I~~~~ e 3;s~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~: w m~~~~~~~~~~~~~~~~~~~~~~~~~ b~~~M f,~~~~~~~~~~~~~~~ 1Rortbwestern ulniversitg Librarp Evanston, Illinois ^3 \ 790I "-4 -:z -)?~- - -, I d~~~~~~~~~~~~~~ I: I E L EME NT S OF GEOM1ETRY AND TRIGONOMWETRY; WITH PRACTICAL APPLICATIONS. B3Y BENJAMIN GREENLEAF, A. M., AUTHOR OF A MATHEMATICAL SERIES. IRVED ELCRTYPE EITMION BOSTO N: PUBLISHED BY ROBERT S. DAVIS & CO. XEW YORK: MASON, BAKER, & PRATT, 142 GRAND STREET. PHILADELPHIA: J. A. BANCROFT &COMPANY. ST. LOUIS: HENDRICKS &CHITTENDEN. CHICAGO: S. C. GRIGGS & CO. NEW COMPREHENSIVE SERIES. An ENqTIRELY NEW MATHEMATICAL COURSE, fully adapted to the best methods of Modern Instructon. GREENLEAF'S NEW PRIMARY ARITHMETIC. GREENLEAF'S NEW INTELLECTUAL ARITHMETIC. GREENLEAF'S NEW ELEMENTARY ARITHMETIC. GREENLEAF'S NEW PRACTICAL ARITHMETIC. GREENLEAF'S NEW ELEMENTARY ALGEBRA. GREENLEAF'S NEW HIGHER ALGEBRA. GREENLEAF'S NEW ELEMENTARY GEOMETRY. GREENLEAF'S ELEMENTS OF GEOMETRY. GREENLEAF'S ELEMENTS OF TRIGONOMETRY. GREENLEAF'S GEOMETRY AND TRIGONOMETRY. 1W Other Boqks -of, a Complete -&-ies, in pparation. KEYS to the PRACTICAL ARITHMETIC, ALGEBRAS. GEcomETRy and TRIGONOMETRY, in separate volumes. jUntered according to, Apt of Xongzw~, iu. the year 1863, by BZNuANI GREENLEAF, RWUA"IhS PRESS: PBTmAn BY B. 0. HOSGETON AND) COMPANT. PREFA CE. THE preparation of this treatise was undertaken at the earnest solicitation of many teachers, who, having used the author's Arithmetics and Algebra with satisfaction, were desirous of seeing his series rendered more nearly complete by the addition of the Elements of Geometry and Trigonometry. That there are peculiar advantages in a graded series of textbooks on the same subject, few, if any, properly qualified to judge, will doubt. The author, therefore, feels justified in introducing this volume to the attention of the public. In the Elements of Geometry, he has followed, in the main, the simple and elegant order of arrangement adopted by Legendre; but in the methods of demonstration no particular authority has been closely followed, the aim having been to adapt the work fully to the latest and most approved modes of instruction. In this respect, there will be found incorporated a considerable number of important improvements. More attention than is usual in elementary works of. this kind has been given to the converse of propositions. In almost all cases where it was possible, the converse of a proposition has been demonstrated. The demonstration of Proposition XX. of the first book is essentially the one given by M. da Cunha in the lincipes Mait, matiques, which has justly been pronounced by the highest mathematical authorities to be a very important improvement in elementary geometry. It has, however, never before been introduced into. a text-book by an American author. PREFACE. The Applications of Geometry to Mensuration, given in the eleventh and twelfth books, are designed to show how the theoretical principles of the science are connected with manifold practical results. The Miscellaneous Exercises, which follow, are calculated to test the thoroughness of the scholar's geometrical knowledge; and sufficient Applications of Algebra to Geometry are given to show the relation existing between these two branches of the mathematics. The Elements of Plane and Spherical Trigonometry present a complete system, theoretical and practical, fully adapted to the wants of advanced classes. The trigonometric functions, in this treatise, have been regarded as ratios, since this improved method has not only now superseded the ancient method in English and French works, but has been approved and adopted generally by the best American mathematicians. Reference, however, is made to whatever is especially valuable in the old method. In the preparation of this work the author has received valuable suggestions from many eminent teachers, to whom he would here express his sincere thanks. Especially would he acknowledge his great obligations to H. B. Maglathlin, A. M., who for many months has been associated with him in his labors, and' to whose experience as a teacher, skill as a mathematician, and ability as a writer, the value of this treatise is largely due. BENJAMIN GREENLEAF. BRLDFORD, Mass., July 25, 1861. NOTICE. A KEY, comprising the Solutions of the Problems contained in this work, is published, for Teachers only; and the same will be mailed, postpaid, to the address ot any Teacher who will forward sixty cents in stamps to the Publishers. CONTEN TS. PLANE GEOMETRY. BO.OK I. PG ELEMENTARYrPRINCIPLES. 7 BOOK II. RATIO AND PROPORTION.. 43 BOOK III. THE CIRCLE, AND THE MEASURE OF ANGLES...55 BOOK IV. PROPORTIONS, AREAS, AND SIMILARITY OF FIGURES..76 BOOK V. PROBLEMS RELATING TO THE PRECEDING BOOKS...118 BOOK VI. REGULAR POLYGONS, AND THE AREA OF THE CIRCLE..142 SOLID GEOMETR'Y. BOOK VII. PLANES. -DIEDRAL AND POLYEDRAL ANGLES...165 BOOK VIII. POLYEDRONS..184 BOOK IX. THE SPHERE, AND ITS PROPERTIES.2114 BOOK X. THE THREE ROUND BODIES.238 vi CONTENTS. PRACTICAL GEOMETRY. BOOK XI. MENSURATION OF' PLANE FIGURES..253 BOOK XII. MENSURATION OF 'SOLIDS..281 BOOK XIII. MISCELLANEOUS EXERCISES....0 01 BOOK XI V. APPLICATION -OF ALGEBRA TO GEOMETRY. il 1 TRI GONOMETRY. BOOK I. LOGARITHMS.2 BOOK II. PLANE TRIGONOMETRY. -DEFINITIONS AND ELEMENTARY PRINCIPLES.1 BOOK III. SOLUTION OF PLANE TRIANGLES....41 BOOK IV. PRACTICAL APPLICATIONS.61 BOOK V. SPHERIC kL TRIGONOMETRY...72 BOOK VI1. APPLICATIONS TO ASTRONOMY AND GEOGRAPHY...105 ELEMENTS OF GEOMETRY. BOOK I. ELEMENTARY PRINCIPLES. DEFINITIONS. 1. GEOMETRY is the science of Position and Extension. The elements of position are direction and distance. The dimensions of extension are length, breadth, and height or thickness. 2. MAGNITUDE, in general, is that which has one or more of the three dimensions of extension. 3. A POINT is that which has position, without magnitude. 4. A LINE is that which has length, without either breadth or thickness. 5. A STRAIGHT LINE, or RIGHT' LINE, is one which has the same A direction in its whole extent; as the line AB. The word line is frequently used alone, to designate a straight line. 6. A CURVED LINE is one which continually changes its direction; C D as the line C D. The word curve is frequently used to designate a curved ELEMENTS OF GEOMETRY. 7. A BROKEN LINE is one which is composed of straighlt lines, not lying in E the same direction; as the line E F. 8. A MIXED LINE is one which is composed of straight lines and of curved lines. 9. A SURFACE is that which has length and breadth, without height or thickness. 10. A PLANE SURFACE, or simply a PLANE, is one in which any two points being taken, the straight line that joins them will lie wholly in the surface. 11. A CURVED SURFACE is one that is not a plane surface, nor made up of plane surfaces. 12. A SOLID, or VOLUME, is that which has length, breadth, and thickness. ANGLES AND LINES. C 13. A PLANE ANGLE, or simply an ANGLE, is the difference in the direction of two lines, which meet at a point; as the angle A. A B The point of meeting, A, is the vertex of the angle, and the lines A B, A C are the sides of the angle. An angle may be designated, not D only by the letter at its vertex, as C, but by three letters, particularly when two.or more angles have the / same vertex; as the angle ACD or C D C B, the letter at the vertex always occupying the middle place. The quantity of an angle does not depend upon the length, but entirely upon the position, of the sides; for the angle remains the same, however tilc lines containing it be increased or diminished. BOOK I. 14. Two straight lines are said to B be perpendicular to each other, when their meeting forms equal adjacent angles; thus the lines A B and C D are perpendicular to each other. C A D Two adjacent angles, as C A B and B A D, have a common vertex, as A; and a common side, as AB. 15. A RIGHT ANGLE is one which C is formed by a straight line and a perpendicular to it; as the angle CAB. A B D 16. An ACUTE ANGLE is one which is less than a right angle; as the angle DE F. E FE An OBTUSE ANGLE is one which is greater than a right angle; as the angle E F G. - F G Acute and obtuse angles have their sides oblique to each other, and are sometimes called oblique angles. 17. PARALLEL LINES are such as, being in the same plane, cannot A meet, however far either way both C D of them may be produced; as the lines AB, CD. 18. When a straight line, as E E F, intersects two parallel lines, as AB, CD, the angles formed A \. B by the intersecting or secant line take particular names, thus- \ INTERIOR ANGLES ON THE SAME C D SIDE are those which lie within the parallels, and on the same F 10 'ELEMENTS OF GEOMETRY. side of the secant line; as the E angles B G H, GHD, and also AGH, GHC. A B EXTERIOR ANGLES ON THE SAME SIDE are those which lie without the parallels, and on the same side C D- \ D of the secant line; as the angles B G E, D H F, and also the angles F AGE, CHF. ALTERNATE INTERIOR ANGLES lie within the parallels, and on different sides of the secant line, but are not adjacent to each other; as the angles B G H, G H C, and also AGH, GHD. ALTERNATE EXTERIOR ANGLES lie without the parallels, and on different sides of the secant line, but not adjacent to each other; as the angles E G B, C H F, and also the angles AGE, DHF. OPPOSITE EXTERIOR and INTERIOR ANGLES lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent to each other; as the angles E GB, GH D, and also E GA, GHC, are, respectively, the opposite exterior and interior angles. PLANM FIGURES. 19. A PLANE FIGURE is a plane terminated on all sides by straight lines or curves. The boundary of any figure is called its perimeter. D 20. When the boundary lines are straight, the space they enclose is C called a RECTILINEAL FIGURE, or POLYGON; as the figure A B C D E. A B 21. A polygon of three sides is called a TRIANGLE; one of four sides, a QUADRILATERAL; 0ne of five, a PENTAGOQO; one of six, a HEXAGON; oni of seven, a E"'AGCON; Ole BOOK I. 11 of eight, an OCTAGON; one of nine, ten, a DECAGON; one of eleven, an twelve, a DODECAGON; and so on. 22. An EQUILATERAL TRIANGLE is one which has its three sides equal; as the triangle A B C. An ISOSCELES TRIANGLE is one which has two of its sides equal; as the triangle D E F. a NONAGON; one of UNDECAGON; ohe of A B/ C D E F G H I L J K A SCALENE TRIANGLE is one which has no two of its sides equal; as the triangle G HI. 23. A RIGHT-ANGLED TRIANGLE is one which has a right angle; as the triangle J K L. The side opposite to the right angle is called the hypothenuse; as the side J L. 24. A1 ACUTE-ANGLED TRIANGLE is one which has three acute angles; as the triangles A B C and D E F, Art. 22. An OBTUSE-ANGLED TRIANGLE is one which has all obtuse angle; as the triangle G H I, Art. 22. Acute-angled and obtuse-angled triangles are also called oblique-angled triangles. 25. A PARALLELOGRAM is a quadrilateral which has its opposite sides parallel. 26. A RECTANGLE is any parallelogram whose angles are right angles; as the parallelogram A B C D. A -- P 12 ELEMENTS OF GEOMETRY. H -- G A SQUARE is a rectangle whose sides are equal; as the rectangle EFGH. E F 27. A RHOMBOID is any parallelo- / gram whose angles are not right angles; as the parallelogram I J K L. A RHOMBUS is a rhomboid whose P/ sides are equal; as the rhomboid MNOP. ]21 --- — / N 28. A TRAPEZOID is a quadrilateral T which has only two of its sides parallel; as the quadrilateral R S T U. R A TRAPEZIUM is a quadrilateral Y X which has no two of its sides parallel; as the quadrilateral V W X Y. D 29. A DIAGONAL is a line joining the vertices of any two angles which \ are opposite to each other; as the \ lines E C and EB in the polygon ABCDE. A - B 30. A BASE of a polygon is the side on which the polygon is supposed to stand. But in the case of the isosceles triangle, it is usual to consider that side the base which is not equal to either of the other sides. 31. An equilateral polygon is one which has all its sides equal. An equiangrular polygon is onle wlhich las BOOK I. 13 all its angles equal. A regular polygon is one which is equilateral and equiangular. 32. Two polygons are mutually equilateral, when all the sides of the one equal the corresponding sides of the other, each to each, and are placed in the same order. Two polygons are mutually equiangular, when all the angles of the one equal the corresponding angles of the other, each to each, and are placed in the same order. 33. The corresponding equal sides, or equal angles, of polygons mutually equilateral, or mutually equiangular, are called homologous sides or angles. AXIOMS. 34. An AXIOM is a self-evident truth; such as, - 1. Things which are equal to the same thing, are equal to each other. 2. If equals be added to equals, the sums will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals, the sums will be unequal. 5. If equals be taken from unequals, the remainders will be unequal. 6. Things which are double of the same thing, or of equal things, are equal to each other. 7. Things which are halves of the same thing, or of equal things, are equal to each other. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. A straight line is the shortest line that can be drawn from one point to another. 11. From one point to another only one straight line can be drawn. 12. Through the same point only one parallel to a straight line can be drawn. 2 ELEMENTS OF GEOMETRY. 13. All right angles are equal to oni another. 14. Magnitudes which coincide throughout their whole extent, are equal. POSTULATES. 35. A POSTULATE is a self-evident problem; such as,1. That a straight line may be drawn from one point to another. 2. That a straight line may be produced to any length. 8. That a straight line may be drawn through a given point parallel to another straight line. 4. That a perpendicular to a given straight line may be drawn from a point either within or without the line. 5. That an angle may be described equal to any given angle. PROPOSITIONS. 86. A DEMONSTRATION is a course of reasoning by which a truth becomes evident. 37. A PROPOSITION is something proposed to be demonstrated, or to be performed. A proposition is said to be the converse of another, when the conclusion of the first is used as the supposition in the second. 38. A THEOREM is something to be demonstrated. 39. A PROBLEM is something to be performed. 40. A LEMMA is a proposition preparatory to the demonstration or solution of a succeeding proposition. 41. A COROLLARY is an obvious consequence deduced from one or more propositions. 42. A SCHOLIUM is a.remark made upon one or more preceding propositions. 48. An HYPOTHESIS is a supposition, made either in the BOOK I. 15 enunciation of a proposition, or in thle oursb of a demonstration. PROPOSITION I.- THEOREM. 44. The adjacent angles which one straight line makes by meeting another straight line, are together equal to two right angles. Let the straight line D C meet E AB, making the adjacent angles D A CD, DCB; these angles to- gether will be equal to two right angles. A B From the point C suppose C E c to be drawn perpendicular to A B; then the angles A C E and E C B will each be a right angle (Art. 15). But the angle A C D is composed of the right angle A C E and the angle E C D (Art. 34, Ax. 9), and the angles E C D and D C B compose the other right angle, E C B; hence tlih angles A C D, D C B together equal two right angles. 45. Cor. 1. If one of the angles ACD, D CB is a right angle, the other must also be a right angle. 46. Cor. 2. All the successive angles, BAC, CAD, DAE, D E A F, formed on the same side C of a straight line, B F, are equal, when taken together, to two right angles; for their sum is equal to B F that of the two adjacent angles, A BAC, CAF. PROPOSITION II.- THEOREM. 47. If one straight line meets two other straight linea at a common point, making adjacent angles, which.tgether are equal to two right angles, the two lines fbrm we and the same straight line. 16 ELEMENTS OF GEOMETRY. Let the straight line D C meet D the two straight lines A C, C B at 'the common point C, making the adjacent angles A C D, D C B to- / gether equal to two right angles; then the lines AC and CB will C... B form one and the same straight line. If C B is not the straight line A C produced, let C E be that line produced; then the line AC E being straight, the sum of the angles A C D and D C E will be equal to two right angles (Prop. I.). But by hypothesis the angles A C D and D C B are together equal to two right angles; therefore the sum of the angles A C D and D C E must be equal to the sum of the angles A C D and D C B (Art. 34, Ax. 2). Take away the common angle A C D from each, and there will remain the angle D C B, equal to the angle D C E, a part to the whole, which is impossible; therefore C E is not the line A C produced. Hence A C and C B form one and the same straight line. PROPOSITION III.- THEOREM. 48. Two straight lines, wzhich have two points common, coincide with each other throughout their whole extent, and form one and the same straight line. Let the two points which are F common to two straight lines be A and B. i The two lines must coincide between the points A and B, for A. E otherwise there would be two B C straight lines between A and B, which is impossible (Art. 84, Ax. 11). Suppose, however, that, on being produced, the lines begin to separate at the point C, the one taking the direc BOOK I. 17 tion C D, and the other C E. From the point C let the line CF be drawn, making, with CA, the right angle A C F. Now, since A C D is a straight line, the angle F C D will be a right angle (Prop. I. Cor. 1); and since A C E is a straight line, the angle F C E will also be a right angle; therefore the angle F C E is equal to the angle F C D (Art. 34, Ax. 13), a part to the whole, which is impossible; hence two straight lines which have two points common, A and B, cannot separate from each other when produced; hence they must form one and the same straight line. PROPOSITION IV.- THEOREM. 49. When two straight lines intersect each other, the opposite or vertical angles which they form are equal. Let the two straight lines A B, C CD intersect each other at the point E; then will the angle AE C A be equal to the angle D E B, and the angle CEB to AED. For the angles AEC, C E B, D which the straight line C E forms by meeting the straight line AB, are together equal to two right angles (Prop. I.); and the angles C E B, B E D, which the straight line B E forms by meeting the straight line C D, are equal to two right angles; hence the sum of the angles A. E C, C E B is equal to the sum of the angles C E B, B E D (Art. 34, Ax. 1). Take away from each of these sums the common angle C E B, and there will remain the angle AE C, equal to its opposite angle, B E D (Art. 34, Ax. 3). In the same manner it may be shown that the angle C E B is equal to its opposite angle, A E D. 50. Cor. 1. The four angles formed by two straight lines intersecting each other, are together equal to four right angles. is ELEMENTS OF OEOMETRY. 51. 'or. 2. All the successive angles, around a comibon point, formed by any number of straight lines meet ing at that point, are together equal to four right angles. PROPOSITION V.- THEOREM. 52. If two triangles have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles will be equal. In the two triangles A BC, DEF, let the side AB be equal to the side DE, the side AC to the side D F, and the B/ angle A to the angle D; then the triangles AB C, D E F will be equal. Conceive the triangle AB C to be applied to the triangle D E F, so that the side A B shall fall upon its equal, D E, the point A upon D, and the point B upon E; then, since the angle A is equal to the angle D, the side AC will take the direction DF. But AC is equal to D F; therefore the point C will fall upon F, and the third side B C will coin'cide with the third side E F (Art. 34, Ax. 11). Hence the triangle A B C coincides with the triangle D E F, and they are therefore equal (Art. 34, Ax. 14). 53. Cor. When, in two triangles, these three parts are equal namely, the side A B equal to D E, the side A C equal to D F, and the angle A equal to D, the other three corresponding parts are also equal, namely, the side B C equal to E F, the angle B equal to E, and the angle C equal to F. PROPOSITION VI.- THEOREM. 54. If two triangles have two angles and the included side in tke one equal to two angles and the included side in the other, each to each, the two triangles will be equal. BOOK.I, 19 Ill the two triangles A D ABC, DEF, let the angle B be equal to the angle E, the angle C to the angle F, and the side B -C E F BC to the side EF; then the triangles A B C, DE F will be equal. Conceive the triangle A B C to be applied to the triangle D E F, so that the side B C shall fall upon its equal, E F, the point B upon E, and the point C upon F. Then, since the angle B is equal to the angle E, the side B A will take the direction E D; therefore the point A will be found somewhere in the line ED. In like manner, since the angle C is equal to the angle F, the line C A will take the direction FD, and the point A will be found somewhere in the line F D. Hence the point A, falling at the same time in both of the straight lines E D and F D, must fall at their intersection, D. Hence the two triangles A B C, D E F coincide with each other, and are therefore equal (Art. 34, Ax. 14). 55. Cor. When, in two triangles, these three parts are equal, namely, the angle B equal to the angle E, the angle C equal to the angle F, and the side B C equal to the side EF, the other three corresponding parts are also equal; namely, the side B A equal to E D, the side C A equal to F D, and the angle A equal to the angle D. PROPOSITION VI. - THEOREM. 56. In an isosceles triangle, the angles opposite the equal sides are equal. Let ABC be an isosceles triangle, in A which the side AB is equal to the side A C; then will the angle B be equal to the angle C. Conceive the angle B A C to be bisected, or divided into two equal parts, by — i 20 ELEMENTS OF GEOMETRY. the straight line AD, making the angle A BAD equal to DAC. Then the two triangles B A D, C A D have the two sides A B, A D and the included angle in the one equal to the two sides AC, A D and the included angle in the other, B i) C each to each; hence the two triangles are equal, and the angle B is equal to the angle C (Prop. V.). 57. Cor. 1. The line bisecting the vertical angle of an isosceles triangle bisects the base at right angles. 58. Cor. 2. Conversely, the line bisecting the base of an isosceles triangle at right angles, bisects also the vertical angle. 59. Cor. 3. Every equilateral triangle is also equiangular. PROPOSITION VIII. -THEOREM. 60. If two angles of a' triangle are equal, the opposite sides are also equal, and the triangle is isosceles. Let A B C be a triangle having the an- A gle B equal to the angle C; then will the side A B be equal to the side A C. D For, if the two sides are not equal, one of them must be greater than the other. Let A B be the greater; then take D B equal to AC the less, and draw CD. B C Now, in the two triangles D B C, A B C, we have D B equal to A C by construction, the side B C common, and the angle B equal to the angle A C B by hypothesis; therefore, since two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each, the triangle D B C is equal to the triangle A B C (Prop. V.), a part to the whole, which is impossible (Art. 34, Ax. 8). Hence the sides AB and A C cannot be unequal; therefore the triangle AB C is isosceles. BOOK I. 21 61. Cor. Therefore every equiangular triangle is equilateral. PROPOSITION IX.- THEOREM. 62. Any side of a triangle is less than the sum of the other two. In the triangle A B C, any one side, C as AB, is less than the sum of the other two sides, A C and C B. For the straight line A B is the shortest line that can be drawn from A B the point A to the point B (Art. 34, Ax. 10); hence the side A B is less than the sum of the sides A C and C B. In like manner it may be proved that the side A C is less than the sum of A B and B C, and the side B C less than the sum of B A and A C. 63. Cor. Since the side AB is less than the sum of A C and C B, if we take away from each of these two unequals the side CB, we shall have the difference between AB and C B less than A C; that is, the difference between any two sides of a triangle is less than the other side. PROPOSITION X. -THEOREM. 64. The greater side of any triangle is opposite the greater angle. In the triangle CAB, let the angle A C be greater than B; then will the side AB, opposite to C, be greater than A C, opposite to B. Draw the straight line C D, making the angle B C D equal to B. Then, in C the triangle B D C, we shall have the side B D equal to D C (Prop. VIII.). But the side A C is less than the sum of A D and D C (Prop. IX.), and the -ELEMENTS OF (N(METRY. -uof A D and D IC is equal to the sum of A D and DB, which is equal to A B; therefore the side A B is greater, than A C. 65. Cor. 1. Therefore the shorter side ig opposite to the less angle. 66. Co~r. 2. In the right-angled triangle the hypothea. nuse is the longest side. PROPOSITION XI. - THEOREMY. 67. The greater angle of apty triangle is. opposite the greater side. In the triangle C AB, suppose"'the A side A B to be greater than A C; then. willt the. angle C, Qpposite to A B, be greater than the angle B, opposite to., For, if the angle C is not greater than C.' B, it must either be equal to it or less. If the angle C were equal to B, then would the side A B be equal to the side A C (Prop. VIII,.), which is contrary, to the hypothesis; and if the angle C were less than B', then would the side A B be- less than A C (Prop. X. Cor. 1), which is also contrary to the hypothesis. Hence, the- angle- C must-be greater than B. 68. Cor. It. follows, therefore, that the less- angle, is. opposite to the ~ shorter side. PPOPOSITION XII.- THEOIJIM. 69. Iffrom a' ty point, within a, triangle., two straight lines are drawn to., t4e, extremities, of either side., titeir somn l be- loss than that- of the other two sides of the BOOK I. Let the two straight lines B O, C 0 A be drawn from the point 0, within the triangle AB C, to the extremities of the side B C; then will the sum of the D two lines B 0 and 0 C be less than the sum of the sides B A and A C. Let the straight line BO be produced till it meets the side A C in the point D; and because one side of a triangle is less than the sum of the other two sides (Prop. IX.), the side 0 C in the triangle C D O is less than the sum of 0 D and D C. To each of these inequalities add B 0, and we have the sum of B O, and 0 C less than the sum of BO, OD, and DC (Art, 34, Ax. 4); or the sum of B 0 and 0 C less than the sum of B D and D C. Again, because the side B D is less than the sum of B A and A D, by adding D C to each, we have the sum of B D and D C less than the sum of B A and A C. But it has been just shown that the sum of B 0 and O C is less than the sum of BD and D C; much more, then, is the sum of B O and O C less than B A and A C. PROPOSITION XIII. - THEOREM. 70. From a point without a straight line, only one perpendicular can be drawn to that line. Let A be the point, and DE the A given straight line; then from the point A only one perpendicular can be drawn to D E. Let it be supposed that we can D c draw two perpendiculars, A B and AC. Produce one of them, as A B, till B F is equal to A B, and join F C. F Then, in the triangles A B C and C B F, the angles,C B A and CB F are both right angles (Prop- I. Cor. 1), the side OB' is common to both, and the side B F is equal to 24 ELEMENTS OF GEOMETRY. the side A B; hence the two triangles A are equal, and the angle B C F is equal to the angle BCA (Prop. V.) But the angle B CA is, by hypothesis, a right angle; therefore B C F must also D C E be a right angle; and if the two adjacent angles, B C A and B C F, are together equal to two, right angles, the F two lines AC and C F must form one and the same straight line (Prop. II.). Whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Art. 34, Ax. 11); hence no more than one perpendicular cal be drawn from the same point to the same straight line. 71. Cor. At the same point C, in the E D line AB, it is likewise impossible to / erect more than one perpendicular to that line. For, if C D and C E were each perpendicular to AB, the angles A BC D, B CE would be right angles; hence the angle B C D would be equal to the angle B C E, a part to the whole, which is impossible. PROPOSITION XIV. THEOREM. 72. If, from a point without a straight line, a perpendicular be let fall on that line, and oblique lines be drawn to different points in the same line;1st. The perpendicular will be shorter than any oblique line. 2d. Any two oblique lines, which meet the given line at equal distances from the perpendicular, will be equal. 3d. Of any two oblique lines, that which meets the given line at the greater distance from the perpendicular will be the longer. BOOK I. 25 Let A be the given point, and A D E tlle given straight line. Draw AB perpendicular to DE, and the oblique lines A E, A C, D< E AD. Produce AB till BF is \ equal to A B, and join C F, DF. First. The triangle B C F is equal to the triangle B C A, for F they have the side C B comlmon, the side A B equal to the side B F, and the angle A B C equal to the angle F B C, botl being right angles (Prop. I. Cor. 1); hence the third sides, CF and AC, are equal (Prop. V. Cor.). But A B F, being a straight line, is shorter than A C F, which is a broken line (Art. 34, Ax. 10); therefore A B, the half of AB F, is shorter than A C, the half of A C F; hence the perpendicular is shorter than any oblique line. Secondly. If B E is equal to B C, then, since A B is common to the triangles, A B E, A B C, and the angles A B E, A B C are right angles, the two triangles are equal (Prop. V.), and the side A E is equal to the side AC (Prop. V. Cor.). Hence tlle two oblique lines, meeting the given line at equal distances from the perpendicular, are equal. Thirdly. The point C being in the triangle AD F, the sum of the lines A C, C F is less than the sum of the sides A, D F (Prop. XII.) But A C has been shown to be equal to C F; and in like manner it may be shown that A D is equal to D F. Therefore A C, the half of the line A C F, is shorter than A D, the half of the line AD F; hence the oblique line which meets the given line the greater distance from the perpendicular, is the longer. 73. Cor. 1. The perpendicular measures the shortest distance of any point from a straight line. 74. Cor. 2. From the same point to a given straight line only two equal straight lines can be drawn. 3 26 ELEMENTS OF GEOMETRY. 75. Cor. 3. Of any two straight lines drawn from a point to a straight lile, that which is not shorter than the other will be longer than aly straight line that can be drawn between them, from the.same point to the same line. PROPOSITION XV. - THEOREM. 76. If from the middle point of a straight line a perpendicular to this line be drawn, — 1st. Any point in the perpendicular will be equally distant from the extremities of the line. 2d. Any point out of the perpendicular will be unequally distant from those extremities. Let D C be drawn perpendicular to D the straight line A B, from its middle point C. First. Let D and E be points, taken / E at pleasure, in the perpendicular, and join DA, DB, and also AE, EB. A B - Then, since A C is equal- to C B, the two oblique Jines D A, D B meet points which are at the same distance from the perpendicular, and are therefore equal (Prop. XIV.). So, likewise, the two oblique lines E A, E B are equal; therefore any point il the perpendicular is equally distant from the extremities A and B. Secondly. Let F be any point out of the perpendicular, and join F A, F B. Then one of those lines must cut the perpendicular, in some point, as E. Join E B; then we have E B equal to E A. But in the triangle F E B, the side F B is less than the sum of the sides E F, E B (Prop. IX.), and since the sum of F E, E B is equal to the sum of F E, E A, which is equal to F A, F B is less than F A. Hence any point out of the perpendicular is at unequal distances from the extremities A and B. 77. Cor. If a straight line have two points, of which * each is equally distant from the extremities of another BOOK I. 27 straight line, it will be perpendicular to that line at its middle point. PROPOSITION XVI. -THEOREM. 78. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angle of the one greater than the included angle of the other, the third side of that which has the greater angle will be greater than the third side of the other. LetABC,DEF D be two triangles, A having the side AB equal to DE, and AC equal to DF, and the angle A greater than D; B C E - G then will the side F BC be greater than EF. Of the two sides D E, D F, let D F be the side which is not shorter than the other; make the angle E D G equal to B A C; and make D G equal to A C or D F, and join EG, GF. Since D F, or its equal D G, is not shorter than D E, it is longer than D H (Prop. XIV. Cor. 3); therefore its extremity, F, must fall below the line E G. The two triangles, A B C and D E G, have the two sides A B, AC equal to the two sides D E, D G, each to each, and the included angle B AC of the one equal to the included angle E D G of the other; hence the side B C is equal to E G (Prop. V. Cor.). In the triangle D F G, since D G is equal to D F, the angle D F G is equal to the angle D G F (Prop. VII.); but the angle DGF is greater than the angle EGF; therefore the angle D F G is greater than E F, and much more is the angle EFG greater than the angle 28 ELEMENTS OF GEOMETRY. E G F. Because the angle E F G in the triangle E F G is greater than E GF, and because the greater side is opposite the greater angle (Prop. X.), tle side E G is greater than E F; and E G has been shown to be equal to B C; hence B C is greater than E F. PROPOSITION XVII. -THEOREM. 79. If two triangles have two sides of the one equal to two sides of the other, each to each, but the third side of the one greater than the third side of the other, the angle contained by the sides of that which has the greater third side will be greater than the angle contained by the sides of the other. Let ABC, DEF be A D two triangles, the side AB equal to DE, and AC equal to DF, and the side CB greater than EF, then will the angle B C E A be greater than D. F For, if it be not greater, it must either be equal to it or less. But the angle A cannot be equal to D, for then the side B C would be equal to E F (Prop. V. Cor.), which is contrary to the hypothesis; neither can it be less, for then the side B C would be less than E F (Prop. XVI.), which also is contrary to the hypothesis; therefore the angle A is not less than the angle D, and it has been shown that is not equal to it; hence the angle A must be greater than the angle D. PROPOSITION XVIII. -THEOREM. 80. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles themselves will be equal. BOOK I. 29 Let the triangles A B C, A D D E F have the side A B equal toDE, A C to DF, and B C to E F; then will the angle A be equal to D, B C E -F the angle B to the angle E, and the angle C to the angle F, and the two triangles will also be equal. For, if the angle A were greater than the angle D, since the sides A B, A C are equal to the sides D E, D F, each to each, the side B C would be greater than E F (Prop. XVI.); and if the angle A were less than D, it would follow that the side B C would be less than E F. But by hypothesis B C is equal to E F; hence the angle A can neither be greater nor less than D; therefore it must be equal to it. In the same manner, it may be shown that the angle B is equal to E, and the angle C to F; hence the two triangles must be equal. 81. Scholium. In two triangles equal to each other, the equal angles are opposite the equal sides; thus the equal angles A and D are opposite the equal sides B C and EF. PROPOSITION XIX. - THEOREM. 82. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let the two right-an- A D gled triangles ABC, DEF, have the hypothenuse A C equal to D F, and the side AB equal to DE; then will the triangle A B C be equal to the triangle D E F. B G C E F The two triangles are evidently equal, if the sides B C and E F are equal (Prop. XVIII.). If it be possible, let 3* 30 ELEMENTS OF GEOMETRY. these sides be unequal, and A D let B C be the greater. Take B G equal to E F, the less side, and join AG. Then, in the two triangles A B G, D E F, the angles B and E are equal, both B G C E F being right angles, the side A B is equal to D E by hypothesis, and the side B G to E F by construction; hence these triangles are equal (Prop. V.); and therefore A G is equal to D F. But by hypothesis D F is equal to A C, and therefore A G is equal to A C. But the oblique line A C cannot be equal to A G, which meets the same straight line nearer the perpendicular AB (Prop. XIV.); therefore B C and E F cannot be unequal, hence they must be equal; therefore the triangles AB C and D E F are equal. PROPOSITION XX. - THEOREM. 83. If a straight line, intersecting two other straight lines, makes the alternate angles equal, the two lines are parallel. Let the straight line E E F intersect the two straight lines A B, CD, G making the alternate an- A -. gles B G H, C H G equal; I -... \... —. K then the lines A B, C D c """ will be parallel. H For, if the lines A B, C D are not parallel, let them meet in some point K, and through 0, the middle point of GH, draw the straight line I K, making IO equal to 0 K, and join H I. Then the opposite angles K 0 G, I O H, formed by the intersection of the two straight lines I K, GH, are equal (Prop. IV.); and the triangles K 0 G, BOOK I. 81 I 0 H have the two sides K, 0 G and the included angle in the one equal to the two sides I 0, OH and the included angle in the other, each to each; hence the angle K G 0 is equal to the angle I H 0 (Prop. V. Cor.). But, by hypothesis, the angle KGO is equal to the angle C HO, therefore the angle 1 H 0 is equal to C H O, so that H I and II C must coincide; that is, the line C D when produced meets I K il two points, I, K, and yet does not form one and the same straight line, which is impossible (Prop. III.); therefore the lines AB, CD cannot meet, consequently they are parallel (Art. 17). NOTE.- The demonstration of the proposition is substantially that given by M. da Cunha in the Principes Mlathe'matiques. This demonstration Young pronounces " superior to every other that has been given of the same proposition "; and Professor Playfair, in the Edinburgh Review, Vol. XX., calls attention to it, as a most important improvement in elementary Geometry. PROPOSITION XXI. -THEOREM. 84. If a straight line, intersecting two other straight lines, makes any exterior angle equal to the interior and opposite ang(le, or makes the interior angles on the same side together equal to two right angles, the two lines are parallel. Let the straight line E F inter- E sect the two straight lines A B, CD, making the exterior angle A \.. B E G B equal to the interior and opposite angle, G H D; then the lines A B, CD are parallel. C l\ D For the angle A G H is equal to the angle E G B (Prop. IV.); F and E G B is equal to D H D, by hypothesis; therefore the angle A G H is equal to the angle GH D; and they are alternate angles; hence the lines A B, C D are parallel (Prop. XX.). 82 ELEMENTS OF GEOMETRY. Again, let the interior angles E on the same side, B G H, G H D, be together equal to two right A B angles; then the lines A B, C D are parallel. \ For the sum of the angles C D BGH, GHD is equal to two right angles, by hypothesis; and F the sum of AGH, B GH is also equal to two right angles (Prop. I.); take away B G H, which is common to both, and there remains the angle G H D, equal to the angle A G H; and these are alternate angles; hence the lines A B, CD are parallel. 85. Cor. If two straight lines are perpendicular to another, they are parallel; thus A B, C D, perpendicular to E F, are parallel. E A B C - D F PROPOSITION XXII. -THEOREM. 86. If a straight line intersects two parallel lines, it makes the alternate angles equal; also any exterior angle equal to the interior and opposite angle; and the two interior angles upon the same side together equal to two right angles. Let the straight line E F inter- E sect the parallel lines A B, C D; the alternate angles A G H, G D are equal; the exterior angle K B E G B is equal to the interior and opposite angle GHD; and the 'C D two interior angles B G H, G H D upon the same side arc together equal to two right anigles. BOOK I. 83 For if the angle A G H is not equal to G H D, draw the straight line K L through the point G, making the angle K GH equal to GHD; then, since the alternate angles GHD, KGH are equal, KL is parallel to CD (Prop. XX.); but by hypothesis A B is also parallel to C D, so tlhat through the same point, G, two straight lines are drawn parallel to C D, wlicl is impossible (Art. 34, Ax. 12). Hence the angles A G H, GH D are not unequal; that is, they are equal. Now, the angle E G B is equal to the angle AGH (Prop. IV.), and A G H has been shown to be equal to G ID; hence E G B is also equal to G H D. Again, add to each of these equals the angle B G I; then the sum of the angles E G B, B G H is equal to the sum of the angles B G H, G H D. But E GB, B G H are equal to two right angles (Prop. I.); hence B GH, GHD are also equal to two right angles. 87. Cor. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other; thus EF (Art. 85), perpendicular to AB, is perpendicular to CD. PROPOSITION XXI II. -THEOREM. 88. If two straight lines intersect a third line, and make the two interior angles on the same side together less than two right angles, the two lines will meet on being produced. Let the two lines KL, CD make E with EF the angles K G H, GH C, togetller less than two right angles; then KL and CD will meet on K being produced. For if they do not meet, they C- D are parallel (Art. 17). But they are not parallel; for then the sum 34 ELEMENTS OF GEOMETRY. of the interior angles K G IT, G II C would be equal to two right angles (Prop. XXII.); but by hypothesis it is less; therefore the lines K L, C D will meet on being produced. 89. Scholium. The two lines K L, C D, on being produced, must meet on the side of E F, on whicl are tlle two interior angles whose sum is less than two right angles. PROPOSITION XXIV. -THEOREM. 90. Straight lines which are parallel to the same line are parallel to each other. Let the straiglt lines A B, C D be each parallel to the line E F; then are they parallel to each other. I Draw G H I perpendicular to C D E F. Then, since A B is parallel to EF, GI will be perpendicular A B to AB (Prop. XXII. Cor.); and since CD is parallel to EF, GI will for a like reason be perpendicular to CD. Consequently A B and C D are perpendicular to the same straight line; hence they are parallel (Prop. XXI. Cor.). PROPOSITION XXV. - THEOREM. 91. Two parallel straight lines are everywhere equally distant from each other. Let AB, CD be two parallel I G straight lines. Through any two C D points in A B, as E and F, draw... the straight lines EG, FH, per- pendicular to A B. These lines F E will be equal to each other. For, if G F be joined, the angles G F E, F G H, considered in reference to the parallels A B, C D, will be alter BOOK I. 35 nate interior angles, and therefore equal to each other (Prop. XXII.). Also, since the straiglit lines E G, F H are perpendicular to the same straight line A B, and consequently parallel (Prop. XXI. Cor.), the angles EGF, G FI, considered in reference to tle parallels E G, F H, will be alternate interior angles, and therefore equal. Hence, the two triangles E F G, F GII, have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, each to each; therefore these triangles are equal (Prop. VI.); hence the side E G, which measures the distance of the parallels AB, CD, at the point E, is equal to the side F I, wlhich measures the distance of the same parallels at tle point F. Hence two parallels are everywhere equally distant. PROPOSITION XXVI. THEOREM. 92. If two angles have their sides parallel, each to each, and lying in the same direction, the two angrles are equal. Let A B C, D E F be two angles, A wrhich have tile side A B parallel to DE, and BC parallel to EF; / then these angles are equal. For produce DE, if necessary, B C till it meets B C in tile point G. H -......... -- F Then, since E F is parallel to G C, the angle D E F is equal to D G C (Prop. XXII.) and since D G is parallel to AB, the angle DGC is equal to ABC; hence the angle DEF is equal to AB C. 93. Scholium. This proposition is restricted to the case where the side E F lies in the same direction with B C, since if F E were produced toward H, the angles D E H, A B C would only be equal when they are right angles. 36 86 ~~~LEMENTS OF GEOMETRY. PROPOSITION XXVH. - TiiEOREM~. 94. If any side of a triangirc be prodluced, tlhe exterior canle is equal to the suynt of the two interior and opposite LI-t A1 B C be a triangle, and A lot one of its sides, B C be produced towards ID; tholl the exterior angle A C ID is equal to the two interior and opposite B D anglesC A B A BC. For, draw E C parallel to the side A 13 then, since A C meets the two parallels A B, E C, the alternate angles BAC, A C E are equal (Pr-op. XXII.). Again, since B 1) meets the two parallels A B, E C, the cxterior angle E C D is equal to the interior and oppo-site angle A B C. But the angle A C E is equal to B A C; thrfrthe whole exterior angle A C ID is equal to the two interior and opposite angles C A B, A 1B C (-Art. '04, Ax. 2). PROPOSITION XXVIII. - THEOREM. 95'. In ever~y triangle the s~m of the three angle is equal to two rig(ht angles. Let A BC be any triangle; A then will the sunm of the angles ABC, C B CA, C AB be equal to two right aingles. For, let the side B C be PAiO- B 1) duced towards ID, making the exterior angle A C ID; then the angle A C ID is equal to C A B and A BC (Prop. XXVII.). To eachi of these equals add the angle A C B, and we shall have the,nnim of BOOK I. 37 A C B and A C D, equal to the sum of A B C, B C A, and C A B. But the sum of A C B and A C D is equal to two right angles (Prop. I.); hence the sum of the three angles A B C, B C A, and C A B is equal to two right angles (Art. 34, Ax. 2). 96. Cor. 1. Two angles of a triangle being given, or merely their sumn, the third will be found by subtracting that sum from two right angles. 97. Cor. 2. If two angles in one triangle be respectively equal to two angles in another, their third angles will also be equal. 98. Cor. 3. A triangle cannot have more than one angle as great as a rightalngle. 99. Cor. 4. And, therefore, every triangle must have at least two acute angles. 100. Cor. 5. In a right-angled triangle the right angle is equal to the sum of the other two angles. 101. Cor. 6. Since every equilateral triangle is also equiangular (Prop. VII. Cor. 3), each of its angles will be equal to two thirds of one right angle. PROPOSITION XXIX.- THEOREM. 102. The sum of all the interior angles of any polygon is equal to twice as many right angles, less four, as the fgure has sides. Let A B C D E be any polygon; then D the sum of all its interior angles, A, B, C, D, E, is equal to twice as many E.C right angles as the figure has sides, less four right angles. \ For, from any point P within the pol- A B ygon, draw the straight lines P A, PB, P C, P D, P E, to the vertices of all the angles, and the polygon will be 4 38 ELEMENTS OF GEOMETRY. divided into as many triangles as it has sides. Now, the sum of the three angles in each of these triangles is equal to two right angles (Prop. XXVIII.); therefore the sum of the angles of all these triangles is equal to twice as many right angles as there are triangles, or sides, to the polygon. But the sum of all the angles about the point P is equal to four right angles (Prop. IV. Cor. 2), which sum forms no part of the interior angles of the polygon; therefore, deducting the sum of the angles about the point, there remain the angles of the polygon equal to twice as many right angles as the figure has sides, less four right angles. 103. Cor. 1. The sum of the angles in a quadrilateral is equal to four right angles; hence, if all the angles of a quadrilateral are equal, each of them is a right angle; also, if three of the angles are right angles, the fourth is likewise a right angle. 104. Cor. 2. The sum of the angles in a pentagon is equal to six right angles; in a hexagon, the sum is equal to eight right angles, &c. 105. Cor. 3. In every equiangular figure of more than four sides, each angle is greater than a right angle; thus, in a regular pentagon, each angle is equal to one and one fifth right angles; in a regular hexagon, to one and one third right angles, &c. 106. Scholiumn. In applying this prop- D osition to polygons which lave re-entrant angles, or angles whose vertices / are directed inward, as B P C, each of these angles must be considered greater than two right angles. But, in order \ B to avoid ambiguity, we shall hereafter A limit our reasoning to polygons with salient angles, or with angles directed outwards, and which may be called convex polygons. Every convex polygon is such that a BOOK I. straight line, however drawn, cannot meet the perimeter of the polygon in more than two points. PROPOSITION XXX.- THEOREM. 107. The sum of all the exterior angles of any polygon, formed by producing, each side in the same direction, is equal to four right angles. Let each side of the polygon A B CD E D be produced in the same direction; then / the sum of the exterior angles A, B, C, D, E, will be equal to four right angles. E For each interior angle, together with its adjacent exterior angle, is equal to two right angles (Prop. 1.); hence the suln of all the angles, both interior and exterior, is equal to twice as many right angles as there are sides to the polygon. But the sum of the interior angles alone, less four right angles, is equal to the same sum (Prop. XXIX.); therefore the sum of the exterior angles is equal to four right angles. PROPOSITION XXXI. -THEOREM. 108. The opposite sides and angles of every parallelogram are equal to each other. Let A B C D be a parallelogram; D C then the opposite sides and angles are equal to each other. Draw the diagonal B D, then, since the opposite sides AB, D C are paral- A B lel, and BD meets them, the alternate angles ABD, BDC are equal (Prop. XXII.); and since AD, B C are parallel, and B D meets them, the alternate angles A D B, D B C are likewise equal. Hence, tle two triangles AD B, D BC have two angles, A B D, A D B, in tlhe one, equal to two angles, BD C, D B C, in the other, eacll to eacll; and since 40 ELEMENTS OF GEOMETRY. the side BD included between these D C equal angles is common to the two triangles, they are equal (Prop. VI.); hence the side AB opposite the angle ____ A D B is equal to the side D C opposite A B the angle D B C (Prop. VI. Cor.); and, in like manner, the side A D is equal to the side B C; hence the opposite sides of a parallelogram are equal. Again, since the triangles are equal, the angle A is equal to the angle C (Prop. VI. Cor.); and since the two angles D B C, A B D are respectively equal to the two angles AD B, BD C, the angle ABC is equal to the angle ADC. 109. Cor. 1. The diagonal divides a parallelogram into two equal triangles. 110. Cor. 2. The two parallels A D, B C, included between two other parallels, A B, C D, are equal. PROPOSITION XXXI.- THEOREM. 111. If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelogram. Let AB C D be a quadrilateral D C having its opposite sides equal; then will the equal sides be parallel, and the figure be a parallelogram. For, having drawn the diagonal A B B D, the triangles A B D, B D C have all the sides of the one equal to the corresponding sides of the other; therefore they are equal, and the angle A D B opposite the side A B is equal to D B C opposite C D (Prop. XVIII. Sch.); hence the side A D is parallel to B C (Prop. XX.). For a like reason, A B is parallel to C D; therefore the quadrilateral A B C D is a parallelogram. BOOK I. 41 PROPOSITION XXXIII. - THEOREM. 112. If two opposite sides of a quadrilateral are equal and parallel, the other sides are also equal and parallel, and the figure is a parallelogram. Let AB CD be a quadrilateral, D C having tlhe sides A B, C D equal and parallel; then will the other sides also be equal and parallel. Draw the diagonal B D; then, since A B A B is parallel to C D, and B D meets them, the alternate angles A B D, B D C are equal (Prop. XXII.); moreover, in the two triangles A B D, D B C, the side B D is commonl; therefore, two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each; hence these triangles are equal (Prop. V.), and the side AD is equal to B C. Hence the angle A D B is equal to D B C, and consequently A D is parallel to B C (Prop. XX.); therefore the figure AB C D is a parallelogram. PROPOSITION XXXIV. -THEOREM. 113. The diagonals of every parallelogram bisect each other. Let A B C D be a parallelogram, D C and A C, DB its diagonals, intersect- /.".... ing at E; then will A E equal E C, and B E equal ED. For, since A B, CD are parallel, A B and B D meets them, tlhe alternate angles C D E, A B E are equal (Prop. XXII.); and since A C meets the same parallels, the alternate angles BAE, ECD are also equal; and the sides AB, CD are equal (Prop. XXXI.). Hence the triangles A B E, C D E have two angles and the in4* 42 ELEMENTS OF GEOMETRY. eluded side in the one equal to two angles and the included side in the other, each to each; hence the two triangles are equal (Prop. VI.); therefore the side A E opposite the angle A B E is equal to C E opposite C D E; hence, also, the sides B E, D E opposite the other equal angles are equal. 114. Scholium. In the case of a rhom- D C bus, the sides A B, B C being equal, the E triangles A E B, E B C have all the sides of the one equal to the corresponding A sides of tle other, and are, therefore, B equal; whence it follows that the angles A E B, B E C are equal. Therefore the diagonals of a rhombus bisect each other at right angles. PROPOSITION XXXV. - THEOREM. 115. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Let AB CD be a quadrilateral, and D C AC, DB its diagonals intersectilg at E;.. then will the figure be a parallelogram. / For, in the two triangles ABE, CDE,.the two sides A E, E B and the included A B angle in tlle ole are equal to the two sides C E, E D and the included angle in tlhe other; hence the triangles are equal, and the side AB is equal to the side CD (Prop. V. Cor.). For a like reason, A D is equal to C B; therefore the quadrilateral is a parallelogram (Prop. XXXII.). BOOK II. RATIO AND PROPORTION. DEFINITIONS. 116. RATIO is the relation, in respect to quantity, which one magnitude bears to another of the same kind; and is the quotient arising from dividing the first by the second. A ratio may be written in the form of a fraction, or with the sign:. Thus the ratio of A to B may be expressed citler by A or )y A: B. 117. The two magnitudes necessary to form a ratio are called tle TERMS of tlhe ratio. The first term is called the ANTECEDENT, and the last, the CONSEQUENT. 118. Ratios of magnitudes may be expressed by numbers, either exactly, or approximately. This may be illustrated by the operation of finding the numerical ratio of two straigllt lines, AB, C D. From the greater line AB cut off a part equal F to tle less C D, as mainy Al - -.,. times as possible; for ex- E ( ample, twice, witll the remainder B E. From the line C D cut off a part equal to the remainder B E as many times as possible; once, for example, with the remainder DF. From tlie first remainder B E, cut off a part equal to the second D F, as many times as possible; once, for example, witll tle remainder B G. 44 ELEMENTS OF GEOMETRY. From the second re- __ mainder D F, cut off a part equal to B G, the third, as many times as A ' possible. Proceed thus till a remainder arises, which is exactly contained a certain number of times in the preceding one. Then this last remainder will be the common measure of the proposed lines; and, regarding it as unity, we shall easily find the values of the preceding remainders; and, at last, those of the two proposed lines, and hence their ratio in numbers. Suppose, for instance, we find G B to be contained exactly twice in F D; B G will be the common measure of the two proposed lines. Let B G equal 1; then will F D equal 2. But E B contains FD once, plus GB; therefore we have EB equal to 3. CD contains E B once, plus F D; therefore we have C D equal to 5. A B contains C D twice, plus E B; therefore we have A B equal to 13. Hence the ratio of the two lines is that of 13 to 5. If the line C D were taken for unity, the line A B would be -1; if A B were taken for unity, C D would be 5%. It is possible that, however far the operation be continued, no remainder may be found which shall be contained an exact number of times in the preceding one. In that case there can be obtained only an approximate ratio, expressed in numbers, more or less exact, according as the operation is more or less extended. 119. When the greater of two magnitudes contains the less a certain number of times without having a remainder, it is called a MULTIPLE of the less; and the less is then called a SUBMULTIPLE, or measure of the greater. Thus, 6 is a multiple of 2; 2 and 3 are submultiples, or measures, of 6. 120. EQUIMULTIPLES, or LIKE MULTIPLES, are those which contain their respective submultiples the same number of BOOK II. 45 times; and EQUISUBMULTIPLES, or LIKE SUBMULTIPLES, are those contained in their respective multiples the same number of times. Thus 4 and 5 are like submultiples of 8 and 10; 8 and 10 are like multiples of 4 and 5. 121. COMMENSURABLE magnitudes are magnituldes of the same kind, which have a common measure, and whose ratio therefore may be exactly expressed ill numbers. 122. INCOMMENSURABLE magnitudes are magnitudes of the same kind, wllicl have no common measure, and whose ratio, therefore, cannot be exactly expressed in numbers. 123. A DIRECT ratio is the quotient of the antecedent by the consequent; an INVERSE ratio, or RECIPROCAL ratio, is the quotient of the consequent by the antecedent, or the reciprocal of the direct ratio. Thus the direct ratio of a line 6 feet long to a line 2 feet long is 1 or 3; and the inverse ratio of a line 6 feet long to a line 2 feet long is A or 1, which is the same as the reciprocal of 3, the direct ratio of 6 to 2. The word ratio when used alone means the direct ratio. 124. A COMPOUND ratio is the product of two or more ratios. Thus the ratio compounded of A: B and C: D is A C AX C B Dor BXD 125. A PROPORTION is an equality of ratios. Four magnitudes are in proportion, when the ratio of the first to the second is the same as that of the third to the fourth. Thus, the ratios of A: B and X: Y, being equal to A X each other, when written A: B = X: Y, or - y, form a proportion. 46 ELEMENTS OF GEOMETRY. 126. Proportion is written not only with the sign -, but, more often, with the sign:: between the ratios. Thus, A: B:: X: Y, expresses a proportion, and is read, The ratio of A to B is equal to the ratio of X to Y; or, A is to B as X is to Y. 127. The first and third terms of a proportion are called the ANTECEDENTS; the second and fourth, the CONSEQUENTS. The first andfourth are also called the EXTREMES, and the second and third the MEANS. Thus, in the proportion A: B:: C: D, A and C are the antecedents; B and D are the consequents; A and D are the extremes; and B and C are the means. The antecedents are called homologous or like terms, and so also are the consequents. 128. All the terms of a proportion are called PROPORTIONALS; and the last term is called a FOURTH PROPORTIONAL to the other three taken in their order. Thus, in the proportion A: B:: C: D, D is the fourth proportional to A, B, and C. 129. When both the means are the same magnitude, either of them is called a MEAN PROPORTIONAL between the extremes; and if, in a series of proportional magnitudes, each consequent is the same as the next antecedent, those magnitudes are said to be in CONTINUED PROPORTION. Thus, if we have A: B:: B: C:: C: D:: D: E, B is a.mean proportional between A and C, C between B and D, D between C and E; and the magnitudes A, B, C, D, E are said to be in continued proportion. 130. When a continued proportion consists of but three terms, the middle term is said to be a MEAN PROPORTIONAL between the other two; and the last term is said to be the THIRD PROPORTIONAL to the first and second. Thus, when A, B, and C are in proportion, A: B:: B: C; in which case B is called a mean proportional between A and C; and C is called the third proportional to A and B. BOOK II. 47 131. Magnitudes are in proportion by INVERSION, or INVERSELY, when each antecedent takes the place of its consequent, and each consequent the place of its antecedent. Thus, let A: B:: C: D; then, by inversion, B: A::: C. 132. Magnitudes are in proportion by ALTERNATION, or ALTERNATELY, when antecedent is compared with antecedent, and consequent with consequent. Thus, let A: B:: D: C; then, by alternation, A: D:: B: C. 133. Magnitudes are in proportion by COMPOSITION, when the sum of the first antecedent and consequent is to the first antecedent, or consequent, as the sum of the second antecedent and consequent is to the second antecedent, or consequent. Thus, let A: B:: C: D; then, by composition, A + B:A:: C-: C, or A + B: B:: C + D: D. 134. Magnitudes are in proportion by DIVISION, when the difference of the first antecedent and consequent is to the first antecedent, or consequent, as the difference of the second antecedent and consequent is to the second antecedent, or consequent. Thus, let A: B:: C: D; then, by division, A-B:A::C-D:C, or A-B:B::C-D:D. PROPOSITION I. THEOREM. 135. If four magnitudes are in proportion, the product of the two extremes is equal to the product of the two means. Let A: B:: C: D; then will A X D - B X C. For, since the magnitudes are in proportion, A C B 48 ELEMENTS OF GEOMETRY. and reducing the fractions of this equation to a common denominator, we have AXD BX C BX D BX D' or, the common denominator being omitted, AX D-B XC, PROPOSITION II.- THEOREM. 136. If the product of two magnitudes is equal to the product qf two others, these four magnitudes form a proportion. Let A X D = B X C; then will A: B:: C: D. For, dividing each member of the given equation by B X D, we have AXD BXC B X D B XD' which, reduced to the lowest terms, gives A C B D Whence A: B:: C: D. PROPOSITION III.-THEOREM. 137. If three magnitudes are in proportion, the product of the two extremes is equal to the square of the mean. Let A: B:: B: C; then will A X C = B2. For, since the magnitudes are in proportion, A B B C' and, by Prop. I., AXC'BXB, or AX C =B2. BOOK II. 49 PROPOSITION IV.- THEOREM. 138. If the product of any two quantities is equal to the square of a third, the third is a mean proportional between the other two. Let A X C = B2; then B is a mean proportional between A and C. For, dividing each member of the given equation by B X C, we have A B B C whence A: B:: B: C. PROPOSITION V. THEOREM. 139. If four magnitudes are in proportion, they will be in proportion when taken inversely. Let A: B:: C: D; then will B: A:: D: C. For, from the given proportion, by Prop. I., we have AXD BXC, or BXC -AXD. Hence, by Prop. II., B: A:: D: C. PROPOSITION VI.- THEOREM. 140. If four magnitudes are in proportion, they will be in proportion when taken alternately. Let A: B:: C: D; then will A: C:: B: D. For, since the magnitudes are in proportion, A C BB D and multiplying each member of this equation by A, we have AXB CB B X C- D xC' 5 50 50 ~~ELEMIENTS OF GEOMETRY. which, reduced to the lowest terms, gives A B whence A:C::B:.D. PROPOSITION VII1. - THEOREM. 141. If four mtagnitudes are in proportion, they witl be in proportion by composition. Let A: B:: C:D; thenr will A + B:A:: C + D:C. For, from the given proportion, by Prop. I., we have B xC = A xD. Adding A X C to each side of this equation, we have A xC~+B xC=~A xC +HA xD, and resolving each member into its factors, (A~+B) X C= (C +D) X A. Hence, by Prop. II., A + B: A: C ~ D: C. PROPOSITION VIII. - THEOREM. 142. If four magnitudes are in proportion, they will be in proportion by division. Let A: B:: C: D;then will A -B: A:: C -D: C. For, from the given proportion, by Prop. I., we have B xC = A xD. Subtracting each side of this equation from A X C, we have A XC - B XC=AXC - A XD) and resolving each member into its factors, (A -B) X C=(C -D) XA. Hence, by Prop. II., A - B: A: C - D: C. BOOK IL.5 51 PROPOSITION IX. - THEOREM. 143. Equimtultiples of two magnitudes have the same ratio as the mcqgnitudes themselves. Let A and B be two magnitudes, and m X A and m X B their equimultiples, then will mi > A:m X B:A:- B. For A x B =.BxA; Multiplying each side of this equation by any number, m, we hiave m x A x B = m x B x A; therefore (m X A) X B=- (m X B) X A. Hence, by Prop. II., m xA:rn X B:A:B. PROPOSITION X. - THEOREM. 144. Magnitudes which are proportional to the sanq proportionals, will be proportional to each other. Let A: B:: E: F,and C':D::E: F; then will A: B: C: D. For, by the given proportions, we have A E C E and -=Therefore, it is evident (Art. 34, Ax. 1), A C j-h j.b Hence A:B::C:D 145. Cor. 1. If two proportions have an antecedent and its consequent the same in both, the remaining terms will be in proportion. 146. Cor. 2. Therefore, by alternation (Prop. VI.), if two proportions have the two antecedents or the two con 52 ELEMENTS OF GEOMETRY. sequents the same in both, the remaining terms will be in proportion. PROPOSITION XI. - THEOREM. 147. If any number of magnitudes are proportional, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let A: B:: C: D E: F; then will A: B:: A + C + E: B + D + F. For, from the given proportion, we have AXDBX C, and AXFBX E. By adding A X B to the sum of the corresponding sides of these equations, we have A X B +A X D +A X F= A X B+B X C+B X E. Therefore, A X (B + D + F) =B X (A + C + E). Hence, by Prop. II., A: B:: A+ C+E: B +D +F. PROPOSITION XII.- THEOREM. 148. If four magnitudes are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. Let A: B:: C: D; then will A + B: A - B:: C + D: C -D. For, from the given proportion, by Prop. VII., we have A + B: A: C + D: C; and from the given proportion, by Prop. VIII., we have A-B:A::C —D:C. Hence, from these two proportions, by Prop. X. Cor. 2, we have A + B: A - B:: C + D: C- D. BOOK II. 6 58 PROPOSITION XIII. - THEOREM. 149. If there be two sets of proportional magnitudes, the products of the corresponding terms will be proportionals. Let A: B: C: D, and E F: G:II; then will A xE:B xF::C xG:D xIH. For, from the first of the giveii proportions, by Prop. I., we have A xD = B xC; and from the second of the given proportions, by Prop. I., we have E xIH= F xG. Multiplying together the corresponding members of these equations, we have A xD xE xH=~B xC xF xG. Hence, by Prop. II., A xE: B xF:: CxG: D xH. PROPOSITION XIV. - THEOREM. 150. If three magnitudes are proportionals, the first wvill be to the third as the square of the first is to the square of the second. Let A: B: B: C; then will A: C: A2: B2. For, from the given proportion, by Prop. III., we have AxC0= B. Multiplying each side of this equation by A. gives A2 X C =Ax B 2. Hence, by Prop. II., A: C A: 2: B2. 54 54 ~~ELEMENTS OF GEOMETRY. PROPOSITION XV. - THEOREM. 151. If four magnitudes are propor'tionals, their like powers and roots will also be proportional. Let A: B:: C: D; then will A':B'::C: D', and AA-: BA:CA:DA. For, from the given proportion, we have A C B D Raising both members of this equation to the nth power, we have and extracting the nth root of each member, we have AkCA Hence, by Prop. II., the last two equations give An: B":: C":D7 and An: BA:: CA: DA. BOOK III. THE CIRCLE, AND THE MEASURE OF ANGLES. DEFINITIONS. 152. A CIRCLE is a plane figure bounded by a curved line, all the I points of whicl are equally distant from a point witlin called the center; as the figure A D B E. D E 153. The CIRCUMFERENCE or PERIPHERY of a circle is its entire bounding line; or it is a curved line, all points of which are equally distant from a point within called the center. 154. A RADIUS of a circle is any straight line drawn from the center to the circumference; as the line CA, CD, or CB. 155. A DIAMETER of a circle is any straight line drawil through tle center, and terminating in both directions in the circumference; as the line A B. All tle radii of a circle are equal; all the diameters are also equal, and each is double the radius. Th 156. An ARC of a circle is any part of the circumference; as the part AD, AE, or EGF. 157. Thle CHORD of an arc is the straight line joining its extremities; tlhus E F is tlle clord of tlhe arc EGF. A B. uE \-F G 56 ELEMENTS OF GEOMETRY. 158. The SEGMENT of a circle is the part of a circle included between an1 arc and its clord; as the surface included between the arc A E G F and the chord E F. 159. The SECTOR of a circle is thle part of a circle included between an arc, and the two radii drawn to tle arc; as the surface included between the two radii CA, CD. 160. A SECANT to a circle is a straight line which meets the circumference in two points, and lies partly within and partly without the circle; as the line A B. D Lk- \ B F G extremities of the the are AD, and C M D 161. A TANGENT to a circle is a straight line which, how far so ever produced, meets the circumferellce in but one point; as the linie CD. The poinlt of meetillg is called the POINT OF CONTACT; as the point 1M. 162. Two circumferences TOUCH eaclh other, whlen thley hlave a poillt of contact without cutting onie an- A B other; thus two circumferences touch each other at the point A, and two at the point B. C 16G3. A STRAIGHT LINE is INSCRIBED in a circle when its extremnities are in the circumference; as the line A B, or B C. 164. An INSCRIBED ANGLE iS one which has its vertex in the circumferencec, and is formed by two chords; as the allc r B C. BOOK III. 57 165. All INSCRIBED POLYGON is one C which las the vertices of all its angles ill tlhe circumference of tlle circle; as tlhe triangle AB C. AB 166. The circle is then said to be CIRCUMSCRIBED about the polygon. D 167. A POLYGON is CIRCUMSCRIBED E about a circle when all its sides are tangents to the circumference; as the polygon A B C D E F. F B A 168. The circle is then said to be INSCRIBED in tlhe polygon. PROPOSITION I. -THEOREM. 169. Every diameter divides the circle and its circumference each into two equal parts. Let AEBF be a circle, and A B F a diameter; then the two parts AEB, AFB are equal. For, if the figure A E B be applied A B to A F B, their cormmon base AB retainillg its position, the curve line A E B must fall exactly on the curve licn AFB; otherwise there would be E points in the one or the other unequally distant from the centre, which is contrary to the definition of the circle (Art. 152). Hence a diameter divides the circle and its circumference into two equal parts. 170. Cor. 1. Conversely, a straight line dividing the circle into two equal parts is a diameter. 58 ELEMENTS OF GEOMETRY. For, let the line AB divide the circle AEBCF into two equal parts; then, if the center is not in A B, let A C be drawn through it, which is c therefore a diameter, and conse- A """, — B quently divides the circle into two equal parts; hence the surface A F C is equal to the surface A F C B, a part E to the whole, which is impossible. 171. Cor. 2. The arc of a circle, whose chord is a diameter, is a semi-circumference, and the included segment is a semicircle. PROPOSITION II. - THEOREM. 172. A straight line cannot meet the circumference of a circle in more than two points. For, if a straight line could meet the circumference AB D, in three points, A, B, D, join each of these points with the center, C; then, since the straight lines C A, CB, A C D are radii, they are equal (Art. 155); hence, three equal straight B lines can be drawn from the same point to the same straight line, which is impossible (Prop. XIV. Cor. 2, Bk. I.). PROPOSITION III.- THEOREM. 173. In the same circle, or in equal circles, equal arcs are subtended by equal chords; and, conversely, equal chords subtend equal arcs. Let A D B and E G F be two equal circles, and let the arc AD be equal to EG; then will the chord AD be equal to the chord E G. BOOK III. 59 For, since the D G diameters A B, EF are equal, A the semicircle AtB E AD B may be applied to the semicircle E G F; and tile curve line AD B will coincide with the curve line EGF (Prop. I.). But, by hypothesis, the arc AD is equal to the arc E G; hence tile point D will fall on G; hence the chord A D is equal to the chord EG (Art. 34, Ax. 11). Conversely, if the chord A D is equal to the chord E G, the arcs A D, E G will be equal. For, if the radii C D, O G are drawn, the triangles A C D, E 0 G, having the three sides of tlle one equal to the three sides of the other, eacll to each, are themselves equal (Prop. XVIII. Bk. I.); therefore the angle A C D is equal to the angle E O G (Prop. XVIII. Seh., Bk. I.). If now the semicircle A D B be applied to its equal E G F, with the radius A C on its equal E 0, since the angles A C D, E 0 G are equal, the radius C D will fall on 0 G, and the point D on G. Therefore tle arcs A D and E G coincide with each other; hence they must be equal (Art. 34, Ax. 14). PROPOSITION IV.- THEOREM. 174. In the same circle, or in equal circles, a greater arc is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc. In the circle of which C is the centre, let tlle arc A B be greater than tlhe arc A D; then will the clord A B be greater tlan the chord AD. Draw tlle radii C.A, CD, and C B. Tlle two sides AC, 60 ELEMENTS OF GEOMETRY. C B in the triangle A C B are equal to tile two A C, C D in tile triangle j BAC D, anid tlhe angle AC B is greater thanl the angle A C D; tlherefore the A third side A B is greater than the tl.ird side A D (Prop. XVI. Bk. I.); hlc;ce tle cllord wlhich subtenids the bgreater arc is tll greater. Conversely, if the chord A B be greater than the chord A D, thl arc A B will be greater than the arc A D. For the triangles A C B, A C D have two sides, A C, C B, in the one, equal to two sides, AC, CD, in the other, while tlhe side A B is greater than the side A D; therefore the angle A C B is greater than the angle A C D (Prop. XVII. Bk. I.); hence the arc A B is greater than the arc A D. 175. Scholium. The arcs here treated of are each less than the semi-circumference. If they were greater, the contrary would be true; in which case, as the arcs increased, the chords would diminish, anld conversely. PROPOSITION V. — THEOREM. 176. In the same circle, or in equal circles, radii which miLake equal anbgwles at the centre intercept equal arcs on the circumference; and, conversely, if the intercepted arcs are equal, the angles made by the radii are also equal. Let A C B and D C E be equal angles imade by radii at the centre of equal circles; then will the I iltercepted arcs A B A D and DE be also equal. First. Since the angles A C B, D C E are equal, the one may be applied to the other; and since tlmeir sides, BOOK III. 61 being radii of equal circles, are equal, the point A will coincide with D, and the point B with E. Therefore the arc A B must also coincide with the arc DE, or there would be points in the one or the other unequally distant from the center, which is impossible; hence the arc A B is equal to the arc D E. Second. If the arcs A B and D E are equal, tle. angles A C B and D C E will be equal. For, if these angles are not equal, let ACB be the greater, and let A C F be taken equal to D C E. From what has been shown, we shall have the arc A F equal to the arc D E. But, by hypothesis, A B is equal to D E; hence A F must be equal to A B, the part to the whole, which is impossible; hence the angle A C B is equal to the angle D C E. PROPOSITION VI.- THEOREM. 177. The radius which is perpendicular to a chord bisects the chord, and also the arc subtended by the chord. Let tlle radius C E be perpendicular to the chord A B; then will C E bisect the chord at D, and the arc AB at E. c Draw the radii CA and C B. Then CA and C B, with respect to -. ) t'le perpendicular C E, are equal A 'B oblique lines drawn to the chord AB; therefore tleir extremities are at equal distances from the perpendicular (Prop. XIV. Bk. I.); hence A D and D B are equal. Again, since the triangle A C B has the sides A C and C B equal, it is isosceles; and the line CE bisects the base A B at right angles; therefore C E bisects also the angle A C B (Prop. VII. Cor. 2, Bk. I.). Since the angles A C D, D C B are equal, the arcs A E, E B are equal 6 62 ELEMENTS OF GEOMETRY. (Prop. V.); hence the radius C E, which is perpendicular to the chord A B, bisects the arc A B subtended by the chord. 178. Cor. 1. Any straight line which joins the centre of the circle and the middle of the chord, or the middle of the arc, must be perpendicular to the chord. For the perpendicular from the centre C passes through the middle, D, of the chord, and the middle, E, of the arc subtended by the chord. Now, aly two of these three points in the straight line C E are sufficient to determine its position. 179. Cor. 2. A perpendicular at the middle of a chord passes through the center of the circle, and through the middle of the arc subtended by the chord, bisecting at the centre the angle which the arc subtends. PROPOSITION VII.- THEOREM. 180. Through three given points, not in the same straight line, one circumference can be made to pass, and but one. Let A, B, and C be any three A points not in the same straight line; one circumference can be made to pass through them, and but one. E Join A B and BC; and bisect.. these straight lines by the perpendiculars D E and F E. Join D F; then, B the angles B D E, B F E, being each a right angle, are together equal to two right angles; therefore the angles E D F, E F D are together less than two right angles; hence D E, F E, produced, must meet in some point E (Prop. XXIII. Bk. I.). Now, since the point E lies in the perpendicular D E, it is equally distant from the two points A and B (Prop. XV. Bk. I.); and since the same point E lies in the per BOOK III. 63 pendicular FE, it is also equally distant from the two points B and C; therefore the three distances, E A, E B, E C, are equal; hence a circumference can be described from the center E passing through the three points A, B, C. Again, the center, lying in the perpendicular DE bisecting the chord AB, and at the same time in the perpendicular FE bisecting the chord B C (Prop. VI. Cor. 2), must be at the point of their meeting, E. Therefore, since there can be but one center, but one circumference can be made to pass through three given points. 181. Cor. Two circumferences can intersect in only two points; for, if they have three points in common, they must have the same center, and must coincide. PROPOSITION VIII. -THEOREM. 182. Equal chords are equally distant from the center; and, conversely, chords which are equally distant from the centre are equal. Let A B and D E be equal chords, and C the center of the circle; and D draw C F perpendicular to A B, and C G perpendicular to DE; then these perpendiculars, which measure A the distance of the chords from the \ F center, are equal. Join C A and C D. Then, in the right-angled triangle C A F, C D G, the hypothenuses C A, C D are equal; and the side A F, the half of A B, is equal to the side D G, the half of D E; therefore the triangles are equal, and C F is equal to C G (Prop. XIX. Bk. I.); hence the two equal chords A B, D E are equally distant from the center. Conversely, if the distances C F and C G are equal, the chords A B and D E are equal. For, in the right-angled triangles A C F, D G, the hypothenuses C A, C D are equal; and the side C F is 64 ELEMENTS OF GEOMETRY. equal to the side C G; therefore the triangles are equal, and A F is equal to D G; hence A B, the double of A F, is equal to D E, the double of D G (Art. 34, Ax. 6). PROPOSITION IX.- THEOREM. 183. Of two unequal chords, the less is the farther from the center. Of the two chords DE and A H, E let A H be the greater; then will D GD E be the farther from the center C. c Since the chord AH is greater A - H than the chord D E, the arc A H is \ " greater than the arc D E (Prop. IV.). Cut off from the arc A H a part, A B, equal D E; draw CF perpendicular to this chord, C I perpendicular to A H, and C G perpendicular to D E. C F is greater than C 0 (Art. 34, Ax. 8), and C 0 than C I (Prop. XIV. Bk. I.); therefore C F is greater tlhan CI. But C F is equal to C G, since the chords A B, D E are equal (Prop. VIII.); therefore, C G is greater than C I; hence, of two unequal chords, the less is the farther from the center. PROPOSITION X.- THEOREM. 184. A straight line perpendicular to a radius at its termination in the circumference, is a tangent to the circle. Let the straight line BD be per- A E pendicular to the radius C A at its B - D tormination A; then will it be a tangent to the circle. Draw from the centre C to BD any other straight line, as C E. Then, since C A is perpendicular to B D, it is shorter than the oblique BOOK III. 65 line C E (Prop. XIV. Bk. I.); hence tile point E is without the circle. The same may be shown of any other point in the line BD, except tlhe point A; therefore B D meets, the circumference at A, and, being produced, does not cut it; hence B D is a tangent (Art. 161). PROPOSITION XI.- THEOREM. 185. If a line is a tangent to a circumference, the radius drawn to the point of contact with it is perpendicular to the tang ent. Let B D be a tangent to the cir- A E cunference, at the point A; then B D will the radius C A be perpendicular to BD. For every point in B D, except A, being without the circumference (Prop. X.), any line C E drawn from tlle center C to BD, at any point other tllhan A, must terminate at E, without the circumference; therefore the radius C A is the shortest line that can be drawn from the center to B D; hence C A is perpendicular to the tangent B D (Prop. XIV. Cor. 1, Bk. I.). 186. Cor. Only one tangent can be drawn through the same point in a circumference; for two lines cannot both be perpendicular to a radius at the same point. PROPOSITION XII.- THEOREM. 1S7. Two parallel straight lines intercept equal arcs of the circumference. First. When the two parallels are secants, as AB, DE. Draw the radius C H perpendicular to A B; and it will also be perpendicular: to D E (Prop. XXII. Cor., Bk. I.); 6* 66 ELEMENTS OF GEOMETRY. therefore the point H will be at the same time the middle of the arc D A H B and of the arc D H E (Prop. VI.); therefore, the arc AH is equal to the arc H B, and the arc D H is equal to the arc H E; hence AH c diminished by D H is equal to H B diminished by H E; that is, the intercepted arcs A D, B E are equal. Second. When of the two parallels, one, as AB, is a secant, and the other, as D E, is a tangent. Draw the radius C H to the point H of contact H. This radius will be D E perpendicular to the tangent DE A- -- -B (Prop. X.), and also to its parallel AB (Prop. XXII. Cor., Bk. I.). c But, since C H is perpendicular to the chord AB, the point H is the middle of the arc A H B; hence the I -- L arcs AH, HB, included between the parallels AB, D E, are equal. Third. When the two parallels are tangents, as D E, IL. Draw the secant A B parallel to either of the tangents, and it will be parallel to the other (Prop. XXIV. Bk. I.); then, from what has been just shown, the arc AH is equal to the arc H B, and also the arc A G is equal to the arc G B; hence the whole arc HA G is equal to the whole arc H B G. It is further evident, since the two arcs H A G, H B G are equal, and together make up the whole circumference, that each of them is a semi-circumference. 188. Cor. Two parallel tangents meet the circumference at the extremities of the same diameter. BOOK III. 67 PROPOSITION XIII. - THEOREM. 189. If two circumferences touch each other externally or internally, their centers and the point of contact are in the same straight line. Let the two circumferences, B whose centers are C and D, touch each other externally in the poilt A; the points C, D, ( c D and A will be all il the same straiglt line. Draw from the point of contact A tle common tangent A B. Then the radius C A of the one circle, and the radius D A of the other, are each perpendicular to A B (Prop. XI.); but there cal be but one straight line drawn tlrougl the point A perpendicular to A B (Prop. XIII. Bk. I.); therefore the points C, D, and A are in one perpendicular; hence they are in one and tlle same straight line. Also, let the two circumferences B touch each other internally in A; then tleir centers, C and D, and the point of contact, A, will be in the same straight line. A C Draw tlle common tangent AB. Then a straiglt line perpendicular to A B, at the point A, on being sufficiently produced, must pass through the two centers C and D (Prop. XI.); but from the same point there can be but one perpendicular; therefore the points C, D, and A are in tlat perpendicular; hence they are in the same strairght line. 190. Cor. 1. When two circumferences toucl each other'externally, the distance between their centers is equal to the sum of their radii. 68 ELEMENTS OF GEOMETRY. 191. Cor. 2. And when two circumferences touch each other internally, the distance between their centers is equal to the difference of their radii. PROPOSITION XIV.- THEOREr. 192. If two circumferences cut each other, the straight line passing through their centers will bisect at right angles the chord which joins the points of intersection. Let two circumferences cut each other at the points A and B; then the straight line passing through the centres C and D will bisect at right angles the chord A B common to the two circles. For, if a perpendicular be erected at the middle of this chord, it will pass through each of tlhe two centers C and D (Prop. VI. Cor. 1). But no more than one straight line can be drawn through two points; hence the straight line C D, passing through the centers, must bisect at right angles the common chord A B. 193. Cor. The straight line joining the points of intersection of two circumferences is perpendicular to the straight line which passes through their centers. PROPOSITION XV.- THEOREI. 194. If two circumferences cut each other, the distance between their centers will be less than the sum of their radii, and greater than their difference. BOOK III. 69 Let two circumferences whose centers are C and D cut each other in the point A, and draw the ( radii CA and DA. Then, in order that the intersection may take place, the triangle C A D must be possible. And in this triangle the side C D must be less than the sum of A C and A D (Prop. IX. Bk. I.); also C D must be greater than the difference between D A and C A (Prop. IX, Cor., Bk. I.). PROPOSITION XVI. -THEOREM. 195. In the same ctrcle, or in equal circles, if two angles at the center are to each other as two whole numbers, the intercepted arcs will be to each other as the same numbers. Let us suppose, for example, that the angles A C B, D C E, at the center of equal circles, are to each other as 7 to 4; or, which amounts to the same thing, that the angle M, which will serve as a common measure, is conC C A B tained seven times in the angle A C B, and four times in the angle D C E. The seven partial angles A C m, m C n, n C p, &c. into which A C B is divided, being each equal to any of the four partial angles into which DCE is divided, each of the partial arcs Am, Inn, np, &c. will 70 ELEMENTS OF GEOMETRY. be also equal to each of the partial arcs Dx, xy, &c. (Prop. V.); therefore the whole arc AB will be to the whole arc D E as 7 to 4. But the same reasoning would apply, if in place of 7 and 4 any numbers whatever were employed; hence, if the ratio of the angles A C B, D C E can be expressed in whole numbers, the arcs AB, DE will be to each other as the angles A C B, D CE. 196. Cor. Conversely, if the arcs A B, D E are to each other as two whole numbers, the angles A C B, D C E will be to each other as the same whole numbers, and we shall have ACB:DCE:: AB: D E. For, the partial arcs A m, m n, &c. and D x, xy, &c. being equal, the partial angles A C m, m C n, &c. and DC x, x Cy, &c. will also be equal. PROPOSITION XVII. - THEOREM. 197. In the same circle, or in equal circles, any two angles at the center are to each other as the arcs intercepted between their sides. Let AC B be the greater, and ACD the less angle; then will the angle A B C be to the angle ACD as the arc A/ A B is to the arc AI AD. Conceive the -less angle to be placed on the greater; then, if the proposition be not true, the angle A C B will be to the angle A C D as the arc A B is to an arc greater or less than A D. Suppose this arc to be greater, and let it be represented by A O; we shall have the angle A C B: angle A C D:: arc A B: arc A 0. Conceive, now, the arc A B to be divided into equal parts, each of which is less BOOK III. 71 than D 0; there will be at least one point of division between D and 0; let I be that point; and join CI. The arcs A B, A I will be to each other as two whole numbers, and, by the preceding proposition, we shall have the angle ACB: angle A C I: arc AB: arc AI. Comparing these two proportions with each other, and observing that the antecedents are the same, we infer that the consequents are proportional (Prop. X. Cor. 2, Bk. II.); hence the angle ACD: angle ACI: arc AO: arc AI. But the arc A 0 is greater than the arc A I; therefore, if this proportion is true, the angle A C D must be greater than the angle A C I. But it is less; hence the angle A C B cannot be to the angle A C D as the arc A B is to an arc greater than AD. By a process of reasoning entirely similar, it may be shown that the fourth term of the proportion cannot be less than A D; therefore it must be A D; hence we have, Angle AC B: angle ACD:: arc AB: arc AD. 198. Scholium 1. Since the angle at the center of a circle, and the arc intercepted by its sides, have such a connection, that, if the one be increased or diminished in any ratio, the other will be increased or diminished in the same ratio, we are authorized to take the one of these magnitudes as the measure of the other. Henceforth we shall assume the arc AB as the measure of the angle A C B. It is to be observed, in the comparison of angles with each other, that the arcs which serve to measure them must be described with equal radii. 199. Scholium 2. Sectors taken in the same circle, or in equal circles, are to each other as their arcs; for sectors are equal when their angles are so, and therefore r. in all respects proportional to their angles. 72 ELEMENTS OF GEOMETRY. PROPOSITION XVIII. -THEOREM. 200. An inscribed angle is mneasured by half the are included between its sides. A Let B A D be an inscribed angle, whose sides include the arc BD; then the angle BAD is measured by half of the arc B D. First. Suppose the center of the circle C to lie within the angle BAD. Draw the diameter AE, lB D and the radii CB, CD. E The angle B C E, being exterior to the triangle AB C, is equal to the sum of the two interior angles C A B, ABC (Prop. XXVII. Bk. I.). But the triangle BAC being isosceles, the angle C A B is equal to A B C; hence, the angle B C E is double BA C. Since B C E lies at the center, it is measured by the arc B E (Prop. XVII. Sch. 1); hence B A C will be measured by half of B E. For a like reason, the angle C A D will be measured by the half of E D; hence B A C and C A D together, or B A D, will be measured by the half of B E and E D, or half B D. Second. Suppose that the center C lies without the angle B A D. Then, drawing the diameter A E, the angle B A E will be measured by the half of B E; and the angle D A E is measured by the half of D E; hence, their difference, BAD, will be measured by the half of BE E_ minus the lalf of ED, or by the D E half of B D. Hence every inscribed angle is measured by the half of the. arc included between its sides. BOOK III. 73 D 201. Cor. 1. All the angles, E B A C B D C, inscribed in the same segment, are equal; because they are all measured by the half B of the same arc, B 0 C. 0 202. Cor. 2. Every angle, BAD, inscribed in a semicircle, is a right angle; because it is measured by half the semi-circumference, BOD; B D that is, by the fourth part of the whole circumference. 203. Cor. 3. Every angle, BAC, o inscribed in a segment greater than A a semicircle, is an acute angle; for it is measured by the half of the arc B 0 C, less than a semi-circumference. And every angle, B 0 C, inscribed in a segment less than a semicircle, \ is an obtuse angle; for it is measured by half of the arc BAC, ~ greater than a semi-circumference. 204. Cor. 4. The opposite angles, A and D, of an inscribed A quadrilateral, A B D C, are together equal to two right angles; for the angle B AC is measured by half the arc BD C, and the angle BDC c is measured by half the arc B A C; D hence the two angles B A C, B D C, taken together, are measured by half the circumference; hence their sum is equal to two right angles. 7 ELEMENTS OF GEOMETRY. PROPOSITION XIX.- THEOREM. 205. The angle formed by the intersection of two chords is measured by half the sum of the two intercepted arcs. Let the two chords A B, C D inter- A C sect each other at the point E; then will the angle D EB, or its equal, A E C, be measured by half the sum of the two arcs D B and A C. Draw A F parallel to D C; then will the arc FD be equal to the arc AC (Prop. XII.), and the an- gle FAB equal to the angle DEB (Prop. XXII. Bk. I.). But the angle FA B is measured by half the arc F D B (Prop. XVIII.); that is, by half the arc D B, plus half the arc FD. Hence, since FD is equal to A C, the angle DEB, or its equal angle A E C, is measured by half the sum of the intercepted arcs D B and AC PROPOSITION XX.- THEOREM. 206. The angle formed by a tangent and a chord is measured by half the intercepted arc. Let the tangent B E form, with D the chord A C, the angle B A C; then B A C is measured by half the arc AMC. From A, the point of contact, M draw the diameter AD. The angle BAD is a right angle (Prop. X.), and is measured by half of B E the semi-circumference A M D (Prop. XVIII.); and the angle D A C is measured by half the arc DC; hence the sum of the angles BAD, D A C, or B A C, is measured by the half of AM D, plus the half of D C; or by half the whole arc A M D C. In like manner, it may be shown that the angle C A E is measured by half the intercepted arc A C. BOOK III. 75 PROPOSITION XXI.- THEOREM. 207. Thle angle formed by two secants is measured by half the difference of the two intercepted arcs. Let A B, AC be two secants A forming the angle B AC; then will that angle be measured by half the difference of the two arcs BEC and D F. Draw D E parallel to AC; then will the arc EC be equal to the arc D F (Prop. XII.); and the B angle B D E be equal to the an- gle B A C (Prop. XXII. Bk. I.). But the angle B D E is measured by half the arc B E (Prop. XVIII.); hence the equal angle B A C is also measured by half the arc B E; that is, by half the difference of the arcs B E C and E C, or, since E C is equal to D F, by half the difference of the intercepted arcs B E C and D F. PROPOSITION XXII. -THEOREM. 208. The angle formed by a secant and a tangent is meas, ured by half the difference of the two intercepted ares. Let the secant AB form, with A the tangent A C, the angle B A; DC then BAC is measured by half the difference of the two arcs BEF and FD. Draw D E parallel to A C; then / will the arc E F be equal to the arc E DF (Prop. XII.), and the angle B BDE be equal to the angle BA C. But the angle B D E is measured by half of the arc B E (Prop. XVIII.); hence the equal angle B A is also measured by half the arc B E; that is, by half the difference of the arcs BEF and EF, or, since EF is equal to DF, by half the difference of the intercepted arcs BEF and DF. BOOK IV. PROPORTIONS, AREAS, AND SIMILARITY OF FIGURES. DEFINITIONS. 209. The AREA of a figure is its quantity of surface, and is expressed by the number of times which the surface contains some other area assumed as a unit of measure. Figures have equal areas, when they contain the same unit of measure an equal number of times. 210. SIMILAR FIGURES are such as have the angles of the one equal to those of the other, each to each, and the sides containing the equal angles proportional. 211. EQUIVALENT FIGURES are such as have equal areas. Figures may be equivalent which are not similar. Thus a circle may be equivalent to a square, and a triangle to a rectangle. 212. EQUAL FIGURES are such as, when applied the one to the other, coincide throughout (Art. 34, Ax. 14). Thus circles having equal radii are equal; and triangles having the three sides of the one equal to the three sides of the other, each to each, are also equal. Equal figures are always similar; but similar figures may be very unequal. 213. In different circles, SIMILAR ARCS, SEGMIENTS, or SECTORS are such as correspond to equal angles at the centres of the circles. BOOK IV. 77 Thus, if the angles A E and E. are equal, the arc A B C will be similar to the arc FG; the segment BDC to the segment FHG, and the sector ABC to the C F G sector E F G. D H 214. The ALTITUDE OF A TRIANGLE A is the perpendicular, which measures the distance of any one of its vertices from the opposite side taken as a base; as the perpendicular AD let fall on tlhe base B C in the triangle B C ABC. 215. The ALTITUDE OF A PARALLEL- E OGRAM is the perpendicular which D C measures the distance between its opposite sides taken as bases; as the / perpendicular E F measuring the dis- A F B tance between the opposite sides, A B, D C, of the parallelogram A B C D. 216. The ALTITUDE OF A TRAPEZOID E is the perpendicular distalce between D - \C its parallel sides; as the distance / measured by the perpendicular E F between the parallel sides, AB, D C, A F B of the trapezoid A B C D. PROPOSITION I. - THEOREM. 217. Parallelograms which have equal bases and equal altitudes are equivalent. Let ABCD, ABEF be two D C F E parallelograms having equal bases ' and equal altitudes; then these parallelograms are equivalent. Let the base of the one paral- A B 7* 78 ELEMENTS OF GEOMETRY. lelogram be placed on that of the other, so that A B shall be the common base. Now, since the two parallelograms are of the same altitude, their upper bases, D C, F E, will be in the same straight line, D C E F, parallel to A B. From the nature of parallelograms D C is equal to A B, and F E is equal to A B (Prop. XXXI. Bk. I.); therefore D C is equal to F E (Art. 34, Ax. 1); hence, if D C and F E be taken away from the same line, D E, the remainders C E and D F will be equal (Art. 34, Ax. 3). But AD is equal to BC and AF to BE (Prop. XXXI. Bk. I.); therefore the triangles D A F, C B E, are mutually equilateral, and consequently equal (Prop. XVIII. Bk. I.). If from the quadrilateral A B E D, we take away the triangle A D F, there will remain the parallelogram ABEF; and if from the same quadrilateral A B E D, we take away the triangle C B E, there will remain the parallelogram A B C D. Hence the parallelograms A B C D, A B E F, which have equal bases and equal altitude, are equivalent. 218. Cor. Any parallelogram is equivalent to a rectangle having the same base and altitude. PROPOSITION II.- THEOREM. 219. If a triangle and a parallelogram have the same base and altitude, the triangle is equivalent to half the parallelogram. Let A B E be a triangle, and D C F E A B CD a parallelogram having the same base, A B, and the same altitude; then will the triangle be equivalent to half A B the parallelogram. Draw AF, FE so as to form the parallelogram ABEF. Then the parallelograms A B C D, A B E F, having the same base and altitude, are equivalent (Prop. I.). But BOOK IV. 79 the triangle A B E is half the parallelogram A B E F (Prop. XXXI. Cor. 1, Bk. I.); hence the triangle A B E is equivalent to half the parallelogram A B C D (Art. 34, Ax. 7). 220. Cor. 1. Any triangle is equivalent to half a rectangle having the same base and altitude, or to a rectangle either having the same base and half of the same altitude, or having the same altitude and half of the same base. 221. Cor. 2. All triangles which have equal bases and altitudes are equivalent. PROPOSITION II.-THEOREM. 222. Two rectangles having equal altitudes are to each other as their bases. LetABCD, AEFD be D F C two rectangles having thle common altitude AD; they are to each other as their.. bases A B, A E. A E B First. Sulppose that the bases A B, A E are commensurable, and are to each other, for example, as the numbers 7 and 4. If A B is divided into seven equal parts, A E will contain four of those parts. At each point of division draw lines perpendicular to the base; seven rectangles will thus be formed, all equal to each other, since they have equal bases and tlle same altitude (Prop. I.). The rectangle A B C D will contain seven partial rectangles,.while A E FD will contain four; hence the rectangle A B C D is to AEFD as 7 is to 4, or as AB is to AE. The same reasoning may be applied, whatever be the numbers expressing the ratio of the bases; hence, whatever be that ratio, when its terms are commensurable, we shall have ABCD: AEFD:: AB: AE. ELEMENTS OF GEOMETRY. Second. Suppose that the bases AB, D F K C A E are incommenlsurable; we slhall still have ABCD: AEFD:: AB: AE. For, if this proportion be not true, the A E I o B first three terms remaining tlie same, the fourth term must be either greater or less than A E. Suppose it to be greater, and that we have ABCD: AEFD:: AB: AO. Conceive A B divided into equal parts, each of which is less than EO. There will be at least one point of division, I, between E and 0. Through this'point, I, draw the perpendicular I K; then the bases A B, A I will be commensurable, and we shall have A B C D: AIKD: AB: AI. But, by hypothesis, we have AB C D: A E F D: A: AO. In these two proportions the antecedents are equal; hence the consequents are proportional (Prop. X. Cor. 2, Bk. 1.), and we have AIKD: AEFD:: AI: AO. But A 0 is greater than A I; tlerefore, if this proportion is correct, the rectangle A E F D must be greater tllan the rectangle A I K D (Art. 125); on the contrary, however, it is less (Art. 34, Ax. 8); therefore the proportion is impossible. Hence, A B C D cannot be to A E F D as A B is to a line greater than A E. In the same manner, it may be shown that the fourth term of the proportion cannot be less than A E; tlerefore it must be equal to A.E. Hence, any two rectangles ABC D, AEFD, having equal altitudes, are to each other as their bases AB, AE. BOOK IV. 81 PROPOSITION IV.-THEOREM. 223. Any two rectangles are to eacn other as the products of their bases multiplied by their altitudes. Let ABCD, AEGF be two H D C rectangles; then will ABCD be to A E G F as A B multiplied by AD is to AE multiplied by AF. E -. B Having placed the two rectangles so that the angles at A are vertiG F cal, produce the sides GE, CD G till they meet il H. The two rectangles A B C D, AEHD, having the same altitude, AD, are to each other as their bases, A B, A E. 11 like manner the two rectalngles A E H D, A E G F, having the same altitude, A E, are to each other as their bases, AD, AF. Hence we have the two proportions, ABCD AEHD:: AB: AE, AEIHD: AEGF:: AD: AF. Multiplying the corresponding terms of these proportions together (Prop. XIII. Bk. II.), and omitting the factor A E H D, which is common to both the antecedent and the consequent (Prop. IX. Bk. II.), we shall have ABCD:AEGF:: ABX AD: AEX AF. 224. Scholium. Hence, we may assume as the measure of a rectangle, tie product of its base by its altitude, provided we understand by this product the product of two numbers, one of which represents tlhe number of linear units contained in the base, the other the number of linear units contained in the altitude. The product of two lines is often used to designate their rectangle; but the term square is used to designate the product of a number multiplied by itself. 82 ELEMENTS OF' GEOMETRY. PROPOSITION V. THEOREM. 225. The area of any parallelogram is equal to the product of its base by its altitude. Let A B C D be any parallelogram, F D E C A B its base, and B E its altitude; tlen will its area be equal to the pro- duct of A B by BE. Draw BE and A F perpendicular A B to A B, and produce C D to F. Then the parallelogram A B C D is equivalent to the rectangle A B E F, which has the same base, AB, and the same altitude, B E (Prop. I. Cor.). But the rectangle A B E F is measured by A B X BE (Prop. IV. Sch.); therefore A B X BE is equal to the area of the parallelogram A B C D. 226. Cor. Parallelograms having equal bases are to each other as their altitudes, and parallelograms having equal altitudes are to each other as their bases; and, in general, parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION VI. - THEOREM. 227. T7ie area of any triangle is equal to the product of its base by half its altitude Let A B C be any triangle, B C its base, A E and A D its altitude; then its area will be equal to the product of B C by half of A D. Draw AE and CE so as to form the parallelogram A B C E; then the triangle B I) A B C is half the parallelogram A B C E, which has the same base B C, and the same altitude A D (Prop. II.); but the area of the parallelogram is equal to B C X AD (Prop. V.); hence the area of the triangle must be. BC X AD, or BC X AD. BOOK IV. 830 228. Cor. Triangles of equal altitudes are to each other as tleir bases, and triangles of equal bases are to each other as their altitudes; and, in general, triangles are to eaclh other as the products of tleir bases and altitudes. PROPOSITION VII. -THEOREnM. 229. The area of any trapezoid is equal to the product of its altitude by half the sum of its parallel sides. Let A B C D be a trapezoid, E F D E C K its altitude, and AB, CD its parallel sides; then its area will be................ equal to the product of E F by half the sum of A B and C D. Through I, the middle point of A F L B the side B C, draw K L parallel to A D; and produce D C till it meet K L. In the triangles I B L, I C K, we have the sides I B, I C equal, by constructionl; the vertical angles L I B, C I K are equal (Prop. IV. Bk. I.); and, since C K and B L are parallel, the alternate angles I B L, IC K are also equal (Prop. XXII. Bk. I.); therefore the triangles I B L, I C K are equal (Prop. VI. Bk. I.); hence the trapezoid A B C D is equivalent to the parallelogram A D K L, and is measured by the product of E F by A L (Prop. V.). But we have AL equal D K; and since the triangles I B L and K C I are equal, the sides B L and C K are equal; therefore the sum of A B and C D is equal to the sum of AL and D K, or twice AL. Hence AL is half the sum of the bases A B, C D; hence the area of the trapezoid A B, C D is equal to the product of the altitude E F by half the sum of the parallel sides AB, C D. Cor. If through I, the middle point of B C, the line IH be drawn parallel to the base AB, the point H will also be the middle point of AD. For, since the figure A H L ELEMENTS OF GEOMETRY. is a parallelogram, as is likewise D 1 I K, their opposite sides being parallel, we have A H equal to I L, and D H equal to IK. But since the triangles B I L, CII K are equal, we have I L equal to I K; lhelce A IH is equal to D H. Now, tile line H I is equal to AL, whicl has been shown to be equal to half the sum of A B and C D; therefore the area of the trapezoid is equal to tie product of E F by H I. Hence, the area of a trapezoid is equal to the product of its altitude by the line connecting the middle points of the sides which are not parallel. PROPOSITION VIII. -THEOREM. 230. If a straight line be divided into two parts, the square described on the whole line is equivalent to the sum of the squares described on the parts, together with twice the rectangle contained by the parts. Let AC be a straight line, divided E H D into two parts, AB, BC, at the point B; then the square described on AC is F G equivalent to the sum of the squares described on the parts A B, B C, together with twice the rectangle contained A B C byAB, BC; that is, AC = -AB2 + B-C2 + 2 AB X BC. On A C describe the square A C D E; take AF equal to A B; draw F G parallel to A C, and B H parallel to A E. The square A C D E is divided into four parts; the first, A B I F, is the square described on A B, since A F was taken equal to A B. Tie second, I G D H, is the square described upon B C; for, since A C is equal to A E, and A B is equal to AF, AC minus AB is equal to AE minus A F, which gives B C equal to E F. But I G is equal to B C, and D G to E F, since the lilnes are parallels; therefore I G D H is equal to tlle square described on B C. BOOK IV. 85 These two parts being taken from the whole square, there remain two rectangles B C G I, E F I H, each of which is measured by A B X B C; hence the square on the whole line AC is equivalent to the squares on the parts AB, BC, together with twice the rectangle of the parts. E H D 231. Cor. The square described on the whole line A C is equivalent to four times the square described on the half I A B. A B C 232. Scholium. This proposition is equivalent to the algebraical formula, (a + b)2= a + 2 a b +. PROPOSITION IX.-THEOREM. 233. The square described on the difference of two straight lines is equivalent to the sum of the squares described on the two lines, diminished by twice the rectangle contained by the lines. Let AB and B C be two lines, and L F G I A C their difference; tlien will the square described on AC be equiva- K E D H lent to the sum of the squares described on A B, B C, diminished by twice the rectangle AB, B C; that is, A C B (AB - B C)2 or C= AB2 BC - 2 A B X BC.. On A B describe the square A B I P; take A E equal to A C; draw C G parallel to B I, H K parallel to A B, and complete the square E F L K. Since AF is equal to AB, and AE to AC, EF is equal to BC, and LF to GI; therefore LG is equal to FI; hence the two rectangles C B I G, G LK D are each 8 ELEMENTS OF GEOMETRY. measured by A B X B C. Take these rectangles from the whole figure A B I L K E, which is equivalent to A B2 + B C2, and there will evidently remain the square ACDE; hence the square on AC is equivalent to the sum of the squares on A B, B C, diminished by twice the rectangle contained by A B, B C. 234. Scholium. This proposition is equivalent to the algebraical formula, (a - b)2 = a2 - 2 a b + b2. PROPOSITION X.-THEOREM. 235. The rectangle contained by the sum and difference of two straight lines is equivalent to the difference of the squares of these lines. F G I Let A B, B C be two lines; then. will the rectangle contained by the D - L sum and difference of A B, B C, be equivalent to the difference of the squares of AB, BC; that is, A C B K 2 ~ -2 (A B + B C) X (AB - B C) = AB2- BC2.C On A B describe the square A B I F, and on A C the square A C D E; produce C D to G; and produce A B until B K is equal to B C, and complete the rectangle AKLE. The base AK of the rectangle is the sum of the two lines A B, B C; and its altitude A E is the difference of the same lines; therefore the rectangle AKLE is that contained by the sum and the difference of the lines A B, B C. But this rectangle is composed of the two parts ABHE and BHLK; and the part BHLK is equal to the rectangle E D G F, since B H is equal to D E, and B K to E F. Hence the rectangle A K L E is equivalent to A B HE plus E DG F, which is equivalent to the dif BOOK IV. 87 ference between the square A B I F described on A B, and D H I G described on B C; hence (AB+ BC) X (AB-BC) = AB2 -B C2. 236. Scholium. This proposition is equivalent to the algebraical formula, (a + b) X (a - b) == a2 - b2 PROPOSITION XI. THEOREM. 237. The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides. Let A B C be a right-angled L triangle, having the right angle at A; then the square described H K on the hypothenuse B C will be \ - equivalent to the sum of the I squares on the sides BA, AC. c On B C describe the square B D B C G F, and on A B, A C the squares ABHL, ACIK; and through A draw AE parallel to BFor C G, and join AF, HC. F E G The angle A B F is composed of the angle A B C, together with the right angle C B F; tle angle C B H is composed of the same angle A B C together with the right angle A B H; therefore the angle A B F is equal to the angle H B C. But we have A B equal to B H, being sides of the same square; and B F equal to B C, for the same reason; therefore the triangles A BF, HB C have two sides and the included angle of the one equal to two sides and the included angle of the other; hence they are themselves equal (Prop. V. Bk. I.). But the triangle A B F is equivalent to half the rectangle B D E F, since they have the same base B F, and the 88 ELEMENTS OF GEOMETRY. same altitude BD (Prop. II. L Cor. 1). The triangle HBC is, in like manner, equivalent II K to half the square A B H L; for.. A. the angles BAC, BAL being I both right, AC and A L form one and the same straiglt line B C parallel to H B (Prop. II. Bk. I.); and consequently the triangle and the square have the same altitude A B (Prop. E G XXV. Bk. I.); and they also have the same base BH; hence the triangle is equivalent to half the square (Prop. II.). The triangle ABF has already been proved equal to the triangle HBC; lence the rectangle B D E F, which is double the triangle A B F,. must be equivalent to the square A B H L, which is double the triangle H B C. In the same manner it may be proved that the rectangle C D E G is equivalent to the square A C I K. But the two rectangles B D E F, C D E G, taken togetler, compose the square B C G F; therefore the square B C G F, described on the hypothenuse, is equivalent to the sum. of the squares A B I L, A C I K, described on the two other sides; that is, BC2 is equivalent to A B + A C2 238. Cor. 1. The square of either of the sides which form the right angle of a right-angled triangle is equivalent to the square of the hypothenuse diminished by the square of the other side; thus, A B is equivalent to B C - A C2. 239. Cor. 2. The square of the hypothenuse is to the square of either of the other sides, as the hypothenuse is to the part of the hypothenuse cut off, adjacent to that side, BOOK IV. 89 by the perpendicular let fall from the vertex of the right angle. For, on account of tile common altitude B F, tle square BC GF is to the rectangle BDEF as the base BC is to hle base B D (Prop. III.); now, the square A B H L has been proved to be equivalent to the rectangle BDEF; therefore we have, BC2: AB2::BC:BD. In like manner, we have, BC2: AC2:: BC: CD. 240. Cor. 3. If a perpendicular be drawn from the vertex of the right angle to the hypothenuse, the squares of the sides about the right angle will be to each other as the adjacent segments of the hypothenuse. For the rectangles B D E F, D C GE, having the same altitude, are to each other as their bases, B D, C D (Prop. III.). But tllese rectangles are equivalent to the squares AB H L, A C I K; therefore we have, AB: AC2:: B D: DC. 241. Cor. 4. The square described on D C the diagonal of a square is equivalent to double the square described on a side. For let A B C D be a square, and A C its diagonal; the triangle AB C being right-angled and isosceles, we llave, A B AC = AB2+ BC = 2 AB2= 2 X ABCD. 242. Cor. 5. Since A C2 is equal to 2 A B2, we have C2: AB:: 2: 1; and, extracting the square root, we have AC: AB:::/2: 1; hence, the diagonal of a square is incommensurable with a side. 8* 90 ELEMENTS OF GEOMETRY. 243. NOTE. -The proposition may also be demonstrated as follows:Let A B C be a right-angled triangle, having the right angle M at A; then the square described on tle hypothenuse B C will N be equivalent to the sum of \ the squares on the sides B A, H A AC. On B C describe the square BCGF, and on AB, AC the B IDC squares ABHL, ACIK; pro- duce FB to N, HL and IK to M; and through A draw E D A parallel to FBN, and meeting ' E G the prolongation of H L in M. Then, since the angles H B A, N B C are both right angles, if the common angle N BA be taken from each of these equals, there will remain the equal angles H B N, A B C; and, consequently, since the triangles H B N, A B C are both right-angled, and have also the sides B H, B A equal, their hypothenuses B N, B C are equal (Prop. VI. Cor., Bk. I.). But B C is equal to B F; therefore B N is equal to B F; hence the parallelograms B A M N, B D E F, of wlicl the common altitude is B D, have equal bases; therefore the two parallelograms are equivalent (Prop. I.). But tlle parallelogram B AM N is equivalelt to the square AB H L, since they have the same base B A, and the same altitude A L; hence the parallelogram B D E F is also equivalent to tle square A B HL. 11 like manner it may be shown that the rectangle D C GE is equivalent to the square A C IK; hence the two rectallgles together, that is, the square B C G F, are equivalent to the sum of the squares A B II L, A C I K. BOOK IV. 91 PROPOSITION XII. - THEOREM. 244. In any triangle, the square of the side opposite an acute angle is less than the sum of the squares of the base and the other side, by twice the rectangle contained by the base and the distance from the vertex of the acute angle to the perpendicular let fall from the vertex of the opposite angle on the base, or on the base produced. Let AB C be any triangle, A A C one of its acute angles, and AD the perpendicular let fall on the base B C, or on B C produced; then, in \ either case, will the square B.. - B ) C D B C of AB be less than the sum of the squares of A C, B C, by twice the rectangle B C X CD. First. When the perpendicular falls within the triangle ABC, we have BD =BC -CD; and consequently, BD2 = B C2 + CD2 2 B C X CD (Prop. IX.). By adding A D to each of these equals, we have BD2 + AD2 = BC2 + CD2 + AD2 — 2 BC X CD. But the two right-angled triangles A D B, A D C give AB2 = B D + AD2, and A C- = C D2 + AD2 (Prop. XI.); therefore, AB2= B-C2+ AC- 2 B C X CD. Secondly. When the perpendicular A D falls without the triangle AB C, we have B D = C D - B C; and consequently, B D2 = C D + B C2 - 2 C D X B C. By adding A D to eacl of these equals, we find, as before, AB2= -C2+ -AC -2 BC X CD. 92 ELEMENTS OF GEOMETRY. PROPOSITION XIII. THEOREM. 245. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the two other sides plus twice the rectangle contained by the one of those sides into the distance from the vertex of the obtuse angle to the perpendicular let fall from the vertex of the opposite angle to that side produced. Let A C B be an obtuse-angled triangle, A having the obtuse angle at C, and let AD D be perpendicular to the base B C produced; then the square of A B is greater than the sum of the squares of B C, A C, by twice the rectangle B C X C D. Since B D is the I sum of the lines B C + C D, we have BD2 BC C + D2 + 2 BC X CD (Prop. VIII.). By adding AD2 to each of these equals, we have BD2 + AD2 = BC2 + D+ 2 AD + 2 BC X C D. But the two right-angled triangles A D B, A D C give AB2= BD2+ AD, and AC 2 =CD- +AD (Prop. XI.); therefore, AB = B C - A C2 + 2 B C X C D. 246. Scholium. The right-angled triangle is the only one in which the sum of the squares of two sides is equivalent to the square of the third; for if the angle contained by the two sides is acute, the sum of their squares will be greater than the square of the opposite side; if obtuse, it will be less. PROPOSITION XIV.- THEOREM. 247. In any triangle, if a straight line be drawn from the vertex to the middle point of the base, the sum of the BOOK IV. 93 squares of the other two sides is equivalent to twice the square of the bisecting, line, together with twice the square of half the base. In any triangle AB C, draw the line A A E from the vertex A to the middle of the base B C; then the sum of tlhe squares of the two sides, A B, A C, is equivalent to twice the square of A E together with twice the square of B E. On B C let fall the perpendicular AD; B E D C then, in the triangle A B E, AB2 =AE2 + EB2 + 2EB X E D (Prop. XIII.), and, in triangle A E C, AC -AE2 + EC2 - 2 E C X ED (Prop. XII.). Hence, by adding the corresponding sides together, observing that since E B and E C are equal, EB2 is equal to E C2, and E B X ED to E C X ED, we have AB2 +A C2 2 AE2+ 2 E B~2 PROPOSITION XV.- THEOREM. 248. In any parallelogram the sum of the squares of the four sides is equivalent to the sum of the squares of the two diagonals. Let A B C D be any parallelogram, D C. the diagonals of which are AC, B D; then the sum of the squares of A B, B C, C D, D A is equivalent to the sum of the squares of A C, B D. A B For the diagonals A C, B D bisect each other (Prop. XXXIV. Bk. I.); hence, in the triangle ABC, AB2 + B Ca 2 AE 2 BE (Prop. XIV.); also, in the triangle A D C, AD2 + DC2 = 2 AE + 2 2DE2. 94 94 ~~ELEMENTS OF GEOMETRY. Hence, by adding the corresponding sides together, and observing that, since B E and D E are equal,n5j' and DE 2 must also be equal, we shall have, AB+BC2+ AD +~DC= 4AE2~4 DE. But 4 K2is the square of 2 A E, or of AC, and 4 IY is the square of 2 D E, or of B D (Prop. VIII. Cor.); hence, BT A2 B C CD +AD2 = AC ~ B D PROPOSITION XVI. - THEOREM. 249. Int any quadrilateral the sum of the squares of the sides is equivalent to the sum of the squares of the diagonals, plus four timtes the square of the strai ght line that joins the middle points of the diagonals. Let A B C D be any quadrilateral, the C diagonals of which are A.C, D B, and B. E F a straight line joining their middle points, E, F; then the sum of the F squares of A B, B C, C D, A D is equivalent toAXC2+BffD2 +w E. AD Join E B and E D; then in the triangle A B C, (Prop. XIV.), and in the triangle A D C, A-D2~CD2=' 2AXE-2+ 2D E. Hence, by adding the corresponding sides, we have A: ' +BC +AD2+CD 2=4AE+2BE+2fBut 4 A2is equivalent to A 2(Prop. V1IJ. Cor.), and 2 B ER2~ 2 D -E2 is equivalent to 4 BE2o + 4 E F2 (Prop. XIV.); hence, AB2+B'C2+A-D +CD C+BD+ 4EF BOOK IV. 95 250. Cor. If the quadrilateral is a parallelogram, the points E and F will coincide; then the proposition will be the same as Prop. XV. 251. Scholiunm. Proposition XV. is only a particular case of this proposition. PROPOSITION XVII.- THEOREM. 252. If a straight line be drawn in a triangle parallel to one of the sides, it will divide the other two sides proportionally. Let ABC be a triangle, and DE a straight line drawn within it parallel to the side B C; then will AD DB:: AE: EC. D Join B E and D C; then the two trian- gles BD E, DEC have the same base, BD E; they have also the same altitude, since the vertices B and C lie in a line parallel to the base; therefore the triangles are equivalent (Prop. II. Cor. 2). The triangles A D E, B D E, having their bases in the same line A B, and having the common vertex E, have the same altitude, and therefore are to each other as their bases (Prop. VI. Cor.); hence ADE: BDE:: AD: DB. The triangles A D E, D E C, whose common vertex is D, have also the same altitude, and therefore are to each other as their bases; hence ADE: DEC:: AE: EC. But the triangles B D E, D E C have been shown to be equivalent; therefore, on account of the common ratio in the two proportions (Prop. X. Bk. II.), AD: DB:: AE: EC. 253. Cor. 1. Hence, by composition (Prop. VII. Bk. 96 ELEMENTS OF GEOMETRY. II.), we have AD + DB: AD:: AE+ EC: AE, or AB:AD::AC: AE; also, AB: BD::AC: EC. 254. Cor. 2. If two or more straight lines be drawn iii a triangle parallel to one of the sides, they will divide the other two sides proportionally. For, in the triangle A B C, since D E A is parallel to B C, by tile theorem, A D: D B:: AE:E C; and, in the triangle / A D E, since F G is parallel to D E, by F the preceding corollary, A D: F D: D/ AE: GE. Hence, since the antece- B C dents are the same in the two proportions (Prop. X. Cor. 2, Bk. II.), F D: D B:: G E C. PROPOSITION XVIII. - THEOREM. 255. If a straight line divides two sides of a triangle proportionally, the line is parallel to the other side of the triangle. Let ABC be a triangle, and D E a A straight line drawn in it dividing the sides AB, AC, so that AD: DB: A E: E C; then will the line DE be parallel D to the side B C. Join BE and D C; then the triangles - C A D E, B D E, having their bases in the same straight line AB, and having a common vertex, E, are to each other as their bases AD, DB (Prop. VI. Cor.); that is, ADE: B D E:: AD: DB. Also, the triangles A D E, D E C, having the common vertex D, and their bases in the same line, are to each other as these bases, A E, E C; that is, ADE:DEC::AE:EC. BOOK IV. '97 But, by hypothesis, A D: D B:: A E: E C; hence (Prop. X. Bk. II.), ADE: BDE:: ADE: DEC; that is, B D E, D E C have the same ratio to AD E; therefore the triangles B D E, D E C lave the same area, and consequently are equivalent (Art. 211). Since these triangles have the same base, D E, their altitudes are equal (Prop. VI. Cor.); hence the line B C, in which their vertices are, must be parallel to D E. PROPOSITION XIX. THEOREM. 256. Thle straight line bisecting any ang'le of a triangle divides the opposite side into parts, which are proportional to the adjacent sides. In any triangle, A B C, let the an- E gle B A C be bisected by the straight.. line AD; then will BD: DC:: AB: AC. Through the point C draw CE C C D B parallel to AD, meeting B A produced in E. Then, since the two parallels AD, EC are met by the straight line A C, the alternate angles D A C, A C E are equal (Prop. XXII. Bk. I.); and the same parallels being met by the straight line B E, the opposite exterior and interior angles B A D, A E C are also equal (Prop. XXII. Bk. I.). But, by hypothesis, the angles DA C, BAD are equal; consequently the angle ACE is equal to the angle A E C; hence the triangle A C E is isosceles, and the side A E is equal to the side A C (Prop. VIII. Bk. I.). Again, since AD, in the triangle E B C, is parallel to E C, we have B D: D C:: A B: A E (Prop. XVII.), and, substituting A C in place of its equal A E, BD: DC:: AB: AC. 9 98 ELEMENTS OF GEOMETRY. PROPOSITION XX.- THEOREM. 257. If a straight line drawn from the vertex of any angle of a triangle divides the opposite side into parts which are proportional to the adjacent sides, the line bisects the angle. Let the straight line AD,. drawn E from the vertex of the angle B A C, in the triangle AB C, divide.the op-. posite side B C, so that BD: D C:: AB: AC; then will the line AD D - bisect the angle B A C. Through the point C draw C E parallel to A D, meeting B A produced in E. Then, by hypothesis, B D: D C:: AB: AC; and since AD is parallel to E C, BD:D C:: AB:AE (Prop. XVII.); then AB: AC:: AB: AE (Prop. X. Bk. II.); consequently A C is equal to A E; hence the angle A E C is equal to the angle A C E (Prop. VII. Bk. I.). But, since C E and A D are parallels, the angle A E C is equal to the opposite exterior angle B A D, and the angle A C E is equal to the alternate angle D A C (Prop. XXII. Bk. I.); hence the angles B A D, D A C are equal, and consequently the straight line A D bisects the angle B AC. PROPOSITION XXI. THEOREM. 258. If the exterior angle formed by producing one of the sides of any triangle be bisected by a straight line which meets the base produced, the distances from the extremities of the base to the point where the bisecting line meets the base produced, will be to each other as the other two sides of the triangle. Let the exterior angle C A E, formed by producing the side B A of the triangle A B C, be bisected by the straight BOOK IV. 99 line A D, which meets the side B C produced in D, then will BD: DC:: AB: AC. A Through C draw C F parallel to AD; then the angle AC F is equal to the alternate angle C AD, B C D and the exterior angle DA E is equal to the interior and opposite angle C F A (Prop. XXII. Bk. I.). But, by hypothesis, the angles CAD, D A E are equal; consequently the angle A C F is equal to the angle C F A; hence the triangle A C F is isosceles, and the side A C is equal to the side A F (Prop. VIII. Bk. I.). Again, since A D is parallel to F C, BD: D C:: BA: A F (Prop. XVII. Cor. 1), and substituting A C in the place of its equal A F, we have BD: DC:: BA: AC. PROPOSITION XXII. -THEOREM. 259. Equiangular triangles have their homologous sides proportional, and are similar. Let the two triangles A B C, D C E F be equiangular; the angle B A C / being equal to the angle C D E, the A' angle A B C to the angle D C E, and the angle A C B to the angle D E C, then the homologous sides will be B C E proportional, and we shall have BC: CE: AB: CD::AC:DE. For, let the two triangles be placed so that two homologous sides, B C, C E, may join each other, and be in the same straight line; and produce the sides BA, ED till they meet in F. Since B C E is a straight line, and the angle B C A is equal to the angle C E D, A C is parallel to F E (Prop. XXI. Bk. I.); also, since the angle A B C is equal to the 100 ELEMENTS OF GEOMETRY. angle D C E, the line B F is parallel F to the line C D. Hence the figure / A C D F is a parallelogram; and, A S consequently, A F is equal to C D, /\ and A C to F D (Prop. XXXI. Bk. I.). B C E In the triangle B E F, since the line A C is parallel to the side FE, we have BC:CE:: BA: AF (Prop. XVII.); or, substituting CD for its equal, A F, B C: C E:: B A: CD. Again, C D is parallel to B F; therefore, B C: C E:: F D: D E; or, substituting A C for its equal F D, BC: CE:: AC: DE. And, since both these proportions contain the same ratio B C: C E, we have (Prop. X. Bk. II.) AC: DE:: BA: CD. Hence, the equiangular triangles B A C, C D E have their.homologous sides proportional; and consequently the two triangles are similar (Art. 210). 260. Cor. Two triangles having two angles of the one equal to two angles of the other, each to each, are similar; since the third angles will also be equal, and the two triangles be equiangular. 261. Scholium. In similar triangles, the homologous sides are opposite to the equal angles; thus the angle A C B being equal to D E C, the side A B is homologous to D C; in like manner, A C and D E are homologous. PROPOSITION XXIII. -THEOREM. 262. Triangles which have their homologous sides proportional, are equiangular and similar. Let the two triangles A B C, D E F have their sides proportional, so that we have BC: EF:: AB: DE:: A C: D F; BOOK IV. 101 then will the triangles D have their angles equal; namely, the angle A equal to the angle D, E F the angle B to the angle E, and the angle C to the angle F. /:\ At the point E, in the B C G straight line E F, make the angle FE G equal to the angle B, and at the point F, the angle E F G equal the angle C; the third angle G will be equal to the third angle A (Prop. XXVIII. Cor. 2, Bk. I.); and the two triangles A B C, E F G will be equiangular. Therefore, by the last theorem, we have BC: E F: A B: E G; but, by hypothesis, we have BC: EF:: AB: DE; hence, E G is equal to D E. By the same theorem, we also have BC: EF:: AC: FG; and, by hypothesis, BC: EP:: AC: DF; hence F G is equal to D F. Hence, the triangles E G F, D E F, having their three sides equal, each to each, are themselves equal (Prop. XVIII. Bk. I.). But, by construction, the triangle E GF is equiangular with the triangle A B C; hence the triangles D E F, A B C are also equiangular and similar. 263. Scholium. The two preceding propositions, together with that relating to the square of the hypothenuse (Art. 237), are the most important and fertile in results of any in Geometry. They are almost sufficient of themselves for all applications to subsequent reasoning, and for the 9* 102 ELEMENTS OF GEOMETRY. solution of all problems; since the general properties of triangles include, by implication, those of all figures. PROPOSITION XXIV. - THEOREM. 264. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing these angles proportional, are similar. Let the two triangles ABC, A D E F have the angle A equal to D the angle D, and the sides containing these angles proportional, so that AB:DE::AC: DF; then G. H the triangles are similar. C Take A G equal D E, and draw GH parallel to BC. The angle AGH will be equal to the angle A B C (Prop. XXII. Bk. I.); and the triangles A G H, A B C will be equiangular; hence we shall have AB: AG:: AC: AH. But, by hypothesis, AB: DE:: AC: DF; and, by construction, A G is equal to D E; hence A H is equal to D F. Therefore the two triangles A G H, D E F, having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are themselves equal (Prop. V. Bk. I.). But the triangle A G H is similar to A B C; therefore D E F is also similar to A B C. PROPOSITION XXV. - THEOREM. 265. Two triangles, which have their sides, taken two and two, either parallel or perpendicular to each other, are similar. Let the two triangles AB C, D E F have the side A B parallel to the side D E, B C parallel to E F, and A C BOOK IV. 103 parallel to D F; these triangles will A D then be similar. For, since the side AB is parallel to the side D E, and B C to E F, the angle ABC is equal to the angle DEF (Prop. XXVI. Bk. I.). Also, since AC is parallel to DF, the angle AC B B C is equal to the angle DF E, and the angle BAC to EDF; therefore the triangles A B C, D E F are equiangular; hence they are similar (Prop. XXII.). Again, let the two triangles A ABC, DEF have the side D E perpendicular to the side/ \G AB, DF perpendicular to AC, and E F perpendicular to B C;D \ these triangles are similar. Produce F D till it meets A C at G; then the angles D G A, D E A of the quadrilateral A E D G are two right angles; and since all tle four angles are together equal to four right angles (Prop. XXIX. Cor. 1, Bk. I.), the remaining two angles, E D G, E A G, are together equal to two right angles. But the two angles E D G, E D F are also together equal to two riglit angles (Prop. I.. I.); hence the angle E D F is equal to EAG or BAC. The two angles, G F C, G C F, in the right-angled triangle F G C, are together equal to a right angle (Prop. XXVIII. Cor. 5, Bk. I.), and the two angles GFC, GFE are together equal to the right angle E F C (Art. 34, Ax. 9); therefore G F E is equal to G C F, or D F E to B C A. Therefore the triangles AB C, DE F have two angles of tlhe one equal to two angles of the other, each to each; hence they are similar (Prop. XXII. Cor.). 266. Scholium. When the two triangles have their sides parallel, the parallel sides are homologous; and when they have them perpendicular, the perpendicular sides are 104 ELEMENTS OF GEOMETRY. homologous. Thus, DE is homologous with AB, D F with AC, and E F with B C. PROPOSITION XXVI. - THEOREM. 267. In any triangle, if a line be drawn parallel to the base, all lines drawn from the vertex will divide the parallel and the base proportionally. In the triangle BAC, let DE A be drawn parallel to the base B C; then will the lines A F, A G, A H, drawn from the vertex, divide the parallel D E, and the base B C, so D J-K that DI:BF::IK:FG::KL:GH. B F G H C For, since D I is parallel to B F, the triangles AD I and A B F are equiangular; and we have (Prop. XXII.), DI:BF::AI:AF; and since I K is parallel to F G, we have in like manner, AI: AF:: IK: FG; and, since these two propositions contain the same ratio, A I: AF, we shall have (Prop. X. Cor. 1, Bk. II.), DI: B F:: I: I FG. In the same manner, it may be shown that IK: FG:: KL: GI:: LE: H C. Therefore the line D E is divided at the points I, K, L, as the base B C is, at the points F, G, H. 268. Cor. If B C were divided into equal parts at the points F, G, H, the parallel D E would also be divided into equal parts at the points I, K, L. PROPOSITION XXVII.- THEOREM. 269. In a right-angled triangle, if a perpendicular is drawn from the vertex of the right angle to the hypothe BOOK IV. 105 nuse, the triangle will be divided into two triangles similar to the given triangle and to each other. In tile right-angled triangle A B C, from the vertex of the right angle B A C, let A D be drawn perpendicu- / lar to the hypothenuse B C; then the triangles B A D, D A C will be simi- lar to the triangle A B C, and to each B other. For the triangles B A D, B A C have the common angle B, the right angle B D A equal to the right angle B A C, and therefore the third angle, B A D, of tile one, equal to the third angle, C, of the other (Prop. XXVIII. Cor. 2, Bk. I.); hence these two triangles are equiangular, and consequently are similar (Prop. XXII.). In the same manner it may be shown that the triangles DAC and B A C are equiangular and similar. The triangles BA D and D A C, being each similar to the triangle B A C, are similar to each other. 270. Cor. 1. Each of the sides containing the right angle is a mean proportional between the hypothenuse and the part of it which is cut off adjacent to that side by the perpendicular from the vertex of the right angle. For, the triangles BAD, B A C being similar, their homologous sides are proportional; hence BD: BA:: BA: BC; and, the triangles D A C, B A C being also similar, DC: AC:: AC: BC; hence each of the sides A B, A C is a mean proportional between the hypothenuse and the part cut off adjacent to that side. 271. Cor. 2. The perpendicular from the vertex of the right angle to the hypothenuse is a mean proportional between the two parts into which it divides the hypothenuse. 106 ELEMENTS OF GEOMETRY. For, since the triangles AB D, AD C are similar, by comparing their homologous sides we have BD: AD:: AD: DC; hence, the perpendicular A D is a mean proportional between the parts D B, D C into which it divides the hypothenuse B C. PROPOSITION XXVIII. THEOREM. 272. Two triangles, having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles. Let the two triangles AB C, A D E A have the angle A in common; then will the triangle ABC be to the tri- D angle AD E as AB X AC toAD X AE. Join B E; then the triangles A B E, B C A D E, having the common vertex E, and their bases in the same line, AB, have the same altitude, and are to each other as their bases (Prop. VI. Cor.); hence ABE: ADE:: AB: AD. In like manner, since the triangles A B C, A B E have the common vertex B, and their bases in the same line, A C, we have ABC: ABE:: AC: AE. By multiplying together the corresponding terms of these proportions, and omitting the common term A B E, we have (Prop. XIII. Bk. II.), ABC:ADE::AB X AC:AD X AE. 273. Cor. If the rectangles of the sides containing the equal angles were equivalent, the triangles would be equivalent. BOOK I V. 107 PROPOSITION XXIX. THEOREM. 274. Similar triangles are to each other as the squares described on their h omologous sides. Let A B C, D E F be two similar A triangles, and let A C, D F be llo- mologous sides; then the triangle A B C will be to the triangle D E F as the square on A C is to the square on DF. For, the triangles being similar, B C E F they have their homologous sides proportional (Art. 210); therefore AB: DE:: AC: DF; and multiplying the terms of this proportion by the corresponding terms of the identical proportion, AC: DF:: AC: D F, we have (Prop. XIII. Bk. II.), AB X AC:DE X DF:: AC2: DF2. But, by reason of the equal angles-A and D, the triangle ABC is to the triangle DEFas AB X AC is to DE X DF (Prop. XXVIII.); consequently (Prop. X. Bk. II.), ABC: DEF:: AC: DF2. Therefore, the two similar triangles A B C, D E F are to each other as the squares described on the homologous sides A C, D F, or as the squares described on any other two homologous sides. PROPOSITION XXX. -THEOREM. 275. Similar polygons may be divided into the same number of triangles similar each to each, and similarly situated. 108 ELEIMENTS OF GEOMETRY. Let ABCDE, C FGIIIK be two similar polygons; B they may be divid-. ed into the same A / F number of triangles similar each E K to each, and similarly situated. From the homologous angles A and F, draw the diagonals A C, A D and F H, F I. The two polygons'being similar, the angles B and G, which are homologous, must be equal, and the sides A B, BC must also be proportional to F G, GH (Art. 210); that is, A B: F G:: B C: G I. Therefore the triangles A B C, F G H have an angle of tlhe one equal to the angle of the other, and the sides containing these angles proportional; hence they are similar (Prop. XXIV.); consequently the angle BC A is equal to tlhe angle G H F. These equal angles being taken from the equal angles B C D, G H I, the remaining angles A C D, F H I will be equal (Art. 34, Ax. 3). But, since the triangles AB C, F G I are similar, we have AC: FH:: BC: GH; and, since the polygons are similar (Art. 210), BC: GH:: CD: HI; hence (Prop. X. Cor. 1, Bk. II.), AC: FIH:: CD: HI. But the terms of the last proportion are the sides about the equal angles A C D, F H I; hence the triangles A C D, F HI are similar (Prop. XXIV.). In the same manner, it may be shown that the corresponding triangles A D E, FIK are similar; hence the similar polygons may be divided into the same number of triangles similar each to each, and similarly situated. 276. Cor. Conversely, if two polygons are composed BOOK IV. 109 of the same number of similar triangles, and similarly situated, the two polygons are similar. For the similarity of the corresponding triangles give the angles A B C equal to F G H, B C A equal to G H F, and A C D equal to F H I; hence, B C ) equal to G H I, likewise C D E equal to H I K, &c. Moreover, we have AB: FG:: BC: GH:: AC: FH::CD: HI,&c. therefore the two polygons have their angles equal and their sides proportional; hence they are similar. PROPOSITION XXXI. - THEOREM. 277. Tle perimeters of similar polygons are to each other as their homologous sides; and their areas are to each other as the squares described on these sides. Let ABCDE, C FGHIK be two H similar polygons; B then their perimr- D G eters are to each A F / other as their ho-/ mologous sides E K A B and F G, B C and G H, &c.; and their areas are to each other as A B2 is to F G, B C2 to GH2, &c. First. Since the two polygons are similar, we have AB: FG:: B C: GH:: CD: HI, &c. Now the sum of the antecedents A-B, B C, CD, &c., which compose the perimeter of the first polygon, is to the sum of the consequents F G, GH, HI, &c., which compose the perimeter of the second polygon, as any one antecedent is to its consequent (Prop. XI. Bk. II.); therefore, as any two homologous sides are to each other, or as AB is to F G. Secondly. From the homologous angles A and F, draw 10 110 ELEMENTS OF GEOMETRY. the diagonals A C, A D and F I, F I. Then, since the triangles A B C, F G H are similar, the triangle A B C: F G H:: AC2:FH2 (Prop. XXIX.); and, since the triangles AC D, F HI are similar, the triangle A C D: FH I:: AC2: FH2. But the ratio AC: F H is common to both of the proportions; therefore (Prop. X. Bk. II.), ABC: FGH:: ACD: FHI. By the same mode of reasoning, it may be proved that ACD: F H I:: A DE: FIK, and so on, if there were more triangles. Therefore the sum of the antecedents A B C, A C D, A D E, which compose the area of the polygon A B C D E, is to the sum of the consequents F G H, F H I, F I K, which compose the area of the polygon F G H I K, as any one antecedent ABC is to its consequent F G H (Prop. XI. Bk. II.), or as A B is to F G2; hence the areas of similar polygons are to each other as the squares described on their homologous sides. 278. Cor. 1. The perimeters of similar polygons are also to each other as their corresponding diagonals. 279. Cor. 2. The areas of similar polygons are to each other as the squares described on their corresponding diagonals. PROPOSITION XXXII.- THEOREM. 280. A chord in a circle is a mean proportional between the diameter and the part of the diameter cut off between one extremity of the chord and a perpendicular drawn from the other extremity to the diameter. Let A B be a chord in a circle, B C a diameter drawn from one extremity of A B, and AD a perpendicular BOOK IV. 111 drawn from the other extremity to B C; then B D: AB:: AB: B C. Join A C; then the triangle B — C A B C, described in a semicircle, is D right-angled at A (Prop. XV1II. Cor. 2, Bk. llI.); and the triangle B A D is similar to the triangle B A C (Prop. XXVII.); hence, we have (Prop. XXVII. Cor. 1), B D: A B::A B: B C; therefore the chord A B is a mean proportional between the diameter B C, and the part, B D, cut off between the extremity of the chord and the perpendicular from the other extremity. 281. Cor. If from aly point, A, in the circumference of a circle, a perpendicular, AD, be drawn to the diameter B C, the perpendicular will be a mean proportional between tlie parts B D, D C into which it divides the diameter. For, joining A B and A C, we have the triangle A B C, right-angled at A, and the triangles BA D, D A C similar to it and to each other (Prop. XXVII.); therefore (Prop. XXVII. Cor. 2), B D: AD:: A D: D C, or, what amounts to the same thing (Prop. III. Bk. II.), BD X D C AD2. Scholium. A part of a straight line cut off by another is called a segment of the line. Thus B D, D C are segments of the diameter B C. PROPOSITIoN XXXIII.-THEOREM. 282. If two chords in a circle intersect each other, the segments of the one are reciprocally proportional to the segnents of the other. 112 ELEMENTS OF GEOMETRY. Let A B, CD be two chords, which intersect each other at E; then will AE:DE:: EC: EB. Join A C and B D. In the triangles / A E C, BED, the angles at E are equal being vertical angles (Prop. A IV. Bk. I.); the angle A is equal to the angle D, being measured by half the same arc, B C (Prop. XVIII. Cor. 1, Bk. III.); for the same reason, the angle C is equal to the angle B; the triangles are therefore similar (Prop. XXII.), and their homologous sides give the proportion, AE: DE:: E: EB. 283. Cor. Hence, AE X EB = DE X EC; therefore the rectangle of tlhe two segments of tle one chord is equal to the rectangle of the two segments of the other. PROPOSITION XXXIV. - THEOREM. 284. If from the same point without a circle two secants be drawn, terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments. Let E B, E C be two secants drawn E from tile point E without a circle, and A/D terminating in the concave arc at the points B and C; then will EB:EC: ED: EA. For, joining A C, B D, the triangles A E C, BE D have the angle E com- B mon; and the angles B and C, being measured by half the same arc, AD, are equal (Prop. XVIII. Cor. 1, Bk. III.); these triangles are therefore similar (Prop. XXII. Cor.), and their homologous sides give the proportion, EBEC E: E: E EA. BOOK IV. 118 285. Cor. Hence, EB X EA=EC X E D; therefore the rectangle contained by the whole of one secant and its external segment is equivalpiit to the rectangle contained by the whole of the other secant and its external segment. PROPOSITION XXXV.- THEOREM. 286. If from a point without a circle there be drawn a tangent terminating in the circumference, and a secant terminating in the concave arc, the tangent will be a mean proportional between the whole secant and its external segment. From the point E let the tangent E E A, and the secant E C, be drawn; D then will E C E A: E A \ ED. For, joining A D and A C, the triangles EAD, EAC have the A angle E common; also, the angle EAD formed by a tangent and a chord has for its measure half the arc A D (Prop. XX. Bk. III.), and the angle C has the same measure; therefore the angle E A D is equal to the angle C; hence the two triangles are similar (Prop. XXII. Cor.), and give the proportion, EC: EA:: EA: ED. 287. Cor. Hence, EA2 = E C X E D; therefore the square of the tangent is equivalent to the rectangle contained by the whole secant and its external segment. PROPOSITION XXXVI. ---THEOREM. 288. If any angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the other two sides is equivalent to the square of the bisecting line plus the rectangle of the segments of the third side..10* 114 ELEMENTS OF GEOMETRY. Let the triangle A B C have the A angle BAC bisected by the straight line A D terminating in the oppo- B - site side BC; then the rectangle B A X AC is equivalent to the square of AD plus the rectangle B D X DC. Describe a circle through the three points A, B, C; E produce A D till A meets the circumference at E, and join CE. The triangles BAD, E A C have, by hypothesis, the angle B A D equal to the angle E A C; also the angle B equal to the angle E, beillg measured by half of the same arc AC (Prop. XVIII. Cor. 1, lk. III.); these triangles arc therefore similar (Prop. XXI. Cor.), and their homologous sides give tlei proportion, BA: AE E:: AD: AC; hence, BA XAC-=AEX AD. But A E is equal to A D + D E, and multiplying each of these equals by A D, we have, AE XD AD=D2 +AD X DE; now, A D X D E is equivalent to BD X DC (Prop. XXXIII. Cor.); hence BA X AC AD2 + BD X DC. PROPOSITION XXXVII.- THEOREM. 289. The rectangle contained by any two sides of a triangle is equivalent to the rectangle contained by the diameter of the circumscribed circle and the perpendicular drawn to the third side from the vertex of the opposite angle. Ill any triangle A B C, let A D be drawn perpendicular to B C; and let E C be tlle diameter of the circle circum BOOK IV. '115 scribed about the triangle; then A will A B X A C be equivalent to E AD X CE. For, joining A E, the angle E A C is a right angle, being inscribed in a semicircle (Prop. XVIII. Cor. 2, B --- Bk. III.); and the angles B and E are equal, being measured by half of the same arc, A C (Prop. XVIII. Cor. 1, Bk. III.); hence the two rightangled triangles are similar (Prop. XXII. Cor.), and give the proportion A B: C E:: AD: A C; hence AB X A C=-CE X AD. 290. Cor. If these equals be multiplied by BC, we shall have AB X AC X BC- CE X AD X BC. But AD X B C is double the area of the triangle (Prop. VI.); therefore the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle. PROPOSITION XXXVIII.- THEOREM. 291. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles of the opposite sides. Let A B C D be any quadrilat- B eral inscribed in a circle, and A C, B D its diagonals; then the rec- tangle A C X B D is equivalent to C the sum of the two rectangles A E AB X CD, AD X BC. For, draw BE, making the angle ABE equal to the angle CB D; to each of these equals add the angle E B D, and we shall have the angle A B D equal to the angle E B C; and the 116 ELEMENTS OF GEOMETRY. angle AD B is equal to the angle BC E, being in the same segment (Prop. XVIII. Cor. 1, Bk. III.); / therefore the triangles A B D, BCE are similar; hence the proportion, A E AD: BD:CE: BC; and, consequently, AD X BC=BDXCE. D Again, since the angle A B E is equal to the angle C B D, and the angle B A E is equal to the angle B D C, being il the same segment (Prop. XVII. Cor. 1, Bk. III.). the triangles A B E, B C D are similar; hence, AB: AE:: BD: CD; and consequently, AB X CD=-AE X BD. By adding the corresponding terms of the two equations obtained, and observing that BD X AE + BD X CE = BD (AE+ CE)-=BD X AC, we have BD X AC=AB X CD +AD X BC. PROPOSITION XXXIX. -THEOREM. 292. The diagonal of a square is incommensurable with its side. Let A B C D be any square, and A C its diagonal; then 'A C is incommensurable with the side AB. To find a common measure, if there be one, we must apply A B, or its equal C B, to CA, as often as it can be done. In order to do this, from the point C as a center, with a radius C B, describe the semicircle E D C. i I A G B FB E, and produce AC to E. It is evident that C B is contained once in AC, BOOK IV. 117 with a remainder AF, which remainder must be compared with B C, or its equal, A B. The angle A B C being a right angle, A B is a tangent to the circumference, and A E is a secant drawn from the same point, so that (Prop. XXXV.) AF: A P: AB: AE. Hence, in comparing A F with A B, the equal ratio of A B to A E may be substituted; but A B or its equal C F is contained twice in A E, with a remainder A F; which remainder must again be compared with A B. Thus, the operation again consists in comparing A F with A B, and may be reduced in the same manner to the comparison of A B, or its equal C F, with A E; which will result, as before, in leaving a remainder A F; hence, it is evident that the process will never terminate; consequently the diagonal of a square is incommensurable with its side. 293. Scholium. The impossibility of finding numbers to express the exact ratio of the diagonal to the side of a square has now been proved; but, by means of the continued fraction which is equal to that ratio, an approximation may be made to it, sufficiently near for every practical purpose. BOOK V. PROBLEMS RELATING TO THE PRECEDING BOOKS. PROBLEM I. 294. To bisect a given straight line, or to divide it into two equal parts. Let A B be a straight line, which it::c is required to bisect. From the point A as a center, with A E B a radius greater than the half of A B, describe an arc of a circle; and from the point B as a center, with the same:D radius, describe another arc, cutting the former in the points C and D. Through C and D draw the straight line C D; it will bisect A B in the point E. For the two points C and D, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle point of AB (Prop. XV. Cor., Bk. I.). Therefore the line C D must divide the line A B into two equal parts at the point E. PROBLEM II. 295. From a given point, without a straight line, to draw a perpendicular to that line. Let A B be the straight line, and let C be a given point without the line. BOOK V. 119 From the point C as a center, and with a radius sufficiently great, describe C an arc cutting the line A B in two points, A and B; then, from the points A and B as centers, with a radius greater than half of A B, describe two arcs cutting each other in D, and draw the straight line C D; it will be the D perpendicular required. For, the two points C and D are each equally distant from the points A and B; hence, the line CD is a perpen. dicular passing through the middle of A B (Prop. XV. Cor., Bk. I.). PROBLEM III. 296. At a given point in a straight line to erect a per, pendicular to that line. Let A B be the straight line, and let C D be a given point in it. In the straight line A B, take the points A and B at equal distances from D; then from the points A and B as A -- B centers, with a radius greater than AD, describe two arcs cutting each other at C; through C and D draw the straight line C D; it will be the perpendicular required. For the point C, being equally distant from A and B, must be in a line perpendicular to the middle of AB (Prop. XV. Cor., Bk. I.); hence C D has been drawn perpendicular to A B at the point D. 297. Scholium. The same construction serves for making a right angle, A D C, at a given point, D, on a given straight line, A B. 120 ELEMENTS OF GEOMETRY. PROBLEM IV. 298. To erect a perpendicular at the end of a gicen straight line. Let A B be the straight line, and B tile end of it at which a perpen- dicular is to be erected. From any point, D, taken without the line A B, with a radius equal to A'. Bthe distance D B, describe an arc... cutting the line A B at tlhe points A and B; through the point A, and the center D, draw the diameter AC. Then through C, where the diameter meets the arc, draw the straight line C B, and it will be the perpendicular required. For the angle A B C, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.). PROBLEM V. 299. At a point in a given straight line to make an angle equal to a given angle. Let A be tlle given E D point, AB the given line, /: ' and EF G the given angle. From the point F as G A B a center, with any radius, describe an arc, GE, terminating in the sides of the angle; from the point A as a centre, with the same radius, describe the indefinite arc B D. Draw the chord GE; then from B as a centre, with a radius equal to G E, describe an arc cutting the arc B D in C. Draw AC, and the angle CAB will be equal to the given angle EF G. For the two arcs, BC and GE, have equal radii and equal chords; therefore they are equal (Prop. III. Bk. BOOK V. 121 III.); hence the angles C A B, E F G, measured by these arcs, are also equal (Prop. V. Bk. III.). PROBLEM VI. 300. To bisect a given arc, or a given angle. First. Let AD B be the given arc C which it is required to bisect. Draw the chord A B; from the centre C draw the line C D perpendicular to A B (Prob. III.); it will bisect the A B arc A D B in the point D. D For C D being a radius perpendicular to a chord A B, must bisect the arc A D B which is subtended by that chord (Prop. VI. Bk. III.). Secondly. Let A C B be the angle winch it is required to bisect. From C as a center, with any radius, describe the arc ADB; bisect this arc, as in the first case, by drawing the line CD; and this line will also bisect the angle ACB. For the angles A C D, B C D are equal, being measured by the equal arcs A D, D B (Prop. V. Bk. III.). 301. Scholium. By the same construction, we may bisect each of the halves A D, D B; and thus, by successive subdivisions, a given angle or a given arc may be divided into four equal parts, into eight, into sixteen, &c. PROBLEM VII. 302. Through a given point, to draw a straight line parallel to a given straight line. Let A be the given point, and A B C D the given straight line. '... From A draw a straight line, C.. AE, to any point, E, in CD. E Then draw AB, making the angle EA B equal to the 11 122 ELEMENTS OF GEOMETRY. angle AEC (Prob. V.); and A....... B AB is parallel to CD. '... For the alternate angles E A B, ' _ "" -... A E C, made by the straight line E D A E meeting the two straight lines A B, C D, being equal, the lines AB and CD must be parallel (Prop. XX. Bk. I.). PROBLEM VIII. 303. Two angles of a triangle being given, to find the third angle. Draw the indefinite straight line C D ABE. At any poilt, B, make the angle ABC equal to one of the given angles (Prob. V.), and the i angle C B D equal-to the other given A B angle; then the angle D B E will be the third angle required. For these three angles are together equal to two right angles (Prop. I. Cor. 2, Bk. I.), as are also the three angles of every triangle (Prop. XXVIII. Bk. I.); and two of the angles at B having been made equal to two angles of the triangle, the remaining angle D B E must be equal to the third angle. PROBLEM IX. 304. Two sides of a triangle and the included angle being given, to construct the triangle. Draw the straight line A B equal to C one of the two given sides. At the point A make an angle, C A B, equal to the given angle (Prob. V.); and take A C equal to the other given side. Join A B B C; and the triangle ABC will be the one required (Prop. V. Bk. I.). BOOK V. 123 PROBLEM X. 305. One side and two angles of a triangle being given, to construct the triangle. The two given angles will either be C both adjacent to the given side, or one adjacent and the other opposite. In the latter case, find the third angle (Prob. VIII.); and the two angles ad- AB jacent to the given side will then be known. In the former case, draw the straight line A B equal to the given side; at the point A, make an angle, B A C, equal to one of the adjacent angles, and at B an angle, ABC, equal to the other. Then the two sides AC, BC will meet, and form with AB the triangle required (Prop. VI. Bk. I.) PROBLEM XI. 306. Two sides of a triangle and an angle opposite one of them being given, to construct the triangle. Draw the indefinite straight C line AB. At the point A make an angle BAC equal to the given angle, and make A C equal to that side which is * - B A E - — B — adjacent to the given angle. Then from C, as a center, with a radius equal to the other side, describe an arc, which must either touch the line A B in D, or cut it in the points E and F, otherwise a triangle could not be formed. When the arc touches A B, a straight line drawn from C to the point of contact, D, will be perpendicular to A B (Prop. XI. Bk. III.), and the right-angled triangle C A J will be the triangle required. When the arc cuts A B in two points, E and F, lying 124 ELEMENTS OF GEOMETRY. on the same side of the point A, draw the straight lines C E, C F; and each of the two triangles C AE, C A F will satisfy the conditions of the problem. If, however, the two points E and F should lie on different sides of the point A, only one of the triangles, as C A F, will satisfy all the conditions; hence that will be the triangle required. 307. Scholium. The problem would be impossible, if the side opposite the given angle were less than the perpendicular let fall from the point C on the straight line AB. PROBLEM XII. 308. The three sides of a triangle being given, to construct the triangle. Draw the straight line AB equal to C one of the given sides; from the point A as a center, with a radius equal to either of the other two sides, describe an arc; from the point B, with a radius / equal to the third side, describe another A B arc cutting the former in the point C; draw the straight lines AC, B C; and the triangle A B C will be the one required (Prop. XVIII. Bk. I.). 309. Scholium. The problem would be impossible, if one of the given sides were equal to or greater than the sum of the other two. PROBLEM XIII. 310. Two adjacent sides of a parallelogram and the included angle being given, to construct the parallelogram. Draw the straight line A B equal D to one of the given sides. At the point A make an angle, B AD, equal to the given angle, and take AD equal to the other given side.. From A B BOOK V. 125 the point D, with a radius equal to A B, describe an arc; and from the point B as a center, with a radius equal to A D, describe another arc cutting the former in the point C. Draw the straight lines CD, CB; and the parallelogram A B C D will be the one required. For the opposite sides are equal, by construction; hence the figure is a parallelogram (Prop. XXXII. Bk. I.); and it is formed with the given sides and the given angle. 311. Cor. If the given angle is a right angle, the figure will be a rectangle; and if the adjacent sides are also equal, the figure will be a square. PROBLEM XIV. 312. A circumference, or an arc, being given, to find the center of the circle. Take any three points, A, B, C, on the given circumference, or arc. Draw the chords A B, BC, and bisect them by the perpendiculars E D E and FE (Prob. I.); the point E, in which these perpendiculars meet, is the center required. For the perpendiculars DE, FE must both pass through the center (Prop. VI. Cor. 2, Bk. III.), and E being the only point through which they both pass, E must be the center. 313. Scholium. By the same construction, a circumference may be made to pass through three given points, A, B, C, not in the same straight line; and also a circumference described in which a given triangle, ABC, shall be inscribed. PROBLEM XV. 314. Through a given point to draw a tangent to a given circle. 11 126 ELEMENTS OF GEOMETRY. First. Let the given point A be in the circumference..A Find the center of the circle, C (Prob. XIV.); draw the radius C A; through the point \ C B A draw AB perpendicular to CA (Prob. IV.); and AB \ - will be the tangent required E (Prop. X. Bk. III.). Secondly. Let the given point B be without the circum'ference. Join the point B and the center C by the straight line B C; bisect BC in D; and from D as a center, with a radius equal to CD or D B, describe a circumference intersecting the given circumference in the points A and E. Draw AB and EB, and each will be a tangent as required. For, drawing C A, the angle CA B, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.); therefore A B is perpendicular to the radius C A at its extremity, A, and consequently is a tangent (Prop. X. Bk. III.). In like manner it may be shown that EB is a tangent. PROBLEM XVI. 315. On a givcen straight line to construct a segment of a circle that shall contain an angle equal to a given angle. Let AB be the given straight line. Through the point B draw the straight line B D, making the angle A B D equal to the given angle; draw B E perpendicular to B D; bisect A B, and from F erect the perpendicular F E. From the point E, where these perpendiculars meet, as a center, A % —~ "./ L/ '.-. G. with the distance EB BOOK V. 127 or E A, describe a circumference, and A C B will be tile segmenlt required. For, since BD is a perpendicular at the extremity of the radius E B, it is a tangent (Prop. X. Bk. III.); and the angle A B D is measured by half the arc A G B (Prop. XX. Bk. III.). Also, the angle A C B, being an inscribed angle, is measured by half the arc A G B; therefore the angle ACB is equal to the angle A B D. But, by construction, the angle ABD is equal to the given angle; hence the segment A C B contains an angle equal to the given angle. 316. Scholium. If tlhe given angle were acute, the center must lie within the segment (Prop. XVIII. Cor. 3, Bk. III.); and if it were right, the center must be in the middle of the line A B, and the required segment would be a semicircle. PROBLEM XVII. 317. To inscribe a circle in any given triangle. Bisect any two of the angles, as A and B, by the straight lines AE and BE, meeting in the point E (Prob. VI.). From the point E let fall tlhe perpen- D E diculars ED, EF, EG (Prolb. II.) on the three sides of the triangle; these perpendiculars A G B will all be equal. For, by construction, we have the angle D A E equal to the angle E A G, and the right angle A D E equal to the right angle AGE; hence the third angle A E D is equal to tlhe third angle A E G. Moreover, A E is common to the two triangles AED, AEG; hence tlhe trianglesthemselves are equal, and ED is equal to EG. In the same manner it may be shown that tile two triangles BEF, 128 ELEMENTS OF GEOMETRY. BE G are equal; therefore E F is equal to E G; hence the three perpendiculars E D, E F, E G are all equal, and if, from the point E as a center, with the radius ED, a circle be described, it must pass through the points F and G. 318. Scholium. The three lines which bisect the angles of a triangle all meet in the center of the inscribed circle. PROBLEM XVIII. 319. To inscribe a circle in a given square. Draw the diagonals AC, D B, and D C from the point E, where the diagonals mutually bisect each other (Prop. / XXXIV. Bk. I.), draw the straight line E F perpendicular to a side of tlhe square. / I From E as a center, with a radius equal to E F, describe a circle, and it will - F Btouch each side of the square. For the square is divided by the diagonals into four equal isosceles triangles; hence, the perpendicular, from the vertex E to the base, is the same in each triangle; therefore the circumference described from the center E, with the radius E F, passes through the extremities of each perpendicular; consequently, the sides of the square are tangents to the circle (Prop. X. Bk. III.). PROBLEM XIX. 320. To find the side of a square which shall be equivalent to the sum of two given squares. Draw the two straight lines A B, B C perpendicular to each other, taking A B equal to a side of one of the given squares, and BC equal to a side of the other. Join AC; this will be the side of the square required. A B BOOK V. 129 For, the triangle A B C being right-angled, the square tlat call be described upon the hypotlenuse A C is equivalent to tlhe sum of the squares that can be described upon tle sides A B and B C (Prop. XI. Bk. IV.). 321. Schlolium. A square may thus be found equivalent to tle sum of aly number of squares; for the construotionl wlich reduces two of them to one, will reduce three of them to two, and these two to one. PROBLEM XX. 322. To find the side of a square which shall be equivalent to the difference of two gi2ven squares. Draw tle two straigllt lines AB, AC C perpendicular to each other, making A C equal to tlle side of tlhe less square. Then fiom C as a center, witll a radius equal to the side of tle other square, describe an arc intersecting A B in tlhe A B point B, and A B will be the side of the required square. For, join BC, and the square tlat can be described upon AB is equivalent to the difference of the squares that can be described on B C and A C (Prop. XI. Cor. 1, Bk. IV.). PROBLEM XXI. 323. To construct a rectan.gle that shall be equivalent to a given triangle. Let AB C be the given triangle. Draw the indefinite straight line C E G D C D parallel to the base A B; bisect A B by the perpendicular EF, and make E G equal to F B. Then, by drawing G B, the rectangle E F B G is equal to the tri- A F B algle ABC. 180 ELEMENTS OF GEOMETRY. For the rectangle EF B G has the same altitude, EF, as the triangle A B C, and half its base (Prop. II. Cor. 1, Bk. IV.). PROBLEM XXII. 324. To construct a triangle that shall be equivalent to a given polygon. Let A B C D E be the given poly- C gon. Draw the diagonal CE, cutting off. D the triangle C D E; through te / point D draw D F parallel to C E,/ and meeting A E produced in F. G A E F Draw C F; and the polygon A B C D E will be equivalent to the polygon A B C F, which has one side less than the given polygon. For the triangles C D E, C F E have the base C E common; they have also the same altitude, since their vertices, D, F, are situated in a line, D F, parallel to tlle base; these triangles are therefore equivalent (Prop. II. Cor. 2, Bk. IV.). Add to each of them the figure A B C E, and the polygon A B C D E will be equivalent to the polygon A B C F. In like manner, the triangle C G A may be substituted for the equivalent triangle A B C, and thus the pentagon A B C D E will be changed into an equivalent triangle GCF. The same process may be applied to every other polygon; for, by successively diminishing the number of its sides, one at each step of the process, the equivalent triangle will at last be found. PROBLEM XXIII. 325. To divide a given straight line into any number of equal parts. BOOK V. 131 Let AB be the given straight E line proposed to be divided into any number of equal parts; for example, six. c Through the extremity A A -: B AD B draw the indefinite straight line AE, making aly angle with A B. Take A C of any convenient length, and apply it six times upon A E. Join the last point of division, E, and the extremity B by the straight line E B; and through thle point C draw C D parallel to E B; then A D will be the sixth part of the line A B, and, being applied six times to A B, divides it into six equal parts. For, since C D is parallel to E B, in the triangle AB E, we have tlie proportion (Prop. XVII. Bk. IV.), AD: A B:: A C: A E. But A C is the sixth part of A E; hence A D is the sixth part of A B. PROBLEM XXIV. 326. To divide a given straight line into parts that shall be proportional to other given lines. I/t A B be tie given straight E line proposed to be divided into I D parts proportional to the given C lines A C CC D, DE. Through the point A draw the - indefinite straigllt line AE, mak- ing any angle with A B. On A E lay off A C, C D, and D E. Join the points E and B by tlhe straiglt line E B, and through the points C and D draw C G and D H parallel to E B; and the line A B will be divided into parts proportional to the given lines. For, since C G and D H are each parallel to EB, wo have the proportion (Prop. XVII. Cor. 2, Bk. IV.), A C: A G:: CD: G H:: D E: H B. 132 ELEMENTS OF GEOMIETRY. PROBLEM XXV. 327. To find a fourth proportional to three given straight lines. Draw the two indefinite straight E lines AB, AE, forming any angle witl each other. c On A B make AD equal to the first of the proposed lines, and A B equal to the second; and on A E A D B make AE equal to the third. Join BE; and through the point D draw D C parallel to B E, and A C will be the fourth proportional required. For, since D C is parallel to B E, we have the proportion (Prop. XVII. Cor. 1, Bk. IV.), AB: AD:: AE: AC. 328. Cor. A third proportional to two given lines, A and B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines, A, B, and B. PROBLEM XXVI. 329. To find a mean proportional between two given straight lines. Draw the indefinite straight line A B. On AB take AC equal to D the first of the given lilies, and C B equal to the second. On A B, as a diameter, describe a semicircle, and A C B at the point C draw the perpendicular C D, meeting the semi-circumference in D; C D will be the mean proportional required. For the perpendicular C D, drawn from a point in the circumference to a point in the diameter, is a mean pro BOOK V. 133 portional between the two segments of the diameter A C, C B (Prop. XXXII. Cor., Bk. IV.); and these segments are equal to the given lines. PROBLEM XXVII. 330. To divide a given straight line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let AB be the given straight.... line.... At the extremity, B, of the line.. ' AB, erect tlhe perpendicular BC,.. equal to the half of A B. From D / the point C as a center, with thle...-, —.....radius C B, describe a circle. A E B Draw A C cutting tile circumference in D; and take A E equal to A D. The line A B will be divided at the point E in the manner required; that is, AB: AE::AE: EB. For A B, being perpendicular to the radius at its extremity, is a tangent (Prop. X. Bk. III.); and if A C be produced till it again meets the circumference, in F, we shall have (Prop. XXXV. Bk. IV.), AF: AB:: AB: AD; hence, by division (Prop. VIII. Bk. II.), AF- AB: AB: AB - AD: AD. But, since the radius is the half of A B, the diameter D F is equal to A B, and consequently A F - A B is equal to AD, which is equal to A E; also, since A E is equal to A D, we have A B- AD equal to E B; heiye, AE: AB:: EB: AD, or AE; and, by inversion (Prop. V. Bk. II.), AB: AE:: AE: EB. 12 134 ELEMENTS OF GEOMETRY. 331. Scholium. This sort of division of the line A B is called division in extreme and mean ratio. PROBLEM XXVIII. 332. Tl7rough a given point in a given angle, to drazo a straight line, which shall have the parts included between that point and the sides of the angle equal to each other. Let E be the given point, and A B C B the given angle. Through the point E draw E F paral- \ lel to BC, make AF equal to BF. \ Through the points A and E draw the straight line A EC, and it will be the A E C line required. For, E F being parallel to B C, we have (Prop. XVII. Bk. IV.), AF: FB:: AE: EC; but A F is equal to F B; therefore A E is equal to E C. PROBLEM XXIX. 333. On a given straight line to construct a rectangle that shall be equivalent to a given rectangle. Let AB be the given straight F E line, and C D E F the given H G rectangle. Find a fourth proportional to the three straight lines AB, CD, DE (Prob. XXV.); A B C ) and let B G be that fourth proportional. The rectangle constructed on A B and B G will be equivalent to the rectangle C D E. For, since AB CD:: DE: BG, it follows (Prop. I. Bk. II.) that AB xBG=CDx DE; BOOK V. 1536 hence, the rectangle A B G H, which is constructed on the line A B, is equivalent to the rectangle C D E F. PROBLEM XXX. 334. To construct a square that shall be equivalent to a given parallelogram, or to a given triangle. First. Let ABCD be the given D C parallelogram, A B its base, and D E its altitude. Find a mean proportional between AB and D E (Prob. XXVI.); and A E B the square constructed on that proportional will be equivalent to the parallelogram AB C D. For, denoting the mean proportional by x y, we have, by construction, AB: xy.: xy:DE; therefore, xy = AB X DE; but A B X D E is the measure of the parallelogram, and x y that of the square; hence they are equivalent. Secondly. Let A B C be the given triangle, BC its base, and AD its altitude. Find a mean proportional between BC and the half of AD, and let xy denote that proportional; the square constructed on x y will be equivalent B D C to the triangle A B C. For since, by construction, B C xy:: xy: ADn, it follows that xy2 BC X 3AD; hence the square constructed on x y is equivalent to the triangle A B C. 136 ELEMENTS OF GEOMETRY. PROBLEM XXXI. 335. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line. Let the straight line AB be equal to the sum of the adjacent sides of D the required rectangle./ \ Upon A B as a diameter describe a semicircle; at the point A, draw A E B A D perpendicular to A B, making A D equal to the side of the given square; then draw the line D C parallel to the diameter A B. From the point C, where the parallel meets the circumference, draw C E perpendicular to the diameter; A E and E B will be the sides of the rectangle required. For their sum is equal to AB; and their rectangle A E X E B is equivalent to the square of C E, or to the square of A D (Prop. XXXII. Cor., Bk. IV.); hence, this rectangle is equivalent to the given square. 336. Scholium. The problem is impossible, when the distance A D is greater than the half the given line A B, for then the line D C will not meet the circumference. PROBLEM XXXII. 337. To construct a rectangle that shall be equivalent to a given square, and the difference.of whose adjacent sides shall be equal to a given line. Let the straight line A B be equal to the difference of the adjacent sides of the required rectangle. Upon A B as a diameter, describe a circle. At the extremity of the diameter, draw the tangent A D, making it equal to the side of the given square. BOOK V. 137 Through the point D and the centre D C draw the secant D C F, intersecting the circumference in E; theln D E \ and D F will be the adjacent sides of tlhe rectangle required. \ For the difference of these lines is A B equal to the diameter E F or AB; / and tle rectangle DE X DF is equal to A D2 (Prop. XXXV. Cor., Bk. IV.); hence it is equivalent to the given square. PROBLEM XXXIII. 338. To construct a square that shall be to a given square as one given line is to another given line. Draw the indefinite line AB, D --- on which take A C equal to one of the given lines, and C B equal A c to the other. Upon A B as a di-............................... ameter, describe a semicircle, and F at the point C draw the perpendicular C D, meeting the circumference in D. Through the points A and B draw the straight lines D E, D F, making the former equal to the side of the given square; and through the point E draw E F parallel to AB; D F will be the side of the square required. For, since E F is parallel to A B, DE: DF:: DA: DB; consequently (Prop. XV. Bk. II.), DE2: DF:: D A2: DB2. But in thle right-angled triangle A D B the square of A D is to tlhe square of D B as the segment A C is to the segment C B (Prop. XI. Cor. 3, Bk. IV.); hence, DIE2: D-F9:: AC: CB. 12* 138 ELEMENTS OF GEOMETRY. But, by construction, D E is equal to the side of the given square; also, A 0 is equal to one of the given lines, and C B to the other; hence, the given square is to that constructed on D F as the one given line is to the other. PROBLEM XXXIV. 339. Upon a given base to construct an isosceles triangle, having each of the angles at the base double the vertical angle. Let A B be the given base. Produce A B to some point C till the D rectangle A C X B C shall be equivalent to the square of A B (Prob. XXXII.); then, with the base A B and sides each equal to A C, construct the isosceles triangle D A B, and the angle A B C A will double the angle D. For, make DE equal to AB, or make A E equal to B C, and join E B. Then, by construction, AD: AB:: AB: AE; for AE is equal to B C; consequently the triangles DAB, B A E have a common angle, A, contained by proportional sides; hence they are similar (Prop. XXIV. Bk. IV.); therefore these triangles are both isosceles, for D A B is isosceles by construction, so that A B is equal to E B; but A B is equal to D E; consequently D E is equal to E B, and therefore the angle D is equal to the angle E B D; hence the exterior angle A E B is equal to double the angle D, but the angle A is equal to the angle AEB; therefore the angle A is double the angle D. PROBLEM XXXV. 340. Upon a given straight line to construct a polygon similar to a given )lyg:,on. BOOK V. 139 Let ABCDE C be the given poly- H gon, and FG the B given straight line. D Draw the diag- F onals AC, AD. At the point F in E K the straiglt line F G, make the angle G F H equal to the angle B A C; and at the point G make the angle F G H equal to the angle A B C. The lines F H, G H will cut each other in H, and F G H will be a triangle similar to ABC. In the same manner, upon F H, homologous to A C, construct the triangle F I H similar to A D C; and upon FI, homologous to AD, construct the triangle FIK similar to A D E. The polygon F G H I K will be similar to A B C D E, as required. For these two polygons are composed of the same number of triangles, similar each to each, and similarly situated (Prop. XXX. Cor., Bk. IV.). PROBLEM XXXVI. 341. Two similarpolygons being given, to construct a similar polygon, which shall be equivalent to their sum or their difference. Let A and B be two homologous sides of the given polygons. Find a square equal to tle sum or to the difference of the A B squares described upon A and B; let x be the side of that square; then will x in the polygon required be the. side which is hom1ologua& to the sides A and B in the givenl polygons. The polygon itself may then be colstrlct.1 o:1 x., by tlhe last problem. 140 ELEMENTS OF GEOMETRY. For similar figures are to each other as the squares of their homologous sides; but the square of the side x is equal to the sum or the difference of the squares described upon the homologous sides A and B; therefore the figure described upon the side x is equivalent to the sum or to the difference of the similar figures described upon the sides A and B. PROBLEM XXXVII. 342. To construct a polygon similar to a given polygon, artd which shall have to it a given ratio. Let A be a side of the given polygon. Find the side B of a square, which is to the square on A in the given ratio of the polygons (Prob. XXXIII.). Upon B construct a polygon similar to the given polygon (Prob. XXXV.), and B will be the polygon required. A For the similar polygons constructed upon A and B have the same ratio to each other as the squares constructed upon A and B (Prop. XXXI. Bk. IV.). PROBLEM XXXVIII. 343. To construct a polygon similar to a given polygon, P, and which shall be equivalent to another polygon, Q. Find M, the side of a square, equivalent to the polygon P, and N, the side of a square equivalent to the polygon Q. Let x be a fourth propor- \ B tional to the three given lines M, N, A B; upon the side x, homologous to A B, describe a polygon similar to the polygon P (Prob. XXXV.); it will also be equivalent to the polygon Q. BOOK V. 141 For, representing the polygon described upon the side x by y, we have P y: A B2: x; but, by construction, AB:x::M:N, or AB2: X::M2: N2; hence, P: y:: M2: N2. But, by construction also, M2 is equivalent to P, and N2 is equivalent to Q; therefore, P:y:: P:Q; consequently y is equal to Q; hence the polygon y is similar to the polygon P, and equivalent to the polygon Q. t. BOOK VI. REGULAR POLYGONS, AND THE AREA OF THE CIRCLE. DEFINITIONS. 344. A REGULAR POLYGON is one which is both 'equilateral and equiangular. 345. Regular polygons may have any number of sides: the equilateral triangle is one of three sides; the square is one of four. PROPOSITION I.-THEOREM. 346. Regular polygons of the same number of sides are similar figures. LetABCDEF, E D GHIKLM, be L K two regular polygons of the same F C number of sides; then these polygons are similar. A B G H For, since the two polygons have the same number of sides, they have the same number of angles; and the sum of all the angles is the same in the one as in the other (Prop. XXIX. Bk. I.). Also, since the polygons are equiangular, each of the angles A, B, C, &c. is equal to each of the angles G, H, I, &c.; hence the two polygons are mutually equiangular. BOOK VI. 143 Again; the polygons being regular, the sides A B, B C, CD, &c. are equal to each other; so likewise are the sides G H, H I, I K, &c. Hence, AB: GH:: BC: H1:: CD: I K, &c. Therefore the two polygons have their angles equal, and their homologous sides proportional; hence they are similar (Art. 210). 347. Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop. XXXI. Bk. IV.). 348. Scholium. The angle of a regular polygon is determined by the number of its sides (Prop. XXIX. Bk. I.). PROPOSITION II. -THEOREM. 349. A circle may be circumscribed about, and another inscribed in, any regular polygon. Let ABCDEFGH be any reg- G F ular polygon; then a circle may be circumscribed about, and another / \ inscribed in it. H Describe a circle whose circum- ference shall pass through the three A / points A, B, C, the center being O; C let fall the perpendicular 0 P from B O to the middle point of the side BC; and draw the straight lines 0 A, 0 B, O C, 0 D. Now, if the quadrilateral 0 P C D be placed upon the quadrilateral OP B A, they will coincide; for the side OP is common, and the angle 0 PC is equal to the angle OP, each being a right angle; consequently the side P C will fall upon its equal, P B, and the point C on B. Moreover, from the nature of the polygon, the angle POD is equal to the angle P BA; therefore C D 'will take the 144 ELEMENTS OF GEOMETRY. direction B A, and C D being equal G F to B A, the point D will fall upon A, and the two quadrilaterals will H FE coincide throughout. Therefore OD is equal to A0, and the circum- ference wlich passes through the three points A, B, C, will also pass through the point D. By the same B C mode of reasoning, it may be shown that the circle which passes through the three vertices B, C, D, will also pass through the vertex E, and so on. Hence, the circumference which passes through the three points A, B, C, passes through the vertices of all the angles of the polygon, and is circumscribed about the polygon (Art. 166). Again, with respect to this circumference, all the sides, A B, B C, C D, &c., of the polygon are equal chords; consequently they are equally distant from the centre (Prop. VIII. Bk. III.). Hence, if from the point 0, as a centre, and with the radius 0 P, a circle be described, the circumference will touch the side B C, and all the other sides of the polygon, each at its middle point, and the circle will be inscribed in the polygon (Art. 168). 350. Scholium 1. The point 0, the common center of the circumscribed and inscribed circles, may also be regarded as the centre of the polygon. The angle formed at the centre by two radii drawn to the extremities of the same side is called the angle at the center; and the perpendicular from the center to a side is called the apothegm of the polygon. Since all the chords A B, B C, C D, &c. are equal, all the angles at the center must likewise be equal; therefore the value of each may be found by dividing four right angles by the number of sides of the polygon. 351. Scholium 2. To inscribe a regular polygon of any number of sides in a given circle, it is only necessary to BOOK VI. 145 divide the circumference into as E - ) many equal parts as the polygon has sides; for the arcs being equal, 0 the chords AB, BC, CD, &c. are F C also equal (Prop. III. Bk. III.); hence likewise the triangles A 0 B, B B 0 C, C 0 D, &c. must be equal, since their sides are equal each to each (Prop. XVIII. Bk. I.); therefore all the angles A B C, B C D, C D E, &c. are equal; hence the figure A B C D E F is a regular polygon. PROPOSITION III. - THEOREM. 352. If from a common center a circle can be circumscribed about, and another circle inscribed within, a polygon, that polygon is regular. Suppose that from the point 0, E D as a center, circles can be circumscribed about, and inscribed in, the polygon ABCDEF; then that F 0 C polygon is regular. For, supposing it to be described, the inner one will touch all the A B sides of the polygon; therefore these sides are equally distant from its center; and consequently, being chords of the outer circle, they are equal; therefore they include equal angles (Prop. XVIII. Cor. 1, Bk. III.). Hence the polygon is at once equilateral and equiangular; consequently it is regular (Art. 344). PROPOSITION IV. -PROBLEM. 353. To inscribe a square in a given circle. Draw two diameters, A C, B D, intersecting each other at right angles; join their extremities, A, B, C, D, and the figure A B C D will be a square. 13 146 ELEMIENTS OF GEOMETRY. For, the angles A 0 B, B 0 C, &c. be- Dc ing equal, the chords AB, BC, &c. are also equal (Prop. III. Bk. III.); and the angles A B C, B C D, &c., being in- \ scribed in semicircles, are right angles (Prop. XVIII. Cor. 2, Bk. III.). Hence A B A B C D is a square, and it is inscribed in the circle ABCD. 354. Cor. Since the triangle A 0 B is right-angled and isosceles, we have (Prop. XI. Cor. 5, Bk. IV.), AB:: AO:: 2:; hence, the side of the inscribed square is to the radius as the square root of 2 is to unity. PROPOSITION V. - THEOREI. 355. The side of a regular hexagon inscribed in a circle is equal to the radius of the circle. Let ABCDEF be a regular ED hexagon inscribed in a circle, the center of which is O; then any Fy side, as B C, will be equal to the C radius 0A. Join B 0; and the angle at the centre, AO B, is one sixth of four A B right angles (Prop. II. Sch. 1), or one third of two right angles; therefore the two other angles, 0 A B, 0 B A, of the same triangle, are together equal to two thirds of two right angles (Prop. XXVIII. Bk. 1.). But A 0 and B 0 being equal, the angles OA B, B A are also equal (Prop. VII. Bk. I.); consequently, each is one third of two right angles. Hence the triangle A 0 B is equiangular; therefore A B, the side of the regular hexagon, is equal to A O, the radius of the circle (Prop. VIII. Cor. Bk. I.). BOOK VI. - 147 356. Cor. 1. To inscribe a regular E hexagon in a given circle, apply the radius, A O, of the circle six times, as a \; chord to the circumference. Hence, / 0 beginning at any point A, and applying \ / - A O six times as a chord to the circum- A C ference, we are brought round to the point of beginning, and the inscribed B figure A B C D E F, thus formed, is a regular hexagon. 357. Cor. 2. By joining the alternate angles. of the inscribed regular hexagon by the straight lines A C, C E, E A, the figure A C E, thus inscribed in the circle, will be an equilateral triangle, since its sides subtend equal arcs, AB C, CD E, E FA, on the circumference (Prop. III. Bk. III.). 358. Cor. 3. Join 0 A, O C, and the figure A B C 0 is a rhombus, for each side is equal to the radius. Hence, the sum of the squares of the diagonals A C, 0 B is equivalent to the sum of the squares of the sides (Prop. XV. Bk. IV.); or to four times the square of the radius 0 B; that is, AC2 + OB2 is equivalent to 4 AB2, or 4 0 B2; and taking away O B from both, there remains A C2 equivalent to 3 O B2; hence AC2: OB2: 3:1, or AC:OB:: 3:1; hence, the side of the inscribed equilateral triangle is to the radius as the square root of 3 is to unity. PROPOSITION VI. - PROBLEM. 359. To inscribe a regular decagon in a given circle. Divide the radius, 0 A, of the given circle, in extreme and mean ratio, at the point M (Prob. XXVII. Bk. V.). 148 ' ELEMENTS OF GEOMETRY. Take the chord A B equal to OM, and AB will be the side of a regular decagon inscribed in the circle. For we have by construction, AO: O:OM:: AM; \ or, since A B is equal to O M, - A: AB: AB: A M. B Draw MB and BO; and the triangles ABO, A MB have a common angle, A, included between proportional sides; hence the two triangles are similar (Prop. XXIV. Bk. IV.). Now, the triangle OA B being isosceles, A MB must also be isosceles, and A B is equal to B M; but A B is equal to O M, consequently M B is equal to M O; hence the triangle M B O is isosceles. Again, the angle A M B, being exterior to the isosceles triangle BMO, is double the interior angle O (Prop. XXVII. Bk. I.). But the angle A MB is equal to the angle M A B; hence the triangle OA B is such, that each of the angles at the base, A B, OB A, is double the angle 0, at its vertex. Hence the three angles of the triangle are together equal to five times the angle 0, which consequently is a fifth part of two right angles, or the tenth part of four right angles; therefore the arc AB is the tenth part of the circumference, and the chord A B is the side of an inscribed regular decagon. 360. Cor. 1. By joining the vertices of the alternate angles A, C, &c. of the regular decagon, a regular pentagon may be inscribed. Hence, the chord A C is the side of an inscribed regular pentagon. 361. Cor. 2. A B being the side of the inscribed regular decagon, let AL be the side of an inscribed regular hexagon (Prop. V. Cor. 1). Join B L; then B L will be the side of an inscribed regular pentedecagon, or regular polygon of fifteen sides. 'For A B cuts off an arc equal to a tenth part of the circumference; and A L subtends an BOOK VI. 149 arc equal to a sixth of the circumference; therefore B L, the difference of these arcs, is a fifteenth part of the circumference; and since equal arcs are subtended by equal chords, it follows that the chord B L may be applied exactly fifteen times around the circumference, thus forming a regular pentedecagon. 362. Scholium. If the arcs subtended by the sides of any inscribed regular polygon be severally bisected, the chords of those semi-arcs will form another inscribed polygon of double the number of sides. Thus, from having an inscribed square, there may be inscribed in succession polygons of 8, 16, 32, 64, &c. sides; from the hexagon may be formed polygons of 12, 24, 48, 96, &c. sides; from the decagon, polygons of 20, 40, 80, &c. sides; and from the pentedecagon, polygons of 30, 60, 120, &c. sides. NOTE. - For a long time the polygons above noticed were supposed to include all that could be inscribed in a circle. In the year 1801, M. Gauss, of G6ttingen, made known the curious discovery that the circumference of a circle could be divided into any number of equal parts capable of being expressed by the formula 2" -+ 1, provided it be a prime number. Now, the number 3 is the simplest of this kind, it being the value of the above formula when the exponent n is 1; the next prime number is 5, and this is contained in the formula. But the polygons of 3 and of 5 sides have already been inscribed. The next prime number expressed by the formula is 17, so that it is possible to inscribe a regular polygon of 17 sides in a circle. The investigations, however, which establish this geometrical fact involve considerations of a nature that do not enter into the elements of Geometry. PROPOSITION VII.- PROBLEM. 363. A regular inscribed polygon being given, to circumscribe a similar polygon about the sanme circle. Let A B C D E F be a regular polygon inscribed in a circle whose centre is O. Through M, the middle point of the arc A B, draw the tangent, G H; also draw tangents at the middle points of 13* 150 ELEMENTS OF GEOMETRY. the arcs B C, CD, &c.; these tangents are parallel to the chords AB, BC, CD, &c. (Prop. XI. \ Bk. III., and Prop. VI. Cor. 1, G A. -D/. Bk. III.), and by their intersec- / tions form the regular circum-: L: scribed polygon GHI, &c. similar H N I to the one inscribed. Since M is the middle point of the arc A B, and N the middle point of the equal arc B C, the arcs B M, B N are halves of equal arcs, and therefore are equal; that is, the vertex, B, of the inscribed polygon is at the middle point of the arc MN. Draw OH; the line OH will pass through the point B. For, the right-angled triangles OM H, ONH, having the common hypothenuse OH, and the side 0 M equal to ON, must be equal (Prop. XIX. Bk. I.), and consequently the angle M 0 H is equal to H 0 N, wherefore the line OH passes through the middle point, B, of the arc M N. In like manner, it may be shown that the line 01 passes through the middle point, C, of the arc N P; and so with the other vertices. Since G H is parallel to A B, and H I to B C, the angle G H I is equal to the angle A B C (Prop. XXVI. Bk. I.); in like manner, H I K is equal to B C D; and so with the other angles; hence, the angles of the circumscribed polygon are equal to those of the inscribed polygon. And, further, by reason of these same parallels, we have GH: AB:: OH: OB, and HI: BC: OI: OB; therefore (Prop. X. Bk. II.), GH: AB:: HI: BC. But A B is equal to B C, therefore G H is equal to H I. For the same reason, HI is equal to I K, &c.; consequently, the sides of the circumscribed polygon are all equal; hence this polygon is regular, and similar to the inscribed one. BOOK VI. 151 364. Cor. 1. Conversely, if the circumscribed polygon G II I K, &c. is given, and it is required, by means of it, to construct a similar ilscribed polygon, draw the straight lines O G, O H, &c. from the vertices of tle angles G, H, I, &c. of the given polygon to tle center; tlle lines will meet the circumference in tlle points A, B, C, &c. Join these points by tlle clords A B, B C, &c., which will form the inscribed polygon. Or simply join the points of contact, M, N, P, &c., by chords, MAN, N P, &c., which likewise would form an inscribed polygon similar to the circumscribed one. 365. Cor. 2. Hence, we may circumscribe about a circle any regular polygon similar to an iiscribed one, and conversely. 366. Cor. 3. It has been shown tlat N I and H M are equal; therefore tlle sum of N II and II M, which is equal to the sum of H M and M G, is equal to H G, one of the equal sides of the polygon. 367. Scholium. From having a circumscribed regular polygon, another having double the number of sides may be readily constructed, by drawing tangents to the points of bisection of the arcs, intercepted by the sides of the proposed polygon, limitilg these tangents by tlose sides. In like manner other circumscribed polygons may be formed; but it is plain that each of the polygons so formed will be less than the preceding polygon, being entirely comprehended in it. PROPOSITION VIII.- THEOREM. 368. Th7e area of a regrular polygon is equivalent to the product of its perimeter by half of the radius of the inscribed circle. Let A B CD E F be a regular polygon, and O the centre of the inscribed circle. From O let tlle straight lines O A, O B, &c. be drawn to 152 ELEMENTS OF GEOMETRY. the vertices of all the angles of E D the polygon, and the polygon will be divided into as many equal triangles as it has sides; and let F O the radii 0M, ON, &c. of the ilnscribed circle be drawn to the centres of the sides of the polygon, or to the points of tangency M, N, &c., and these radii are perpendicular to the sides respectively (Prop. XI. Bk. III.); therefore the radius of the circle is equal to the altitude of the several triangles. Now, the triangle A 0 B is measured by the product of AB by half of 0 M (Prop. VI. Bk. IV.); the triangle O B C by the product of B C by half of O N. But O M is equal to 0 N; hence the two triangles taken together are measured by the sum of A B and B C by half of 0 M. In like manner the measure of the other trialgles may be found; hence, the sum of all the triangles, or the whole polygon, is equal to the sum of the bases A B, B C, &c., or the perimeter of the polygon, multiplied by half of OM, or half the radius of the inscribed circle. PROPOSITION IX. -THEOREM. 369. The perimeters of two regular polygons, having the same number of sides, are to each other as the radii of the circumscribed circles, and, also, as the radii of the inscribed circles; and their areas are to each other as the squares of those radii. Let A B be a side of one polygon, O 0 the center, and consequently 0 A / the radius of the circumscribed circle, and 0 M, perpendicular to A B, the radius of the inscribed circle. / Let G H be a side of the other poly- gon, C the center, C G and C N the A B G H radii of the circumscribed and the inscribed circles. BOOK VI. 153 The perimeters of tlle two polygons are to eacl other as the sides AB and GH (Prop. XXXI. Bk. IV.), but tlle angles A and G are equal, being each half of the angle of the polygon; so also are the angles B and H; hence, drawing 0 B and C H, tlle isosceles triangles A B O, GH C are similar, as are likewise the right-angled triangles A I 0, G N C; hence AB: GH:: AO: GC:: MO: NC. Hence, the perimeters of the polygons are to each other as the radii A O, G C of the circumscribed circles, and, also, as the radii M O, N C of the inscribed circles. The surfaces of these polygons are' to each other as the squares of the homologous sides A B, G H (Prop. XXXI. Bk. IV.); they are thlerefore to each other as the squares of A O, G C, the radii of tlle circumscribed circles, or as the squares of 0 M, C N, the radii of the inscribed circles. PROPOSITION X. - PROBLEM. 370. The surface of a regular inscribed polygon, and that of a similar circumscribed polygon, being given; to find the surfaces of regular inscribed and circumscribed polygons having double the number of sides. Let AB be a side of the given E P M inscribed polygon; E F, parallel to A B, a side of the circumscribed. polygon, and C the centre of the i circle. Draw the cllord A M, and tle tangents AP, BQ; then AM M will be a side of the inscribed polygon, having twice tle number of C sides; and P Q, the double of P M, will be a side of the similar circumscribed polygbn. Let A, then, be the surface of the inscribed polygon whose side is AB, B that of the similar circumscribed polygon; A' the surface of the polygon whose side is A M, 154 ELEMENTS OF GEOMETRY. B' that of the similar circumscribed polygon: A and B are given; we have to find A' and B'. First. The triangles A C D, A C M, whose common vertex is A, are to each other as their bases C D, C M (Prop. VI. Cor., Bk. IV.); they are likewise as the polygons A and A'; hence E P M Q F ~,, C A: A':: CD: C M. Again, the triangles C A M, C M E, whose common vertex is M, are to each other as their bases C A, C E; they are likewise to each other as the polygons A' and B; hence A': B:: CA: CE. But, since A D and M E are parallel, we have, CD: CM:: CA: CE; hence A: A':: A': B; hence, the polygon A' is a mean proportional between the two given polygons. Secondly. The altitude C M being common, the triangle CPM is to the triangle CPE as PM is to PE; but since CP bisects the angle MC E, we have (Prop. XIX. Bk. IV.), PM: PE:: CM: CE:: CD: CA:: A: A'; hence CPM: CPE:: A: A'; and, consequently, CPM: CPM +CPE or CME: A:A + A'. But C M P A or 2 C MP and C M E are to each other as the polygons B' and B; hence B': B:: 2 A: A + A'; which gives 2A B A + A7 BOOK VI. 155 or, the polygon B' is equal to the quotient of twice the product of the given polygons divided by the sum of the inscribed polygons. Thus, by means of the polygons A and B, it is easy to find tle polygons A' and B', which have double the number of sides. PROPOSITION XI.- THEOREM. 371. A circle being given, two similar polygons can always be formed, the one circumscribed about the circle, the other inscribed in it, which shall differ from each other by less than any assignable surface. Let Q be the side of a square I less than the given surface. H Bisect A C, a fourth part of. the circumference, and tlhen bi- sect the half of this fourtlL, and G A o - L so proceed until an arc is found Q whose chord AB is less than Q. As this arc must be an ex- act part of the circumference, if we apply the chords A B, B C, &c., each equal to AB, the last will terminate at A, and tlere will be inscribed in the circle a regular polygon, A B C D E, &c. Next describe about the circle a similar polygon, G H I K L, &c. (Prop. VII.); and the difference of these two polygons will be less than the square of Q. Find the center, 0; from the points G and H draw the straight lines G O, H 0, and they will pass through the points A and B (Prop. VII.). Draw also OM to the point of tangency, M; and it will bisect A B in N, and be perpendicular to it (Prop. VI. Cor. 1, Bk. III.). Produce AO to E, and draw BE. Let P represent the circumscribed polygon, and p the inscribed polygon. Then, since these polygons are similar, they are as the squares of tlhe homologous sides G H, 156 ELEMENTS OF GEOMETRY. A B (Prop. XXXI. Bk. IV.); but the triangles G OH, AOB are similar (Prop. XXIV. Bk. IV.); hence they are to each other as the squares of the homologous sides O G and O A (Prop. XXIX. Bk. IV.); therefore P:p:: 0G2: A2 or'-M2. Again, the triangles 0 G M, E A B, having their sides respectively parallel, are similar; therefore P:p: G2: fO::E2 I-:BE; and, by division, P: P - p:: A-E2: E2 - E B2 or AB2. But P is less than the square described on the diameter A E; therefore P —p is less than tile square described on AB, that is, less than the given square Q. Hence, the difference between the circumscribed and inscribed polygons may always be made less than any given surface. 372. Cor. Since the circle is obviously greater than any inscribed polygon, and less thlan any circumscribed one, it follows that a polygon may be inscribed or circumscribed, which will differ from the circle by less than any assignable magnitude. PROPOSITION XII. - PROBLEM. 373. To find the approximate area of a circle whose radius is unity. Let the radius of the circle he 1, and let the first inscribed and circumscribed polygons be squares; the side of the inscribed square will be 2- (Prop. IV. Cor.), and that of the circumscribed square will be equal to the diameter 2. Hence the surface of the inscribed square is 2, and that of the circumscribed square is 4. Let, therefore A= 2, and B= 4. Now it has been proved, in Proposition X., that the surface of the inscribed octagon, or, as it has been represented, A', is a meall proportional / BOOK VI.15 157 between the two squared A and B, so that A,' = V8=f 2.8284271; and it has also been proved, in the same proposition, that the circumscribed octagon, represented by B', 2A x B. so that B' = 3.3137085. The A~~_+ -A/ ~ 2 +-V 8 inscribed and the circumscribed octagons being thus determined, we can easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put A 2.8284271, B - 3.3137085; and we shall find A' = Wy A x B = 3.0614674, and B' = 2 A x B - 3.1825979. A + Al In like manner may be determined the area of polygons of sixteen sides, and thence the area of polygons of thirtytwo sides, and so on till we arrive at an inscribed an~d a circumscribed polygon differing so little from each other, and consequently from the circle, that the difference shall be less than any assignable magnitude (Prop. XI. Cor.). The subjoined table exhibits the area, or numerical expression for the surface, of these polygons, carried on till they agree as far as the seventh place of decimals. Number of sides. Inscribed Polygons. 4.... 2.0000000 8.... 2.8284271 16....3.0614674 32....3.1214451 64....3.1365485 128....3.1403311 256....3.1412772 512....3.1415138 1024....3.1415729 2048....3.1415877 4096....3.1415914 8192....3.1415923 16384 3.1415925 32768....3.1415926 14 Circumscribed Polygons. 4.0000000 3.3137i085 3.1825979 3.1517249 3.1441148 3.1422236 3.1417504 3.1416321 3.1416025 3.14159551 3.1415933 3.1415928 3.1415927 3.1415926 158 ELEMENTS OP GEOMETRY. It appears, therefore, that the inscribed and circumscribed polygons of 32768 sides differ so little from each other that the numerical value of each, as far as seven places of decimals, is absolutely the same; as the circle is between the two, it cannot, strictly speaking, differ from either so much as they do from each other; so that the number 3.1415926 expresses the area of a circle whose radius is 1, correctly, as far as seven places of decimals. Some doubt may exist, perhaps, about the last decimal figure, owing to errors proceeding from the parts omitted; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place. 374. Cor. Since the inscribed and circumscribed polygons are regular, and have the same number of sides, they are similar (Prop. I.); tlerefore, by increasing the number of the sides, the corresponding polygons formed will approach to an equality with the circle. Now if, by continual bisections, the polygons formed shall have their number of sides indefinitely great, each side will become indefinitely small, and the inscribed and circumscribed polygons will ultimately coincide with each other. But when they coincide with each other, they must each coincide with the circle, since no part of al inscribed polygon can be without the circle, nor can any part of a circumscribed one be within it; hence, the perimeters of the polygons must coincide with the circumference of the circle, and be equal to it. 375. Scholium. Every circle, therefore, may be regarded as a polygon of an infinite number of sides. NOTE. -This new definition of the circle, if it does not appear at first view to be very strict, has at least the advantage of introducing more simplicity and precision into demonstrations. (Cours de Geomgtrie Elementaire, par Vincent et Bourdon.) BOOK VI. 159 PROPOSITION XIII. - THEOREM. 376. The circumferences of circles are to each other as their radii, and their areas are to each other as the squares of their radii. Let C denote the circumference of one of A B the circles, R - its radius OA, A its area; and \\ let C' denote the circumference of the other circle, r its radius 0 B, A' its area; then will C: C': R: r, and A: A':: R2: r2. Inscribe within the given circles two regular polygons of the same number of sides; and, whatever be the number of sides, the perimeters of the polygons will be to each other as the radii O A and O B (Prop. IX.). Now, conceive the arcs subtending the sides of the polygon to be continually bisected, forming other inscribed polygons, until polygons are formed of an indefinite number of sides, and therefore having perimeters coinciding with the circumference of the circumscribed circles (Prop. XII. Cor.); and we shall have C: C':: R: r. Again, the areas of the inscribed polygons are to each other as 0 A2 to 0 B2 (Prop. IX.). But when the number of sides of the polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles; hence we shall have A: A':: R2: r2. ELEMENTS OF GEOMETRY. 377. Cor. 1. The circumferences of circles are to each other as twice their radii, or as their diameters. For, multiplying the terms of the second ratio in the first proportion by 2, we have C: C':: 2 R: 2 r. 378. Cor. 2. The areas of circles are to each other as the squares of their diameters. For, multiplying the second ratio of the second proportion by 4, or 2 squared, we have A: A':: 4 R2: 4r2. PROPOSITION XIV.- THEOREM. 379. Similar arcs are to each other as their radii; and similar sectors are to each other as the squares of their radii. Let A B, DE be similar arcs; AC B, DOE, similar A B D E sectors; and denote the radii CA and OD by R and r; then will AB: DE::R:r, c o and ACB:DOE:: R2: r2. For, since the arcs are similar, the angle C is equal to the angle O (Art. 213). But the angle C is to four right angles as the arc AB is to the whole circumference described with the radius CA (Prop. XVII. Sch. 2, Bk. III.); and the angle 0 is to four right angles as the arc D E is to the circumference described with the radius 0 D. Hence, the arcs A B, D E are to each other as the circumferences of which they form a part. But these circumferences are to each other as their radii, C A, 0 D (Prop. XIII.); therefore Arc A B: Arc DE:: R: r. By like reasoning, the sectors A C B, D O E are to each BOOK VI. 161 other as the whole circles of which they are a part; and these are as the squares of their radii (Prop. XIII.); therefore Sector A C B: Sector D E:: R2: r2. PROPOSITION XV.- THEOREM. 380. The area of a circle is equal to the product of the circumference by half the radius. Let C denote the circumference of the circle, whose center is O, R its / P. radius OA, and A its area; then will // \ A-=C X R. For, inscribe in the circle any reg- ular polygon, and from the center draw OP perpendicular to one of the sides. The area of the polygon, whatever be the number of sides, will be equal to its perimeter multiplied by half of 0 P (Prop. VIII.). Conceive the arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides; its perimeter will be equal to the circumference of the circle (Prop. XII. Cor.), and 0 P be equal to the radius O A; therefore the area of the polygon is equal to that of the circle; hence A - C X R. 381. Cor. 1. The area of a sector is equal to the product of its arc by half of its radius. For, let C denote the circumference of the circle of which'the sector D OE is a part, R its radius O D, and A its area; then we shall have (Prop. XVII. Sch. 2, Bk. III.), Sector D O E: A: Arc D E: C; 1l* 162 ELEMENTS OF GEOMETRY. hence, since equimultiples of two magnitudes have the same ratio as the magnitudes themselves (Prop. IX. Bk. II.), Sector D OE: A:: Arc D E X ~ R: C X. R. But A, or the area of the whole circle, is equal to C X f R; hence, the area of the sector D 0 E is equal to the arc D E X R. 382. Cor. 2. Let the circumference of the circle whose diameter is unity be denoted by 7 (which is called pi), the radius by R, and the diameter by D; and the circumference of any other circle by C, and its area by A. Then, since circumferences are to each other as their diameters (Prop. XIII. Cor. 1), we shall have, C: D:: n: 1; therefore C= D 7r 2RX r. Multiplying both numbers of this equation by ~ R, we have C X R R2X I, or A = R2 X 7; that is, the area of a circle is equal to the product of the square of its radius by the constant number n. 383. Cor. 3. The circumference of every circle is equal to the product of its diameter, or twice its radius, by the constant number n. 384. Cor. 4. The constant number n denotes the ratio of the circumference of any circle to its diameter; for C 385. Scholiumn 1. The exact numerical value of the ratio denoted by nr can be only approximately expressed. The approximate value found by Proposition XII. is 3.1415926; but, for most practical purposes, it is sufficiently accurate to take r == 3.1416. Tile symbol n is the first letter of the Greek word rrepltperpov, perimetron, wlhicll signifies circumference. BOOK VI. 168 386. Scholium 2. The QUADRATURE OF THE CIRCLE is the problem which requires the finding of a square equivalent in area to a circle having a given radius. Now, it has just been proved that a circle is equivalent to the rectangle contained by its circumference and half its radius; and this rectangle may be changed into a square, by finding a mean proportional between its length and its breadth (Prob. XXVI. Bk. V.). To square the circle, therefore, is to find the circumference when the radius is given; and for effecting this, it is enough to know the ratio of the circuinference to its radius, or its diameter. But this ratio has never been determined except approximately; but the approximation has been carribd so far, that a knowledge of the exact ratio would afford no real advantage whatever beyond that of the approximate ratio. Professor Rutherford extended the approximation to 208 places of decimals, and Dr. Clausen to 250 places. The value of Tr, as developed to 208 places of decimals, is 3.14159265358979323846264338327950288419716939937 5105820974944592307816406286208998628034825342717 0679821480865132823066470938446095505822317253594 0812848473781392038633830215747399600825931259129 40183280651744. Such an approximation is evidently equivalent to perfect correctness; the root of an imperfect power is in no case more accurately known. PROPOSITION XVI. - PROBLEM. 387. To divide a circle into any number of equal parts by means of concentric circles. Let it be proposed to divide the circle, whose center is 0, into a certain number of equal parts, - three for instance, -by means of concentric circles. Draw the radius AO; divide AO into three equal parts, A B, B C, C O. Upon A O describe a semi-circumference, 164 ELEMENTS OF GEOMETRY. and draw the perpendiculars, B E, C D, meeting that semi-circumference il the points E, D. Join ~ S 0 E, 0 D, and with these lines as \ radii from the center, O, describec A circles; these circles will divide the given circle into the required number of equal parts. For join A E, A D; then the angle A D O, being in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.); hence the triangles D A 0, D C 0 are similar, and consequently are to each other as the squares of their homologous sides; that is, DAO:DCO::: A2: OD2; but DAO: DCO:: OA: 0C; hence OA 2: D2::OA: 0C; consequently, since circles are to each other as the squares of their radii (Prop. XIII.), it follows that the circle whose radius is OA, is to that whose radius is OD, as OA to 0 C; that is to say, the latter is one third of the former. In the same manner, by means of the right-angled triangles E A O, E B O, it may be proved that the circle whose radius is 0 E, is two thirds that whose radius is 0 A. Hence, the smaller circle and the two surrounding annular spaces are all equal. NOTE. —This useful problem was first solved by Dr. Hutton, the justly distinguished English mathematician. BOOK VII. PLANES. - DIEDRAL AND POLYEDRAL ANGLES. DEFINITIONS. A 388. A STRAIGHT line is perpendicular to a plane, when it is M perpendicular to every straight line which it meets in that plane. Conversely, the plane, in the /N same case, is perpendicular to the line. The foot of the perpendicular is the point in which it meets the plane. Thus the straight line A B is perpendicular to the plane MN; the plane M N is perpendicular to the straight line A B; and B is the foot of the perpendicular A B. 389. A line is parallel to a plane when it cannot meet the plane, however far both of them may be produced. Conversely, the plane, in the same case, is parallel to the line. 390. Two planes are parallel to each other, when they cannot meet, however far both of them may be produced. 391. A DIEDRAL ANGLE is an M angle formed by the intersection of two planes, and is measured by the inclination of two straight lines drawn from any A. — - -*- N — point in the line of intersection, perpendicular to that line, one being drawn in each plane. B ELEMENTS OF GEOMETRY. The line of common section AM is called the edge, and the two planes are called the faces, of the diedral angle. Thus the two planes A B M, A —. — -— L N A B N, whose line of intersection is A B, form a diedral angle, of which the line AB B is the edge, and the planes A BM, A BN are the faces. 392. A diedral angle may be acute, right, or obtuse. If the two faces are perpendicular to each other, the angle is right. 393. A POLYEDRAL ANGLE is S an angle formed by the meeting at one point of more than two plane angles, which are not in the same plane. The common point of meeting A.-. C of the planes is called the vertex, each of the plane angles a face, 1 and the line of common section of any two of the planes an edge of the polyedral angle. Thus the three plane angles A S B, B S C, C S A form a polyedral angle, whose vertex is S, whose faces are the plane angles, and whose edges are the sides, AS, BS, CS, of the same angles. 394. A polyedral angle formed by three faces is called a triedral angle; by four faces, a tetraedral; by five faces, a pentaedral, &c. PROPOSITION I. - THEOREM. 395. A straight line cannot be partly in a plane, and partly out of it. For, by the definition of a plane (Art. 10), a straight BOOK VII. 167 line which has two points in common with a plane lies wholly in that plane. 396. Scholium. To determine whether a surface is a plane, apply a straight line in different directions to that surface, and ascertain whether the line throughout its whole extent touches the surface. PROPOSITION II. - THEOREM. 397. Two straight lines which intersect each other lie in the same plane and determine its position. Let A B, A C be two straight lines A which intersect each other in A; then these lines will be in the same plane. Conceive a plane to pass through AB, and to be turned about AB, B until it pass through the point C; then, the two points A and C being in this plane, the line A C lies wholly in it (Art. 10). Hence, the position of the plane is determined by the condition of its containing the two straight lines A B, A C. 398. Cor. 1. A triangle, A B C, or three points, A, B, C, not in a straight line, determine the position of a plane. E 399. Cor. 2. Hence, also, two parallels, AB, CD, determine the position of a plane; for, A drawing the secant E F, the plane of the two straight lines C D A B, E F is that of the parallels AB, CD. F PROPOSITION III.- THEOREM. 400. If two planes cut each other, their common section is a straigkt line. 168 ELEMENTS OF GEOMETRY. Let the two planes A B, C D cut each other, and let E, F be two points in their C B common section. Draw the straight linle E F. Now, since the points E and F are in the plane A B, and also in the plane CD, the straight line E F, joining E and F, must be wholly in each plane, or is conmon to both of them. Therefore, tlhe common section of the two planes AB, A 1) CD is a straight line. PROPOSITION IV. THEOREM. 401. If a straight line is perpendicular to each of two straight lines, at their point of intersection, it is perpendicular to the plane in which the two lines lie. Let tle straight line AB be A perpendicular to each of the straight lines C D, E F, at B, the point of their intersection, and M MN the plane in which the lines / / C D, EF lie; then will A B be /E l / perpendicular to the plane MN. N Through the point B draw any straight line, B G, in the plane M N; and through any point G draw D G F, meeting the lines C D, E F in such a manner that D G shall be equal to GF (Prob. XXVIII. Bk. V.). Join AD, AG, AF. The line D F being divided into two equal parts at tile point G, the triangle D B F gives (Prop. XIV. Bk. IV.) B F2 + BD = 2 BG2+ 2 GF2. The triangle D A F, in like manner, gives A F2 + A D2 = 2 AG2 + 2 G F2 Subtracting the first equation from tile second, and ob BOOK VII. 169 serving that the triangles A B F, A B D, each being rightangled at B, give A F2- BF2 =A B and AD2. B D2 = B, we shall have AB2 +AB2 AG2 -2 BG2. Therefore, by taking the halves of both members, we have AB =AG -BG2, or AG =AB2 +B G2; hence, the triangle A B G is right-angled at B, and the side A B is perpendicular to B G. In the same manner, it may be shown that A B is perpendicular to any other straight line il the plane MN, which it may meet at B; therefore AB is perpendicular to the plane MN (Art. 388). 402. Scholium. Thus it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot, in a plane, but it always must be so whenever it is perpendicular to two straight lines drawn in the plane; which shows the accuracy of the first definition (Art. 388). 403. Cor. 1. The perpendicular AB is shorter than any oblique line A G; therefore it measures the shortest distance from the point A to the plane M N. 404. Cor. 2. From any given point, B, in a plane, only one perpendicular to that plane can be drawn. For if there could be two, conceive a plane to pass through them, intersecting the plane MN in B G; the two perpendiculars would then be perpendicular to the straight line B G at the same point, and in the same plane, which is impossible (Prop. XIII. Cor., Bk. I.). It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane. For, suppose A B, A G to be two such perpendiculars, then the triangle A B G will have two right angles, A B G, A G B, which is impossible (Prop. XXVIII. Cor. 3, Bk. I.). 15 ELEMENTS OF GEOMETRY. PROPOSITION V. - THEOREM. 405. Oblique lines drawn from a point to a plane at equal distances from a perpendicular drawn from the same point to it, are equal, and of two oblique lines unequally distant from the perpendicular, the more remote is the longer. Let A B be perpendicular A to the plane MN; and AC, AD, AE be oblique lines, from the point A, meeting / the plane at equal distances, /.. BC, BD, BE, from the per- / ' F pendicular; and A F a line meeting the plane more remote from the perpendicular; then will A C, A D, A E be equal to each other, and A F be longer than A C. For, the angles A B C, A B D, A B E being right angles, and the distances B C, B D, B E being equal to each other, the triangles A B C, A B D, A B E have in each an equal angle contained by equal sides; consequently they are equal (Prop. V. Bk. I.); therefore, the hypothenuses, or the oblique lines A C, A D, A E, are equal to each other. In like manner, since the distance B F is greater than BC, or its equal BE, the oblique line AF must be greater than AE, or its equal AC (Prop. XIV. Bk. I.). 406. Cor. All the equal oblique lines A C, AD, A E, &c. terminate in the circumference of a circle, C D E, described from B, the foot of the perpendicular, as a center; therefore, a point, A, being given out of a plane, the point B, at which the perpendicular let fall from it would meet that plane, may be found by taking upon the plane three points, C, D, E, equally distant from the point A, and then finding the center of the circle which passes through these points; this centre will be the point B required. BOOK VII. 171 407. Scholium. The angle A C B is called the inclination of the oblique line A C to the plane M N; which inclination is evidently equal with respect to all such lilies, A C, A D, A E, as are equally distant from the perpendicular; for all the triangles A C B, AD B, A E B, &c. are equal to each other. PROPOSITIN VI.- THEOREM. 408. If from the foot of a perpendicular a straight line be drawn at right angles to any straight line of the plane, and a straight line be drawn from the point of intersection to any point of the perpendicular, this last line will be perpendicular to the line of the plane. Let A B be perpendicular to the plane MN, and B D a straight line drawn through B, cutting at right angles the \ straight line C E in the plane; draw the straight line A D from the point of intersection, D, to M any point, A, in the perpendicular A B; and AD will be perpendicular to CE. For, take DE equal to D C, and join BE, BC, AE, AC. Since D E is equal to D C, the two right-angled triangles B D E, B D C are equal, and the oblique line B E is equal to B C (Prop. V. Bk. I.); and since B E is equal to B C, the oblique line A E is equal to A C (Prop. V. Bk. I.); therefore the line AD has two of its points, A and D, equally distant from tlhe extremities E and C; hence, A D is a perpendicular to E C, at its middle point, D (Prop. XV. Cor., Bk. I.). 409. Cor. It is also evident tllat C E is perpendicular to the plane of tlle triangle A B D, si1nce C E is perpendicular at the same time to the two straight lines A D and B D (Prop. IV.). 172 ELEMENTS OF GEOMETRY. PROPOSITION VII.- THEOREM. 410. If a straight line is perpendicular to a plane, every plane which passes through that line is also perpendicular to the plane. Let A B be a straight line perpendicular to the plane M N; then will any plane, A C, passing through A B, be perpendicular to MN. M.... For, let C D be the inter- section of the planes Aa, / E- D MN; in the plane MN draw E F, through the point B, perpendicular to C D; then the line A B, being perpendicular to the plane M N, is perpendicular to each of the two straight lines C D, E F (Art. 388). But the angle A B E, formed by the two perpen. diculars A B, E F to their common section, C D, measures the angle of the two planes A C, MN (Art. 391); hence, since that angle is a right angle, the two planes are perpendicular to each other. 411. Cor. When three straight lines, as A B, CD, E F, are perpendicular to each other, each of those lines is perpendicular to the plane of the other two, and the three planes are perpendicular to each other. PROPOSITION VIII.- THEOREM. 412. If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their common section, will be perpendicular to the other plane. Let A C, MN be two planes perpendicular to each other, and let the straight line AB be drawn in the plane A C perpendicular to the common section C D; then will A B be perpendicular to the plane M N. BOOK VII. For, in the plane MN, draw E F, through the point A B, perpendicular to CD; then, since the planes AC, M N are perpendicular, the angle M, A B E is a right angle (Art. /. 391); therefore the line AB E D is perpendicular to the two straight lines C D, E F, at the point of their intersection; hence it is perpendicular to their plane, MN (Prop. IV.). 413. Cor. If the plane A C is perpendicular to the plane M N, and if at a point B of the common section we erect a perpendicular to the plane M N, that perpendicular will be in the plane AC. For, if not, there may be drawn ill the plane A C a line, A B, perpendicular to the common section CD, which would be at the same time perpendicular to the plane M N. Hence, at the same point B there would be two perpendiculars to the plane MN, which is impossible (Prop. IV. Cor. 2). PROPOSITION IX. THEOREM. 414. If two planes which cut each other are perpendicular to a third plane, their common section is perpendicular to the same plane. Let the two planes C A, D A, which cut each other in the straight line A B, be each perpendicular to the plane MN; then will their common section A B be per-. pendicular to MN. For, at the point B, erect N a perpendicular to the plane M N; that perpendicular must be at once in the plane C A and in the plane D A (Prop. VIII. Cor.); hence, it is their common section, A B. 15* ELEMENTS OF GEOMETRY. PROPOSITION X. -THEOREM. 415. If one of two parallel straight lines is perpendicular to a plane, the other is also perpendicular to the same plane. Let AB, CD be two parallel A straight lines, of which A B is per- I pendicular to the plane MN; then M will C D also be perpendicular to it. For, pass a plane through the B / parallels A B, CD, cutting the plane... D M N in the straight line BD. In N the plane M N draw the straight line E F, at right angles with BD; and join AD. Now, E F is perpendicular to the plane A B D C (Prop. VI. Cor.); therefore the angle CD E is a right angle; but the angle C D B is also a right angle, since A B is perpendicular to B D, and C D parallel to A B (Prop. XXII. Cor., Bk. I.); therefore the line CD is perpendicular to the two straight lines E F, B D; hence it is perpendicular to their plane, M N (Prop. IV.). 416. Cor. 1. Conversely, if the straight lines A B, C D are perpendicular to the same plane, M N, they must be parallel. For, if they be not so, draw, through the point D, a line parallel to A B; this parallel will be perpendicular to the plane M N; hence, through the same point D more than one perpendicular may be erected to the same plane, which is impossible (Prop. IV. Cor. 2). 417. Cor. 2. Two lines, A and B, parallel to a third, C, are parallel to each other; for, conceive a plane perpendicular to the line C; the lines A and B, being parallel to C, will be perpendicular to the same plane; hence, by the preceding corollary, they will be parallel to each other. The three lines are supposed to be not in the same plane; otherwise the proposition would be already demonstrated (Prop. XXIV. Bk. I.). BOOK VII. PROPOSITION XI. -THEOREM. 418. If a straight line without a plane is parallel to a line within the plane, it is parallel to the plane. Let tlhe straight line A B, with- A B out the plane M N, be parallel to I: the line C D il that plane; thenl M/ -- will A B be parallel to the plane / i MN. /c -- D Conceive a plane A B C D to pass tllrough tlle parallels AB, CD. Now, if the line A B, which lies in the plane A B C D, could meet the plane MN, it could only be in some point of the line C D, the common section of the two planes; but the line A B cannot meet C D, since they are parallel (Art. 17); tlerefore it will not meet tlhe plane M N; hence it is parallel to that plane (Art. 389). PROPOSITION XII. -THEOREM. 419. If two planes are perpendicular to the samne straight line, they are parallel to each other. Let the planes AM N, M A PQ, be each perpendic- N/..^ ---- '.j N ular to the strailght line 0. --- A B; then will they be parallel to each other.. p For, if they can meet,.. —........- on being produced, let O be one of their corn- Q mon points; and join O A, O B. The line A B, which is perpendicular to the plane MI N, is perpendicular to the straight line 0 A, drawn through its foot in that plane (Art. 388). For the same reason, AB is perpendicular to B 0. Therefore 0 A and 0 B are two perpendiculars let fall from tile same point, 0, upon the same straight line, A B, which is impossible (Prop. XIII. Bk. I.). ELEMENTS OF GEOMETRY. Therefore, the planes M N, P Q cannot meet on being produced; hence they are parallel to each other. PROPOSITION XIII. - THEOREM. 420. If two parallel planes are cut by a third plane, the two intersections are parallel. Let the two parallel planes MN and P Q be cut by the plane E F G H, and let their intersections with it be E F, G H; then E F is parallel to G H. For, if the lines E F, G H, lying in the same plane, were not parallel, they would meet each F M P G other on being produced; therefore the planes M N, P Q, in which those lines are situated, would also meet, which is impossible, since these planes are parallel. PROPOSITION XIV. THEOREM. 421. A straight line which is perpendicular to one of two parallel planes, is also perpendicular to the other plane. Let M N, P Q be two parallel planes, and AB a straight line perpendicular to the plane M N; then A B is also perpendicular to the plane P Q. Draw any line, B C, in the plane P Q; and through the lines A B, B C, conceive a plane, A B C, to / B f................. P/. D *.......... Q pass, intersecting the plane M N in A D; the intersection AD will be parallel to B C (Prop. XIII.). But the line A B, being perpendicular to the plane M N, is perpendicular to the straight line A D; consequently it will be perpendicular to its parallel BC (Prop. XXII. Cor., Bk. I.). BOOK VII. 177 Hence the line A B, being perpendicular to any line, B C, drawn through its foot in the plane P Q, is consequently perpendicular to the plane P Q (Art. 388). PROPOSITION XV.- THEOREM. 422. Parallel straight lines included between two parallel planes are equal. Let EF, GH be two parallel straight planes, included between E -- / two parallel planes, MN, PQ; N then E F and G H are equal. For, through the parallels EF, p G H conceive the plane E F GH /.-' / to pass, intersecting the parallel F planes in E G, F H. Tile intersections E G, F H are parallel to each other (Prop. XIII.); and E F, G H are also parallel; consequently the figure E F G H is a parallelogram; hence E F is equal to GH (Prop. XXXI. Bk. I.). 423. Cor. Two parallel planes are everywhere equidistant. For, if E F, G H are perpendicular to the two planes MN, P Q, they will be parallel to each other (Prop. X. Cor. 1); and consequently equal. PROPOSITION XVI. -THEOREM. 424. If two angles not in the same plane have their sides parallel and lying in the same direction, these angles will be equal, and their planes will be parallel. Let B A C, E D F be two tri-, angles, lying in different planes, B H / MN and PQ, having their sides - G - /N parallel and lying in the same I [ direction; then the angles BAC, P/ iD EDF will be equal, and their E.\F / planes, MN, PQ, be parallel..Q 178 ELEMENTS OF GEOMETRY. For, take AB equal to ED, and AC equal to DF; and join B C / BC, EF, BE,AD, CF. Since --. /N A B is equal and parallel to ED, the figure ABED is a /.i D j 7 parallelogram (Prop. XXXIII. E/....\V / Bk. I.); therefore A D is equal Q and parallel to B E. For a similar reason, C F is equal and parallel to A D; hence, also, B E is equal and parallel to C F; hence the figure B C F E is a parallelogram, and the side B C is equal and parallel to E F; therefore the triangles B A C, E D F have their sides equal, each to each; hence the angle B A C is equal to the angle E D F. Again, the plane B A C is parallel to the plane E D F. For, if not, suppose a plane to pass through the point A, parallel to ED F, meeting the lines BE, C F, in points different from B and C, for instance G and H. Then the three lines GE, AD, HF will be equal (Prop. XV.). But the three lines B E, A D, C F are already known to be equal; hence B E is equal to G E, and H F is equal to C F, which is absurd; hence the plane RA C is parallel to the plane E D F. 425. Cor. If two parallel planes M N, P Q, are met by two other planes, A B E D, A C F D, the angles B A C, E D F, formed by the intersections of the parallel planes, are equal; for the intersection AB is parallel to ED, and A C to D F (Prop. XIII.); therefore the angle B A C is equal to the angle E D F. PROPOSITION XVII. - THEOREM. 426. If three straight lines not in the same plane are equal and parallel, the triangles formed by joining the extremities of these lines will be equal, and their planes will be parallel. Let B E, A D, C F be three equal and parallel straight lines, not in the same plane, and let B A C, E D F be two BOOK VII. 179 triangles formed by joining the M — extremities of these lines; then will these triangles be equal, N and their planes parallel. For, since B E is equal and 1) parallel to A D, the figure E A B E D is a parallelogram; Qhence, the side A B is equal and parallel to D E (Prop. XXXIII. Bk. I.). For a like reason, the sides BC, EF are equal and parallel; so also are A C, D F; hence, the two triangles B A C, E D F, having their sides equal, are themselves equal (Prop. XVIII. Bk. I.); consequently, as shown in tile last proposition, their planes are parallel. PROPOSITION XVIII.- THEOREM. 427. If two straight lines are cut by three parallel planes, they will be divided proportionally. Let the straight line A B meet M the parallel planes, M N, P Q, R S, / A at the points A, E, B; and the straight line C D meet the same E. \G.. planes at the points C, F, D; then Q will AE:EB::CF:FD. -- Draw the line AD, meeting the /. D/s plane P Q in G, and draw A C, E G, B D. Then the two parallel planes PQ, R S, being cut by the plane A B D, the intersections E G, B D are parallel (Prop. XIII.); and, in the triangle AB D, we have (Prop. XVII. Bk. IV.), AE: EB:: AG: GD. In like manner, the intersections A C, G F being parallel, in the triangle A D C, we have AG: GD:: CF:FD; ELEMENTS OF GEOMETRY. hence, since the ratio A G: G D is common to both proportions, we have AE: EB:: CF: FD. PROPOSITION XIX.- THEOREM. 428. The sum of any two of the plane angles which form a triedral angle is greater than the third. The proposition requires dem- S onstration only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Let the triedral angle whose A/ \ vertex is S be formed by the three plane angles A S B, A S C, B SC; and suppose the angle A S B to be greater than either of the other two; then the angle A S B is less than the sum of the angles A S C, B S C. In the plane A SB make the angle B SD equal to BS C; draw the straight line AD B at pleasure; make S C equal S D, and draw A C, B C. The two sides B S, S D are equal to the two sides B S, S C, and the angle B S D is equal to the angle B S C; therefore the triangles B S D, B S C are equal (Prop. V. Bk. I.); hence the side B D is equal to the side B C. But AB is less than the sum of AC and BC; taking B D from the one side, and from the other its equal, B C, there remains A D less than AC. The two sides A S, S D of the triangle A S D, are equal to the two sides A S, S C, of the triangle A S C, and the third side A D is less than the third side A C; hence tle angle A S D is less than the angle A S C (Prop. XVII. Bk. I.). Adding B S D to one, and its equal, B S C, to the other, we shall have the sum of ASD, B S D, or A S B, less than the sum of ASC, BSC. BOOK VII. 181 PROPOSITION XX. - THEOREM. 429. The sum of the plane angles which form any polyedral angle is less than four right angles. Let the polyedral angles whose vertex is S be formed by any number of plane angles, A S B, B S C, C S D, &c.; the sum of all these plane angles is less than four right angles. /..' Let the planes forming the poly- A - - \ ---- D edral angle be cut by any plane, ABCDEF. From any point, 0, B C in this plane, draw the straight lines A 0, B O, C O, D O, E 0, F O. The sum of the angles of the triangles A SB, BSC, &c. formed about the vertex S, is equal to the sum of the angles of an equal number of triangles A 0 B, B O C, &c. formed about the point O. But at the point B the sum of the angles A B O, 0 B C, equal to A B C, is less than the sum of the angles AB S, S B C (Prop. XIX.); in the same manner, at the point C we have the sum of B C O, C D less than the sum of B C S, S C D; and so with all the angles at the points D, E, &c. Hence, the sum of all the angles at the bases of the triangles whose vertex is 0, is less than the sum of all the angles at the bases of the triangles whose vertex is S; therefore, to make up the deficiency, the sum of the angles formed about the point 0 is greater than the sum of the angles formed about the point S. But the sum of the angles about the point 0 is equal to four right angles (Prop. IV. Cor. 2, Bk. I.); therefore the sum of the angles about S must be less than four right angles. 430. Scholium. This demonstration supposes that the polyedral angle is convex; that is, that no one of the faces would, on being produced, cut the polyedral angle; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude. 16 182 ELEMENTS OF GEOMETRY. PROPOSITION XXI. -THEOREM. 431. If two triedral angles are formed by plane angles which are equal each to each, the planes of the equal angles will be equally inclined to each other. Let the two triedral an- S T gles whose vertexes are S and T, have the angle ASC equal to DTF, the angle A S B equal to D T E, and the angle B S C equal to A. -. ETF; then will the incli- -. p - nation of the planes A S C, AS B be equal to that of the planes D T F, D T E. For, take S B at pleasure; draw B perpendicular to the plane A S C; from the point 0, at which the perpendicular meets the plane, draw 0 A, 0 C, perpendicular to S A, S C; and join AB B B C. Next, take TE equal S B; draw E P perpendicular to the plane D TE; from the point P draw P D, P F, perpendicular respectively to T D, TF; and join DE, EF. The triangle S A B is right-angled at A, and the triangle T D E at D; and since the angle A S B is equal to D T E, we have S B A equal to T E D. Also, SB is equal to T E; therefore the triangle S A B is equal to T D E; hence S A is equal to T D, and A B is equal to D E. In like manner it may be shown that S C is equal to T F, and B C is equal to E F. We can now show that the quadrilateral A S CO is equal to the quadrilateral D T FP; for, place the angle AS C upon its equal D TF; since S A is equal to T D, and S C is equal to T F, the point A will fall on D, and the point C on F; and, at the same time, A 0, which is perpendicular to S A, will fall on D P, which is perpendicular to T D, and, in like manner, C O on F P; wherefore the point 0 will fall on the point T' and A 0 will be equal to D P. ROOK VII. But the triangles A 0 B, D P E are right-angled at 0 and P; the hypotenuse AB is equal to D E, and the side A 0 is equal to D P; hence the two triangles are equal (Prop. XIX. Bk. I.); and, consequently, the angle OA B is equal to the angle PD E. The angle O A B is tie inclination of the two planes A SB, AS C; and the angle PD E is that of the two planes D T E, D TF; hence, those two inclinations are equal to each otller. 432. Scholium 1. It must, however, be observed, tllat the angle A of the right-angled triangle 0 A B is properly the inclination of the two planes A S B, A S C only when the perpendicular B 0 falls on the same side of S A with S C; for if it fell on the other side, the angle of the two planes would be obtuse, and joined to the angle A of the triangle 0 A B it would make two right angles. But, in the same case, the angle of the two planes D T E, D T F would also be obtuse, and joined to the angle D of the triangle D P E it would make two right angles; and the angle A being thus always equal to the angle D, it would follow in the same manner that the inclination of the two planes A S B, A S C must be equal to that of the two planes D TE, D TF. 433. Scholium 2. If two triedral angles are formed by three plane angles respectively equal to each other, and if at the same time the equal or homologous angles are similarly situated, the two angles are equal. For, by the proposition, the planes which contain the equal angles of the triedral angles are equally inclined to each other. 434. Scholium 3. When the equal plane angles forming the two triedral angles are not similarly situated, these angles are equal in all their constituent parts, but, not admitting of superposition, are said to be equal by symmetry, and are called symmetrical angles. BOO K VIII. POLYEDRONS. DEFINITIONS. 435. A POLYEDRON is a solid, or volume, bounded by planes. The bounding planes are called the faces of the polyedron; and the lines of intersection of the faces are called the edges of the polyedron. 436. A PRISM is a polyedron having two of its faces equal and parallel polygons, and the other faces parallelo- F I grams. The equal and parallel polygons are called the bases of the prism, and the / parallelograms its lateral faces. The E lateral faces taken together constitute A / the lateral or convex surface of the B C prism. Thus the polyedron A B C D E-K is a prism, having for its bases the equal and parallel polygons A B C D E, F G H I K, and for its lateral faces the parallelograms ABGF, BCHG, &c. The principal edges of a prism are those which join the corresponding angles of the bases; as A F, B G, &c. 4837. The altitude of a prism is a perpendicular drawn from any point in one base to the plane of the other. 438. A RIGHT PRISM is one whose principal edges are perpendicular to the planes of its bases. Each of the BOOK VIII. edges is then equal to the altitude of the prism. Every other prism is oblique, and has each edge greater than the altitude. 439. A prism is triangular, quadrangular, pentangular, hexangular, &c., according as its base is a triangle, a quadrilateral, a pentagon, a hexagon, &c. H 440. A PARALLELOPIPEDON is a prism E G whose bases are parallelograms; as the F prism ABCD-lH. The parallelopipedon is rectangular when all its faces are rectangles; as the.-. parallelopipedon A B C D - H. A C B E H 441. A CUBE, or REGULAR HEXAEDRON, is a rectangular parallelopipedon having F G all its faces equal squares; as the paral- A............ lelopipedon A B C D -. H. B C 442. A PYRAMID is a polyedron of S which one of the faces is any polygon, and all the others are triangles meeting at a common point. The polygon is called the base of the pyramid, the triangles its lateral faces, and the point at which the triangles meet A its vertex. The lateral faces taken to- B gether constitute the lateral or convex surface of the pyramid. Thus the polyedron A B C D E - S is a pyramid, having for its base the polygon A B C D E, for its lateral faces the triangles A S B, B S C, C S D, &c., and for its vertex the point S. 16* ELEMENTS OF GEOMETRY. 443. The ALTITUDE of a pyramid is a perpendicular drawn from the vertex to the plane of the base. 444. A pyramid is triangular, quadrangular, &c., according as its base is a triangle, a quadrilateral, &c. 445. A RIGHT PYRAMID is one whose base is a regular polygon, and the perpendicular drawn from the vertex to the base passes through the centre of the base. In this case the perpendicular is called the axis of the pyramid. 446. The SLANT HEIGHT of a right pyramid is a line drawn from the vertex to the middle of one of the sides of the base. 447. A FRUSTUM of a pyramid is the part of the pyramid included between the base and a plane cutting the pyramid parallel to the base. 448. The ALTITUDE of the frustum of a pyramid is the perpendicular distance between its parallel bases. 449. The SLANT HEIGHT of a frustum of a right pyramid is that part of the slant height of the pyramid which is intercepted between the bases of the frustum. 450. The Axis of the frustum of a pyramid is that part of the axis of the pyramid which is intercepted between the bases of the frustum. 451. The DIAGONAL of a polyedron is a line joining the vertices of any two of its angles which are not in the same face. 452. SIMILAR POLYEDRONS are those which are bounded by the same number of similar faces, and have their polyedral angles respectively equal. 453. A REGULAR POLYEDRON is one whose faces are all equal and regular polygons, and whose polyedral angles are all equal to each other. BOOK VIII. 187 PROPOSITION I. -THEOREM. 454. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Let A B C D E-K be a right prism; K then will its convex surface be equal F,, to the perimeter of its base, AB+BC-+-CD+DE+EA, I multiplied by its altitude A F. For, the convex surface of the prism A is equal to the sum of the parallelo- grams AG, B H, CI, DK, EF (Art. B C 436). Now, the area of each of those parallelograms is equal to its base, AB, BC, CD, &o., multiplied by its altitude, AF, B G, CH, &c. (Prop. V. Bk. IV.). But the altitudes A F, B G, C H, &c. are each equal to A F, the altitude of the prism. Hence, the area of these parallelograms, or the convex surface of the prism, is equal to (A B + B C + C D + DE + E A) X AF; or the product of the perimeter of the prism by its altitude. 455. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases. PROPOSITION II. THEOREM. 456. In every prism, the sections formed by parallel planes are equal polygons. Let the prism ABC DE-K be intersected by the parallel planes N P, S V; then are the sections N OP Q R, S T V X Y equal polygons. For the sides S T, N 0 are parallel, being the intersections of two parallel planes with a third plane A B G F ELEMENTS OF GEOMETRY. (Prop. XIII. Bk. VII.); these same K sides S T, NO, are included between I tlhe parallels N S, 0 T which are sides F \ of the prism; hence NO is equal to Y/ S T. For like reasons, the sides P, s /... PQ, QR, &c. of the section N P Q R, are respectively equal to the sides N, 7 V TV, VX, XY, &c. of the section ST V X Y; and since the equal sides are at the same time parallel, it follows that the angles NOP, OP Q, &c. A c of the first section are respectively B equal to the angles S T V, T V X of the second (Prop. XVI. Bk. VII.). Hence, the two sections N 0 P Q R, S T V X Y, are equal polygons. 457. Cor. Every section made in a prism parallel to its base, is equal to that base. PROPOSITION III. THEOREM. 458. Two prisms are equal, when the three faces which form a triedral angle in the one are equal to those which form a triedral angle in the other, each to each, and are similarly situated. Let the two prisms K Y ABCDE-K and \ LMOPQ-Y have R the faces which form / the triedral angle B equal to the faces. which form the tri- A E L edral angle M; that is, the base ABCDE B C M 0 equal to the base L M N 0 P Q, the parallelogram A B G F equal to the parallelogram L MS R, and the parallelogram B C H G equal to MOTS; then the two prisms are equal. BOOK VIII. 189 For, apply the base K y ABCDE to the equal base LMOPQ; I v then, the triedral an- H gles B and M, being equal, will coincide, since the plane an- A E D L Q gles which form these triedral angles are B C M 0 equal each to each, and similarly situated (Prop. XXI. Sch. 2, Bk. VII.); hence the edge B G will fall on its equal M S, and the face B H will coincide with its equal MT, and the face BF with its equal MR. But the upper bases are equal to their corresponding lower bases (Art. 436); therefore the bases F G H I K, R S T V Y are equal; hence they coincide with each other. Therefore H I coincides with T V, I K with V Y, and K F with Y R; and consequently the lateral faces coincide. IHence the two prisms coincide throughout, and are equal. -459. Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to LM, and the altitude B G to M S, the rectangle A B G F is equal to the rectangle L M S R; so, also, the rectangle B G H C is equal to M S T O; and thus the three faces which form the triedral angle B, are equal to the three faces which form the triedral angle M. Hence the two prisms are equal. PROPOSITION IV. -THEOREM. 460. In every parallelopipedon the opposite faces are equal and parallel. Let A B C D - H be a parallelopipedon; then its opposite faces are equal and parallel. The bases A B C D, E F GH are equal and parallel (Art. 436), and it remains only to be shown that the same is ELEMENTS OF GEOMETRY. true of any two opposite lateral faces, as B C G F, ADHE. Now, since the base - A BC D is a parallelogram, the side A D is equal and parallel to B C. For G a similar reason, AE is equal and par- A. allel to BF; hence the angle DA E is equal to the angle C B F (Prop. XVI. \ Bk. VII.), and the planes DA E, CBF B C are parallel; hence, also, the parallelogram B C G F is equal to the parallelogram AD H E. In the same way, it may be shown that the opposite faces A B F E, D C G H are equal and parallel. 461. Cor. Any two opposite faces of a parallelopipedon may be assumed as its bases, since any face and the one opposite to it are equal and parallel. PROPOSITION V.-THEOREM. 462. The diagonals of every parallelopipedon bisect each other. Let AB CD-H be a parallelo- E H pipedon; then its diagonals, as B H, D F, will bisect each other. - For, since B F is equal and par- allel to D H, the figure B F H D is A. - -- a parallelogram; hence the diago- \ - nals B H, D F bisect each other at B C the point 0 (Prop. XXXIV. Bk. I.). In the same manner it may be shown that the two diagonals A G and C E bisect each other at the point O; hence the several diagonals bisect each other. 463. Scholium. The point at which the diagonals mutually bisect each other may be regarded as the centre of the parallelopipedon. BOOK VIII. 191 PROPOSITION VI.- THEOREM. 464. Any parallelopipedon may be divided into two equivalent triangular prisms by a plane passing through its opposite diagonal edges. Let any parallelopipedon, A B C D-H, G be divided into two prisms, A B C- G, AD C-G, by a plane, A C G E, passing H through opposite diagonal edges; then p F will the two prisms be equivalent..... —'N Through tie vertices A and E, draw E. the planes A K L M, E N 0 P, perpen- D / dicular to the edge A E, and meeting M\ / BF, C G, DH, the three other edges. -- of the parallelopipedon, in the points K, L, M, and il N, O, P. The sections A K L M, E N 0 P are equal, since they are formed by planes perpendicular to the same straight lines, and hence parallel (Prop. II.). They are parallelograms, since the two opposite sides of the same section, A K, L M, are the intersections of two parallel planes, A B F E, D C GH, by the same plane, A K L M (Prop. XIII. Bk. VII.). For a like reason, the figure A M PE is a parallelogram; so, also, are A K N E K L N L P, L M P the other lateral faces of the solid A K L M - P; consequently, this solid is a prism (Art. 436); and this prism is right, since the edge AE is perpendicular to the plane of its base. This right prism is divided by the plane ALOE into the two right prisms A K L- O, A M L -0, which, having equal bases, A K L, A M L, and the same altitude, A E, are equal (Prop. III. Cor.). Now, since A E H D, A E P M are parallelograms, the sides D H, M P, being each equal to A E, are' equal to each other; and taking away the common part, D P, there remains D M equal to H P. In the same manner it may be shown that C L is equal to GO. 192 ELEMENTS OF GEOMETRY. Conceive now E P 0, the base of the G solid EPO-G, to be applied to its equal A M L, the point P falling upon H M, and the point 0 upon L; the edges -- -:. '1F GO, HP will coincide with their equals.... N C L, DM, since they are all perpen- E C dicular to the same plane, AK L M. D /! \i Hence the two solids coincide through- M —... M.. out, and are therefore equal. To each \ of these equals add the solid AD C-P, and the riglit prism AML-O is equivalent to the prism ADC-G. In the same manner, it may be proved that the right prism AKL-O is equivalent to the prism ABC-G. The two riglt prisms A K L - 0, A M L - 0 being equal, it follows that two triangular prisms, A B C - G, A D C - G, are equivalent to each other. 465. Cor. Every triangular prism is half of a paralleloi ' pipedon having the same triedral angle, with the same edges. PROPOSITION VII. -THEOREM. 466. Two parallelopipedons, having. a common lower base, and their upper bases in the same plane and between the same parallels, are equivalent to each other. Let the two parallelo- H G M L pipedons A G, AL have the common base ABCD, I and their upper bases, EFGH, IKLM, in the same plane, and be- D -. / tween the same paral- \ lels, EK, ILL; then the A parallelopipedons will be equivalent. BOOK VIII. 193 There may bc tlree cases, according as E I is greater or less than, or equal to, E F; but the demonstration is the same for each. Since AE is parallel to B F, and H E to GF, the plane angle AEI is equal to B FK, H EI to GFK, and HEA to G F B. Of these six plane angles, the three first form the polyedral angle E, the three last the polyedral angle F; consequently, since these plane angles are cqua:l cach to each, and similarly situated, the polyedral angles, E, F, must be equal. Now conceive tlhe prism A E I- M to be applied to tlhe prism B F K - L; the base A E I, being placed upon the base B F K, will coincide with it, since they are equal; and, since the polyedral angle E is equal to the polyedral angle F, the side E H will fall upon its equal, F G. But the base A E I and its edge E H determine the prism A E I - M, as the base B F K and its edge F G determine the prism B FK-L (Prop. III.); hence the two prisms coincide throughout, and therefore are equal to each other. Take away, now, from the whole solid A E L C, the prism AEI-M, and there will remain the parallelopipedon AL; and take away from the same solid A L the prism BFK-L, and there will remain the parallelopipedon A G; hence the two parallelopipedons A L, A G are equivalent. PROPOSITION VIII. — THEOREM. 467. Two parallelopipedons having' the same base and the same altitude are equivalent. Let the two parallelopipedons A G, A L have the common base A B C D, and the same altitude; then will the two parallelopipedons be equivalent. For, the upper bases EFGH, IKLM being in the same plane, produce the edges E F, H G, L K, I M, till by their intersections they form the parallelogram N 0 P Q; this parallelogram is equal to either of the bases I L, E G, and 17 194 ' ELEMENTS OF GEOMETRY. is between the same par- Q p H G allels; hence N O P Q is equal to the common.......... --- base ABCD, and is M ' / / parallel to it. K/ / Now, if a third parallelopipedon be conceived, which, with the same lower base A B C D, has/ / for its upper base NOPQ, c this third parallelopipe- A l B don will be equivalent to the parallelopipedon A G, since the lower base is the same, and the upper bases lie in the same plane and between the same parallels, G Q, F N (Prop. VII.). For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon A L; hence the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent. PROPOSITION IX.-THEOREM. 468. Any oblique parallelopipedon is equivalent to a rectangular parallelopipedon having the same altitude and an equivalent base. Let AG be any paral- H G lelopipedon; then A G will be equivalent to a / rectangular parallelopip- edon having the same / altitude and an equivalent base. From the points A, B, C, D, draw A,B, B.. / C L, D M, perpendicular c to the lower base, and A equal in altitude to A G; there will thus be formed the BOOK VIII. 195 parallelopipedon AL, equivalent to AG (Prop. VIII.), and having its lateral faces, AK, B L, &c., rectangular. Now, if the base A B C D is a rectangle, AL will be a rectangular parallelopipedon equivalent to A G. But if ABC D is not a rectangle, MQ LP draw A 0, B N, each perpendicular to C D; also 0 Q, N P, each perpendicular K to the base; then we shall have a rectangular parallelopipedon A B N 0- Q. For, by construction, the bases A B N O, I K P Q are rectangles; so, also, are the lateral faces, the edges A I, 0 Q, D. N &c. being perpendicular to the plane A B of the base; therefore the solid A P is a rectangular parallelopipedon. But the two parallelopipedons A P, A L may be considered as having the same base, A B K I, and the same altitude, AO; hence they are equivalent. Hence the parallelopipedon A G, which was shown to be equivalent to the parallelopipedon AL, is also equivalent to the rectangular parallelopipedon A P, having the same altitude, A I, and a base, A B N 0, equivalent to the base A BC D. PROPOSITION X.-THEOREM. 469. Two rectangular parallelopipedons, which have the same base, are to each other as their altitudes. Let the two parallelopipedons A G, A L have the same base, A BC D; then they are to each other as their altitudes, AE, AI. First. Suppose the altitudes AE, AI are to each other as two whole numbers; for example, as 15 is to 8. Divide A E into 15 equal parts, of which A I will contain 8. Through x, y, z, &c., the points of division, conceive planes to E H 0.. G j B C 196 ELEMENTS OF GEOMETRY. pass parallel to the common base. E H These planes will divide the solid A G into 15 small parallelopipedons,all equal F G to each other, having equal bases and equal altitudes; equal bases, since every section, as I K L M, parallel to the base L A BC D, is equal to that base (Prop. y II.), and equal altitudes, since the alti- A.. tudes are the equal divisions Ax, x y, y z, &c. But of those 15 equal parallel- B C opipedons, 8 are contained in AL; hence the parallelopipedon A G is to the parallelopipedon A L as 15 is to 8, or, in general, as the altitude A E is to the altitude A I. Secondly. If the ratio of A E to A I cannot be exactly expressed by numbers, we shall still have the proportion, Solid A G Solid AL:: A E: AI. For, if this proportion is not correct, suppose we have Solid A G: Solid A L:: A E: A O greater than A I. Divide A E into equal parts, each of which shall be less than I O; there will be at least one point of division, m, between I and 0. Let P represent the parallelopipedon, whose base is A B C D, and altitude A m; since the altitudes A E, A m are to each other as two whole numbers, we shall have Solid AG: P:: AE: Am. But, by hypothesis, we have Solid A G: Solid A L A A E: A 0; hence (Prop. X. Cor. 2, Bk. II.), Solid AL: P:: AO: Am. But A 0 is greater than Am; hence, if the proportion is correct, the parallelopipedon A L must be greater than P. On the contrary, however, it is less; consequently the solid A G cannot be to the solid A L as the line A E is to a line greater than A I. '... BOOK VIII. 197 - By the same mode of reasoning, it may be shown that the fourth term of the proportion cannot be less than A I; therefore it must be equal to AI. Hence rectangular parallelopipedons, having the same base, are to each other as their altitudes. PROPOSITION XI. -THEOREM. 470. Two rectangular parallelopipedons, having the same altitude, are to each other as their bases. Let the two rectanI E H gular parallelopipedons _- A G, AK have the same L altitude, A E; then they are to each other as their bases. G Place the two solids so that their faces, B E,! O E, may have the com- mon angle B A E; pro- M - ---...-. duce the plane ONKL... till it meets the plane N DCGH il PQ; we shall thus have a third B C parallelopipedon, A Q, which may be compared with each of the parallelopipedons A G, A K. The two solids, A G, A Q, having the same base, A E H D, are to each other as their altitudes A B, A O (Prop. X.); in like manner, the two solids A Q, A K, having the same base, A O L E, are to each other as their altitudes AD, AM. Hence we have the two proportions, Solid A G: Solid A Q:: A B A O, Solid A Q: Solid A K:: AD: AM. Multiplying together the corresponding terms of these 17* 198 ELEMENTS OF GEOMETRY. proportions, and omitting, in the result, the common factor Solid A Q, we shall have, Solid AG: Solid AK:: AB X AD: AO X AM. But A B X A D measures the base A B C D (Prop. IV. Sch., Bk. IV.); and AO X AM measures the base A M N 0; hence two rectangular parallelopipedons of the same altitude are to each other as their bases. PROPOSITION XII. - THEOREM. 471. Any two rectangular parallelopipedons are to each other as the product of their bases by their altitudes. Let AG, AZ be two E H rectangular parallelopipedons; then they are \K. _1 \ to each other as the Q product of their bases, I h G ABCD, AMNO, by x their altitudes, A E, AX. \. Place the two solids Z -- i so that their faces, B E, I 0 X, may have the con- M l mon angle BAE; pro- '............. duce the planes neces- sary for completing the third parallelopipedon, B C A K, having the same altitude with the parallelopipedon A G. By the last proposition, we shall have Solid A G: Solid AK:: ABCD: AMNO. But the two parallelopipedons A K, A Z, having the same base, A M N O, are to each other as their altitudes, A E, A X (Prop. X.); hence we have Solid AK: Solid A Z:: A E: A X. Multiplying together the corresponding terms of these BOOK VIII. 199 proportions, snd omitting, in the result, the common factor Solid A K, we shall have Solid A G: SolidAZ::ABCD X AE: AMNO X AX. Hence, any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes. 472. Scholium 1. We are consequently authorized to assume, as the measure of a rectangular parallelopipedon, the product of its base by its altitude; in other words, the product of its three dimensions. But by the product of two or more lines is always meant the product of the numbers which represent them; those numbers themselves being determined by the particular linear unit, which may be assumed as the standard. It is necessary, therefore, iln comparing magnitudes, that the measuring unit be the same for each of the magnitudes compared. 473. Scholium 2. The measured magnitude of a solid, or volume, is called its volume, solidity, or solid contents. We assume as the unit of volume, or solidity, the cube, each of whose edges is the linear unit, and each of whose faces is the unit of surface. PROPOSITION XIII. -THEOREM. 474. The solid contents of a parallelopipedon, and of any other prism, are equal to the product of its base by its altitude. First. Any parallelopipedon is equivalent to a rectangular parallelopipedon having the same altitude and an equivalent base (Prop. IX.). But the solid contents of a rectangular parallelopipedon are equal to the product of its base by its altitude; therefore the solid contents of any parallelopipedon are equal to the product of its base by its altitude. Second. Any triangular prism is half of a parallelopipedon, so constructed as to have the same altitude, and a 200 ELEMENTS OF GEOMETRY. base twice as great (Prop. VI.). But the solid contents of the parallelopipedon are equal to the product of its base by its altitude; lhence, that of the triangular prism is also equal to the product of its base, or half that of the parallelopipedon, by its altitude. Third. Any prism may be divided into as many triangular prisms of the same altitude, as there are triangles in the polygon taken for a base. But the solid contents of each triangular prism are equal to the product of its base by its altitude; and, since the altitude is the same in each, it follows that the sum of all these prisms is equal to the sum of all the triangles taken as bases multiplied by the common altitude. Hence the solid contents of aly prism are equal to the product of its base by its altitude. 475. Cor. When any two prisms have the same altitude, the products of the bases by the altitudes will be as the bases (Prop. IX. Bk. II.); hence, prisms of the same altitude are to each other as their bases. For a like reason, prisms of the same base are to each other as their altitudes. PROPOSITION XIV.- THEOREM. 476. Similar prisms are to each other as the cubes of their homologous edges. Let ABC-E, FHI-M E be two similar prisms; these prisms are to each / M other as the cubes of their homologous edges, AB \ K and FH. For, from D and K, ho- B A B A H F mologous angles of the two prisms, draw the perpendiculars D N, K 0, to the bases A B C, F H I. Take A K' equal to F K, and join A N. BOOK VIII. 201 Draw K' 0' perpendicular to A N in the plane A N D, and K' O will be perpendicular to the plane A B C, and equal to K O, the altitude of the prism F H I - M. For, conceive the triedral angles A and F to be applied the one to the other; the planes containing them, and therefore the perpendiculars K' O', K O, will coincide. Now, since the bases A B C, F H I are similar, we have (Prop. XXIX. Bk. IV.), Base AB C: Base F H I:: AB2: FH2; and, because of the similar triangles D A N, K F O, and of the similar parallelograms D B, K H, we have DN:KO::DA: KF::AB: FH. Hence, multiplying together the corresponding terms of these proportions, we have Base A B C X D N: Base F HI X KO: A B3: F 3. But the product of the base by the altitude is equal to the solidity. of a prism (Prop. XIII.); hence Prism AB C - E Prism F H I-M:: AB3: FH3. PROPOSITION XV.- THEOREM. 477. The convex surface of a right pyramid is equal to the perimeter of its base, multiplied by half the slant height. Let AB CDE-S be a right pyra- mid, and S M its slant height; then the convex surface is equal to the perimeter AB+ BC + CD +DE+EA mul- tiplied by j S M. / The triangles SAB, SBC, S CD, &c. are all equal; for the sides AB, A' —/ \ BC, CD, &c. are equal (Art. 445), and \ D the sides S A, S B, S C, &c., being ob- lique lines meeting the base at equal 202 ELEMENTS OF GEOMETRY. distances from a perpendicular let fall from the vertex S to the center of the base, are also equal (Prop. V. Bk. VII.). Hence, these triangles are all equal (Prop. XVIII. Bk. I.); and the altitude of each is equal to the slant height S M. But the area of a triangle A is equal to the product of its base mul- M tiplied by half its altitude (Prop. VI. Bk. LV.). Hence, the areas of the triangles SAB, SBC, SCD, &c. are equal to the sum of the bases A B, B C, C D, &c. multiplied by half the common altitude, S M; that is, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height. 478. Cor. The lateral faces of a right pyramid are equal isosceles triangles, having for their bases the sides of the base of the pyramid. PROPOSITION XVI.-THEOREM. 479. If a pyramid be cut by a plane parallel to its base,1st. The edges and the altitude will be divided proportionally. 2d. The section will be a polygon similar to the base. Let the pyramid A B C D E - S, whose S altitude is SO, be cut by a plane, G H I K L, parallel to its base; then will the edges S A, S B, S C, &c., with 1 the altitude SO, be divided proportion- I ally; and the section GHI K L will be similar to the base A B C D E... First. Since the planes A B C, GHI A C are parallel, their intersections A B, G H, by the third plane S AB, are B parallel (Prop. XIII. Bk. VII.); hence BOOK VIII. 208 the triangles S A B, SGH are similar (Prop. XXV. Bk. IV.), and we have SA: SG:: SB: SH. For the same reason, we have SB: SH:: SC: SI; and so on. Hence all the edges, S A, S B, S C, &c., are cut proportionally in G, H, I, &c. The altitude S 0 is likewise cut in the same proportion, at the point P; for B 0 and H P are parallel; therefore we have SO: SP:: SB: SH. Secondly. Since G H is parallel to AB, H I to B C, I K to C D, &c. the angle G H I is equal to A B C, the angle H I K to B C D, and so on (Prop. XVI. Bk. VII.). Also, by reason of the similar triangles S A B, S G H, we have AB: GH:: SB: SH; and by reason of the similar triangles S B C, S H I, we have SB: SH:: BC: HI; hence, on account of the common ratio S B: S H, AB: G H:: BC: H I. For a like reason, we have BC: HI:: CD:IK, and so on. Hence the polygons AB C DE, GH I K L have their angles equal, each to each, and their homologous sides proportional; hence they are similar. 480. Cor. 1. If two pyramids have the same altitude, and their bases in the same plane, their sections made by a plane parallel to the plane of their bases are to each other as their bases. Let A B C D E - S, M N 0-S be two pyramids, having the same altitude, and their bases in the same plane; and let G II I K L, P Q R be sections made by a plane parallel 204 ELEMENTS OF GEOMETRY. to the plane of their bases; then these sections are to each other as the bases ABCDE, MNO. L For, the two polygons c/ \R ABCDE, GHIKL be- ing similar, their surfaces / D.. \ are as the squares of the A homologous sides AB, GH A\ / M -.. (Prop. XXXI. Bk. IV.). N But AB: GH:: SA: SG. Hence, ABCDE: GHIKL:: S A2: S G2. For the same reason, MNO: PQR:: SM: SP2 But since G H I K L and P Q R are in the same plane, we have also (Prop. XVIII. Bk. VII.), SA: SG:: SM: SP; hence, ABCDE: GHIKL:: MNO: PQR; therefore the sections G H I K L, P Q R are to each other as the bases A B C D E, M N O. 481. Cor. 2. If the bases A B C D E, M N 0 are equivalent, any sections, G H I K L, P Q R, made at equal distances from those bases, are likewise equivalent. PROPOSITION XVII. - THEOREM. 482. The convex surface of a frustum of a right pyramid is equal to half the sum of the perimeters of its two bases, multiplied by its slant height. Let A B C D E- L be the frustum of a right pyramid, and MN its slant height; then the convex surface is equal to the sum of the perimeters of the two bases A B C D E, G H I K L, multiplied by half of M N. BOOK VIII. 205 F-or the upper base G H I K L is similar L to the base A B C D E (Prop. XVI.), and K A B C D E is a regular polygon (Art. 445); hence the sides GH, HI, I, K K L, and L G are all equal to each other. The angles GAB, ABH, H B C, &c. 1IKI are equal (Prop. XV. Cor.), and the A D edges A G, B H, C I, &c. are also equal (Prop. XVI.); therefore the faces A H, B C B I, C K, &c. are all equal trapezoids (Art. 28), having a common altitude, M N, the slant height of the frustum. But the area of either trapezoid, as A H, is equal to k (AB + GH) X AMN (Prop. VII. Bk. IV.); hence the areas of all the trapezoids, or the convex surface of frustum, are equal to half the sum of the perimeters of the two bases multiplied by the slant height. PROPOSITION XVIII. THEOREM. 483. Triangular pyramids, having equivalent bases and the same altitude, are equivalent. T S S' HX7F D.A^p A' EA A C, B B' Let A B C - S, A' B' C' - S' be two triangular pyramids, having equivalent bases, A B C, A' B' C', situated in the same plane; and let them have the same altitude, AT; then these pyramids are equivalent. For, if the two pyramids are not equivalent, let A' B C' - S' be the smaller, and suppose A X to be the 18 206 ELEMENTS OF GEOMETRY. altitude of a prism, which, having A B C for its base, is equal to their difference. T S S' A Al El'. GA: I GC t B B' Divide the altitude AT into equal parts, each less than A X; through each point of division pass a plane parallel to the plane of the base, thus forming corresponding sections in the two pyramids, equivalent each to each, namely, D E F to D E' F', G H I to G' HI' I, &c. Upon the triangles A B C, D E F, G H I, &c., taken as bases, construct exterior prisms, having for edges the parts A D, D G, G K, &c. of the edge S A; in like manner, on the bases D'E'F', G'H' I', &c. in the second pyramid, construct interior prisms, having for edges the corresponding parts of S' A'. It is plain that the sum of all the exterior prisms of the pyramid A B C-S is greater than this pyramid; and also that the sum of all the interior prisms of the pyramid A'B'C'-S' is less than this pyramid. Hence, the difference between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. Now, beginning with the bases ABC, A'B'C', the second exterior prism, D E F - G, is equivalent to the first interior prism, D'E'F'-A', since they have equal altitudes, and their bases, DEF, D'E'F', are equivalent. For a like reason, the third exterior prism, GHI - K, and the second interior prism, G' H' I'- D', are equivalent; and so BOOK VIII. 207 on to the last in each series. Hence, all the exterior prisms of the pyramid A B C - S, excepting the first prism, A B C-D, have equivalent corresponding ones in the interior prisms of tlle pyramid A B C' - S. Therefore tlie prism A B C- D is the difference between the sum of all the exterior prisms of the pyramid A B C-S, and the sum of the interior prisms of the pyramid A'B' C'-S'. But the difference between tlese two sets of prisms has been proved to be greater thall that of the two pyramids, which latter difference we supposed to be equal to the prism A B C - X. Hence, the prism A B C - D must be greater than the prism A B C - X, which is impossible, since they have the same base, A B C, and the altitude of the first is less than A X, the altitude of the second. Hence, the supposed inequality between the two pyramids cannot exist; therefore the two pyramids A B C- S, A' B' C'- S', having the same altitude and equivalent bases, are themselves equivalent. PROPOSITION XIX. THEOREM. 484. Every triangular pyramid is a third part of a triangular prism having the same base and the same altitude. Let A B C-F be a triangu- E D lar pyramid, and A B C- D E F a triangular prism of the same base and the same altitude; then tle pyramid is one third of the prism. Cut off the pyramid AB C-F from the prism, by the plane A FAC; there will remain the A solid A C D E- F, which may be considered as a quadrangular pyramid, whose vertex is F, B and whose base is the parallelogram A C D E. Draw the 208 ELEMENTS OF GEOMETRY. diagonal C E, and pass the plane E D F C E, which will cut tlhe quadrangular pyramid into two tri- angular ones, ACE-F, EDC-F. These two triangular pyramids \ have for their common altitude the perpendicular let fall from F on the plane A CD E; they j -. have equal bases, since the triangles A C E, C D E are halves of the same parallelogram; hence the two pyramids A CE-F, B C D E - F are equivalent (Prop. XVIII.). But the pyramid C D E - F and the pyramid A B C - F have equal bases, A B C, D E F; they have also the same altitude, namely, the distance between the parallel planes ABC, D E F; hence the two pyramids are equivalent. Now, the pyramid C D E - F has been proved equivalent to AC E-F; hence the three pyramids ABC-F, CDE-F, A C E-F, which compose the whole prism A B C - D E F, are all equivalent; therefore, either pyramid, as ABC-F, is the third part of the prism, which has the same base and the same altitude. 485. Cor. 1. Every triangular prism may be divided into three equivalent triangular pyramids. 486. Cor. 2. The solidity of a triangular pyramid is equal to a third part of the product of its base by its altitude. PROPOSITION XX.-THEOREM. 487. The solidity of every pyramid is equal to the product of its base by one third of its altitude. Let A B C D E - S be any pyramid, whose base is AB C D E, and altitude SO; then its solidity is equal to ABCDE X j SO. BOOK VIII. 209 Draw the diagonals A C, A D, and pass the planes S A C, S A D through these diagonals and the vertex S; the polygonal pyramid ABCDE-S will be divided into several triangular pyra- / mids, all having the same altitude, S O. / E But each of these pyramids is measured A | by the product of its base, B A C, CAD,. ---- D D A E, by a third part of its altitude, SO (Prop. XIX. Cor. 2); hence, the B C sum of these triangular pyramids, or the polygonal pyramid A B C D E - S, will be measured by the sum of the triangles B A C, C A D, D A E, or the polygon AB C D E, multiplied by one third of S O; hence, every pyramid is measured by the product of its base by one third of its altitude. 488. Cor. 1. Every pyramid is the third part of the prism which has the same base and the same altitude. 489. Cor. 2. Pyramids leaving the same altitude are to each other as tleir bases. 490. Cor. 3. Pyramids having the same base, or equivalent bases, are to each other as their altitudes. 491. Cor. 4. Pyramids are to each other as the products of their bases by their altitudes. 492. Scholium. The solidity of alny polyedron may be found by dividing it into pyramids, by passing planes through its vertices. PROPOSITION XXI. -THEOREM. 493. A frustum of a pyramid is equivalent to the sum of three pyramids, having for their common altitude the altitude of the frustum., and whose bases are the two bases of the frustum and a mean proportional between them. First. Let A B C - D E F be the frustum of a pyramid, whose base is a triangle. Pass a plane through the points 18* 210 ELEMENTS OF GEOMETRY. A, E, C; it cuts off the triangular D F pyramid A B C- E, whose altitude is that of the frustum, and whose base, A B C, is the lower base of.. the frustum. Pass another plane througll tle points D, E, C; it cuts / \ off the triangular pyramid D E F- C, A whose altitude is that of tle frus-\ i tum, and whose base, D E F, is the G / upper base of tle frustum. There now remains of the frustum the pyramid A C D - E. Draw E G parallel to AD; join C G and D G. Then, since E G is parallel to A D, it is parallel to the plane A C D (Prop. XI. Bk. VII.); and the pyramid A C D - E is equivalent to the pyramid A C D - G, siice they have the same base, A C D, and their vertices, E and G, lie in the same straight line parallel to the common base. But the pyramid A C D- G is the same as the pyramid AG C- D, whose altitude is that of the frustum, and whose base, A G C, as will be proved, is a mean proportional between the bases A B C and D E F. The two triangles A G C, D E F have the angles A and D equal to each other (Prop. XVI. Bk. VII.); hence we have (Prop. XXVIII. Bk. IV.), / /^? AGC:DEF::AGX AC:DEXDF; but since A G is equal to D E, AGC: DEF::.C: DF. We have, also (Prop. VI. Cor., Bk. IV.), ABC: AGC::: AB: AGorD E. But the similar triangles A B C, D E F give AB: DE:: AC: DF; hence (Prop. X. Bk. II.), AB C: AGC::AGC: DEF; BOOK VIII. 211 that is, the base AG C is a mean proportional between the bases A B C, D E F of the frustum. Secondly. Let G H I K L - M N 0 P Q be the frustum of a pyramid, whose base is any polygon. Let ABC-S be S a triangular pyramid T having the same alti- / tude, and an equiva- M O lent base, with any polygonal pyramid, /L \ A —.......C...GHIKL-T; these pyramids are equiva- B lent (Prop. XX. Cor. 3.) The bases of the two pyramids may be regarded as situated in the same plane, in which case the plane MNOPQ produced will form in the triangular pyramid a section, D E F, at the same distance above the common plane of the bases; and therefore the section D E F will be to the section MNOPQ as the base ABC is to the base G H I K L (Prop. XVI. Cor. 1); and since the bases are equivalent, the sections will be so likewise. Hence, the pyramids M N 0 P Q - T, D E F - S, having the same altitude and equivalent bases, are equivalent. For the same reason, the entire pyramids G'H I K L - T, A B C - S are equivalent; consequently, the frustums GHIKLMNOPQ, ABC-DEF, are equivalent. But the frustum A B C -D E F has been shown to be equivalent to the sum of three pyramids having for their common altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between them. Hence the proposition is true of the frustum of any pyramid. PROPOSITION XXII. -THEOREM. 494. Similar pyramids are to each other as the cubes of their homologous edges. 212 ELEMENTS OF GEOMETRY. Let ABC-S and S S DEF-S be two sim- ilar pyramids; these pyramids are to each other as the cubes of D - F their homologous edges / AB and DE, or BC A i and EF, &c. ' For, the two pyra- \ mids being similar, the B homologous polyedral angles at the vertices are equal (Art. 452); hence the smaller pyramid may be so applied to the larger, that the polyedral angle S shall be common to both. In that case, the bases AB C, D EF will be parallel; for, since the homologous faces are similar, the angle S D E is equal to S A B, and S E F to S B C; hence the plane A B C is parallel to the plane D E F (Prop. XVI. Bk. VII.). Then let SO be drawn from the vertex S perpendicular to the plane A B C, and let P be the poilt where this perpendicular meets the plane D E F. From what has already been shown (Prop. XVI.), we shall have SO: SP:: SA: SD:: AB: DE; and consequently, SO: SP::AB: DE. But the bases A B C, D E F are similar; hence (Prop. XXIX. Bk. IV.), ABC: DEF::A B: DDE. Multiplying together the corresponding terms of these two proportions, we have AB C X SO:-DEF X SP: AB3: D~E3. Now, A B C X S 0 represents the solidity of the pyramid A B C - S, and D E F X > SP that of the pyramid D E F-S (Prop. XX.); hence two similar pyramids are to each other as the cubes of their homologous edges. BOOK VIII. 213 PROPOSITION XXIII. - THEOREM. 495. There can be no more than five regular polyedrons. For, since regular polyedrons have equal regular polygons for their faces, and all their polyedral angles equal, there cal be but few regular polyedrons. First. If the faces are equilateral triangles, polyedrons may be formed of them, having each polyedral angle contained by three of these triangles, forming a solid bounded by four equal equilateral triangles; or by four, forming a solid bounded by eight equal equilateral triangles; or by five, forming a solid bounded by twenty equal equilateral triangles. No others can be formed with equilateral triangles. For six of these angles are equal to four right angles, and cannot form a polyedral angle (Prop. XX. Bk. VII.). Secondly. If the faces are squares, their angles may be arranged by threes, forming a solid bounded by six equal squares. Four angles of a square are equal to four right angles, and cannot form a polyedral angle. Thirdly. If the faces are regular pentagons, their angles may be arranged by threes, forming a solid bounded by twelve equal and regular pentagons. We can proceed no farther. Three angles of a regular hexagon are equal to four right angles; three of a heptagon are greater. Hence, there can be formed no more than five regular polyedrons, - three with equilateral triangles, one with squares, and one with pentagons. 496. Scholium. The regular polyedron bounded by four equilateral triangles is called a TETRAEDRON; the one bounded by eight is called an OCTAEDRON; the one bounded by twenty is called an ICOSAEDRON. The regular polyedron bounded by six equal squares is called a HEXAEDRON, or CUBE; and the one bounded by twelve equal and regular pentagons is called a DODECAEDRON. BOOK IX. THE SPHERE, AND ITS PROPERTIES. DEFINITIONS. 497. A SPHERE is a solid, or volume, bounded by a curved surface, all points of which are equally distant from a point within, called the center. The sphere may be con- D ceived to be formed by the revolution of a semicircle, D A E, about its diameter, D E, which remains fixed. A -- 498. The RADIUS of a sphere is a straight line drawn from the center to any point in surface, as the line C B. The DIAMETER, or Axis, of a sphere is a line passing through the center, and terminated both ways by the surface, as the line D E. Hence, all the radii of a sphere are equal; and all the diameters are equal, and each is double the radius. 499. A CIRCLE, it will be shown, is a section of a sphere. A GREAT CIRCLE of the sphere is a section made by a plane passing through the center, and having the center of the sphere for its center; as the section AB, whose center is C. 500. A SMALL CIRCLE of the sphere is any section made by a plane not passing through the center. 501. The POLE of a circle of the sphere is a point in the BOOK IX. 215 surface equally distant from every point in the circumference of the circle. 502. It will be shown (Prop. V.) that every circle, great or small, has two poles. 503. A PLANE is TANGENT to a sphere, when it meets the sphere in but one point, however far it may be produced. 504. A SPHERICAL ANGLE is the difference il the direction of two arcs of great circles of the sphere; as A E D, formed by the arcs E A, D E. It is the same as the angle resulting from passing two planes through those arcs; as the angle formed on the edge EF, by the planes EAF, EDF. A C FD 505. A SPHERICAL TRIANGLE is a portion of the surface of a sphere bounded by three arcs of great circles, each arc being less than a semi-circumference; as A E D. These arcs are named the sides of the triangle; and the angles which their planes form with each other are the angles of the triangle. 506. A spherical triangle takes the name of rig'ht-angled, isosceles, equilateral, in the same cases as a plane triangle. 507. A SPHERICAL POLYGON is a portion of the surface of a sphere bounded by several arcs of great circles. 508. A LUNE is a portion of the surface of a sphere com- EF prehended between semi-cir- /1 ' cumferences of two great cir-... cles; asA I G B D F. Ct 509. A SPHERICAL WEDGE, / or UNGULA, is that portion of a sphere comprehended between 216 ELEMENTS OF GEOMETRY. two great semicircles having a A common diameter. 510. A ZONE is a portion of / the surface of a sphere cut off I. by a plane, or comprehended C< D.. between two parallel planes; as EIFK-A, or CGDHEIFK. 511. A SPHERICAL SEGMENT is a portion of the sphere cut off by a plane, or comprehended between two parallel planes. 512. The ALTITUDE of a ZONE or of a SPHERICAL SEGMENT is the perpendicular distance between the two parallel planes which comprehend the zone or segmellt. In case the zone or segment is a portion of the sphere cut off, one of the planes is a tangent to the splere. 513. A SPHERICAL SECTOR is a solid described by the revolution of a circular sector, in the same manner as the semicircle of whichl it is a part, by revolving round its diameter, describes a sphere. 514. A SPHERICAL PYRAMID is a portion of the sphere comprehended between the planes of a polyedral angle whose vertex is the center. The base of the pyramid is the spherical polygon intercepted by the same planes. PROPOSITION I. THEOREM. 515. Every section of a sphere made by a plane is a circle..Let A B E be a section made by a plane in the sphere whose center is C. From the centre, C, draw CD perpendicular to' the plane A B E; and draw the lines C A, C B, C E, to different points of the curve A B E, which bounds tle section. ^.d r BOOK IX. Z1l The oblique lines C A, C B, C E are equal, being radii of the sphere; A/'..I.._^ D therefore they are equally distant from thle perpendicular, C D (Prop. V. Cor., Bk. VII.). Hence, thec, lines DA, D B, D E, and, in like / manner, all the lines drawn from D to the boundary of the section, \__ are equal; and therefore the section ABE is a circle whose center is D. 516. Cor. 1. If the section passes through the center of the sphere, its radius will be the radius of the sphere; hence all great circles are equal. 517. Cor. 2. Two great circles always bisect each other. For, since the two circles have the same center, their common intersection, passing through the center, must be a common diameter bisecting both circles. 518. Cor. 3. Every great circle divides the sphere and its surface into two equal parts. For if the two hemispheres were separated, and afterwards placed on the coinmon base, with their convexities turned the same way, the two surfaces would exactly coincide. 519. Cor. 4. The center of a small circle, and that of the sphere, are in a straight line perpendicular to the plane of the small circle. 520. Cor. 5. Small circles arc less according to their distance from the center; for, the greater the distance C D, the smaller the chord A B, the diameter of the small circle A B E. 521. Cor. 6. The arc of a great circle may be made to pass through any two points on the surface of a sphere; for the two given points and the center of the sphere determine the position of a plane. If, however, the two given points bo the extremities of a diameter, these two points 19 218 ELEMENTS OF GEOMETRY. and the center would be in a straight line, and any number of great circles may be made to pass through the two given points. PROPOSITION II. -THEOREM. 522. Any one side of a spherical triangle is less than the sum of the other two. Let AB C be any spherical triangle; A - then any side, as A B, is less than the sum of the other two sides, A C, B C. For, draw the radii O A, 0 B, O C, and the plane angles A B, AO C, C OB will form a triedral angle, O. The angles AOB, AOC, C OB will be measured by AB, AC, BC, the O side of the spherical triangle. But each of the three plane angles forming a triedral angle is less than the sum of the other two (Prop. XIX. Bk. VII.). Hence, any side of a spherical triangle is less than the sum of the other two. PROPOSITION III. -THEOREM. 523. The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the tuo given points. Let ABD be the arc of the B great circle which joins the points A A and D; then the line A B D is the shortest path from A to D on the surface of the sphere. For, if possible, let the shortest path on the surface from A to D pass through the point C, out of the arc of the great circle A B D. Draw A C, D C,.arcs of great circles, and take D B equal to D C. Then in the spherical triangle A B D C the side A B D is less than the sum of the sides A C, ID C (Prop. II.); and BOOK IX. subtracting the equal D B and D C, there will remain A B less than A C. Now, the shortest path, on the surface, from D to 0, whether it is the arc D C, or any other line, is equal to the shortest path from D to B; for, revolving D C about the diameter which passes through D, the point C may be brought into the position of the point B, and the shortest path from D to C be made to coincide with the shortest path from D to B. But, by hypothesis, the shortest path from A to D passes through C; consequently, the shortest path on the surface from A to C cannot be greater than that from A to B. Now, since A B has been proved to be less than AC, the shortest path from A to C must be greater than that from A to B; but this has just been shown to be impossible. Hence, no point of the shortest path from A to D can lie out of the arc A B D; consequently, this arc of a great circle is itself the shortest path between its extremities. 524. Cor. The distance between any two points of surface, on the surface of a sphere, is measured by the arec df a great circle joining the two points. PROPOSITION IV.- THEOREM. 525. The sum of all the sides of any spherical polygon is less than the circumference of a great circle. Let A B C D E be a spherical polygon; E then the sum of the sides A B, B C, C D, / D &c. is less than the circumference of a A great circle. \B For, from 0, the center of the sphere, draw the radii OA, OB, OC, &c., and the plane angles AOB, BOC, COB, \ &c. will form a polyedral angle at O. Now, the sum of the plane angles which Q 220 ELEMENTS OF GEOMETRY. form a polyedral angle is less than four right angles (Prop. XX. Bk. VII.). Hence, the sum of the arcs A B, B C, C D, &c., which measure these angles, and bound the spherical polygon, is less than the circumference of a great circle. 526. Cor. The sum of the three sides of a spherical triangle is less than the circumference of a great circle, since a triangle is a polygon of three sides. PROPOSITION V. - THEOREM. 527. The extremities of a diameter of a sphere are the poles of all circles of the sphere whose planes are perpendicular to that diameter. Let D E be a diameter per- D pendicular to AHB, a great circle of a sphere, and also to p/...r the small circle FIG; then r.... \ D and E, the extremities of. this diameter, are the poles of A j these two circles. H For, since D E is perpendic- ular to the plane A H B, it is perpendicular to all the straight lines, A C, H C, B C, &c., drawn through its foot in this plane; hence, all the arcs D A, D H, D B, &c. are quarters of the circumference. So, likewise, are all the arcs E A, EH, E B, &c.; hence the points D and E are each equally distant from all the points of the circumference, AHB; consequently D and E are poles of that circumference (Art. 501). Again, since the radius D C is perpendicular to the plane AH B, it is perpendicular to the parallel plane FIG; hence it passes through 0, the center of the circle F I G (Prop. I. Cor. 4). Hence, if the oblique lines D F, D I, D G, &c. be drawn, these lines will be equally distant from BOOK IX. 221 the perpendicular D0, and will themselves be equal (Prop. V. Bk. VII.). But the chords being equal, the arcs are equal; hence the point D is a polo of the small circle FIG; and, for like reasons, the point E is the other pole. 528. Cor. 1. Every arc of a great circle, D H, drawn from a point in the arc of a great circle, A H B, to its pole, is a quarter of the circumference, and is called a quadrant. This quadrant makes a right angle with the are A II. For, the line D C being perpendicular to the plane A I C, every plane D H C passing through the line D C is perpendicular to the plane A H C (Prop. VII. Bk. VII.); hence the angle of those planes, or the angle AH D, is a right angle (Art. 506). 529. Cor. 2. To find the pole of a given arc, A H, draw the indefinite arc HD perpendicular to AH, and take HD equal to a quadrant; the point D will be one of the poles of the arc A II D; or at each of the two points A and H, draw the arcs AD and H D perpendicular to AH; the point of their intersection, D, will be the pole required. 530. Cor. 3. Conversely, if the distance of the point D from each of the points A and H is equal to a quadrant, the point D will be the pole of the arc A H; and the angles D A H, A H D will be right. For, let C be the center of the sphere, and draw the radii C A, C D, C H. Since the angles A C D, H C D are right, the line C D is perpendicular to the two straight lines C A, C H; hence it is perpendicular to their plane (Prop. IV. Bk. VII.). Hence the point D is the pole of the arc A H; and consequently the angles D A II, A H D are right angles. 531. Scholium. A circle may be described on the surface of a sphere with the same facility as on a plane surface. For instance, by turning the arc D F, or any other line extending to the same distance, round the point D the 19* 222 ELEMENTS OF GEOMETRY. extremity, F, will describe the small circle F I G; and by turning the quadrant D F A round the point D, its extremity, A, will describe the great circle A H B. PROPOSITION VI. - THEOREM. 532. A plane perpendicular to a radius, at its termination in the surface, is tangent to the sphere. Let ADB be a plane per- A D E B pendicular to a radius, C D, at / F its termination, D; then the plane A D B is a tangent to the sphere..... For, draw from the center, C C, any other straight line, C E, to the plane, ADB. Then, since C D is perpendicular to the plane, it is shorter than the oblique line C E; hence the radius C F is shorter than C E; consequently.the point E is without the sphere. The same may be shown of any other point in the plane A D B, except the point D; hence the plane can meet the sphere in but one point, and therefore is a tangent to the sphere (Art. 503). 533. Scholium. In the same manner, it may be proved that two spheres are tangent to each other, when the distance between their centers is equal to the sum or the difference of their radii; in which case the centers and the point of contact lie in the same straight line. PROPOSITION VII. -THEOREM. 534. The angle formed by two arcs of great circles is equal to the angle formed by the tangents of those arcs at the point of their intersection, and is measured by the arc of a great circle described from its vertex as. a pole, and intercepted between its sides, produced if necessary. BOOK IX. 228 Let B A C be an angle formed A F by tile two arcs AB, A C; then will it be equal to the angle E E A F, formed by the tangents. A E, AF, and it is measured (... by B C, the arc of a great circle 0 - described from the vertex A as a pole. For the tangent AE, drawn in the plane of the arc A B, is D perpendicular to the radius A 0 (Prop. X. Bk. III.); and the tangent A F, drawn in the plane of the arc A C, is perpelidicular to the same radius A O. Hence the angle E A F is equal to tlle angle of the planes A 0 B, A O (Art. 391); which is that of the arcs A B, A C. Also, if the arcs A B, A C are both quadrants, the lines 0 B, 0 C will be perpendicular to AO, and the angle B O C will be equal to the angle of the planes A O B, AO C; hence the arc B C is the measure of the angle of these planes, or the measure of the angle C A B. 535. Cor. 1. The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles, and included between tljeir sides; hence it is easy to make an angle of thils kind equal to a given angle. 536. Cor. 2. Vertical angles, B such as AOC and BOD, are equal; for each of them is equal to the angle formed by \ the two planes A 0 B, C D. C It is also evident that the two adjacent angles, A O C, COB, taken together, are A equal to two right angles. 224 ELEMENTS OF GEOMETRY. PROPOSITION VIII.- THEOREM. 537. If from the vertices of any spherical triangle, as poles, arcs of great circles are described, a second triangle is formed, whose vertices will be poles to the sides of the first triangle. Let A B C be any spher- D ical triangle; and from the vertices, A, B, C, as poles, let the arcs E F, A FD, DE be described, and a second triangle, DEF, is formed, whose E B vertices, D, E, F, will be F poles to the sides of the triangle ABC. H For, the point A being the pole of the arc E F, the distance AE is a quadrant; tle point C being the pole of the arc D E, the distance C E is also a quadrant; hence the point E is at the distance of a quadrant from each of the points A and C; hence it is the pole of the arc A C (Prop. V. Cor. 3). In like manner, it may be shown that D is the pole of the arc B C, and F that of the arc A B. 538. Scholium. Hence the triangle A B C may be described by means of D E F, as D E F may be by means of A B C. Spherical triangles thus described are said to be polar to each other, and are called polar or supplemental triangles. PROPOSITION IX. - THEOREM. 539. Each of the angles of a spherical triangle is measured by a semi-circumference minus the side lying opposite to it in the polar triangle. Let A B C be a spherical triangle, and D E F a triangle polar to it; then each of the angles of A B C is measured BOOK IX. 22X by a semi-circumference D minus the side lying opposite to it in D E F. For, produce the sides A B, A C, if necessary, till they meet EF in G and H. The point A being the B pole of the arc GH, the \ angle A will be measured by that arc (Prop. VII.). H But, E being the pole of A H, tlle arc E H is a quadrant; and F being the pole of A G, F G is a quadrant. Hence, E H and G F together are equal to a semi-circumference. Now, the sum of E H and G F is equal to the sum of E F and G H; hence the arc G 11, which measures the angle A, is equal to a semi-circumference milus the side E F. In like manner, the angle B will be measured by a semicircumference minus D F; and the angle C by a semicircumference minus D E. 540. Cor. This property must be reciprocal in the two triangles, since they are polar to eacl other. The angle D, for example, of the triallgle D'E F, is measured by the arc I K; but the sum of I K and B C is equal to the sum of I C and B K, which is equal to a semi-circumference; hence tile arc I K, the measure of D, is equal to a semicircumference minus B C. In like manner, it may be shown tlat E is measured by a semi-circumference minus A C, and F by a semi-circumference minus A B. PROPOSITION X.-THEOREM. 541. Tle sum of the angles in any spherical trianles is less than six right angles, and greater than two. First. Every angle of a spherical triangle is less than two right angles; hence, the sum of the three is less than six right angles. 226 ELEMENTS OF GEOMETRY. Secondly. The measure of each angle of a spherical triangle is equal to the semi-circumference minus the corresponding side of the polar triangle (Prop. IX.); hence, the sum of the three is measured by three semi-circumforences minus the sum of the sides of the polar triangle. Now, this latter sum is less than a circumference (Prop. IV. Cor.); therefore, taking it away from three semicircumferences, the remainder will be greater than one semi-circumference, which is the measure of two right angles; hence, the sum of the three angles of a spherical triangle is greater than two right angles. 542. Cor. 1. The sum of the angles of a spherical triangle is not constant, like that of the angles of a rectilineal triangle. It varies between two right angles and six, without ever arriving at either of these limits. Two given angles, therefore, do not serve to determine the third. 543. Cor. 2. A spherical triangle may have two, or even three right angles, or obtuse angles. 544. Scholium. If a spherical triangle has two right angles, it is said to be bi-rectangular; and if it has three right angles, it is said to be tri-rectangular, or quadrantal. The quadrantal triangle is evidently contained eight times in the surface of the sphere. PROPOSITION XI. - THEOREM. 545. If around the vertices of any two angles of a given spherical triangle, as poles, the circumferences of two circles be described, which shall pass through the third angle of the triangle, and then if through the other point in which these circumferences intersect, and the vertices of the first two angles of the triangles, arcs of two great circles be drawn, the triangle thus formed will have all its parts equal to those of the given triangle, each to each. BOOK IX. 227 Let A B C be the given spherical A triangle, and CED, DFC arcs described about the vertices of any O two of its angles, A and B, as poles; / then will the triangle AD B lave C all its parts equal to those of A BC. D For, by construction, the side AD ^ is equal to A C, D B is equal to B C, and A B is common; hence tlhe two B triangles have their sides equal, eacll to each. We are now to show that tlhe angles opposite these equal sides are also equal. If the center of the sphere is supposed to be at 0, a triedral angle may be conceived as formed at O by tlhe three plane angles A O B, A OC, B 0 C; also, another triedral angle may be conceived as formed by the three plane angles A OB, AOD, BOD. Now, since the sides of the triangle AB C are equal to those of the triangle AD B, the plane angles forming the one of these triedral angles are equal to the plane angles forming the other, each to each. Therefore the planes, in which the equal angles lie, are equally inclined to each other (Prop. XXI. Bk. VII.); hence, all tlhe angles of the spherical triangle DAB are respectively equal to those of the triangle C A B; namely, D A B is equal to B A C, D B A to A B C, and A D B to A C B; hence, the sides and angles of the triangle A D B are equal to the sides and the angles of the triangle A C B, each to each. 546. Scholium. The equality of these triangles is not, however, an absolute equality, or one of superposition; for it would be impossible to apply them to each other exactly, unless they were isosceles. The equality here meant is that by symmetry; therefore the triangles AC B, A D B are termed symmetrical triangles. 228 ELEMENTS OF GEOMETRY. PROPOSITION XII.- THEOREM. 547. If two triangles on the same sphere, or on equal spheres, are mutually equilateral, they are mutually equiangular; and their equal angles are opposite to equal sides. Let ABC, A B D be two triangles on A the same sphere, or on equal spheres, having the sides of tlhe one respectively equal to those of the other; tllenl tlhe angles opposite to the equal sides, C il the two triangles, are equal. D For, with three given sides, A B, AC, B C, there can be constructed only two triangles, A C B, A B D, and these triangles will be equal, each to each, in the magnitude of all tleir parts (Prop. XI.). Hence, these two triangles, which are mutually equilateral, must be either absolutely equal, or equal by symmetry; in either case they are mutually equiangular, and the equal angles lie opposite to equal sides. PROPOSITION XIII. -THEOREM. 548. If two triangles on the sante sphere, or on equal spheres, are mutually equiangular, they are mutually equilateral. Let A and B be the two given triangles; P and Q, their polar triangles. Since the angles are equal in the triangles A and B, the sides will be equal in the polar triangles P and Q (Prop. IX.). But since the triangles P and Q are mutually equilateral, they must also be mutually equiangular (Prop. XII.); and, the angles being equal in the triangles P and Q, it follows that the sides are equal in their polar triangles A and B. Hence, the triangles A and B, which are BOOK IX. 229 mutually equiangular, are at the same time mutually equilateral. PROPOSITION XIV.- THEOREM. 549. If two triangles on the same sphere, or on equal spheres, have two sides and the included angle in the one equal to two sides and the included angle in the other, each to each, the two triangles are equal in all their parts. Il the two triangles ABC, DEF, let the side A B be equal to the side D E, the side A C to the side D F; and the angle BA C to tlle angle E D F; then the triangles will be equal in all tleir parts. A C B D O/)F G /FE E Let the triangle D E G be symmetrical with tile triangle D E F (Prop. XI. Scll.), having the side E G equal to E F, the side G D equal to F D, and the side E D common, and consequently the angles of the one equal to those of the other (Prop. XII.). Now, the triangle AB C may be applied to the triangle D E F, or to D E G symmetrical with D E F, just as two rectilineal triangles are applied to each other, when they have an equal angle included between equal sides. Helice, all the parts of the triangle A B C will be equal to all the parts of the triangle D E F, each to each; that is, besides the three parts equal by hypothesis, we shall have the side B C equal to E F, the angle A B C equal to D E F, and the angle A C B equal to D F E. 550. Cor. If two triangles, A B C, D E F, on the same sphere, or on equal spheres,.have two angles and the included side in the one equal to two angles and the included side in the other, each to each, the two triangles are equal in all their parts. 20 230 ELEMENTS OF GEOMETRY. For one of these triangles, or the triangle symmetrical with it, may be applied to the other, as is done in the corresponding case of rectilineal triangles. PROPOSITION XV.- THEOREM. 551. In every isosceles spherical triangle, the angles op. posite the equal sides are equal; and, conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. Let ABC be an isosceles spherical A triangle, in which the side A B is equal o to the side A C; then will the angle B be equal to the angle C. For, if the arc AD be drawn from the vertex A to the middle point, D, of the base, the two triangles A B D, A C D B C will have all the sides of the one re- D spectively equal to the corresponding sides of the other, namely, A D common, B D equal to D C, and A B equal to A C; hence their angles must be equal; consequently, the angles B and C are equal. Conversely. Let the angles B and C be equal; then will the side A C be equal to A B. For, if A C and A B are not equal, let A B be the greater of the two; take B 0 equal to A C, and draw O C. The two sides B 0, B C in the triangle B 0 C are equal to the two sides A B, B C in the triangle B A C; the angle O B C, contained by the first two, is equal to A C B, contained by the second two. Hence, the two triangles B 0 C, BAC have all their other parts equal (Prop. XIV. Cor.); hence the angle 0 C B is equal to AB C. But, by hypothesis, the angle A B C is equal to A C B; hence we have 0 C B equal to AC B, which is impossible; therefore A B cannot be unequal to A C; consequently the sides A B, A C, opposite the equal angles B and C, are equal. BOOK IX. 552. Cor. The angle B A D is equal to D A C, and the angle B D A is equal to A D C; the last two are therefore right angles; hence the arc drawn from.the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to the base, and bisects the vertical angle. PROPOSITION XVI. -THEOREM. 553. In a spherical triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side. In the triangle AB C, A let the angle A be greater than B; then will the side. \ BC, opposite to A, be greater than A C, opposite to B. Take the angle BAD B equal to the angle B; then, in the triangle AB D, we shall have the side AD equal to DB (Prop. XV.). But the sum of AD plus D C is greater than A C; hence, putting D B in the place of AD, we shall have the sum of D B plus D C, or B C, greater than AC. Conversely. Let the side B C be greater than AC; then the angle B A C will be greater than A B C. For, if B A C were equal to AB C, we should have BC equal to AC; and if B A C were less than A B C, we should then have, as has just been shown, B C less than A C. Both of these results are contrary to the hypothesis; hence the angle BAC is greater than ABC. PROPOSITION XVII. -THEOREM. 554. If two triangles on the same sphere, or on equal spheres, are mutually equilateral, they are equivalent. 232 ELEMENTS OF GEOMETRY. Let ABC, DEF be two D A triangles, having tlhe three sides of the one equal to the three sides of the other, each / to each, namely, AB to D E, F/..-.p O~'." — C AC to DF, and CB to EF; then their triangles will be equivalent. E B Let 0 be the pole of the small circle passing through the three points A, B, C; draw the arcs 0 A, OB, 0 C, and they will all be equal (Prop. V. Sch.). At the point F make the angle D F P equal to ACO; make the arc F P equal to CO; and draw DP, EP. The sides D F, F P are equal to the sides A C, C 0, and the angle D F P is equal to the angle A C 0; hence tll3 two triangles D FP, A CO are equal in all their parts (Prop. XIV.); hence the side D P is equal to A O, and the angle D P F is equal to A 0 C. In the triangles D F E, A B C, the angles D F E, A C B, opposite to the equal sides D E, A B, are equal (Prop. XII.). Taking away the equal angles D F P, A CO, there will remain the angle P F E, equal to 0 C B. The sides P F, F E are equal to the sides 0 C, C B; hence the two triangles F P E, C 0 B are equal in all their parts (Prop. XIV.); hence the side PE is equal to 0 B, and the angle F P E is equal to C 0 B. Now, the triangles D F P, A C 0, which have the sides equal, each to each, are at the same time isosceles, and may be applied the one to the other. For, having placed O A upon its equal P D, the side 0 C will fall on its equal P F, and thus the two triangles will coincide; consequently they are equal, and the surface D P F is equal to A 0 C. For a like reason, the surface F P E is equal to C 0 B, and the surface D P E is equal to A 0 B; hence we have BOOK IX. 283 AOC +COB- AOB DPF+FPE -DPE, or, A B C =D E F. Hence the two triangles A B C, D E F are equivalent. 555. Cor. 1. If two triangles on the same sphere, or on equal spheres, are mutually equiangular, they are equivalent. For in that case the triangles will be mutually equilateral. 556. Cor. 2. Hence, also, if two triangles on the same sphere, or on equal spheres, have two sides and the included angle, or have two angles and the included side, in the one equal to those in the other, the two triangles are equivalent. 557. Scholium. The poles O and P might lie within the triangles A B C, D E F; in which case it would be requisite to add the three triangles DPF, FPE, DPE together, to form the triangle D E F; and in like manner to add tlhe three triangles A O C, C B, A O B together, to form the triangle A B C; in all other respects the demonstration would be tile same. PROPOSITION XVII. - THEOREM. 558. The area of a lune is to the surface of the sphere as the angle of the lune is to four right angles, or as the arc which measures that angle is to the circumference. Let A C B D be a lune upon A a sphere whose diameter is i A B; tlen will the area of the luile be to the surface of the..... sphere as the angle D 0 C to C...... four riglt angles, or as the arc D e0 D C to tlle circumference of a great circle. For, suppose the arc CD to B be to the circumference C D E F in the ratio of two whole numbers, as 5 to 48, for example. 20* 2384 ELEMENTS OF GEOMETRY. Then, if the circumference A CD EF be divided into 48 equal parts, C D will contain 5 of them; and if the pole. A be joined with the several C....... points of division by as many D quadrants, we shall have 48 triangles on the surface of the hemisphere ACDEF, all equal, since all their parts are equal. Hence, the whole sphere must contain 96 of these triangles, and the lune A C B D 10 of them; consequently, the lune is to the sphere as 10 is to 96, or as 5 to 48; that is, as the arc C D is to the circumference. If the arc C D is not commensurable with the circumference, it may still be shown, by a mode of reasoning exemplified in Prop. XVI. Bk. III., that the lune is to the sphere as C D is to the circumference. 559. Cor. 1. Two lunes on the same sphere, or on equal spheres, are to each other as the angles included between their planes. 560. Cor. 2. It has been shown that the whole surface of the sphere is equal to eight quadrantal triangles (Prop. X. Sch.). Hence, if the area of a quadrantal triangle be represented by T, the surface of the sphere will be represented by 8T. Now, if the right angle be assumed as unity, and the angle of the lune be represented by A, we have, Area of the lune: 8 T A: 4, which gives the area of lune equal to 2 A X T. 561. Cor. 3. The spherical ungula included by the planes A C B, A D B, is to the whole sphere as the angle D 0 C is to four right angles. For, the lunes being equal, the spherical ungulas will also be equal; hence, two spherical ungulas on tie same sphere, or on equal spheres, BOOK IX. are to each other as the angles included between their.planes. PROPOSITION XIX.- THEOREM. 562. If two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed is equivalent to a lune, whose angle is equal to the angle formed by the circles. Let the great circles B A D, A C A E intersect on the surface of a hemisphere, AB D E; then will the sum of the opposite triangles, B A, D A E, B be equal to a lune whose angle is DAE. For, produce the arcs AD, AE till they meet in F; and the arcs BAD, ADF will each be a semi-circumference. Now, if we take away A D from both, we shall have D F equal to B A. For a like reason, we have E F equal to C A. D E is equal to B C. Hence, the two triangles BAC, DEF are mutually equilateral; therefore they are equivalent (Prop. XVII.). But the sum of the triangles D E F, D A E is equivalent to the lune A D F E, whose angle is D A E. PROPOSITION XX.-THEOREM. 563. The area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles, multiplied by the quadrantal triangle. Let A B C be a spherical triangle; its area is equal to the excess of the sum of its angles, A, B, C, above twq right angles multiplied by the quadrantal triangle. For produce the sides of the triangle AB C till they 2836 ELEMENTS OF GEOMETRY. meet the great circle D E F G H I, F drawn without the triangle. The E two triangles ADE, A G H are / together equivalent to the lune whose angle is A (Prop. XIX.), A and whose area is expressed by 2 A X T (Prop. XVIII. Cor. 2). \ Hence we have D. ADE+AGH=-2AXT; and, for a like reason, BGF +BID= 2B T, and CIH+CFE= 2C XT. But the sum of these six triangles exceeds the hemisphere by twice the triangle A B C; and the hemisphere is represented by 4 T; consequently, twice the triangle A B C is equivalent to 2A X T- +2B X T + 2C X T - 4T; therefore, once the triangle A B C is equivalent to (A + B + C- 2) X T. Hence the area of a spherical triangle is equal to the excess of the sum of its three angles above two right angles multiplied by the quadrantal triangle. 564. Cor. If the sum of the three angles of a spherical triangle is equal to three right angles, its area is equal to the quadrantal triangle, or to an eighth part of the surface of the sphere; if the sum is equal to four right angles, the area of the triangle is equal to two quadrantal triangles, or to a fourth part of the surface of the sphere, &c. PROPOSITION XXI. -THEOREM. 565. The area of a spherical polygon is equal to the excess of the sum of all its angles above two rig'ht angles taken as many times as the polygon has sides, less two, multiplied by the quadrantal triangle. BOOK IX. 237 Let A B C D E be any spherical C polygon. From one of the vertices, A, draw the arcs AC, AD to the / opposite vertices; the polygon will \ \ be divided into as many spherical \ triangles as it has sides less two. B But the area of each of these trian- A gles is equal to the excess of the sum of its three angles above two right angles multiplied by the quadrantal triangle (Prop. XX.); and the sum of the angles in all the triangles is evidently the same as that of all the angles in the polygon; hence the area of the polygon A B C D E is equal to the excess of the sum of all its angles above two right angles taken as many times as the polygon has sides, less two, multiplied by the quadrantal triangle. 566. Cor. If the sum of all the angles of a spherical polygon be denoted by S, the number of sides by n, the quadrantal triangle by T, and the right angle be regarded as unity, the area of the polygon will be expressed by S-2 (n-2) X T= (S- 2 +4) X T. Ct. C (s-:z T Z. f X - SC -- 2' '., BOOK X. THE THREE ROUND BODIES. DEFINITIONS. 567. A CYLINDER is a solid, which may be described by the revolution of a rectan- AN gle turning about one of its sides, which remains immovable; as the solid described by the rectangle A B C D revolving about its side AB. -.. The BASES of the cylinder are the circles B described by the sides, AC, BD, of the D revolving rectangle, which are adjacent to the immovable side, AB. The AXIS of the cylinder is the straight line joining the centres of its two bases; as the immovable line A B. The CONVEX SURFACE of the cylinder is described by the side C D of the rectangle, opposite to the axis A B. 568. A CONE is a solid which may be described by the revolution of a right- angled triangle turning about one of its perpendicular sides, which remains im- / movable; as the solid described by the right-angled triangle A B C revolving about its perpendicular side A B. The BASE of the cone is the circle described by the revolution of the side B B C, which is perpendicular to the im- C movable side. BOOK X. The CONVEX SURFACE of a cone is described by the hypothenuse, A C, of the revolving triangle. The VERTEX of the cone is the point A, where the hypothenuse meets the immovable side. The AXIS of the cone is the straight line joining the vertex to the centre of the base; as the line A B. The ALTITUDE of a cone is a line drawn from the vertex perpendicular to the base; and is the same as the axis, AB. The SLANT HEIGHT, or SIDE, of a cone, is a straight line drawn from the vertex to the circumference of the base; as the line A C. 569. The FRUSTUM of a cone is the part of a cone included between the FA base and a plane parallel to the base; as the solid C D- F. The AXIS, or ALTITUDE, of the frus —.. tum, is the perpendicular line A B ji- C D eluded between the two bases; and the SLANT HEIGHT, or SIDE, is that portion of the slant height of the cone which lies between the bases; as F C. 570. SIMILAR CYLINDERS, or CONES, are those whose axes are to each other as the radii, or diameters, of their bases. 571. The sphere, cylinder, and cone are termed the THREE ROUND BODIES of elementary Geometry. PROPOSITION I. -THEOREM. 572. The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude. Let A B CD E F- G be a cylinder, whose circumference is the circle A B CD EF, and whose altitude is the line A G; then its convex surface is equal to ABCDEF multiplied by AG. 240 ELEMENTS OF GEOMETRY. In the base of the cylinder inscribe any regular polygon, A B C D E F, and G on this polygon construct a right prism of the same altitude with the cylinder. The prism will be inscribed in the con- vex surface of the cylinder. The convex surface of this prism is equal to the / - E; D perimeter of its base multiplied by its A D altitude, A G (Prop. I. Bk. VIII.). B C Conceive now the arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides; its perimeter will then be equal to the circumference of the circle ABCDEF (Prop. XII. Cor., Bk. VI.); and thus the convex surface of the prism will coincide with the convex surface of the cylinder. But the convex surface of the prism is always equal to the perimeter of its base multiplied by its altitude; hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude. 573. Cor. 1. If two cylinders have the same altitude, their convex surfaces are to each other as the circumferences of their bases. 574. Cor. 2. If H represent the altitude of a cylinder, and R the radius of its base, then we shall have the circumference of the base represented by 2 R X n (Prop. XV. Cor. 3, Bk. VI.), and the convex surface of the cylinder by 2 R X X H. PROPOSITION II. THEOREM. 575. The solid contents of a cylinder are equal to the product of its base by its altitude. Let A B C D E F - G be a cylinder whose base is the circle A B C D E F, and whose altitude is the line A G; then its solid contents are equal -to the product of ABCDEF by AG. BOOK X. 241 In the base of the cylinder inscribe any regular polygon, A B C D E F, and G. I on this polygon construct a right prism of the same altitude with the cylinder. Tlhe prism will be inscribed in the convex surface of the cylinder. The solid contents of this prism are equal to E i D the product of its base by its altitude A (Prop. XIII. Bk. VIII.). B C Conceive now the number of the sides of the polygon to be indefinitely increased, until its perimeter coincides with the circumference of the circle A B C D E F (Prop. XII. Cor., Bk. VI.), and the solid contents of the prism will equal those of the cylinder. But the solid contents of the prism will still be equal to the product of its base by its altitude; hence the solid contents of the cylinder are equal to the product of its base by its altitude. 576. Cor. 1. Cylinders of the same altitude are to each other as their bases; and cylinders of equal bases are to each other as their altitudes. 577. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their radii (Prop. XIII. Bk. VI.), and the cylinders being similar, the radii of their bases are to each other as their altitudes (Art. 570); therefore the bases are as the squares of the altitudes; hence, the products of the bases by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. 578. Cor. 3. If the altitude of a cylinder be represented by H, and the area of its base by R2 X, (Prop. XV. Cor. 2, Bk. VI.), the solid contents of the cylinder will be representedby R2 X n X tH. 21. 242 ELEMENTS -OF GEOMETRY. PROPOSITION III. THEOREM. 579. The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height. LetA B C D E F-S be a cone S whose base is the circle A B C D E F, and whose slant height is the line S A; then its convex surface is equal to A B C D E F multiplied by /; S A. In the base of the cone inscribe / any regular polygon, A B C D E F, A D and on this polygon construct a reg- I ular pyramid having the same ver- B C tex, S, with the cone. Then a right pyramid will be inscribed in the cone. From S draw S H perpendicular to B C, a side of the polygon. The convex surface of the pyramid is equal to the perimeter of its base, multiplied by half its slant. height, SH (Prop. XV. Bk. VIII.). Conceive now the: arcs subtending the sides of the polygon to be continually bisected, until a polygon is formed having an indefinite number of sides; its perimeter will equal the circumference of the circle A B C D E F; its slant height, S H, will equal that of the cone, and its convex surface coincide with the convex surface of the cone. But the convex surface of every right pyramid is equal to the perimeter of its base, multiplied by half the slant height; hence the convex surface of the cone is equal to the circumference of its base multiplied by half its slant height. 580. Cor. If S A represent the slant height of a cone, and R the radius of the base, then, since the circumference of the base is represented by 2 R X n (Prop. XV. Cor. 3, Bk. VI.), the convex surface of the cone will be represented by 2R X n SA, equal to n X R X SA. BOOK x. 243 PROPOSITION IV. THEOREM. 581. The convex surface of a frustunt of a cone is equal to half the sum of the circumference of the two bases multiplied by its slant height. Let A B C D E F -M be the frustum. M L of a cone, and A G its slant height; G then the convex surface is equal to / 1I \ half the sum of tile circumferences of /I / the two bases ABCDEF, GFIIIKLM,. \ multiplied by A G. A.E A D For, inscribe in the bases of the frus-. tum two regular polygons of the same B C number of sides, having their sides parallel, each to each. Draw the straight lines A G, B H, C I, &c., joining the vertices of the corresponding angles, and these liles will be the edges of the frustum of a pyramid inscribed in the frustum of the cone. The convex surface of the frus.tum of the pyramid is equal to half the sum of tle perimeters of the two bases multiplied by its slant height, ON (Prop. XVII. Bk. VIII.). Conceive now the number of sides of the inscribed polygons to be indefinitely increased; the perimeters of the polygons will then coincide with the circumferences of the circles A B C D E F, G H I K L M; and the slant height, O N, of the frustum of the pyramid, will equal tle slant height, A G, of the frustum of the cone; and the surfaces of tile two firstums will coincide. But the convex surface of every frustum of a riglht pyramid is equal to half the sum of the perimeters of its two bases, multiplied by its slant height; hence, the convex surface of the frustum of the cone is equal to half the sum of the circumference of its two bases multiplied by half its slant heiglit. 582. Cor. Through R, the middle point -of the side KD% 2A44 ELEMENTS OF GEOMETRY. draw the diameter R ST, parallel to the Gr diameter A QD, and the straight ilnes RU, KV, parallel to the axis PQ. T/-s \ Then, since D R is eq.ual to R K, D U is equal to UV (Prop. XVII. Cor. 2, i Bk. IV.); hence, the radius S R is equal A -D to half.the sum of the radii Q D, P K. But the circumferences of circles being to each other as tlheir radii (Prop. XIII. Bk. VI.), the circumference of the section of which S R is the radius is equal to half tlhe sum of the circumferences of which Q D, PK are tlle radii; lence, the convex surface of a frustum of a cone is equal to the slant hleight multiplied by the circumference of a section at equal distances between the two bases. PROPOSITION V. - THEOREM. 583. The solidity of a cone is equal to the product of its base by one third of its altitude. Let ABCDEF-S be a cone, S whose base is A B C D E F, and altitude S I.; then its solidity is equal to ABCDEF X S H. A In the base of the cone inscribe any regular polygon, AB C D E F, and on this polygon construct a reg- ular pyramid, having the same vertex, A D S, with the cone. Then a right pyra13 C mid will be inscribed in the cone; and its solidity will be equal to the product of its base Ly one third of its altitude (Prop. XX. Bk. VIII.). Conceive, now, tle number of sides of the polygon to be indefinitely increased, and its perimeter will become equal to the circumference of the cone; and the pyramid will exactly coincide with the cone. But the solidity of every right pyramid is equal to the product of the' base by one BOOK -X..~24 third of its altitude; hence, the solidity of a cone is equal to tlhe product of its base by one third of its altitude. 584. Cor. 1. A cone is the third of a cylinder having the same base and the same altitude; hence it follows,1. Tllat cones of equal altitudes are to each other as their bases; 2. That cones of equal bases are to each other as their altitudes; 3. That similar cones are as the cubes of the diameters of their bases, or as tile cubes of their altitudes. 585. Cor. 2. If the altitude of a cone be represented by H, ahd the radius of its base by R, the solidity of the cone will be represented by R2 X 7 X 1H, or X R2 X H. PROPOsITION VI. THEOREII. 586,. The solidity of thefrustum of a cone is equivalent to the sum of three cenes, having for their common altitude the altitude of the frustum, and whose bases are the two bases of th e frustum, and a mean proportional between them. Let A B C ) E F - M be the frus- M L tum of a cone; then will its solidity be equivalent to th1e sum of three cones having the same altitude as the frustum, and whose bases are-. i the two bases of the frustum, and a mean proportional between them. II F For, inscribe in the two bases of A. the frustum two regular polygons ~ - lhaving tile same number of sides, B C and alaviilg their sides, parallel, each to cachl.. Let the Vertic6s of 'the corresponding angles be joined:by the straight lineS B H, C I, &., 'and there is.in1sribed in 1te 21* 246 ELEMENTS OF GEOMETRY. frustum of the cone the frustum of a regular pyramid. The solidity of the frustum of this pyramid is equivalent to the sum of three pyramids, having for their colnmon altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between them. Conceive now the number of the sides of the polygons to be indefinitely increased; and the bases of the frustum of the pyramid will equal the bases of the frustum of the cone; and the two frustums will coincide. Hence the frustum of a cone is equivalent to the sum of three cones, having for their common altitude the altitude of the frustum, and whose bases are the two bases of the frustum, and a mean proportional between them. PROPOSITION VII.- THEOREM. 587. If any regular semi-polygon be revolved about a line passing through the center and the vertices of oppo-site angles, the surface described will be equal to the product of its axis by the circumference of its inscribed circle. Let the regular semi-polygon A B C D E F A be revolved about A F as an axis; then B. G the surface. described by the sides A B, K BC, CD, &c. will equal the product of C'/ —; — H A F by the inscribed circle. 0 For, from the vertices B, C, D, E of the D --- semi-polygon, draw B G, C H, D M, E N, perpendicular to the axis A F; and from E N the center, 0, draw 01 perpendicular to one of the sides; also draw I K perpendicular to A F, anl B L perpendicular to C H. Now 0 I is the radius of the inscribed circle (Prop. II. Bk. VI.); and the surface described by the revolution of a side, B C, of a regular polygon, is equal to B C multiplied ty the circumference, I K (Prop. IV. Cor.). BOO K X. 247 The two triangles O I K, B C L, having their sides perpendicular to each other, are similar (Prop. XXV. Bk. IV.); therefore, BC: BL or GH: O: I: I Circ. 0: Circ. IK. Hence (Prop. I. Bk. II.), BC X Circ. IK= GH X Circ. OI; that is, the surface described by B C is equal to the product of the altitude G H by the circumference of the inscribed circle. The same may be shown of each of the other sides; hence, the surface described by all the sides taken together is equal to the product of the sum of the altitudes AG, GH H HM, MN, NF, by the circ. O, or to the product of the axis A F by the circ. 0 I. PROPOSITION VIII.- THEOREM. 588. The surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let ABCDEF be a semicircle in A which is inscribed any regular semi-poly- B G gon; from the center, 0, draw 0I per-. pcndicular to one of the sides. C - If now the semicircle and the semi- 0 polyon be revolved about the axis A F, D M thl, surface described by the semicircle will be the surface of a sphere (Art. 497), and that described by the semi-polygoi will be equal to the product of its axis, A F, by the circumference, 0I (Prop. VII.); and the same is true, whatever be the number of sides of the polygon. Conceive the number of sides of the semi-polygon to be made, by continual bisections, indefinitely great; then its perimeter will coincide with the semi-circumference A B C D E F, and the perpendicular O will be equal to the radius O A; hence, the surface of the sphere is equal 248 ELEMENTS OF GEOMETRY. to the product of the diameter by the A circumference of a great circle. B 589. Cor. 1. The surface of a sphere y. is equal to the area of four of its great circles. 0 For the area of a circle is equal to the D" M product of the circumference by half the E< radius, or one fourth of the diameter (Prop. XV. Bk. VI.). 590. Cor. 2. The surface of a zone or sement is equal to the product of its altitude by the circunrference of a great circle. For the surface described by the sides B C, C D of the inscribed polygon is equal to the product of the altitude G M by the circumference of the inscribed circle O I. If, now, the number of the sides of an inscribed polygon be indefinitely increased, its perimeter will equal the circle, and BC, CD will coincide with the arc B CD; consequently, the surface of the zone described by the revolution of B C D is equal to the product of its altitude by the circumference of a great circle. In like manner, the same may be proved true of a segment, or a zone having but one base. 591. Cor. 3. The surfaces of two zones, or segments upon the same sphere, are to each other as their altitudes; and any zone or segment is to the surface of the sphere as the altitude of that zone or segment is to the diameter. 592. Cor. 4. If the radius of a sphere is represented by R, and its diameter by D, its surface will be represented by 47X R2, or 7X D2. 593. Cor. 5. Hence, the surfaces of spheres are to each other as the squares of their radii or diameters. 594. Cor. 6. If the altitude of a zone or segment is . - OOK X -I - S4 represented by H, the surface of a zone or segment will be represented by 2nXRX H, or n XDXH. PROPOSITION IX. -THEOREM. 595. The solidity of a sphere is equal to the product of its surface by one third of its radius. For a sphere may be regarded as composed of an indefinite number of pyramids, each having for its base a part of the surface of the sphere, and for its vertex the center of the sphere; consequently, all these pyramids have the radius of the sphere as their common altitude. Now, the solidity of every pyramid is equal to the pro duct of its base by one third of its altitude (Prop. XX. Bk. VIII.); hence, the sum of the solidities of these pyramids is equal to the product of the sum of theirbases by one third of their common altitude. But the sum of their bases is the surface of the sphere, and their common altitude its radius; consequently, the solidity of the sphere is equal to the product of its surface by one third of its radius. 596. Cor. 1. The solidity of a spherical pyramid or sector is equal to the product of the polygon or zone which forms its base, by one third of the radius. For the polygon or zone forming the base of the spherical pyramid or sector may be regarded as composed of an indefinite number of planes, each serving as a base to a pyramid, having for its vertex the centre of the sphere. 597. Cor. 2. Spherical pyramids, or sectors of the same sphere or of equal spheres, are to each other as their bases. 598. Cor. 3. A spherical pyramid or sector is to the sphere of which it is a part, as its base is to the surface of the sphere. 599. Cor. 4. Hence, spherical sectors upon the same 250 ELEMENTS OF GEOMETRY. sphere are to each other as the altitudes of the zones forming their bases (Prop. VIII. Cor. 3); and any spherical sector is to the sphere as the altitude of the zone forming its base is to the diameter of the sphere. 600. Cor. 5. If the radius of a sphere is represented by R, its diameter by D, and its surface by S, its solidity will be represented by S X -R - 4 7 X R2 X - R= R 4 X R3 or Q'i X D3. 601. Cor 6. Hence, the solidities of spheres are to cacl other as the cubes of their radii. 602. Cor. 7. If the altitude of tle zone which forms the base of a sector be represented by I, the solidity of the sector will be represented by 2 r X R X H X R= 7 X R2 X H. 603. Scholium. The solidity of the spller- A ical segment less than a hemisphere, and of B one base, formed by the revolution of a portion, AB C, of a semicircle about the radius O A, is equivalent to the solidity of the spherical sector formed by A B, less the E solidity of the cone formed by 0 B C. The solidity of the spherical segment greater than a hemisphere, and of one base, formed by tlhe revolution of A D E, is equivalent to tlhe solidity of tle. spherical sector formed by A 0 D, plus the solidity of tllc cone formed by O D E. The solidity of the spherical segment of two bases formed by the revolution of C B D E about the axis A F, is equivalent to the solidity of the segment formed by. A D E, less the solidity of the segment formed by A B C. PROPOSITION X. — THEOREM. 604. The surface of a sphere is equivalent to the convex bwface of ithe circumseribed cylinder, and t is two thirds BOOK X. 9251 of the zwhole surface of the cylinder; also, the solidity of the sphere is two thirds of that of the circumscribed cylinder. Let A BFI be a great circle of tile sphiere; DEGHI tile circum- l - H scribed square; then, if the semi-, circle A B F and tile semi-square. - A D E F be revolved about the di- B......... I ameter A F the semicircle will describe a sphere, and the semi- square a cylinder circumscribillg F the sphere. The convex surface of tlle cylinder is equal to the circumference of its base multiplied by its altitude (Prop. I.). But tlhe base of the cylinder is equal to the great circle of the sphere, its diameter E G being equal to thle diameter B I, and the altitude D E is equal to the diameter AF; hence, tlhe convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter. This measure is tlhe same as that of the surface of the sphere (Prop. VIII.); hence, the surface of the sphere is equal to tlhe convex surface of the circumscribed cylinder. But the surface of the sphere is equal to four great circles of the sphere (Prop. VIII. Cor. 1); hence, the convex surface of the cylinder is also equal to four great circles; and adding the two bases, each equal to a great circle, the whole surface of the circumscribed cylinder is equal to six great circles of the sphere; hence, the surface of the sphere is % or * of the wlole surface of the circumscribed sphere. In the next place, since the base of the circumscribed cylinder is equal to a great circle of tile splere, and its altitude to the diameter, the solidity of the cylinder is equal to a great circle multiplied by its diameter (Prop. II.). But the solidity of the splere is equal to its sur 252 ELEMENTS OF GEOMETRY. face, or four great circles, multiplied by one third of its radius (Prop. IX.), which is the same as one great circle multiplied by 4 of the radius, or by 2 of the diameter; hence, the solidity of the sphere is equal to i- of that of the circumscribed cylinder. 605. Cor. 1. Hence the sphere is to the circumscribed cylinder as 2 to 3; and their solidities are to each other as their surfaces. 606. Cor. 2. Since a cone is one third of a cylinder of the same base and altitude (Prop. V. Cor. 1), if a cone has the diameter of its base and its altitude each equal to the diameter of a given sphere, the solidities of the cone and sphere are to each other as 1 to 2; and the solidities of the cone, sphere, and circumscribing cylinder are to each other, respectively, as 1, 2, and 3. BOO K XI. APPLICATIONS OF GEOMETRY TO THE MENSURATION OF PLANE FIGURES. DEFINITIONS. 607. MENSURATION OF PLANE FIGURES is the process of determining the areas of plane surfaces. 608. The AREA of a figure, or its quantity of surface, is determined by the number of times the given surface contains some other area, assumed as the unit of measure. 609. The MEASURING UNIT assumed for a given surface is called the superficial unit, and is usually a square, taking its name from the linear unit forming its side; as a square whose side is 1 inch, 1 foot, 1 yard, &c. Some superficial units, however, have no corresponding linear unit; as the rood, acre, &c. 610. TABLE OF LINEAR MEASURES. 12 Inches make 1 Foot. 3 Feet " 1 Yard. 5. Yards " - 1 Rod or Pole. 40 Rods " 1 Furlong. 8 Furlongs " 1 Mile. ' Also, 7i^ Inches " 1 Link. 25 Links " 1 Rod or Pole. 100 Links " 1 Chain. 10 Chains " 1 Furlong. 8 Furlongs " 1 Mile. NOTE. — For other linear measures, see National Arithmetic, Art. 133, 134, 136. 22 264 ELEMENTS OF GEOMETRY. 611. TABLE OF SURFACE MEASURES. 144 Square Inches make 1 Square Foot. 9 Square Feet " 1 Square Yard. 301 Square Yards " 1 Square Rod or Pole. 40 Square Rods " 1 Rood. 4 Roods " 1 Acre. 640 Acres " 1 Square Mile. Also, 625 Square Links " 1 Square Rod. 16 Square Rods " 1 Square Chain. 10 Square Chains " 1 Acre. 612. Since an acre is equal to 10 chains, or 100,000 links, square chains may be readily reduced to acres by pointing off one decimal place from the right, and square links by pointing off five decimal places from the right. PROBLEM I. 613. To find the area of a PARALLELOGRAM. Multiply the base by the altitude, and the product will be the area (Prop. V. Bk. IV.). EXAMPLES. D C 1. What is the area of a square, A B C D, whose side is 25 feet? 25 X 25 625 feet, Ans. 2. What is the area of a square field whose. A B side is 35.25 chains? Aiis. 124 A. 1 R. 1 P. 3. How many square feet of boards are required to lay. a floor 21 ft. 6 in, square? 4. Required the area of a square farm, whose side is 3,525 links. 5. What is the area of the rectangle D C A BC D, whose length, AB, is 56 feet, and whose widthi AD, is 37 feet.?; 56 X 37 = 2,072 feet, Ans. --- I A BA BOOK- XI. 255 6. How many square feet in a plank, of a rectangular form, which is 18 feet long and 1 foot 6 inches wide? 7. How many acres in a rectangular garden, whose sides are 326 and 153 feet? Ans. 1 A. 23 P. 61 yd. 8. A rectangular court 68 ft. 3 il. long, by 56 ft. 8 ill. broad, is to be paved with stones of a rectangular form, each 2 ft. 3 in. by 10 in.; how many stones will be required? Ans. 2,0622 stones. 9. Required the area of the rhomboid A B C D, of which the side A B D E C is 354 feet, and the perpendicular distance, E F, between A B and the oppo- site side C D, is 192 feet. A F B 354 X 192 = 67,968 feet, Ans. 10. How many square feet in a flower-plat, in the form of a rhombus, whose side is 12 feet, and the perpendicular distance between two opposite sides of which is 8 feet? 11. How many acres in a rhomboidal field, of which the sides are 1,234 and 762 links, and the perpendicular distance between the longer sides of which is 658 links? Ans. 8 A. 19 P. 4 yd. 6j ft. PROBLEM II. 614. The area of a SQUARE being given, to find the side. Extract the square root of the area. ~ Scholium. This and the two following problems are the converse of Prob. I. IXAMPLES. 1. What is the side of a square containing 625 square feet? __. 625.- 25 feet, the side required. 2. The area of a square farm is 124 A. 1 R. 1 P.; how: many links in length is its side?.. 3.A certain corn-field. ia.the form. of... a g.qare.jtais 256 ELEMENTS -OF GEOMETRY. 15 A. 2 R. 20 P. If the corn is planted on the margin, 4 hills to a rod in length, how many hills are there on the margin of the field? Ans. 800 hills. PROBLEM III. 615. The area of a RECTANGLE and either of its sides being given, to find the other side. Divide the area by the given side, and the quotient will be the other side. EXAMPLES. 1. The area of a rectangle is 2,072 feet, and the length of one of the sides is 56 feet; what is the length of the other side? 2072 - 56 = 37 feet, the side required. 2. How long must a rectangular board be, which is 15 inches in width, to contain 11 square feet? 3. A rectangular piece of land containing 6 acres is 120 rods long; what is its width? Ans. 8 rods. 4. The area of a rectangular farm is 266 A. 3 R. 8P., and the breadth 46 chains; what is the length? Ans. 58 chains. PROBLEM IV. 616. The area of a RHOMBOID or RHOMBUS and the length of the base being given, to find the altitude; or the area and the altitude being given, to find the base. Divide the area by the length of the base, and the quotient will be the altitude; or divide the area by the altitude, and the quotient will be the length of the base. -".......: EXAMPLES. 1. The area of a rhomboid is 67,968 square feet, and the length of the side taken as its base 354 feet; what is the altitude? 67,968 -- 354 = 192 feet, the altitude required. 2. The area.,of a piece of land in the form of a rhombus BOOK XI. 26T7 is 69,452 square feet, and the perpendicular distance between two of its opposite sides is 194 feet; required the length of one of the equal sides. Ans. 358 ft. 3. On a base 12'feet il length it is required to find the altitude of a rhomboid containing 968 square feet. 4. The area of a rhomboidal-shaped park is 1A. 3R. 34P. 5. yd.; and the perpendicular distance between the two shorter sides is 96 yards; required the length of each of these sides? Ans. 18 rods. PROBLEM V. 617. Tile diagonal of a SQUARE being given, to find the area. Divide the square of the diagonal by 2, and the quotient will be the area. (Prop. XI. Cor. 4, Bk. IV.) EXAMPLES. D C 1. The diagonal, A C, of the square A B C D, is 30 feet; what is the area? 302 = 900; 900 - 2 = 450 square feet, [the area required. A B 2. The diagonal of a square field is 45 chains; how many acres does it contain? 3. The distance across a public square diagonally is 27 rods; what is the area of the square? PROBLEM VI. 618. The area of a SQUARE being given, to find the diagonal. Extract the square root of double the area. Scholium. This problem is the converse of the last. EXAMPLES. 1. The area of a square is 450 square feet; what is its diagonal? 450 X 2 = 900; V/900 = 30 feet, the diagonal required. 22* 2658 ELEMENTS.OF GEOMETRY. 2. The area of a public square is 4 A. 2 R. 9 P.; what is the distance across it diagonally? 3. The area of a square farm is 57.8 acres; what is the diagonal in chains? Alls. 34 chains. PROBLEM VII. 619. The sides of a RECTANGLE being given, to cut off a given area by a line parallel to either side. Divide the given area by the side which is to retain its length or width, and the quotient will be the length or width of the part to be cut off. (Prop. IV. Sch., Bk. IV.) EXAMPLES. 1. If the sides of a rectangle, ABCD, D F C are 25 and 14 feet, how wide an area, E BC F, to contain 154 square feet, can be cut off by a line parallel to the side A D? A E B 154 + 14 = 11 feet, the width required. 2. A farmer has a field 16 rods square, and wishes to cut off from one side a rectangular lot containing exactly one acre; what must be the width of the lot? 3. A carpenter sawed off, from the end of a rectangular plank, in a line parallel to its width, 5 square feet. From the remainder lie then sawed off, in a line parallel to the length, 8 square feet. Required the dimensions of the part still remaining, provided the original dimensions of the plank were 20 feet by 15 inches. Ans. 16 feet by 9 inches. 4. The length of a certain rectangular lot is 64 rods, and its width 50 rods; how far fiom the longer side must a parallel line be drawn to cut off an area of 4 acres, and how far from the shorter side of the remaining portion to cut off 5 acres and 2 roods? How many acres will remain after the two portions are cut off? BOOK XI. 259 PROBLEM VIII. 620. To find the area of a TRIANGLE, the base and alti. tude being given. Multiply the base by half the altitude (Prop. VI. Bk. IV.). 621. Scholium. The same result can be obtained by multiplying the altitude by half the base, or by multiplying together the base and altitude and taking half the product. EXAMPLES. 1. Required the area of the triangle A A B C, whose base, B C, is 210, and altitude, AD, is 190 feet. 190 210 X -2- = 19,950 square feet, the [area required. B --- I 2. A piece of land is in the form of a right-angled triangle, having the sides about the right angle, the one 254 and-the other 136 yards; required the area in acres. Ans. 3 A. 2 R. 10 P.: 29yyd. 3. Required the number of square feet in a triangular board whose base is 27 inches and altitude 27 feet. 4. What is the area of a triangle whose base is 15.75 chains, and the altitude 10.22 chains?... 5. What is the area of a triangular field whose base is 97 rods, and the perpendicular distance from tle base t'. the opposite angle 40 rods? Ans. 12 A. 20 P1 PROBLEM IX. 622. To find the area of a TRIANGLE, the three sides being given. From half the sum of the three sides subtract each ..260 260 ~~ELEMENTS OF GEOMETRY. side; multiply the half sumt and the three remainders tog-ether, and the square root of the product will be the area required. For, let A B C be a triangle whose three c sides, A B, B C, A C, are given, but not the altitude C D, and let the side B C be represented by a, AC by b, and AB by c. Now, since A. is an acute angle of the triangle A B C, we have (Prop. XII. Bk. -IV.), A I) B a2=-b2~c2-2cX.AD, or AD= - _a 2c ilence, in the right-angled triangle A DO, we have'.(Prop. XI. Cor. 1, Bk. JY.),' CD2~~b __(b2 +c2- a2)2 4 M c2-( (b+c2 - 2)2 C D2 b ~ c2 4c and, by extracting the square root, CD - V4 bj c.- (bJ + 0 - Opy But the area of the triangle -A B C -is equivalent to the product of ~c by half of C D (Prob. VIII.); hence The expression 4 b2 c2 - (b2+ ~ 2 a2)2, being the difference of two squares,. can be decomposed into (2b c +b2 +c-a') X (2 b -b21c2~a2) Now, the first of these factors may be transformed to (b + c)2 - a', and consequently may be resolved into (b +c +a) X (b + c -a); and the second is the same thing as a2 - (b -c)2, which -is equal to (a~ b-c)..X (a -b + c). WVe have then, 4 b2c2- (b2~ +c2-a 2) 2=.- (a+ b +c) X (b-f+c-a) X(a.+ c.-b) X(a +bc) BOOK'XL. 261 261 Let S represent half the sum of the three sides of the triangle; then a+ b+c = 2 S; b + c- a = 2(S-a); a~c-b=~2 (S -b); a+b-c= 2 (S-c)-;hence A B Ci 16 S(S -a) X (S-=b) X (S:- c), whlichl being reduced, gives as the area, of the triangle, as given above, _ _ _ _ _ _ _ _ _ _ _ V S(S -a) x (Sb) X (S -c). EXAMPLES. C 1. What is the area of a triangle, -A BC, whose sides, A B, B C, C A, are 40, 30, atnd 50 feet? A 13 30 +40 + 50 + 2 -60, half the sum of the three sides. 60 - 30 =30, first remainder. 60-40 =20, second remainder. 60 -50 =10, third remainder. 60 X 30 x 20 x. 10 =360,000'; V3601000 600, square feet, the area required. 2. How many' -square feet in a triangular floor,whe sides are -15, 16, -and 21 feet? 3. Required the area of a triangular field whose sides a~r 8465,ad 423- links. Ans. l A.'iR. 20P. 4yd. 1.6 -ft.. 4-. Required the area of an equilateral triangle,~ of whic~h each side-is 15 yards. 5. What is the area of a garden in the -form -of a -parallelogram, whose sides are 432 and 263 feet,'and a diagonal 342 fe~et? Ans. 2IA.` 10 P. 11.-46 yd., — 6. Rqie teae of anl isosceles triangle, WhOse base is 25 and each of its. equal sides 40 rods..7. What is the area of a rhomodl ilwhose sides are 57 and 83 rods, and -the. diagonal 127 rods? ins. 22 A. S R. 21 P. 26 yd. 6 it 262 ELEMENTS 'OF GEOMETRY. PROBLEM X. 623. Any two sides of a RIGHT-ANGLED TRIANGLE being given, to find the third side. To the square of the base add the square of the perpendicular; and the square root of the sum will give the hypothenuse (Prop. XI. Bk. IV.). From the square of the hypothenuse subtract the square of the given side, and the square root of the difference will be the side required (Prop. XI. Cor. 1, Bk. IV.). EXAMPLES. 1. The base, AB, of the triangle AB C is 48 feet, and the perpendicular, B C, 36 feet; what is the hypothenuse? 482 + 362 = 3600;. 3600.= 60 feet, [the hypothenuse required. A 1 2. The hypothenuse of a triangle is 53 feet, and the perpendicular 28 feet; what is the base? 3. Two ships sail from the same port, one due-west 50 miles, and the other due south 120 miles; how far are they-apart? Ans. 130 miles. 4. A rectangular common is 25 rods long and. 20 rods wide; what is the distance across it diagonally? 5. If a house is 40 feet long and 25 feet wide, with a pyramidal-shaped roof 10 feet in height, how long is a rafter which reaches from the vertex of the roof to a corner of the building? 6. There is a park in the form of a square containing 10 acres; how many rods less is the distance from the centre to each corner, than the length of the side of the square? Ans. 11.716 rods. PROBLEM XI. 624. The sum of the hypothenuse and: perpendicular r.... *~~~~~~~. ' d ~.'. ..-:* BOoK: XI..: - 268 and the base of a RIGHT-ANGLED TRIANGLE being given, to find the hypothenuse and the perpendicular...To the square- of the sum add the square of the lbase, and-divide the amount by twice the sum of the hypothenuse and perpendicular, and the quotient will be the 'hypothe nuse. From the -sum of. the hypothenuse and perpendicular subtract the hypothenuse, and the remainder will be. the. perpendicular. 625. Scholium. This problem may be regarded as equivalent to the sum of two numbers and the difference of their squares being given, to find the numbers (National Arithmetic, Art. 553). NOTE. - The learner should be required to give a geometrical demonstration of the problem, as an exercise in the application of principles. EXAMPLES. 1. The sum of the hypothenuse and the perpendicular of a right-angled triangle is 160 feet, and the base 80 feet;.required the hypothenuse and the perpendicular. Ans. Hypothenuse, 100 ft.; perpendicular, 60 ft. 1602 + 802= 32,000; 32,000-(160 X 2) = 100; 160 100 = 60. 2. Two ships leave the same anchorage; the one, sailing due north, enters a port 50 miles from the place.of departture, aid the other, sailing due east, also enters a port, but by sailing thence in a direct course enters the port of the first; now, allowing that the second passed over, in all, 90 miles, how far apart are the two ports? 3. A tree 100 feet high, standing perpendicularly on a horizontal plane, was broken by the wind, so that, as it fell, while the part broken off remained in contact with the upright portion, the top reached the ground 40 feet' from the foot of the tree; what is the.length of each-part? Ans. The.part broken off, 58 ft.; the upriglt, 42 ft. 264 ELEMENTS OF GEOMETRY. PROBLEM XII. 626. The area and the base of a TRIANGLE being given, to find the altitude; or the area ild altitude being given, to find the base. Divide double the area by the base, and the quotient will be the altitude; or divide double the area by the altitude, and the quotient will be the' base. 627. Scholium. This problem is the converse of Prob. VIII. EXAMPLES.. 1. The area of a triangle is 1300 square feet, and the base G5 feet; what is the altitude? 1300 X 2 = 2600; 2600 - 65 = 40 ft., altitude required. 2. The area of a right-angled triangle is 17,272 yards, of which one of the sides about the right angle is 136 yards; required the other perpendicular side. 3. The area of a triangle is 46.25 chains, and the altitude 5.2 chains'; what is the base? 4, A triangular field contains 30 A. 3 R. 27 P.; one of its sides is 97 rods; required the perpendicular distance from the opposite angle to that side. Ans. 102 rods. PROBLEM XIII, 628. To find the area of a TRAPEZOID. Multiply half the sum of its parallel sides by its altitude (Prop. VII. Bk. IV.). EXAMPLES. D E C 1. What is the area of the trapezoid A B C D, whose parallel sides, A B, ) C, are 32 and 24 feet, and the alti- tude, EF, 20 feet? A F B 82 + 24 56; 56 -- 2= 28; 28 X 20 == 560 Eq. ft., [the area required. BOOK XI. 265 2. How many square feet in a board in the form of a trapezoid, whose width at one end is 2 feet 3 inches, and at tle other 1 foot 6 inches, the length being 16 feet? 3. Required the area of a garden in the form of a trapezoid, wlose parallel sides are 786 and 473 links, and the perpendicular distance between them 986 links. Ans. 6A. 33 P. 3yd. 4. How many acres in a quadrilateral field, having two parallel sides 83 and 101 rods in length, and whicl are distant from each other 60 rods? PROBLEM XIV. 629. To find the area of a REGULAR POLYGON, the perimeter and apothegm being given. Multiply the perimeter by half the apothegm, and the product will be the area (Prop. VIII. Bk. VI.). 630. Scholium. This is in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the center to all the angles, then finding their areas, and taking their sum. EXAMPLES. 1. Required the area of a regu- E D lar hexagon, A B C D E F, wlose sides, A B, B C, &c. are each 15 yards, and the apothegm, O M, 13 F( C yards. 15X6= 90; 90- -= 585yd., [the area required. A M B 2. What is the area of a regular pentagon, whose sides are each 25 feet, and the perpendicular from the center to a side 17.205 feet? 3. A park is laid out in the form of a regular heptagon, whose sides are each19.263chains; and the perpendicular 23 266 ELEMENTS OF GEOMETRY. distance from the center to each of the sides is 20 chains. How many acres does it contain? Ans. 134 A. 3 R. 14 P. PROBLEM XV. 631. To find the area of a REGULAR POLYGON, its side or perimeter being given. Multip'y the square of the side of the polygon by the area of a similar polygon whose side is unity or 1 (Prop. XXXI. Bk. IV.). 632. A TABLE OF REGULAR POLYGONS WHOSE SIDE IS 1. NAMES. AREAS. NAMES. AREAS. Triangle, 0.4330127 Octagon, 4.8284271 Square, 1.0000000 Nonagon, 6.1818242 Pentagon, 1.7204774 Decagon, 7.6942088 Hexagon, 2.5980762 Undecagon, 9.3656399 Heptagon, 3.6339124 Dodecagon, 11.1961524 < _________ _________ The apothegm of any regular polygon whose side is 1 being ascertained, its area is computed readily, by Prob. XIV. EXAMPLES. 1. Required the area of an equilateral triangle, whose side is 100 feet. 100 =- 10,000; 10,000 X 0.4330127 = 4330.127 square [feet, the area required. 2. What is the area of a regular pentagon, whose side is 37 yards? 3. How many acres in a field in the form of a regular undecagon, whose side is 27 yards? Ans. 1 A. 1 R. 25 P. 21 yd. 2.7 ft. BOOK XI. 267 4. What is the area of an octagonal floor, whose side is 15 ft. 6 in.? 5. How many acres in a regular nonagon, whose perimeter is 2286 feet? Ans. 9 A. 24 P. 28 yd. PROBLEM XVI. 633. To find the side of any REGULAR POLYGON, its area being given. Divide the given area by the area of a similar polygon whose side is 1, and the square root of the quotient will be the side required. 634. Scholium. This problem is the converse of Prob. XV. EXAMPLES. 1. The area of an equilateral triangle is 4330.127 square feet; what is its side? 4330.127 -.4330127 = 10,000; V/10,000 = 100 feet, [the side required. 2. The area of a regular hexagon is 1039.23 feet; what is its side? 3. The area of a regular decagon is 7 P. 18yd. 5 ft. 128.55 in.; what is its side? Ans. 16 ft. 5 in. PROBLEM XVII. 635. To find the area of an IRREGULAR POLYGON. Divide the polygon into triangles, or triangles and trapezoids, and find the areas of each of them separately; the sum of these areas will be the area required. 636. Scholium. When the irregular polygon is a quadrilateral, the area may be found by multiplying together the diagonal and half the sum of the perpendiculars drawn from it to the opposite angles. 268 ELEMENTS OF GEOMETRY. EXAMPLES. ' 1. Required the area of the irregular C pentagon A B C D E, of which the diag- onal AC is 20 feet, and AD 36 feet; \ B D and the perpendicular distance from A^' the angle B to A C is 8 feet, from C to AD 12 feet, and from E to AD 6 feet. E 20 X. =80; 36 X -1 — 216; 36 X = 108; 80 + 216 + 108 = 504 sq. ft., the area required. 2. What is the area of a trapezium, whose diagonal is 42 feet, and the two perpendiculars from the diagonal to the opposite angles are 16 and 18 feet? 3. In an irregular hexagon, A B C D E F, are given the sides A B 536, B C 498, C D 620, D E 580, E F 398, and A F 492 links, and the diagonals A C 918, C E 1048, and A E 652 links; required tile area. Ans. 6 A. 2 R. 9 P. 23 yd. 8.4 ft. 4. In measuring along one side, AB, of a quadrangular field, A B C D, that side and the perpendiculars let fall on it from two opposite corners measured as follows: A B 1110,- A E 110, A F 745, D E 352, C F 595 links. What is the area of the field? Ans. 4 A. 1 R. 5 P. 24 yd. 5. In a four-sided rectilineal field, A B C D, on account of obstructions, there could be taken only the followilg measures: the two sides B C 265 and A D 220 yards, tle diagonal A C 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, A E 100, and C F 70 yards. Required the area in acres. PROBLEM XVIII. 637. To find the circumference of a CIRCLE, when tlie diameter is given, or the diameter when the circumference,is given. Multiply the diameter by 3.1416, and the product will be the circumference; or, divide the circumference by BOOK XI. 3.1416, and the quotient will be the diameter (Prop. XV. Cor. 3, Bk. VI.). 638. Scholium. The diameter may also be found by multiplying the circumference by.31831, the reciprocal of 3.1416. E EXAMPLES. 1. The diameter, A B, of the circle AE BF is 100 feet; what is its circumference? A 100 X 3.1416 =314.16 feet, the [circumference required. F 2. Required the circumference of a circle whose diameter is 628 links. Ans. 1 fur. 38 rd. 5 yd. 1.56 in. 3. If the diameter of the earth is 7912 miles, what is its circumference? 4. Required the diameter of a circular pond whose circumference is 928 rods. Ans. 7 fur. 15 rd. 2 yd. 5.55 in. 5. The circumference of a circular garden is 1043 feet; what is its radius? Ans. 10 rd. 1 ft. PROBLEM XIX. 639. To find the length of an arc of a circle containing any number of degrees, the radius or diameter being given. Multiply the number of degrees in the given arc by 0.01745, and the product by the radius of the circle. For, when the diameter of a circle is 1, the circumference is 3.1416 (Prop. XV. Sch. 1, Bk. VI.); hence, when the radius is 1, the circumference is 6.2832; which, divided by 360, the number of degrees into which every circlb is supposed to be divided, gives 0.01745, the length of the arc of 1 degree, when the radius is 1. 640. Scholium. Each of the 360 degrees of a circle, 23 270 ELEMENTS OF GEOMETRY. marked thus, 360~, is divided into 60 minutes, marked thus, 60', and each minute into 60 seconds, marked thus, 60" (National Arithmetic, Art. 143). EXAMPLES. 1. What is the length of an arc, AD, containing 60~ 30' on the circumference of a circle whose radius, AC, is 100 feet? C 60~ 30' == 60.5~; 60.5 X 0.01745 = 1.055725; 1.055725 X 100 105.5725 ft., arc required. A _ D 2. Required the length of an arc of 31~ 15', the radius being 12 yards. 3. Required the length of an arc of 12~ 10', the diameter being 20 feet. Ans. 2.1231 feet. 4. What is the length of an arc of 57~ 17' 441", the radius being 25 feet? Ans. 25 feet. PROBLEM XX. 641. To find the area of a circle. Multiply the circumference by half the radius (Prop. XV. Bk. VI.); or, multiply the square of the radius by 3.1416 (Prop. XV. Cor. 2, Bk VI.). 642. Scholium. Multiplying the circumference by half the radius is the same as multiplying the circumference and diameter together, and taking one fourth of the product. Now, denoting the circumference by c, and the diameter by d, since c = 3.1416 X d (Prob. XVIII.), we have (d X 3.1416 X d) 4 = d2 X 0.7854 = the area of a circle. Again, since d-= c- 3.1416 (Prob. XVIII.), we have c + 3.1416 X c + 4 = c2 + 12.5664, which is, by taking the reciprocal of 12.5664, equal to c2 X 0.07958 =- the area of tle circle. Hence the area of the circle may also be found by multiplying the square of the diam BOOK XI. 271 eter by 0.7854; or by multiplying the square of the circumference by 0.07958. EXAMPLES. 1. The circumference of a circle is 314.16 feet, and its radius 50 feet; what is its area? 314.16 X -50 = 7854 feet, the area required. 2. If the circumference of a circle is 355 feet, and its diameter 113 feet, what is the area? 3. What is the area of a circular garden, whose radius is 281k links? Ans. 2 A. 1 R. 38 rd. 9 yd. 5 ft. 4. A horse is tethered in a meadow by a cord 39.25075 yards long; over how much ground call le graze? 5. Required the area of a semicircle, the diameter of the whole circle being 751 feet. Ans. 5A. 13P. 16yd. PROBLEM XXI. 643. To find the DIAMETER or CIRCUMFERENCE, the area being given. Divide the area by 0.7854, and the square root of the quotient will be the diameter; or, divide the area by 0.07958, and the square root of the quotient will be the circumference. 644. Scholium. This problem is the converse of Prob. XX. EXAMPLES. 1. The area of a circle is 314.16 feet; what is the diameter? 314.16 - 0.7854 = 400; ^400 = 20 feet, the diameter [required. 2. What must be thie length of a cord to be used as a radius in describing a circle which shall contain exactly 1 acre? 8. The area of a circular pond is 6 A. 1 R. 27 P. 18.2 yd.; wlat is the circumference? Ans. 625 yd. 272 ELEMENTS OF GEOMETRY. 4. The area of a circle is 7856 feet; what is the circumference? 5. The length of a rectangular garden is 32, and its width 18 rods; required the diameter of a circular garden having the same area. Ans. 27 rd. 1 ft. 4 in. PROBLEM XXII. 645. To find the area of a SECTOR of a circle. Multiply the arc of the sector by half of its radius (Prop. XV. Cor. 1, Bk. VI.); or, As 3600 are to the degrees in the arc of the sector, so is the area of the circle to the area of the sector. EXAMPLES. 1. Required the area of a sector, DE, whose arc is 80 feet, and its radius, 0 E, 70 feet. 80 X 7 = 2800 square feet, the area [required. 2. Required the area of a sector, of which the arc is < and the radius 112 yards. 3. Required the area of a sector, of which the angle 137~ 20', and the radius 456 links. Ans. 2 A. 1 R. 38 P. 21.92yd. 90 is PROBLEM XXIII. 646. To find the area of a SEGMENT of a circle. Find the area of the sector having the same arc with the segment, and also the area of the triangle formed by the chord of the segment and the radii of the sector. Then, if the segment is less than a semicircle, take the difference of these areas; but if greater, take their sum. 647. Scholium. When the height of the segment and BOOK XI. 273 the diameter of the circle are given, the area may be readily foiuid by means of a table of segm-enits, by dividingt the height by the diamneter, and looking- in the table for the quotient in the column of heights, and taking out, in the next column on the right hand, the corresponding area; which, multiplied by the square of the diameter, will give the area required. When the quotient cannot be exactly found in the table, proportions may be instituted so as to find the area bctween the iiext higrher and the next lower, inl the same ratio that the given height varies from the next h1igher and lower heigyhts. 648. TABLE OF SEGMENTS. 1.1 Seg. -~Seg..~Seg. Seg. -~Seg. q~Area. ~3Area. Area. Area. Ar-ea..01.00133.11.04701.21.11990.311.20738.41.30319.02.00375 112.05339 I.22.12811 I.32.21667.42.31304.03:00687:.13.06000 1.23.13646.33.22603.43.32293.04.01054:.14.06683.24.14494!.34.23547.44.33284.05.01468 1.15.07387.25.15354.35 24498.45.34278.06.01924 1.16.08111.26.16226.36' 25455.46.35274.07.02417.17.08853.27:171091:37.26418.47.36272.08.02944.18.09613.28.18002.38, 27386.48.37270.09.03502.19.10390.29.18905.39':28359.49.38270.10.04088.20.11182.30-.19817 40.29337 I.50.39270 The segments in the table are those of a circle-whose diameter is 1, and the first column contains the corresponding heights divided by the diameter. The method of calcuflating the areas of segments from the- elements in the table depends upon the principle' that simnilar plane figures are to each other as the squares of their like linear dimensions. - EXAMPLES. 1. What is the area of the se(gmuent A B E, its arc A. E B.being - 73.74',. its chord. A B being' J 2 feet, and ELEMENTS OF GEOMIETRY. the radius, C B, of the circle 10 feet? 0.7854 X 202 - 314.1G, area of circle; then 3G0~: 73.740::314.16: G4.3504, area of sector A E B C; and, by Problem IX., 48 is the area of the trian- A' glc A B C; 64.3504 —48 = 16.3504 feet, the area required. 2. Required the area of a segment whose ]leigllt is 18, and the diameter of the circle 50 feet. IS -- 50 -.36; to whicll the corresponding area in thll table is.25455;.25455 X 502 = 636.375, area required. 3. Required the area of a segment whose arc is 100~, chord 153.208 feet, and tlhe diameter of the circle 200 feet. 4. What is the area of a segment whose height is 4 feet, and the radius 51 feet? Ans. 106 feet. 5. Required the area of a segment, the arc being 160~, chord 196.9616 feet, and the radius of the circle 100 feet. PROBLEM XXIV. 649. To find the area of a CIRCULAR ZONE, or the space included between two parallel chords and their intercepted arcs. From the area of the whole circle subtract the areas of the segments on the sides of the zone. EXAMPLES. 1. What is the area of a zone whose chords are each 12 feet, subtending each an are of 73.74~, when the radius of the circle is 10 feet? Area of the whole circlo by Prob. XX. = 314.16; area of each segment by Prob. XXIII. = 16.3504; 1G.3504 X 2 = 32.7008 = area of both segments; 314.16 - $2.7008 2981.4592, the area required. BOOK XI. 27 5 2. What is the area of a circular zone whose longer chord is 20 yards, subtending an arc of 60~, and the shorter chord 14.66 yards, subtending an arc of 43~, the diameter of the circle beinig 40 yards? 3. A circle whose diameter is 20 feet is divided into three parts by two parallel chords; one of the segments cut off is 8 feet in height, and the other 6 feet; what is the area of the circular zone? Ans. 117.544 ft. PROBLEM XXV. 650. To find the area of a CRESCENT. Find the difference of the areas of the two se.gments formed by the arcs of the crescent and its chord. EXAMPLES. 1. The arcs AC B, AE B, of C circles having the same radius, 50 rods, intersecting, form the crescent A C B E; the height, / D C, of the segment A C B is 60 A _ -- B rods, and the height, D E, of the segment A B E is 40 rods; what is the area of the crescent? The area of the segment A C B, by Prob. XXIII., is 4920.3 rods, and that of the segment A B E is 2933.7 rods; 4920.3 - 2933.7 = 1986.6 rods, the area of the crescent. 2. If the arc of a circle whose diameter is 24 yards intersects a circle whose diameter is 20 yards, forming a crescent, so that the height of the segment of the first circle is 5.072 yards, and that of the segment of the second circle is 8 yards, what is the area of the crescent? PROBLEM XXVI. 651. To find the area of a CIRCULAR RING, or the space included between two concentric circles. Find the areas of the two circles separately (Prob. XX.). and take the difference of these areas; or sub 276 ELEMENTS OF GEOMETRY. tract the square of the less diameter from the square of the greater, and multiply their difference by 0.7854 (Prob. XX. Scll.). EXAMPLES. 1. Required the area of the ring formed by two circles whose diameters are 30 and 50 feet. 50' - -02 = 1400; 1400 X 0.7854 = 1099.56 sq. feet, [the area of tlhe ring. 2. What is the area of a ring formed by two circles whose radii are 36 and 24 feet? 3. A circular park, 256 yards il diametr, lhas a carriage-way running around it 29 feet wide; what is tleo area of the carriage-way? Ans. 1 A. 2 R. 26 P. 21.5 yd. PROBLEM XXVII. 652. The diameter or circumference of a CIRCLE being given, to find the side of an EQUIVALENT SQUARE. Multiply the diameter by 0.8862, or the circumference by 0.2821; the product in either case will be the side of an equivalent square. For, since 0.7854 is the area of a circle whose diameter is 1 (Prob. XX. Sch.), the square root of 0.7854, which is 0.8862, is the side of a square which is equivalent to a circle whose diameter is 1. Now when tlle circumference is 1, the side of an equivalent square must have the same ratio to 0.8862 as the diameter 1 las to its circumference 3.1416 (Prop. XV. Cor. 4, Bk. VI.); and 0.8862- 3.1416 gives 0.2821 as the side of the equivalent square when the circumference is 1. EXAMPLES. 1. The diameter of a circle is 120 feet; what is the side of an equivalent square? 120 X 0.8862 = 106.344 feet, the side required. BOOK XI. 277 2. The circumference of a circle is 100 yards; what is the side of an equivalent square? Ans. 28.21 yd. 3. There is a circular floor 30 feet in diameter; what is the side of a square floor containing the same area? 4. If 500 feet is the circumference of a circular island, what is the side of a square of equal area? Ans. 141.05 ft. PROBLEM XXVIII. 653. The diameter or circumference of a CIRCLE being given, to find the side of the INSCRIBED SQUARE. Multiply the diameter by 0.7071, or the circumference by 0.2251; the product in either case will be the side of the inscribed square. For 0.7071 is the side of the inscribed square when the diameter of the circumscribed circle is 1, since the side of the inscribed square is to the radius of the circle as the square root of 2 to 1 (Prop. IV. Cor., Bk. VI.); consequently, the side is to the diameter, or twice the radius, as half the square root of 2 is to 1, and half the square root of 2 is 0.7071, approximately. Now, the ratio of the diameter of a circle to the side of its inscribed square being as 1 to 0.7071, and the ratio of the circumference of a circle to its diameter as 3.1416 to 1, the ratio of the inscribed square is to the circumference of the circle as 0.7071 to 3.1416; and 0.7071 + 3.1416 gives 0.2251 as the side of the inscribed square when the circumference is 1. EXAMPLES. 1. The diameter, AC, of a circle is Dc 110 feet; what is the side, A B, of the inscribed square? 110 X 0.7071 - 77.781 feet, the side [required. 24 278 ELEMENTS OF GEOMETRY. 2. The circumference of a circle is 300 feet; what is the side of the inscribed square? Ans. 67.53 ft. 3. A log is 36 inches in diameter; of how many inches square can a stick be hewn from it? 4. There is a circular field 1000 rods in circuit; what is the side of the largest square that can be described in it? Ans. 225.10 rods. PROBLEM XXIX. 654. The diameter or circumference of a CIRCLE being given, to find the side of an INSCRIBED EQUILATERAL TRIANGLE. Multiply the diameter by 0.8660, or the circumference by 0.2757; the product in either case will be the side of the inscribed equilateral triangle. For 0.8660 is the.side of the inscribed equilateral triangle when the diameter of the circumscribed circle is 1, since the side of the inscribed equilateral trihngle is to the radius of the circle as the square root of 3 is to 1 (Prop. V. Cor. 3, Bk. VI.); consequently, tle side is to the diameter, or twice the radius, as half the square root of 3 is to 1, and half the square root of 3 is 0.8660, approximately. Also, since the ratio of the circumference of a circle to its diameter is as 3.1416 to 1, tle side of the inscribed equilateral triangle, when the circumference is 1, equals 0.8660 - 3.1416, or 0.2757. EXAMPLES. 1. Required the side of an equilateral triangle that may be inscribed in a circle 101 feet in diameter. 101 X 0.8660 = 87.4660 feet, the side required. 2. Required the side of an equilateral triangle that may be inscribed in a circle 80 rods in circumference. Ans. 22.05 rods. 3. Required the side of the largest equilateral triangular beam that can be hewn from a piece of round timber 86 incles in diameter. BOOK XI. 279 4. Required the side of an equilateral triangle that can be inscribed ill a circle 251.33 feet il circumference. 5. How much less is tile area of an equilateral triangle that call be inscribed in a circle 100 feet in diameter, than the area of the circle itself? Ans. 4606.4 sq. ft. THE ELLIPSE. 655. An ELLIPSE is a plane figure bounded by a curve, from any point of which tlle sum of tlle distances to two fixed points is equal to a straight line drawn tlrough those two points, and terminated both ways by the curve. Thus AD B C is an ellipse. The two fixed points G and H are called thefoci. The longest diameter, AB, of the ellipse is called its major or transverse axis, and its shortest diameter, CD, is called its minor or D conjugate axis. 656. The AREA of -an ellipse is a mean proportional between the areas of two circles whose diameters are the two axes of the ellipse. This, however, can only be well demonstrated by means of Analytical Geometry, a branch of the mathematics with which the learner here is not supposed to be acquainted. PROBLEM XXX. 657. To find tile area of an ELLIPSE, the major and minor axes being given. Multiply the axes together, and their product by 0.7854, and the result will be the area. For A B2 X 0.7854 expresses the area of a circle whose diameter is A B, and C D2 X 0.7854 expresses the area of a circle whose diameter is C D; and the product of these two areas is equal to A B2 X C D2 X 0.78542, which is 280 ELEMENTS OF GEOMETRY. equal to the square of A X CD X 0.7854; hence, A B X C D X 0.7854 is a mean proportional between the areas of the two circles wlose diameters are A B and C D (Prop. IV. Bk. II.); consequently it measures tile area of an ellipse whose axes are AB and CD (Art. 656). EXAMPLES. 1. Required the area of an ellipse, of which the major axis is 60 feet, and the minor axis 40 feet. 60 X 40 X 0.7854 = 1884.96 sq. ft., the area required. 2. What is tlle area of all ellipse whose axes are 75 and 35 feet? 3. Required the area of an ellipse whose axes are 526 and 354 inches. Ans. 112 yd. 7 ft. 84.62 in. 4. How many acres in all elliptical pond whose semiaxes are 436 and 254 feet? Ans. 7 A. 3R. 37 P. 27yd. 7ft. BOOK XII. APPLICATIONS OF GEOMETRY TO TIFE MENSURATION OF SOLIDS. DEFINITIONS. 658. MENSURATION OF SOLIDS, or VOLUMES, is the process of determining their contents. The SUPERFICIAL CONTENTS of a body is its quantity of surface. The SOLID CONTENTS of a body is its measured magnitude, volume, or solidity. 659. The UNIT OF VOLUME, or SOLIDITY, is a cube, whose faces are each a superficial unit of the surface of the body, and whose edges are each a linear unit of its linear dimensions. 660. TABLE OF SOLID MEASURES. 1728 Cubic 27 " 4492o, 32,768,000 Also, 231 " 2G8 " 2150, " 128 " Inches make 1 Cubic Foot Feet Feet Rods " 1 " Yard. " 1 " Rod. " 1 " Mile. " 1 Liquid Gallon. " 1 Dry Gallon. " 1 Bushel. " 1 Cord. Inches Inches Inches Feet PROBLEM I. 661. To find the surface of a RIGHT PRISM. Multiply the perimeter of the base by the altitude, and the product will be the CONVEX surface (Prop. I. Bk. 21 282 ELEMENTS OF GEOMETRY. VIII.). To this add the areas of the two bases, and the result will be the ENTIRE surface. EXAMPLES. F 1. Required the convex surface of a pentangular prism, having eacl side ofG 1 its base, A B C D E, equal to 2 feet, and its altitude, A F, equal to 5 feet. E.-. 2 X 5 = 10; 10 X 5 = 50 square feet, ^\ / [the convex surface required. B C 2. The altitude of a hexangular prism is 12 feet, two of its faces are eacl 2 feet wide, tlree are each 2} feet wide, and the remaining face is 9 inches wide; what is the convex surface of the prisln? 3. Required tlle entire surface of a cube, the length of each edge being 25 feet. 4. Required, in square yards, the wall surface of a rectangular room, whose height is 20 feet, width 30 feet, and length 50 feet. Ans. 355k sq. yd. PROBLEMI II. 662. To find the solidity of a PRISM. Multiply the area of its base by its altitude, and the product will be its solidity (Prop. XIII. Bk. VIII.).,. EXAMPLES. 1. Required the solidity of a pentangular prism, having each side of its base equal to 2 feet, and its altitude equal to 5 feet. 22 X 1.72048 = 6.88192; 6.88192 X 5 = 34.40960 cubic [feet, the solidity required. 2. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of whose base are 3, 4, and 5 feet. Anls. 60. 3. A slab of marble is 8 feet long, 3 feet wide, and 6 inches thick; required its solidity. BOOK XII. 283 4. There is a cistern in the form of a cube, whose edge is 10 feet; what is its capacity in liquid gallons? Ans. 7480.519 gallons. 5. Required the solid contents of a quadrilateral prism, the length being 19 feet, the sides of the base 43, 54, 62, and 38, and the diagonal between the first and second sides, 70 inches. Ans. 306.047 Cu. ft. 6. How many cords in a range of wood cut 4 feet long, the range being 4 feet 6 inches high and 160 feet long? PROBLEM III. 663. To find the surface of a RIGHT PYRAMID. Multiply the perimeter of the base by half its slant height, and the product will be the CONVEX surface (Prop. XV. Bk. VIII.). To this add the area of the base, and the result will be the ENTIRE surface. 664. Scholiunz. The surface of an oblique pyramid is found by taking the sum of the areas of its several faces. S EXAMPLES. 1. Required the convex surface of a pentangular pyramid, A B C D E - S, each side of whose base, A B C D E, is 5 feet, and whose slant height, S M, is /. 20 feet. Ai D 5 X 5 25; 25 X - - 250 square M [feet, the surface required. B C 2. What is the entire surface of a triangular pyramid, of which the slant height is 18 feet, and each side of the base 42 inches? Ans. 99.804 sq. ft. 3. Required the convex surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet.4. What is the entire surface of a quadrangular pyramid, the sides of the base being 40 and 30 inches, and the slant height upon the greater side 20.04, and upon the less side 20.07 feet? Ans. 125.308 ft. 284 ELEMENTS OF GEOMETRY. PROBLEM IV. 665. To find the surface of a FRUSTUM OF A RIGHT PYRAMID. Multiply half the sum of the perimeters of its two bases by its slant height, and the product will be the CONVEX surface (Prop. XVII. Bk. VIII.); to this add the areas of the two bases, and the result will be the ENTIRE surface. EXAMPLES. 1. What is the entire surface of a rectangular frustum whose slant height is 12 feet, and the sides of whose bases are 5 and 2 feet? 5X4=-20; 2X4=-8; 20+8- 28; 28-x12=-168; 52 + 22 = 29; 168 + 29 = 197 sq. ft., area required. 2. Required the convex surface of a regular hexangular frustum, wliose slant height is 16 feet, and the sides of whose bases are 2 feet 8 inches and 3 feet 4 inches. 3. What is the entire surface of a regular pentangular frustum, whose slant height is 11 feet, and the sides of whose bases are 18 and 34 inches? Ans. 136.849 sq. ft. PROBLEM V. 666. To find the solidity of a PYRAMID. Multiply the area of its base by one third of its altitude (Prop. XX. Bk. VIII.). EXAMPLES. 1. Required the solidity of a pentangular pyramid, AB CD E-S, each side of whose base, A B C D E, is 5 feet, and whose altitude, S 0, is 15 feet. 5" X 1.7205 — 43.0125; 43.0125 X -- - =215.0575 cu. ft., the solidity required. s it,.A. D B C BOOK XII. Ao.3 4b00 2. What is the solidity of a hexangular pyramid, the altitude of which is 9 feet, and each side of the base 29 inches? 3. What is the solidity of a square pyramid, each side of whose base is 30 feet, and whose perpendicular height is 25 feet? Ans. 7500. 4. Required the solid contents of a triangular pyramid, the perpendicular eighllt of which is 24 feet, and tie sides of the base 34, 42, and 50 inches. Ans. 39.2054 cu. ft. PROBLEM VI. 667. To find the solidity of a FRUSTUM OF A PYRAMID. Add together the areas of the two bases and a mean proportional between them, and multiply that sum by one third of the altitude of the frustum (Prop. XXI. Bk. VIII.). EsXAMPLES. 1. Required tlhe solidity of tie frustum of a quadrangular pyramid, the sides of whose bases are 3 feet and 2 feet, and whose altitude is 15 feet. 3X3=9;2X2=4; V /9X4=6(Prop.IV.Bk.II.); (9 + 4 + 6) X -J = 95 cu. ft., solidity required. 2. How many cubic feet in a stick of timber in the form of a quadrangular frustum, the sides of whose bases are 15 incies and 6 inches, and whose altitude is 20 feet? 3. Required the solid contents of a pentangular frustum, whose altitude is 5 feet, each side of whose lower base is 18 inches, and each side of whose upper base is 6 inches. Ans. 9.319 cu. ft. 4. Required the solidity of the frustum of a triangular pyramid, the altitude of which is 14 feet, the sides of the lower base 21, 15, and 12, and those of the upper base 14, 10, and 8 feet. Ans. 868.752 cu. ft. 286 ELEMENTS OF GEOMETRY. THE WEDGE. 668. A WEDGE is a polyedron bounded by a rectangle, called the base of the wedge; by two trapezoids, called the sides, which meet in an edge parallel to the base; and by two triangles, called the ends of the wedge. Thus AB C D-GH is G H a wedge, of which ABCD is the rectangular base; / ABHG, DCHG, tletra- pezoidal sides, which meet / in the edge GII; and ADG, B C H, the triangular ends. B The altitude of a wedge is the perpendicular distance from its edge to the plane of its base; as G P. PROBLEM VII. 669. To find the solidity of a WEDGE. Add the length of the edge to twice the length of the base; multiply the sum by one sixth of the product of the altitude of the wedge and the breadth of the base. For, let L equal AB, the length of the base; I equal GH, the length of the edge;. b equal B C, the breadth / / of the base; and 1 equal D- --.,c. / P G, the height of the \ wedge. Then L -I I -- AB - GH = A M. Now, if the length of the base and the edge be equal, the polyedron is equal to half a parallelopipedol having the same base and altitude (Prop. VI. Bk. VIII.), and its solidity will be equal to I b 1 h (Prop. XIII. Bk. VIII.). If the length of the base is greater than that of the edge, let a section, M N G, be made parallel to B C H. BOOK XII.28 -287 This section will divide the whole wedge into the quadrangular pyramid AAM N D - G, and the triangrular prism B CH- G. The solidity of AMAlN D-G is equal to I bh AX (L-l1) (Prob.. Y.); and the solidity of B C H - G is equal to ~-b I h; hence the solidity of the whole wedge is equal to I -bhII+ }b h X(L -1)= b h 3 +-l+ bIt2 Lb bh 21= - bh X (2 L + 1). But, if the length of the base is less thaii that of the edge, the solidity of the wedge will be equal to the prism less the pyramid; or to jbhl- ~3b ItX (I- L) = b bt 3i- b h 2 1~.b h2L =bhItX (2 L +l). EXAMPLES. 1. Required the solidity of a wedge, the edge of which is 10 inches, the sides of 'the base 12 inches and 6 inches,. and the altitude 14 inches. 10 +(12 X2)=~34;34 X 14 X6~ 476 cu. in. thle 6 [solidity required. 2. What 'is the solidity of a wedge, of which the edge is 24 inches, the sides of the base 36 inches and 9 inches, and the altitude 22 inches? 3. How many solid feet in'a wedge, of which the sides of the base are 35 inches and 15 inches, the length of the edge 55 inches, and the altitude 17 3 inches? Atis. 3 cu. ft. 1751 CU. ill. RECTANGULAR PRISMOID. 670. A RECTANGULAR PRISMOID is a polyedron bounded by two rectangles, called the bases of the prismoid, and by fou'r trapezoids called the lateral faces of the prismoid. The altitude of a prismoid is the perpendicular distance between its ba'ses. OQ8 ELEMENTS OF GEOMETRY. PROBLEM1 VIII. 671. To find the solidity of a RECTANGULAR PRTSMOTD. Add the area of the two bases tofour intes the area of a parallel section at equal distances firont the bases; multiply the sum by one sixth, of the altitude. Let L and B be the length and breadth n ~ of the lower base, I and b the length and breadth of the upper base, M and m the length and breadth of the parallel section.......M..... equidistant from the bases, and ht the L altitude of the prisinoid. If a plane be passed through the opposite edges L and 1, the prismoid will be divided into two wedges, havinga for bases the bases of the prisnoid, and for edges L and I. The solidity of these wedges, which compose the prismold, is (Prob. VII.), B h X (2 L + I) + bh ItX (.(2 I+ L)= Iht(2 B L + B I+2bl+ b L). But M being equally distant from L and 1, 2 M = L ~ I, and 2 m - B + b (Prop. VII. Cor., Bk. IV.); consequently, 4 M m = (L +l1) X (B + -b)= B L B I + bL + b 1. Substituting 4 M m for its base, in the preceding oqunation, we have, as the expression of the solidity of a prismoid, ~h (BL +-b I + 4 Mm). 672. Scholium. This demonstration applies to prismoids of other forms. For, whatever be the form of the two bases, there may be inscribed in each such a number of small rectangles that the sum of them in each base shall differ less from that base than any assignable quantity; so that the sum of the rectangular prismoids that may be BOOK XII. 289 constructed on these rectangles will differ from the given prismoid by loss thlan ally assignable quantity. EXAMPLES. 1. Required the solidity of a prismoid, the larcger base of which is 30 inclies by 27 inches, tile smallcr base 2I inches by 18 inches, and the altitude 48 inclhcs. 30 X- 27-= 810; 24 X 18 = 432; + 24 X..27... X 4 - 2430; (810 + 432 + 2430) X 4 = 29,376 cu. i:i. 17 cu. ft., the solidity required. 2. What is tlhe solidity of a stick of timber, wh]lose largcr end is 24 inclhes by 20 inches, the smaller end 16 inches by 12 inches, and the lenlgth 18 feet? 3. What is the solidity of a block, whose enlds are respectively 30 by 27 inches and 24 by 18 inches, and whose length is 36 inches? 4. What is the capacity in gallons of a cistern 47- inches deep, whose inside dimensions are, at the top 81. and 55 inches, and at the bottom 41 and 29} inches? Ans. 546.929 gall. PROBLEM IlX. 673. To find the surface of a REGULAR POLYEDRON. Multiply the area of one of the faces by the number of faces; or multiply the square of one of the edges of tile polyedron by the surface of a similar polyedron whose edges are 1. For, since the faces of a regular polyedron are all equal, it is evident that the area of one face multiplied by the number of faces will give the area of the whole surface. Also, since tile surfaces of regular polyedrons of the same name are blounded by the 'same number of similar polygolns (Prop. I. Bk. VI.), their surfaces are to each other as the squares of the edges of the polyedroils (Prop. I. Cor., Bk. VI.),-...2 25 290 ELEMENTS OF GEOMIETRY. 674. TABLE OF SURFACES AND SOLIDITIES OF POLYEDRONS WHOSE EDGE IS 1. NAMES. NO. OF FACES. SURFACES. SOLIDITIES. Tetraedron, 4 1.7320508 0.1178511 Iexaedron, 6 6.0000000 1.0000000 Octaedron, 8 3.4641016 0.4714045 Dodecaedron, 12 20.6457288 7.6631189 Icosaedron, 20 8.6602540 2.1816950 The surfaces in the table are obtained by multiplying the area of one of the faces of the polyedron, as given in Art. 632, by the number of faces. EXAMPLES. 1. What is the surface of an octaedron whose edge is 16 inches? 162 X 3.4641016 = 886.81 sq. in., the area required. 2. Required the surface of an icosaedron whose edge is 20 inches. 3. Required the surface of a dodecaedron whose edge is 12 feet. Ans. 2972.985 sq. ft. PROBLEM X. 675. To find the solidity of a REGULAR POLYEDRON. Multiply the surface by one third of the perpendicular distance from the center to one of the faces; or multiply the cube of one of the edges by the solidity of a similar polyedron whose edge is 1. For any regular polyedron may be divided into as many equal pyramids as it has faces, the common vertex of the pyramids being the center of the polyedron; hence, the solidity of the polyedron must equal the product of the areas of all its faces by one third the perpendicular distance from the center to each face of the polyedron. BOOK XII. 291 Also, since similar pyramids are to cach other as tle cubes of their homologous edges (Prop. XXII. Bk. VIII.), two polyedrons containing the same number of similar pyramids are to cach other as the cubes of their edges; hence, tile solidity of a polycdron whose edge is 1 (Art. 673), may be used to measure other similar polycdrons. EXAMPLES. 1. Required the solidity of an octaedron whose edge is 16 inches. 168 X 0.4714045 = 1930.8728 cu. in., solidity required. 2. What is the solidity of a tctraedron whose edge is 2 feet? 3. Required the solidity of an icosacdron whose edge is 15 inches. Aiis. 7363.2206 cu. in. PROBLEM XI. 676. To find the surface.of a CYLINDER. Multiply the circumference of its base by its altitude, and the product will be the CONVEX surface (Prop. I. Bk. X.). To this add the areas of its two bases, and the result will be the ENTIRE surface. EXAMPLES. 1. What is the entire surface of a cylin- der, the altitude of which, A B, is 10 feet, and the circumference of the base 20 feet? 10 X 20= 200; 202 X 0.07958 X 2....-. 63.264; 200 + 63.264 263.264 sq. ft., B the surface required. 2. Required the convex surface of a cylinder wllose altitude is 16 feet, and the circumference of whose base is 21 feet. 3. What is the entire surface of a cylinder whose altitude is 10 inches, and whose circumference is 4 feet? 292 ELEMENTS OF GEOMETRY. 4. I-Iow many times must a cylinder 5 feet 3 inches long, and 21 inches in diameter, revolve, to roll an acre? Ans. 1509.18 times. PROBLEM XII. 677. To find the solidity of a CYLINDER. Multiply the area of the base by the altitude, and the product will be the solidity (Prop. II. Bk. X.). EXA.MPLES. 1. What is the solidity of a cylinder, whose altitude is 10 feet, and the circumference of whose base is 20 feet? 202 X 0.07958 X 10 = 318.32 cu. ft., solidity required. 2. Required the solidity of a cylindrical log, whose length is 9 feet, and the circumference of whose base is 6 feet. Ans. 25.7831 cu. ft. 3. The Winchester bushel is a liollow cylinder 18} inches in diameter, and 8 inches deep; what is its capacity in cubic inches?..! PROBLEM XIII. 678. To find the surface of a CONE. Multiply the circumference of the base by half the slant height (Prob. III. Bk. X.), and the product will be the convex surface. To this add the area of the base, and the result will be the entire surface. EXAMPLES. 1. What is the convex surface of a cone, whose slant height is 28 feet, and the circumference of whose base is 40 feet? 40 X - = 560 sq. ft., the surface required. 2. Required the -entire surface of a cone, whose slant heighllt is 14 feet, and the circumference of wlhose base is 92 inches. - BOOK XII. 293 3. What is the surface of a cone, whose slant height is 9 feet, and tle diameter of wlose base is 36 inches? 4. How many yards of canvas are required for the covering of a conical tent, tlhe slant height of which is 30 feet, and the circumference of the base 900 feet? Ans. 1500 sq. yd. PROBLEM XIV. 679. To find tle surface of a FRUSTUM OF A CONE. Multiply half the sum of the circumferences of its two bases by its slant hezight, and the product will be the convex surface (Prop. IV. Bk. X.). To this add the area of its bases, and the result will be the entire surface. 680. Scholium. The convex surface of a frustum of a cone may also be found by multiplying the slant height by the circumference of a section at equal distances between the two bases (Prop. IV. Cor., Bk. X.). EXAMPLES. 1. Required the convex surface of a frustum of a cone, whose slant lheight is 20 feet, and the circumferences of whose bases are 30 feet and 40 feet. 30 -+ 40 2 + X 20 = 700 sq. ft., the surface required. 2. Required the surface of a frustum of a cone, the diameters of the bases being 43 inches and 23 inches, and the slant heiglit 9 feet. 3. What is the convex surface of a frustum of a cone, of which a section equidistant from its two bases is 24 feet in circumference, the slant height of the frustum being 19 feet? 4. From a cone the circumference of whose base is 10 feet, and whose slant heigllt is 30.fcet, a cone has been cut off, whllose slant llcight is 8 feet. VWhat is tile convex surface of the frustum? Ans. 139* sq. ft. 25* 294 ELEMENTS OF GEOMETRY. PROBLEM XV. 681. To find the solidity of a CONE. Multiply the area of its base by one third of its altitude, and the product will be the solidity (Prop. V. Bk. X.). EXAMPLES. 1. What is tle solidity of a cone whose altitude is 42 feet, and the diameter of whose base is 10 feet? 102 X 0.7854 X -4 = 1099.56 cu. ft., solidity required. 2. Required tile solidity of a cone wlose altitude is 63 feet, and the radius of wlose base is 12 feet 6 inches. 3. How many cubic feet in a conical stick of timber, wlose length is 18 feet, the diameter at tie larger end being 42 inches? Ans. 57.7269 cu. ft. PROBLEM XVI. 682. To find the solidity of the FRUSTU1M OF A CONE. Add together the areas of the two bases and a mean proportional between them, and multiply that sum by one third of the altitude of the frustum; and the result will be the solidity required (Prop. VI. Bk. X.). EXAMPLES. 1. What is the solidity of a frustunm of a cone, CD EF, whose altitude, A B, is 21 feet, and the area of whose bases, FE, CD, are 80 square feet / and 300 square feet? (80 + 300 + /80X 300) X -- = B 3732.96 cu. ft., solidity required. 2. Required the solidity of a frustum of a cone, the diameters of the bases being 38 and 27 inches, and the altitude 11 feet. 3. If a cask, which is two equal frustums of cones joined together at the larger bases, have its bung diameter 28 BOOK XII. 295 inches, the head diameter 20 incies, and length 40 inches, how many gallons of wine will it hold? Ans. 79.06. PROBLEM XVII. 683. To find the surface of a SPHERE. Multiply the diameter by the circumference of a great circle of the sphere (Prop. VIII. Bk. X.); or multiply the area of one great circle of the sphere by 4 (Prop. VIII. Cor 1, Bk. X.); or multiply 3.1416 by the square of the diameter (Prop. VIII. Cor. 4, Bk. X.). EXAMPLES. 1. What is the surface of a sphere, whose diameter, ED, is 40 feet, and wlhose circumference, A E B D, is 125.664? 125.664 X 40 = 5026.56 sq. [ft., the surface required. D A B E 2. Required the surface of a sphere whose diameter is 30 inches. 3. What is the surface of a globe whose diameter is 7 feet and circumference 21.99 feet? Ans. 153.93. 4. How many square miles of surface has the earth, its diameter being 7912 miles? PROBLEM XVIII. 684. To find the surface of a ZONE or SEGMENT OF A SPHERE. Multiply the altitude of the zone or segment by the circumference of a great circle of the sphere (Prop. VIII. Cor. 2, Bk. X.); or multiply the product of the diameter and altitude by 3.1416 (Prop. VIII. Cor. 6, Bk. X.). 296 ELEMENTS OF GEOMETRY. EXAMPLES. 1. What is the surface of a segment of a sphere, the altitude of tle segment beilg 10 fet, and the diameter of the sphere 50 feet? 50 X 10 X 3.1416 = 1570.80 sq. ft., surface required. 2. The altitude of a segment of a spliere is C8 inches, and the circumference of the sphere is 25 feet; what is the surface of the segment? 3. Required the surface of a zone or segment, the diameter of the sphere being 72 feet, and tlhe altitude of the zone 24 feet. Ans. 5428.6848 sq. ft. 4. If the eartl be regarded as a perfect sphere whose axis is 7912 miles, and tile part of tlle axis corresponding to each of tlhe frigid zones is 327.192848, to each of tile temperate zones 2053.468612, and to tlhe torrid zone 3150.67708 miles; what is tlle surface of each zone? Ans. Eacl frigid zone 8132797.39568; eacl temperate zone 51041592.99898; torrid zone 78314115.07768 miles. PROBLEM XIX. 685. To find the solidity of a SPHERE. Multiply the surface of the sphere by one third of its radius (Prop. IX. Bk. X.); or multiply the cube of the diameter of the sphere by 0.5236 (Prop. IX. Cor. 5, Bk. X.). EXAMPLFS. 1. What is the solidity of a sphere whose diameter is 40 inches? 40s X 0.5236 = 33510.4 cu. in., the solidity required. 2. Required the solidity of a globe whose circumfcrcnce is 60 inches. 3. What is the solidity of tle moon in cubic miles, supposing it a perfect sphere with a diameter of 2160 miles? 4. Required the solidity of the earth, supposing it to be a perfect sphere, whose diameter is 7912 miles. Ans. 259332805349.80493 cu. miles. U OOK~ XII. 29 &d.097 PROBLEM XX. 686. To find the surface of a SPHERICAL POLYGON. Fromn the sumn of all the angles subtract the product of two rio/ht ankc1es by the numnber of sides lcss twoo; divide the remnainder by 900, and miultiply the quotient by one eig-hth of the surface of the sphere; an(d the result will be the surface of the spherical polygon (Prop. XX. Bk. 1X.). E-XA-MPLES9. 1. Required thec surface of a spherical polygon hiaving five sides, described on a sphiere whiose diameter is 100 feet, the suni of the angles being 720 degrees. 2 X 900 X (5 -2) =540; (7200" - 400) ~900=~-2; 1002 X 3.46 346;2x7 7834 sq. ft., the surface required. 2. Whiat is the surface of a triangle on a sphere whose diameter is 20'feet, the angles being 1500, 900, and. 540? PROBLEM XXI. 687. To find the solidity of a SPHERICAL PYRAMID or SECTOR. Multiply the area of the polygon or zone whichl forms the base of the pyramid or sector by one third of the radius (Prop. IX. Cor. 1, Bk. X.); or multiply the altitude of the base by the square of the radius, and that product by 2.0944 (Prop. IX. Cor. 7, Bk. X.). EXAMPLES. 1. Required the solidity of a E spherical sector, A C B E, the al-.......... titude, E D, of the Zone forming A __ its base being 5 feet, and the radius, C B, of the sphere being 12 feet. C 5 X 24 X 3.1416 =376.992; 876.992 x =1507.968 cu. ft., the solidity required. 298 ELEMENTS OF GEOMETRY. 2. What is the solidity of a sphlrical pyramid, the area of its base being 364 square fect, and the diameter of tho sphere 60 feet? 3. Required the solidity of a spherical sector, whose base is a zone 16 inches in altitude, in a sphere 3 feet in diameter. 4. What is the solidity of a spherical sector, whose base is a zone 6 feet in altitude, in a sphere 18 feet in diaimeter? Ans.. 1017.88 cu. ft. PROBLEM XXII. 688. To find the solidity of a SPHERICAL SEGMENT. When the segment is LESS than a hemisphere, from the solidity of the spherical SECTOR whose base is the zone of the segment, take the solidity of the cone whose vertex is the center of the sphere, and whose base is the circular base of the segment; but when the segment is GREATER than a hemisphere, take the sum of these solidities (Prop. IX. Sch., Bk. X.). 689. Scholium. If the segment has two plane bases, its solidity may be found by taking the difference of the two segmnents which lie on the same side of its two bases (Prop. IX. Sch., Bk. X.). EXAMPLES. 1. What is the solidity of a E segment, ABE, whose altitude, ED, is 5 feet, cut from a sphere A^. D whose radius, C E, is 20 feet? Tle altitude of the cone A B C is equal to C E -ED, or 20-5, which is equal to 15 feet; and the radius of its base is equal to /C A2 -CD2, or V 20- 152~, which is equal to 13.23; consequently the diameter A B is equal to 26.46 feet; 5 X 202 X 2.0944 = 4188.8 BOOK XII. 299 cubic feet, the solidity of the sector A CT) E (Prob. XXI.); 26.462 X 0.7'54 X -'s — 2946.99 cubic feet, the solidity of the cone A B - C (Prob. XV.); 4188.8 - 2946.99 = 1241.81 cubic feet, the solidity of the segment A B E required. 2. Required the solidity of a segment, whose altitude is 57 inches, the diameter of the sphere being 153 inches. 3. What is tlle solidity of a spherical segment, whose altitude is 13 feet, and the diameter of the sphere 33 feet 6 inches? 4. Required the solidity of tlhe segments of the earth which are bounded severally by its five zones, the eartl's diameter being 7912 miles, and tlhe part of tile diameter corresponding to each of the frigid zones being 327.19, to each temperate zone 2053.47, and to tlhe torrid zone 3150.68. Ans. Each frigid zone 1293793463.32, each temperate zone 55013912318.45, and the torrid zone 146717393786.26 cubic miles. THE SPHEROID. 690. A SPHEROID is a solid wlich many be described by the revolution of an ellipse about one of its axes, wlich remains immovable. An oblate spheroid is one described by the revolution of the ellipse about its minor or conjugate axis. A prolate spheroid is one described by the revolution of the ellipse about its major or transverse axis. PROBLEM XXIII. 691. To find the solidity of a SPHEROID. Multiply the square of the axis of revolution by the fixed axis, and that product by 0.52C6. A full demonstration of this, which is based upon the principle that a spheroid is two thirds of its circumscribing 800 ELEMENTS OF GEOMETRY. cylinder, would require a knowledge of Conic Sections, or of the Differential and Integral Calculi, with neither of which is the learner here supposed to be acquainted. The relation, however, of the spheroid to its circumscribing cylinder, is that which the sphere sustains to its circumscribing cylinder (Prop. X. Bk. X.).. Now the area of the base of the cylinder is found by multiplying tle square of the axis of revolution by 0.7854, and the solidity of the cylinder by multiplying that product by the fixed axis (Prop. II. Bk. X.). But the solidity of the spheroid is only two thirds of tliat of the cylinder; hence, to obtain the solidity of tlle former, instead of multiplying by 0.7854, we mutst use a factor only two thirds as large, which will be 0.5236. X AMPIES. 1. What is the solidity of the ob- C late spheroid ACBD, whose fixed axis, C D, is 30 inches, and the axis A.\.... B of revolution, A B, 40 inches. 402 X 30 X 0.5236 = 25132.8 cubic D inches, the solidity required. 2. Required the solidity of a prolate spheroid, whose fixed axis is 50 feet, and the axis of revolution 36 feet. 3. What is the solidity of a prolate, and also of an oblate spleroid, tlle axes of each being 25 anld 15 inches? Ans. Prolate, 2945.25 cu. in.; oblate, 4908.75 cu. in. 4. What is the solidity of a prolate, and also of an oblate spheroid, the axes of each being 3 feet 6 inches and 2 feet 10 inches? 5. Required the solidity of the earth, its figure being that of an oblate spheroid whose axes are 7925.3 and 7898.9 miles. Ans. 259774584886.834 cubic miles. BOOK XIII. MISCELLANEOUS GEOMETRICAL EXERCISES. 1. IF the opposite angles formed by four lines meeting at a point are equal, these lines form but two straight lines. 2. If the equal sides of an isosceles triangle are produced, the two exterior angles formed witl the base will be equal. 3. The sum of any two sides of a triangle is greater tlan the third side. 4. If from any point witlin a triangle two straight lines are drawn to the extremities of either side, they will ilclude a greater angle tlan that contained by the other two sides. 5. If two quadrilaterals have the four sides of the one equal to the four sides of the other, each to each, and the angle included by any two sides of tie one equal to the angle contained by the corresponding sides of the other, the quadrilaterals are themselves equal. 6. The sum of the diagonals of a trapezium is less than tlhe sum of any four lines which can be drawn to the four angles from any point within the figure, except from the intersection of the diagonals. 7. Lines joining the corresponding extremities of two equal and parallel straight lines, are themselves equal and parallel, and the figure formed is a parallelogram. 8. If, in the sides of a square, at equal distances from the four angles, points be taken, one in each side, the straight lines joining these points will form a square. 26 302 ELEMENTS OF GEOMETRY. 9. If one angle of a parallelogram is a right angle, all its angles are righlt angles. 10. Any straight line drawn through the middle point of a diagonal of a parallelogram to meet the sides, is bisected in that point, and likewise bisects the parallelogram. 11. If four magnitudes are proportionals, the first and second may be multiplied or divided by the same magnitude, and also the third and fourth by the same magnrlitude, and the resulting magnitudes will be proportionals. 12. If four magnitudes are proportionals, the first and third may be multiplied or divided by the same magnitude, and also the second and fourth by the same magnitude, and the resulting magnitudes will be proportionals. 13. If there be two sets of proportional magnitudes, the quotients of the corresponding terms will be proportionals. 14. If any two points be taken in tlhe circumference of a circle, the straight line joining them will lie wholly witlin the circle. 15. The diameter is tile longest straight line that can be inscribed in a circle. 16. If two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel. 17. If a straight line be drawn to touch a circle, and be parallel to a chord, the point of contact will be the middle point of the arc cut off by that chord. 18. If two circles cut eacl other, and from either point of intersection diameters be drawn, the extremities of these diameters and the other point of intersection will be in the same straight line. 19. If one of the equal sides of an isosceles triangle be the diameter of a circle, tlle circumference of the circle will bisect tlie base of tlle triangle. 20. If the opposite angles of a quadrilateral be together equal to two right angles, a circle may be circumscribed about the quadrilateral. BOOK XIII. 803 21. Parallelograms which hlave two sides and the included angle equal in each, are tlhemlsclves equal. 22. Equivalent triangles upon the same base, and upon the same side of it, are between the same parallels. 23. If the middle points of the sides of a trapczoid, which are not parallel, be joined by a straight line, tlat lile will be parallel to each of the two parallel sides, and be equal to half their sum. 24. If, in opposite sides of a parallelogram, at equal distances from opposite angles, points be takel, one il each side, the straight line joining these points will bisect the parallelogram. 25. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude with, and equivalent to, the given triangle. 26. If the sides of tile square described upon the hypothenuse of a riglt-angled triangle be produced to meet tle sides (produced if necessary) of the squares described upon the other two sides of the triangle, the triangles thus formed will be similar to the given triangle, and two of them will be equal to it. 27. A square circumscribed about a given circle is double a square inscribed in the same circle. 28. If the sum of the squares of the four sides of a quadrilateral be equivalent to the sum of the squares of the two diagonals, the figure is a parallelogram. 29. Straight lines drawn from the vertices of a triangle, so as to bisect the opposite sides, bisect also the triangle. 30. The straight lines which bisect the three angles of a triangle meet in the same point. 31. The area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle. 32. If the points of bisection of the sides of a given triangle be joined, the triangle so formed will be one fourth of the given triangle. 83. To describe a square upon a given straight line. 304 ELEMENTS OF GEOMETRY. 34. To find in a given straight line a point equally distant from two given points. 35. To construct a triangle, the base, one of the angles at the base, and the sum of the other two sides being given. 36. To trisect a right angle. 37. To divide a triangle into two parts by a line drawn parallel to a side, so that these parts shall be to each other as two given straight lines. 38. To divide a triangle into two parts by a line drawn perpendicular to tlie base, so that these parts shall be to each other as two given lines. 39. To divide a triangle into two parts by a line drawn from a given point in one of the sides, so that tie parts shall be to each other as two given lines. 40. To divide a triangle into a square number of equal triangles, similar to each other and to the original triangle. 41. To trisect a given straight line. 42. To inscribe a square in a given right-angled isosceles triangle. 43. To inscribe a square in a given quadrant. 44. To describe a circle that shall pass through a given point, have a given radius, and touch a given straight line. 45. To describe a circle, the centre of which shall be in the perpendicular of a given riglt-angled triangle, and the circumference of which shall pass through the right angle and touch the hypothenuse. 46. To describe three circles of equal diameters whicl shall touch each other, and to describe another circle which slall touch the three circles. 47. If, on the diameter of a semicircle, two equal circles be described, and in the curvilinear space included by tle three circumferences a circle be inscribed, its diameter will be to that of tie equal circles in tlle ratio of two to three. 48. If two points be taken in tle diameter of a circle, BOOK XIII. 380 equidistant from the centre, the sum of tlhe squares of two lines drawn from these points to- ally point in the circumference will always be the same. 49. Given the vertical angle, and the radii of the inscribed and circumscribed circles, to construct the triangle. 50. If a diagonal cuts off three, five, or aly odd number of sides from a regular polygon, the diagonal is parallel to one of the sides. 51. The area of a regular hexagon inscribed in a circle is double that of an equilateral triangle inscribed in the same circle. 52. The side of a square circumscribed about a circle is equal to the diagonal of a square inscribed in the same circle. 53. To describe a circle equal to half a given circle. 54. A regular duodecagon is equivalent to three fourths of the square constructed on the diameter of its circum-,scribed circle; or is equal to the square constructed on the side of the equilateral triangle inscribed in the same circle. 55. If semicircles be described on the sides of a rightangled triangle as diameters, the one described on the hypothenuse will be equal to the sum of the other two. 56. If on the sides of a triangle inscribed in a semicircle, semicircles be described, the two crescents thus formed will together equal the area of the triangle. 57. If the diameter of a semicircle be divided into any number of parts, and on them semicircles be described, their circumferences will together be equal to the circumference of tle given semicircle. 58. To divide a circle into any number of parts, which shall all be equal in area and equal in perimeter, and not have tile parts in the form of sectors. 59. To draw a straight line perpendicular to.a plane, from a given point above the plane. 60. Two straight lines not in the same plane being 26* 306 ELEMENTS OF GEOMETRY. given in position, to draw a straight line which shall be perpendicular to tllhem both. 61. The solidity of a triangular prism is equal to the product of the area of either of its rectangular sides as a base multiplied by half its altitude on that base. 62. All prisms of equal bases and altitudes are equal in solidity, whatever be the figure of their bases. 63. The convex surface of a regular pyramid exceeds the area of its base in the ratio that the slant height of the pyramid exceeds the radius of the circle inscribed in its base. 64. If from any point in the circumference of the base of a cylinder, a straight line be drawn perpendicular to the plane of the base, it will be wholly in the surface of the cylinder. 65. A cylinder and a parallelopipedon of equal bases and altitudes are equivalent to each other. 66. If two solids have the same height, and if their sections made at equal altitudes, by planes parallel to the bases, have always the same ratio which the bases have to one another, the solids have to one another the same ratio which their bases have. 67. The side of the largest cube that can be inscribed in a sphere, is equal to the square root of one third of the square of the diameter of the sphere. 68. To cut off just a square yard from a plank 14 feet 3 inches long, and of a uniform width, at what distance from the edge must a line be struck? Ans. 7f- in. 69. How much carpeting a yard wide will be required to cover the floor of an octagonal hall, whose sides are 10 feet each? 70. The perambulator, or surveyilg-wheel, is so constructed as to turn just twice in the length of a rod; what is its diameter? Ans. 2.626 ft. 71. What is the excess of a floor 50 feet long by ~0 broad, above two others, each of half its dimensions? BOOK XIII. 307 72. The four sides of a trapezium are 13, 13.4, 24, and 18 feet, and the first two contain a right angle. Required the area. Ans. 253.38 sq. ft. 73. If an equilateral triangle, whose area is equal to 10,000 square feet, be surrounded with a walk of uniform width, and equal to the area of the inscribed circle, what is the widtl of the walk? Ans. 11.701 ft. 74. A right-angled triangle has its base 16 rods, and its perpendicular 12 rods, and a triangle is cut off from it by a line parallel to its base, of which the area is 24 rods. Required the sides of that triangle. Ans. 8, 6, and 10 rods. 75. There is a circular pond whose area is 5028f square feet, in the middle of which stood a pole 100 feet high; now, the pole having been broken off, it was observed that the top portion resting on the stump just reached the brink of the pond. What is the height of the piece left standing? Ans. 41.9968 ft. 76. The area of a square inscribed in a circle is 400 square feet; required the diagonal of a square circumscribed about the same circle. 77. The four sides of a field, whose diagonals are equal, are known to be 25, 35, 31, and 19 rods, in a successive order; what is the area of the field? Ans. 4 A. 1 R. 38- p. 78. The wheels of a chaise, each 4 feet higl, in turning withi a ring, moved so that the outer wheel made two turns while the inner made one, and their distance from one another was 5 feet; what were the circumferences of the tracks described by tllem? Ans. Outer, 62.8318 ft.; inner, 31.4159 ft. 79. The girt of a vessel round the outside of the loop is 22 inches, and the hoop is 1 inch thick; required the true girt of the vessel. 80. If one of the Egyptian pyramids is 490 feet high, having eacl slant side an equilateral triangle and the base a square, what is the area of tie base? Ans. 11 A. 3 rd. 2231 ft. 308 ELEMENTS OF GEOMETRY. 81. An ellipse is surroun:.dedl 1,y a wall 1- illnllcs thick; its axes are 840 links and (112 links; reqllired thl qualtity of ground enclosed, and the quantity occupied by the wall. Ans. 4 A. 6 rd. enclosed, and 1760.49 sq. ft., area occupied by the wall. 82. There is a meadow of 1 acre in the form of a square; what must be the length of the rope by which a horse, tied equidistant from each angle, can be permitted to graze over the entire meadow? 83. A gentleman has a rectangular garden, whose lengtl is 100 feet and breadth 80 feet; what must be the uniform width of a walk half-way round the same, to take up just half the garden? Ans. 25.9688 ft. 84.. Two trees, 100 feet asunder, are placed, the one at the distance of 100 feet, and the other 50 feet from a wall; what is the distance that a person must pass over in running from one tree to touch the wall, and then to the other tree, the lines of distance making equal angles with tile wall? Ans. 173.205 ft. 85. There is a rectangular park 400 feet long and 300 feet broad, all round which, and close by the wall, is a border 10 feet broad; close by the border there is a walk, and also two others, crossing each Other and the park at right angles, in the middle of the garden. The walks are all of one breadth, and their area takes up one tenth of the whole park; required the breadth of the walks. Ans. 6.2375 ft. 86. A farmer borrowed a cubical pile of wood, which measured 6 feet every way, and repaid it by two cubical piles, of which the sides were 3 feet each; what part of the quantity borrowed lias lie returned? 87. A board is 10 feet long, 8 inches in breadth at the greater end, and 6 inches at the less; how much must be cut off from the less end to make a square foot? Ans. 23.2493 in. 88. A piec of timber is 10 feet long, each side of tle BOOK XIII. greater base 9 inches, and each side of tlhe less 6 inches; how much must be cut off from the less end to contain a solid foot? Ans. 3.39214 ft. 89. What must be the inside dimensions of a cubical box to hold 200 balls, each 2i inches in diameter? 90. Near my house I intend making a hexagonal or sixsided seat around a tree, for which I have procured a pine plank 16~ feet long and 11 inches broad; what must be the inner and outer lengths of each side of the seat, that there may be the least loss in cutting up the plank? -Ans. 26.64915 in. inner, and 39.35085 in. outer length. 91. Required the capacity of a tub in the form of a frustum of a cone, of which the greatest diameter is 48 inches, the inside length of the staves ~0 inches, and the diagonal between the farthest extremities of the diameters 50 inches. Ans. 165.34 gals. 92. The front of a house is of such a height, that, if the foot of a ladder of a certain length be placed at the distance of 12 feet from it, tlhe top of the ladder will just reach to the top of the house; but if the foot of the ladder be placed 20 feet from the front, its top will fall 4 feet below the top of the house. Required the height of the house, and the length of the ladder. Ans. 34 feet, the height of the building; 36.0555 feet, the length of the ladder. 93. A sugar-loaf in form of a cone is 20 inches lhighl; it is required to divide it equally among three persons, by cctions parallel to the base; what.is the height of each part? Ans. Upper 13.8672, next 3.6044, lowest 2.5284 in. 94. Within a rectangular court, whose length is four cliains, and breadth three chains, there is a piece of water in tlie form of a trapezium, whose opposite angles are in a direct line with those of the court, and the respective distances of the angles of the one from those of the other are 20, 25, 40, and 45 yards, in a successive order;i' equired the area of the water. Ans. 960 sq. yd. 310 ELEMENTS OF GEOMETRY. 95. What will the diameter of a sphere be, whcl its solidity and the area of its surface are expressed by the same numbers? Ails. 6. 96. There is a circular fortification, which occupies a quarter of an acre of ground, surrounded by a ditch coinciding with the circumference, 24 feet wide at bottom, 26 at top, and 12 deep; how much water will fill the ditch, if it slope equally on both sides? Ans. 18.5483.25 cu. ft. 97. A father, dying, left a square field containing "0 acres to be divided among his five sons, in such a manner that the oldest son may have 8 acres, the second 7, the third 6, the fourth 5, and the fifth 4 acres. Now, the division fences are to be so made that the oldest son's share shall be a narrow piece of equal breadth all around the field, leaving the remaining four shares in the form of a square; and in like manner for each of the other shares, leaving always the remainders in form of squares, one within another, till the share of the youngest be the inllermost square of all, equal to 4 acres. Required a side of each of the enclosures. Ans. 17.3205,14.8324, 12.2474, 9.4868, and 6.3246 chains. 98. Required the dimensions of a cone, its solidity being 282 inches, and its slant height being to its base diameter as 5 to 4. Ans. 9.796 in. the base diameter; 12.246 in. the slant height; and 11.223 in. the altitude. 99. A gentleman lhas a piece of ground in form of a square, the difference between whose side and diagonal is 10 rods. He would convert two thirds of the area into a garden of an octagonal form, but would have a fish-pond at the centre of the garden, in the form of an equilateral triangle, whose area must equal five square rods. Required the length of each side of the garden, and of each side of the pond. Ans. 8.9707 rods, each side of the garden, and 3.398 rods, each side of the pond. BOOK XIV. APPLICATIONS OF ALGEBRA TO GEOMETRY. 692. WHEN it is proposed to solve a geometrical prolb lem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required. Represent the known parts by the first letters of the alphabet, and the required parts by the last letters. Then, observing the geometrical relations that the parts of the figure have to each other, make as many independent equations as there are unknown quantities introduced, and the solution of these equations will determine the unknown quantities or required parts. To form these equations, however, no definite rules can be given; but the best aids may be derived from experience, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem. PROBLEM I. 693. In a right-angled triangle, having given the lhy pothenuse, and the sum of the other two sides, to determine these sides. C Let AB C be the triangle, right-angled at B. Put A C = a, the sum AB / + B C =s, AB = x, and B C =. 312 ELEMENTS OF GEOMETRY. Then, x + y s. and (Prop. XI. Bk. Iv.),.IC2 + V2 _ a2. From the first equastoli,; =s -- y. Substitute in second equation this value of x, - 2 sy+ 2 y2- a2. Or, 2 y2- 2 s y a2 - s, Or, y2 S y a2 --- 2. By completing the square, y2 - y S2 - a2 -- s, Extracting sq. root, y - s ~: / a' - j s, Or, y = - s ~ ^ a -- s2. If A C 5, and the sum AB + B C =7, y=4 r 3, and x=- 3 or 4. PROBLEM II. 694. Having given the base aln perpendicular of a triangle, to find the side of an inscribed square. Let ABC be the triangle, C and l- E F G the inscribed square. PutAB= b,CD=a, F and GF or GH DI=x; then will CI CD-DI= a-x., Since the triangles AB C, A H D E B G F C are similar, AB: CD:: GF: CI, or: a:: x:.a -- x. Hence, ab - b= ax, ab ora, x = BOOK XIV. 313 that is, the side of the inscribed square is equal to the product of the base by the altitude, divided by their sum. PROBLEM III. 695. Having given the lengths of two straight lines drawn from the acute angles of a right-angled triangle to the middle of the opposite sides, to determine those sides. Let A B C be the given triangle, A and AD, BE the given lines. Put A D=a, B E b, CD or. CB=-x, and CE or CA-=y; then, since C D2 + C A2 = A D2, and E2 + C B2 = B E2, we have x2 +- 4 y2 a2 and y' + 4 x2 b2. B D C By subtracting the second equation from four times the first, 15 y2 4 a2 b2 or, y 4_4a-. b2; by subtracting the first equation from four times the second, 15 x = 4 b2 - a, or, x= 4b -a2; s 15 which values of x and y are half the base and perpendiculars of the triangle. PROBLEM IV. 696. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within to the three sides, to determine these sides. 27 314 314 ~~ELEMENTS OF GEOMETRY. Let A B C be the equilateral trian- C gle, and D E, D F, D G the given perpendiculars from the point ID. Draw F D A, D B, D C to the vertices of the, 'D E three angles, and let fall the perpendicular, C H, on the base, A B. Put DE~a, DF b DG-c, A H1G B and A H or B H, half the side of the equilateral triangle, =-X. Then ACor BC=2x, anidCH=V/AC2 -All2 -v4-x2 - V3x= V3. Now, since the area of a triangle is equal to the product of half 'its base by its altitude (Prop. VI. Bk. IV.), The triangle A CB = AB X CH=1 x xN X 3xV 3. ABD=~,AB xD G~xxc =cX. BCD=iBCXDE~xXa =ax. ACD=JACXDF~xXb =bx. But the three triangles A B D, B C ID, A C ID are together equal to the triangle A C B. Hence, X2 A/3 == a x ~ b x ~ c x = x (a + b ~ c), or, xV/3=a +b c; 2aabb~c Hence each side, or 2 x=2(a~b+c vf/ 697. Cor. Since the perpendicular, C H, is equal to XV3 it is equal to a + b + c; that is, the whole perpen~dicular of an equilateral triangle is equal to the sum of all the perpendiculars let fall from any point in the tri.. angle to each of its sides. PROBLEM V. 698. To determine the radii of three equal circles described within and tangent to a given circle, and also tangent to each other. BOOK XIV. 315 Let A F be the radius of the given circle, and B E the radius of one of the equal circles described within it. Put AF = a, and BE - x; then each side of the equilateral triangle, BCD, formed by joining the centers of the required circles, will be represented by 2 x, and its altitude, C E, by V4 x2 x2, or x 4v3. The triangles B C E, A B E are similar, since the angles B C E and A B E are equal, each being half as great as one of the angles of the equilateral triangle, and the angle B E C is common. Hence, CE: BE:: BC:AB, or x 3: x:: 2 x: AB, and A B 2 2/3 But AB + BF - AF; 2x hence, - + x = a, 1J3 or 2 x + x V3 = a /3, or (2 + V/) x =a a 3. Hence, 2+3 - 2.547 a X 0.4641. PROBLEM VI. 699. In a right-angled triangle, having given the base, and the sum of the perpendicular and hypothenuse, to find these two sides. PROBLEM VII. 700. In a rectangle, having given the diagonal and perimeter, to find the sides. 316 ELEMENTS OF GEOMETRY. PROBLEM VIII. 701. In a right-angled triangle, having given the base, and the difference between the hypothenuse and perpendicular, to find both these two sides. PROBLEM IX. 702. Having given the area of a rectangle inscribed in a given triangle, to determine the sides of the rectangle. PROBLEM X. 703. In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle. PROBLEM XI. 704. In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base, to find the sides of the triangle. PROBLEM XII. 705. In a triangle, having given the two sides about the vertical angle together with the line bisecting that angle, and terminating in the base, to find the base. PROBLEM XIII. 706. To determine a right-angled triangle, having given the perimeter and the radius of its inscribed circle. PROBLEM XIV. 707. To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides. PROBLEM XV. 708. To determine a right-angled triangle, having given the hypothenuse, and the side of the inscribed square. ~ BOOK XIV. 317 PROBLEM XVI. 709. In a right-angled triangle, having given the perimeter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse, to determine the triangle, that is, its sides. PROBLEM XVII. 710. To determine a right-angled triangle, having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. PROBLEM XV1II. 711. To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides. PROBLEM XIX. 712. To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides. PROBLEM XX. 713. In a triangle, having given all the three sides, to find the radius of the inscribed circle. PROBLEM XXI. 714. To determine a right-angled triangle, having given the side of the inscribed square, and the radius of the inscribed circle. PROBLEM XXII. 715. To determine a triangle, having given the base, the perpendicular, and the rectangle of the two other sides. 27* 818 ELEMENTS OF GEOMETRY. PROBLEM XXIII. 716. To determine a right-angled triangle, having given the hypothenuse, and the radius of the inscribed circle. PROBLEM XXIV. 717. To determine a right-angled triangle, having given the hypothenuse and the difference between a side and the radius of the inscribed circle. PROBLEM XXV. 718. To determine a triangle, having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle. PROBLEM XXVI. 719. There are two stone pillars in a garden, whose perpendicular heights are 20 and 30 feet, and the distance between them 60 feet. A ladder is to be placed at a certain point in the line of distance, of such a length, that it may just reach the top of both the pillars. What is the length of the ladder, and how far from each pillar must it be placed? Ans. 39.5899 feet, length of the ladder; 341 feet, distance of the foot of the ladder from the bottom of the lower pillar; and 25~ feet, distance of the foot of the ladder from the bottom of the higher pillar. PROBLEM XXVII. 720. There is a cistern, the sum of the length and breadth of which is 84 inches, the diagonal of the top 60 inches, and the ratio of the breadth to the depth as 25 to 7. What are its dimensions, provided it has the form of a rectangular parallelopipedon? Ans. Length 48 inches; width 36 inches; depth 10.08 inches. BOOK XIV. 319 PROBLEM XXVIII. 721. The three distances from an oak, growing in an open plain, to the three visible corners of a square field, lying at some distance, are known to be 78, 59.161, and 78 poles, in successive order. What are the dimensions of the field, and its area? Ans. Side of the square 24 rd.; area 3 A. 2 R. 16 rd. PROBLEM XXIX. 722. There is a house of three equal stories in height. Now a ladder being raised against it, at 20 feet distance from the foot of the building, reaches the top; whilst another ladder, 12 feet shorter, raised from the same point, reaches only to the top of the second story. What is the height of the building? Ans. 41.696 ft. PROBLEM XXX. 723. The solidity of a cone is 2513.28 cubic inches, and the slant side of a frustum of it, whose solidity is 2474.01, is 19.5 inches. Required the dimensions of the cone. Ans. Altitude 24 inches; base diameter 20 inches. PROBLEM XXXI. 724. Within a rectangular garden containing just an acre of ground, I have a circular fountain, whose circumference is 40, 28, 52, and 60 yards distant from the four angles of the garden. From these dimensions, the length and breadth of the garden, and likewise the diameter of the fountain, are required. Ans. Length 94.996 yds.; width 50.949 yds.; diameter of the fountain 20 yds. PROBLEM XXXII. 725. There is a vessel in the form of a frustum of a cone, standing on its lesser base, whose solidity is 8.67 feet, the depth 21 inches, its greater base diameter to that 320 ELEMENTS OF GEOMETRY. of the lesser as 7 to 5, into which a globe had accidentally been put, whose solidity was 2- times the measure of its surface. Required the diameters of the vessel and of the globe, and how many gallons of water would be requisite just to cover the latter within the former. Ans. 35 and 25 inches, top and bottom diameters of the frustum; 15 inches, diameter of the globe; and 34.2 gallons, the water required. PROBLEM XXXIII. 726. Three trees, A, B, C, whose respective heights are 114, 110, and 98 feet, are standing on a horizontal plane, and the distance from A to B is 112, from B to C is 104, and from A to C is 120 feet. What is the distance from the top of each tree to a point in the plane which shall be equally distant from each? Ans. 126.634 ft. PROBLEM XXXIV. 727. A person possessed a rectangular meadow, the fences of which had been destroyed, and the only mark left was an oak-tree in the east corner; he however recollected the following particulars of the dimensions. It had once been resolved to divide the meadow into two parts by a hedge running diagonally; and lie recollected that a segment of the diagonal intercepted by a perpendicular from one of the corners was 16 chains, and the same perpendicular, produced 2 chains, met the other side of the meadow. Now the owner has bequeathed it to four grandchildren, whose shares are to be bounded by the diagonal and perpendicular produced. What is the area of the meadow, and what are the several shares? Ans. Area of the whole meadow, 16 acres; shares, 1 R. 24 rd.; 1 A. 2R. 16 rd.; 6 A. 1 R. 24 rd.; 7 A. 2 R. 16 rd. THE END E LE ME NT S or PLANE AND SPHERICAL TRIGONOMETRY; WITH PRACTICAL APPLICATIONS. I1' TRIGONOMETRY. BOOK I. LOGARITHMS. 1. THE LOGARITHM of a number is the exponent of the power to which a given fixed number must be raised in order to produce the first number. 2. The BASE of the system is the fixed number. 3. The base, in the common system of logarithms, is 10. Hence, since 10~ = 1, 0 is the logarithm of 1; 101 = 10, 1 " i 10; 10W 100, 2 " " 100; 10 = 1,000, 3 " " 1,000; 10 -= 10,000, 4 " " 10,000; &c., &c. It thus appears that, in the common system, the logarithm of every number between 1 and 10 is some number between 0 and 1; that is, a proper fraction. The logarithm of every number between 10 and 100 is some number between 1 and 2; that is, 1 plus a fraction. The logarithm of every number between 100 and 1,000 is some number between 2 and 3; that is, 2 plus a fraction; and so on. 4. By means of negative exponents the application of logarithms may be extended, in the common system, to numbers less than 1. Thus, since BOOK I. 3 10-' - 0.1, -1 is the logarithm of 0.1; 10-2 0.01, 2 " 0 9.01; 10-3 0.001, - 3 " 0.001; 10-4 0.0001, -4 " " 0.0001; &c., &c. From this it appears that the logaritlm of every number between 1 and 0.1 is some number between 0 and -1; that is, -1 plus a fraction. The logarithm of every number between 0.1 and 0.01 is some number between -1 and -2; that is, -2 plus a fraction. The logarithm of every number between 0.01 and 0.001 is some number between -2 and -3; that is, -3 plus a fraction; and so on. 5. In the common system, as the logarithms of all numbers which are not exact powers of 10 are incommensurable with those numbers, their values can only be obtained approximately, and are expressed by decimals. 6. The integral part of any logarithm is called the CHARACTERISTIC, and the decimal part is sometimes called the MANTISSA. 7. The characteristic of the logarithm of ANY NUMBER GREATER TIAN UNITY, is one less than the number of integralfigures in the given number. For it has been shown (Art. 3) that the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of 1000 is 3, and so on. 8. The characteristic of the logarithm of ANY DECIMAL FRACTION is a negative number, and is one more than the number of ciphers between the decimal point and the first significant figure. For it has been shown (Art. 4) that the logarithm of 0.1 is -1, of 0.01 is -2, of 0.001 is -3, and so on. NOTE. -In general, whether the given number be integral, fractional, or mixed, the characteristic of the logarithm of any number expressed decimally is the distance of the first, or left-hand, significant figure from the units'place, being positive when that figure is on the left of the units' place, and negative when on the right. GENERAL PROPERTIES OF LOGARITHMS. 9. The logarithm of a PRODUCT is equal to the sum of the logarithms of its factors. 4 TRIGONOMETRY. For let Mland Nbe any two numbers, x and y their respective logarithms, and a the base of the system. Then, by definition (Art. 1), we have M — a, N -ay. Multiplying equations, member by member, we have MlN' _ a' ay - ax +. Therefore, log (MX N) -x-+y = log -M+log NV 10. The logarithm of a QUOTIENT is equal to the logarithm of the dividend diminished by that of the divisor. For, by Art. 9, we have M = ax, N = a". Dividing the first equation by the second, member by member, we have 31r a~ ~ a — a Y. N y - a a Therefore, log (N- x-y = log l- log N 11. The logarithm of any POWER of a number is equal to the product of the logarithm of the number by the exponent of the power. For let m be any number, and take the equation (Art. 9) M- a, then, raising both sides to the mth power, we have Mim = (ax)m - a. Therefore, log (MJ") = x m = (log M) X m. 12. The logarithm of the ROOT of any number is equal to the logarithm of the number divided by the index of the root. For, let n be any number, and take the equation (Art. 9) M- ax, then, extracting the nth root of both sides, we have vM=^=az BOOK I. 5 Therefore, log (3M' ) =! 13. Hence, by means of logarithms, we can perform multiplication by addition, and division by subtraction; also, we can raise a number to any power by a single multiplication, and extract any root of a number by a single division. 14. All numbers, integral, fractional, or mixed, having the same succession of significant figures, have logarithms with the same decimal part. For since the logarithm of 10 is 1, the product of any number by 10 will have a logarithm increased by 1; and, likewise, the quotient of any number divided by 10 will have a logarithm diminished by 1; and, 1 being an integer, the logarithms will differ only in their characteristics. Thus, the logarithm of 235 is 2.371068 6C" " 2350 " 3.371068 "i " 23.5 " 1.371068 "< " 2.35 " 0.371068 i" ".235 " 1.371068 <(" ".0235 " 2.371068 15. The negative sign placed over the characteristic indicates that it alone is negative, the decimal part being always positive. TABLE OF LOGARITHMS. 16. A Table of Logarithms usually contains all the whole numbers between 1 and a given number, with their logarithms. The accompanying table contains the logarithms of all numbers from 1 up to 10,000, calculated to six places of decimals. 17. In the table, the characteristics of the logarithms of the first 100 numbers are inserted; but for all other numbers the decimal part only of the logarithms is given, while the characteristic is left to be supplied by inspection, according to the principles already furnished (Art. 7, 8). 18. The numbers are in the column headed N, and their logarithms, or the decimal parts of their logarithms, are opposite 6 TRIGONOMETRY. on the same line. When the first two figures of the decimal are the same for several successive logarithms, they are not repeated for each, but, being used once, are then left to be supplied., 19. In the column headed D are the mean or average differences of the ten logarithms against which they are placed. To FIND THE LOGARITHM OF ANY NUMBER. 20. When the given number is any integer of ONE or TWO figures. Look on the first page of the table, and opposite the given number will be found the logarithm with its characteristic. Thus, the logarithm of 63 is 1.799341; "< " ~98 " 1.991226. 21. When the given number is any integer of THREE FIGURES. Look in the table for the given number, and opposite the same, in the column headed 0, will be found the decimal part of the logarithm, to which must be prefixed 2 as the characteristic (Art. 7). Thus, the logarithm of 110 is 2.041393; <(" " 817 " 2.912222. 22. When the given number is any integer of FOUR figures, either with or without ciphers annexed. Find the first three figures of the given number in the column headed N, and, opposite to them, in the column headed by the fourth figure, will be found the decimal part of the logarithm; to which the characteristic, as determined by Art. 7, must be prefixed. Thus, the logarithm of 4901 is 3.690285; "( " < 9677 " 3.985741; " " 31250 " 4.494850. 23. When the given number is any integer of FIVE or MORE figures. Find the logarithm of the first four figures as in Art. 22. BOOK I. ' 7 regarding the others as ciphers annexed; then take, from the col umn headed D, the number on the same horizontal line with the decimal part of the logarithm, and multiply it by the remaining figures of the given number; reject from the right of the product thus obtained as many figures as there were in the multiplier, and add what is left to the decimal part of the logarithm already found; and the sum will be the required logarithm. Thus, if it be required to find the logarithm of 93192: the logarithm of 93190 is 4.969369 Dif. from col. D, 47 9.4 2 " (" 93192 " 4.969378. 9.4 This process is based upon the supposition that the differences of logarithms are proportional to the differences of their corresponding numbers, which is not strictly correct, yet sufficiently exact for practical purposes. When the figure or figures rejected from the right of the product, considered decimally, exceed in value.5, the right-hand figure of what is left to be added must be increased by 1, to insure greater accuracy in the result. 24. When the given number is any DECIMAL FRACTION, or any mixed number expressed decimally. Find the decimal part of the logarithm of the number, as if it were an integer, and prefix the proper characteristic (Arts. 7 and 8). Thus, the logarithm of 93.192 is 1.969378; "( " 4526.375 " 3.655751; " ".0006801 " 4.832573. 25. When the given number is any COMMON FRACTION. Reduce the given fraction to a decimal, and find its logarithm, as in Art. 24; or, since a fraction is an expression of division, subtract the logarithm of the denominator from the logarithm of the numerator, and the difference will be the logarithm of the fraction (Art. 10). Thus, the logarithm of f, or.75, is 1.875061. Or, gTRIGONOMETRY. the logarithm of the numerator, 3, is 0.477121 " " " denominator, 4, " 0.602060 " " " fraction, " 1.875061 TO FIND THE NUMBER CORRESPONDING TO ANY LOGARITHM, 26. When the given logarithm is WITHIN the limits of the table. Look in the column headed 0, for the first two or three figures of the logarithm, neglecting the characteristic, and, if all the figures of the logarithm are found in that column, the corresponding number will be on the same horizontal line, in the column headed N. If, however, the decimal part of the logarithm be not exactly found in the column headed 0, look for it in one of the nine following columns, and the first three figures of the corresponding number will be on the same horizontal line in the column headed.N, and the fourth will be at the head of the column in which the logarithm was found. Make the number correspond with the characteristic given, if necessary, by pointing off decimals or annexing ciphers (Arts. 7 and 8). Thus, the number corresponding to the logarithm 3.146128 is 1400; c "it 4 '0.370143 " 2.345; " " " 2.907680 ".08085. 27. When the given logarithm is NOT WITHIN the limits of the table. From the decimal part of the given logarithm subtract the decimal part of that next less; annex to their difference two or more ciphers, and divide by the number, in the column headed D, opposite the decimal part taken from the table. Annex the result to the number corresponding to the lesser logarithm, and point off according to the characteristic, as before. It sometimes happens, in dividing by the tabular difference, that there are not as many figures in the quotient as there are ciphers annexed to the dividend. In such a case, supply the deficiency, as in the division of decimals, by prefixing a cipher or ciphers to the quotient before annexing. BOOK I. 9 This process, like its converse (Art. 23), is based upon the supposition that the differences of logarithms are proportional to the differences of their corresponding numbers. NOTE. The number corresponding to a given logarithm, when obtained by the use of a table calculated to six decimal places, is reliable only to the sixth figure, and sometimes that figure of an answer is not strictly correct. Let it be required to find the number corresponding to the logarithm 2.633356. The dec. part of the given log. is.633356 " " log. next less is.633266, correspon. num., 4298 Their difference is 90.00 -~- ~ 89 Difference from column D is 101 Logarithm 2.633356 has for its corresponding number 429.889 The number corresponding to the logarithm 3.441049 is 2760.89 6" " " 2.497935 is.0314728 " " " c " 2.436811 is 273.408 ARITHMETICAL COMPLEMENT. 28. The arithmetical complement of any logarithm is the difference between it and 10. Thus, the arithmetical complement of the logarithm of 41, is 10-log 41 = 10-1.612784= 8.387216. 29. The arithmetical complement of a logarithm may be readily found, from the table, by subtracting each figure of the logarithm from 9, excepting the last significant figure at the right, which must be taken from 10; for this is equivalent to subtracting the logarithm from 10. 30. 7Te difference of two logarithms is equal to the sum of the logarithm to be diminished and the arithmetical complement of the other, less 10. For let a be any logarithm, and b a logarithm to be subtracted from it; then their difference will be a - b. Now the arithmetical complement of b is 10-b; adding 10-b to a, we have a + 10-b; subtracting 10, we have 2 10 'TRIGONOMET'RY. a+ 0-l-b-10- a-b; the same result as before. 31. Hence, since an arithmetical complement added makes the result 10 too great, a corresponding allowance must be made in any operation in which arithmetical complements of logarithms are used. MULTIPLICATION BY LOGARITHMS. 32. Add the logarithms of the factors; and the sum will be the logarithm of their product (Art. 9). The term sum, here used, is to be understood in its algebraic signification. Therefore, since the characteristic alone of a logarithm is negative (Art. 15), whatever there is to be carried from the decimal part, in the operation, must either be added to a positive characteristic, or subtracted from one that is negative. Also, when the characteristics of the logarithms are not either all positive or all negative, the difference between their sums must be taken, and the sign of the greater prefixed. Ex. 1. Multiply 120, 101, and.015 together. Log 120 2.079181 " 101 - 2.004321 ".015 - 2.176091 Product = 181.8..... 2.259593 Here, the 2 cancels a positive 2, so we have but 2 to set down. 2. Multiply 3.26 by.0085. Ans..02771. 3. Multiply 6651 by 108. Ans. 718308. DIVISION BY LOGARITHMS. 33. Subtract the logarithm of the divisor from the logarithm of -the dividend, and the difference will be the logarithm of their quotient (Art. 10). Or, Add the arithmetical complement of the logarithm of the divisor to the logarithm of the dividend, and the sum, less 10, will be the logarithm of the quotient (Art. 31). The term difference, here used, is to be understood in its BOOK I. 11 algebraic signification. Therefore, the sign of the characteristic of the divisor must be changed; and then, if the characteristics of the divisor and dividend have the same sign, their sum must be taken, but when of different signs, their difference, with the sign of the greater, for the characteristic of the logarithm of the quotient. Also, if 1 is carried from the decimal part, it must be regarded as positive, and jnust be united with the characteristic of the divisor before it is changed. Ex. 1. Divide 850 by.093. FIRST OPERATION. SECOND OPERATION. Log 850 = 2.929419 Log 850 = 2.929419 ".093 - 2.968483 ".093 ar. co.= 11.031517 Quot. 9139.8..... 3.960936 Quot. 9139.8...3.960936 Here, in the first operation, 1 carried from the decimal part to the 2 changes it to 1, which being taken from 2, leaves 3 to set down; and, in the second operation, 10 is taken from the sum of the characteristics (Art. 31). 2. Divide 2625 by 125. Ans. 21. 3. Divide.02771 by.0085. Ans. 3.26. 4. Divide 117.1347 by 5.062. Ans. 23.14. 5. Find the 4th term of the proportion, 219: 63.05:: 378. Log 378 - 2.577492 " 63.05 - 1.799685 " 219 ar. co. - 7.659556 4th term = 108.826 2.036733 6. Find the 4th term of the proportion, 720: 196:: 155.5. Ans. 42.33. INVOLUTION BY LOGARITHMS. 34. Multiply the logarithm of the given number by the exponent of the power to which the number is to be raised; and the product will be the logarithm of the required power (Art 11). Since the exponent of any power is positive, a negative char 12 TRIGONOMETRY. acteristic multiplied by it will give a negative result; but that which is to be carried from the decimal part will be positive; therefore, their difference will be the characteristic of the product. Ex. 1. Required the square, or second power, of 31. Log 31 - 1.491362 2 Ans. 961 2.982724 2. Required the cube, or third power, of.25. Log.25 = 1.397940 3 Ans. 0.015625 2.193820 8. Required the tenth power of.64. Ans..0115292. EVOLUTION BY LOGARITHMS. 35. Divide the logarithm of the given number by the index of the root; and the quotient will be the logarithm of the required root (Art. 12). When the characteristic of the logarithm is negative, and does not contain the given divisor without a remainder, we may increase the characteristic by any number that will make it exactly divisible, provided we prefix an equal positive number to the decimal part of the logarithm. Ex. 1. Required the square, or second root, of 1296. Log 1296 - 3.112605 (Log 1296)-. 2 1.556303 Ans. 36. 2. Required the cube, or third root, of.00048. Log.00048 = 4.681241 (Log.00048) - 3 2.893747 Ans..078297. Here, the negative characteristic 4 not being exactly divisible by 3, it is increased by 2 to make it so, and then the 2 borrowed is restored, by regarding 2 as prefixed to the decimal part. j 3. Required the fourth root of.434296. Ans..811794. 4. What is the tenth root of 2? Ans. 1.0718. BOOK II. PLANE TRIGONOMETRY. DEFINITIONS AND ELEMENTARY PRINCIPLES. 36. TRIGONOMETRY is the science which treats of methods of computing angles and triangles. 37. PLANE TRIGONOMETRY treats of methods of computing plane angles and triangles. 38. The MAGNITUDE OF ANGLES is represented by numbers expressing how many times they contain a certain angle fixed upon as the unit of angular measure. For this purpose a right angle is generally divided into 90 equal parts called degrees, each degree into 60 equal parts called minutes, each minute into 60 equal parts called seconds; then an angle is expressed by the number of degrees, minutes, seconds, and decimal parts of a second, which it contains. 39. Degrees, minutes, and seconds, are marked by the symbols ~, ', /; thus, to represent 16 degrees, 9 minutes, 23.5 seconds, we write 16~ 9' 23".5. 40. Since angles at the centre of a circle are to each other as the arcs of the circumference intercepted between their sides (Geom., Prop. XVII. Bk. III.), these arcs may be regarded a:s the measures of the angles, and the number of units of arc intercepted on the circumference may be used to express both the arc and the corresponding angle. 41. A degree of arc is zeu of a circumference; a minute, j^, of a degree; a second, -7l of a minute; and these arcs subtend angles of a degree, a minute, and a second, respectively, at the centre. 2* 14 TRIGONOMETRY. 42. For simplifying calculations, the radius employed in measuring angles, being constant, is taken at an assumed value of unity, as the linear unit of measure. 43. Since the value of the constant ratio of the circumference to the diameter of a circle, represented by n, is 3.14159 (Geom., Prop. XV. Sch. 1, Bk. VI.), if the radius of a circle is denoted by r, its circumference is 2 nr, where n = 3.14159. Hence, as r is taken as unity, any number of degrees may be expressed as a multiple or fractional part of n. Thus 360~ - 2 n, 180~ =n, 900 =f, and 30~0. 44. The COMPLEMENT OF AN ANGLE, or arc, is the remainder obtained by subtracting the angle or arc from 90~. Thus the complement of 45~ is 45~, and the complement of 31~ is 59~. When an angle, or arc, is greater than 90~, its complement is negative. Thus the complement of 127~ is - 37~. Since the two acute angles of a right-angled triangle are together equal to a right angle, they are complements one of the other. 45. The SUPPLEMENT OF AN ANGLE, or are, is the remainder obtained by subtracting the angle or arc from 180~. Thus the supplement of 110~ is 70~. When the angle is greater than 180~, its supplement is negative. Thus the supplement of 200~ is-20~. Since the three angles of any triangle are together equal to two right angles, any one of them is a supplement of the sum of the other two. TRIGONOMETRIC FUNCTIONS. 46. TRIGONOMETRIC FUNCTIONS are the quantities by which angles are subjected to computation. These we shall consider, in accordance with the best modern authorities, as ratios formed by comparing the sides of a rightangled triangle, and thus capable of comparison one with another by means of their geometrical properties. These ratios have received the special names of sine, tangent, secant, cosine, cotangent, and cosecant. There are also sometimes employed the quantities termed versed fine, coversed sine, and suversed sine. BOOK 11. -1 P. o 47. The SINE Of an angle is the ratio of the op'posite side to the liypothenuse. B Thus, in any right-angled triangle, A B 0', if the sides be denoted by p, b, h, we shallP have, A b 48.. The TANGENT of an angle is the ratio of the opposite side to thea~jcent side. Thus, tan A - -P tan B —b (2) 49. The SECANT of an angle is the ratio of the hypothenuse to the ad'jacent side. Thus, seecA~ seecB=:- 50. The COSINE, COTANGENT, and COSECANT of an angle are respectively the SINE, TANGENT, and SECANT of its complement. Hence, since the acute angles of a right-angled triangle are complements one of the other (Art. 44), we have, according to the definitions, cos A= sin B — Cos B==sin A-lu h' cot A =tan B- cot B =tai A- P (4) A b cosec A = sec B-, cosec B = ec A A o _p p b hA 51. Since-is the reciprocal of -P fb o, and )K of 6, we see that the' cosecant, cotangent, and cosine of an angle are respectively the reciprocals of the sine, tangent, and secant of the angle. That is, cos A= cot A 1 oe A- 1= sec A' tan A coe sin A' ci i osec A ta A cot A e cos A' isin A cosec A = 1, cos A4 sec A-i 1 tan A cot A = 1.J 16 TRIGONOMETRY. 52. If the cosine of A be subtracted from unity, the remainder is called the versed sine of A; if the sine of A be subtracted from unity, the remainder is called the coversed sine of A; an:l if thel cosine of A be added to unity, the sum is called the suversed sine of A. Hence, vers A 1 -cos A, covers A 1 sin A,! (6) suvers A - 1 + cos A. 53. The values of trigonometric ratios remain the same so long as the angle continues the same. Let BA C be any angle; in A B take any point, B, and draw BC perpendicular to A C; also take any other point, B', and draw B' C' perpendicular to A C. Then, since B' B the triangles A B C, A B' C' are similar, their sides have to one another the same ratio (Geom., Art. 210), and therefore sin A, tan A, A C- &c. will have the same values, whether A B C or A B' C' be the triangle by the sides of which they are expressed. It is also evident that their values would change with a change. of the angle. Hence, The. trigonometric ratios determine the angles, and conversely; that is, any determinate values being given for the one, determinate values can be found for the other. 54. The terms sine, tangent, secant, &c., were formerly * considered to befunctions of an arc, and denoted certain trigonometric lines. Thus, let 0 be the centre of any circle, A A" its diameter, and ABany arc; draw the radius 0 A' at right angles to AA", and draw tan- At cot. T' gents to the circle at the points A D' and A'; produce OB to meet the first tangent in T and the second tan- /, s. A gent in T'; draw B D perpendicular "\ to OA, and B D' perpendicular to OA '. Then, by the old definitions, the lines of the figure are considered to * " The modern method has now completely superseded the ancient method in English works." — Todunter's Trigonometry, p. 49. BOOK IL I7 - be the functions of the arc A B. BD is the sine of the arc A B, 0 D its cosine, A T its tangent, A' T' its cotangent, 0 T its secant, 0 T' its cosecant, A D its versed sine, A' D' its coversed sine, and A "ID its suversed sine. Also the line joining A and B is the chord of the arc A B. That is, in the circle whose radius is unity;The SINE of an arc, or of the angle measured by that arc, is the perpendicular let fall from one extremity of the arc, upon the diameter passing through the other extremity. The COSINE is the distance from the centre to the foot of the sine. The trigonometric TANGENT is that part of the tangent touching. one extremity of the arc, which is intercepted between that extremity and the radius produced passing through the other extremity. The SECANT is that part of the radius produced which is intercepted between the centre and the tangent. The VERSED SINE is that part of the diameter intercepted between the foot of the sine and the origin of the arc. The COTANGENT, COSECANT, and COVERSED SINE are tangent, secant, and versed sine, respectively, of the complement of an arc or angle. The cosine is also equal to the sine of the complement, as OD D'B. The SUVERSED SINE is that part of the diameter which remains after taking away the versed sine, or it is the versed sine of the supplement. 55. If the radius of the circle be unity, the numerical value of the sine and other trigonometric functions is the same in both the old and new systems, for sin A O B B sin A B =BD. But 0 B is the radius of the circle, and denoting it by r, we have sin A B sin A B = sin A 0B X r, sin A B s - B; r and making radius = 1, we have sin A B - sin A 0 B. (7) In like manner it may be shown, that similar results hold for 18 TRIGONOMETRY. all the other trigonometric functions. Hence any formula expressed in the old system may be immediately converted into a formula expressed in the new system, by supposing the radius of the circle to be equal to unity. 56. The sine, cosine, tangent, and cotangent constitute the primary class of trigonometric ratios, as they are by far most frequently used; and the others form a subordinate class, the employment of which is occasionally attended with convenience. They are collected, for more ready reference, in the following TABLE. 1. sin A - b 2. cos A - A 3. tan AJ - 4. cot A = - P h 5. sec A — 6. cosec A = - P 7.- 2 vers A - 1 - 8. - covers A = 1- 9. vers 9.t 4uvers A = 1 + 1. sin A. cosec A 2. cos A = sec A 3. tanA 1 cot A 4. cotA 1 tan A 5. sec A - cos A 6. cosec A = 1 sin A 7. vers A - 1- cos A 8. covers A = 1 - sin A 9. suvers A = 1 - cos A OTHER RELATIONS BETWEEN TRIGONOMETRIC FUNCTIONS OF THE SAME ANGLE. 57. To find the COSINE of an angle by means of its sine. From the right-angled triangle ABC B (Geom., Prop. XI. Bk. IV.) we have h d b s o h2 e h. Dividing both sides of the equation by hs A b 0 BOOK IL. gives 1,+ b or, by definition (Art. 47, 50)., sin' A +cos,2A=1 8 therefore co A = 1 - sin2 A, (9) and cosA =VI -sin 2A (10) in which "sin 2A" denotes "1the square of the sine of A." 58. To find the SINE, of an angle by means of its cosine. Since, by (8), sin2 A ~ cos' A 1 sin2A=1Cos2 A, (11) and sin A = -COS2 A. (12) 59. To find the TANGENT and COTANGENT of an angle by means of the sine and cosine. By (2) we have tan A p and sin p _b p Cos k b' therefore tan A -'in A (13). cos A Then, by Art. 51, cotan A - cos A (4 sin A'(4 60. To find the SECANT and COSECANT of an angrie by means of the tangent. From the right-angled triangle A B C B we have Dividing both sides of the equation by MyA -, gives AllbPC or, by definitions (Art. 48, 49), seeA = 1 +tan" A. (15) 20 TRIGONOMETRY. Again, since h2 = p2 +b19, oQr, by definitions (Art. 50), coseC2'A= 1 + cot2 A.' (16) 61. In general, any one of the six trigonometric ratios of an angle being given, the relations expressed by the foregoing formulae will enable us to find the value of all the rest. These are termed the elementary frrmulce, and are collected in the following TABLE. 1. sin 2A + cos2 A 1 2. sin2A - 1- cos2A. 3. cos2 A = 1 sin2 A. tn -sin A ~ A cos A ta -cos A 5. ctA=sin A -6. sec2A=1+ tan 2A. 7. cosec2A=1~ +cot2A., RELATIONS BETWEEN TRIGONOMETRIC FUNCTIONS 'OF DIFFERENT ANGLES. 62. To find tke SINE, and COSINE of the sum of two ang les b means of their -sines and cosines. Let the two angles be CO0D, DOE f. In the line O E take any point, E, draw EF perpendicular to '. O C, and -ED perpendicular to O D.E Draw D G perpendicular to EF, and D C perpendicular to OC. The trian-. gles G E D and CO0D have their sides perpendicular, hence they are similar (Geom., Prop. XXV. Bk. IV.), and the ____ angles D E G and CO0D are equal. 0 P Let a=COD =DEG, and b-=DOER Then a + b=COE, EF GF+EG' DC EG and sin (a + i)= —E= OEk —! —E+ BOOK H. 21 DC or, substituting for the ratios of which it is formed, D C OD OD X O =sina cosl E G and in like manner, for EG ED E XOE Cos a sin 1b, we have sin (a +b) =sina cos b +cos asin b. (17) Again, cos (a +b)=? _00 D or, substituting for 2 2 the ratios of which it is formed, OE 00 OD co~s a cos DG and in like manner, for U DC DE sin a sin b DEX OE we have cos (a +) =cosa cos b - sin asin b. (18) 63. Yo find the SINE, and cosiNE: of thC DIFFERENCE Of tWO angles b~y means of their si~nes and cosines..Let the two angles be F CD, D CEB. In the line C E take any point, E, draw '" E F perpendicular to O F, and E D f perpendicular to C D. Draw D C perpendicular to FE produced, and D C perpendicular to C F. The triangles C BED and C C D have their sides perpendicular, hence they are similar. - (Geom., Prop. XXV. Bk. IV.), and the angles D B C and COCD are equal. Let a= CODDE6!, and b~=DOE. Then a-b=FOE, 3 22 *TRIGONOMETRY. U 2k an sn(a- ) E F GF- GE DC- GE and sn~a-b=~1E_ OE 0E 0E' or, substituting for ' _the ratios of which it is formed, - sin a cosb G E and in like manner, for-0E GE ED E D _U__ -cos a sinb we have sin (a - ) = sin a cos b -cos a sin b. (19) Again, cos (a -b)= OF OE+ F-OC+D or, substituting for ~2E the ratios of which it is formed, 00 OD DXCO ~7 oa Cos b, and in like manner, for D0 E -xsin a sn6 we have cos (a - b) =cos a cos b~ sin a sinl'. (20) 64. The four formula3 last established apply to arcs as well as angles, and may be considered the fundamental formuke of subsequent analysis. They are brought together in the following TABLE. 1. sin (a +b) = sin a cos b ~ cos a sin 6. 2. cos (a +b) =cos acos b-sin a sin b. 8. sin (a-b) = sin acos b-cos asin b. 4. cos (a-b = cos a cos b+ sin a sin 6. BOOK II. 23 SIGNS OF THE TRIGONOMETRIC FUNCTIONS. 65. If on any line, straight or curved, different distances be measured from a fixed point of origin, the distances which have contrary situations may by convention be introduced into our calculations, by affecting the quantities representing their magnitudes by contrary signs. Let O be a fixed point in any line, A A", and suppose we have to determine the situations of other points in this line with respect to O. The position of any point in the line will be known if we know the distance of the point from 0, and also know on which side of O the O point lies. Now it is found convenient A A to adopt the following convention: distances measured in one direction from O along the line will be denoted by positive quantities, and distances measured in the contrary direction from O will be denoted by negative quantities. Thus, for example, suppose that distances measured from O towards the right are denoted by positive numbers, and let A be a point, the distance of which from O is denoted by 2 or + 2; then if A' be a point situated just as far to the left of 0:as A is to the right, the distance of A" from 0 will be denoted bly -2. In like manner, if distances originating in A A", and taken along 0A', or only parallel to OA', when measured upwards be denoted by positive quantities, on being measured downwards they will be denoted by negative quantities. 66. A similar convention may con- Ai veniently be adopted with respect to angles. Let any line, 0 B, revolve from the A position OA, in one direction round 0, 0 forming the angle B O A, and let this angle be denoted by a positive quantity; then, if the line OB revolve, A 24 TRIGONOMETRY. from the position 0A, round 0 in the contrary direction, form. ing the angle B 0 A, this angle may be denoted by a negative quantity. Thus, for example, if each of the angles A OB and A O B is two ninths of a right angle, and we denote the former by 20~ or -+20~, the latter may be denoted by -20~. The direction of the positive distances is quite indifferent; but, being once fixed, the negative distances must lie in the contrary direction. 67. The representation of angles as the measure of the revolution of a line, turning in a plane about one of its own points, leads to the consideration of angles, not only greater than two right angles, but of all possible magnitudes. Thus, when the line 0 B, starting from the initial position 0 A, has passed A", or made more than half a revolution, we have described an angular magnitude of more than 180~; and when it has passed on to A, we have an angular magnitude of 360~. If it now contin- BA ues to revolve in the same direction till A < A it arrives again at B, we have an an- ' gular magnitude of 360~+ 20~= 380~, and thus, we may conceive of angles of all magnitudes. In like manner negative angles of all magnitudes may be formed by the describing line 0 B revolving from O A, but in a contrary direction. 68. The algebraic signs of the trigonometric functions can be readily fixed in the mind by being represented geometrically. Thus, B 3 Let the extremity of a revolving line, starting from the initial position OA, describe the positive arc - A A B less than 90~, A Bi between Dl 0 D 90~ and 180~, AA'B' between 180~ and 270~, and AA'A"B" between 270~ and 360~. Then, according to B the definitions of Art. 54, BD, B'D', BOOK II. 25 BI'yD, and B"'D are the sines, and O D and O D'are the cosines, of the angles measured by the arcs terminating in each of the four quadrants. As all the functions of an angle less than 900 are considered positive, their direction will fix the signs for other quadrants. It will be seen (Art. 65) that the sines are above the diameter A A", and positive, in the first and second quadrants, but below, and negative, in the third and fourth quadrants. Also (Art. 65), the cosines are to the right of the diameter A' A"', and positive, in the first and fourth quadrants, but to the left, and negative, in the second and third quadrants. sin A As tan A = A (13), the tangent must be positive when the sine and cosine have the same sign, and negative when they have unlike signs. Hence the tangents are positive in the first and third quadrants, and negative in the second and fourth quadrants, a result which may also be obtained by noticing whether the tangents must run above or below the origin A, to meet the secant. The cotangent of any angle or arc always has the same algebraic sign as its tangent, the secant the same as the cosine, and the cosecant the same as the sine; for they are reciprocals (Art, 51). The versed sine, coversed sine, and suversed sine, since they are referred to the origins of their arcs, A, A', and A", as fixed points, instead of the centre 0, can have but one direction, and therefore are always positive. By comparing the sine B"'D and the cosine 0D of the negative. arc A B1" (Art. 66) with those of the equal positive arc A B, it will be seen that the cosines are identical, and consequently the secants are equal; but the sines, and, consequently, the tangents, cotangents, and cosecants, have unlike algebraic signs. The functions of the arc A B1", terminating in the first negative quadrant, are the same as those of the arc A A'A 'Be", terminating in the fourth positive quadrant. The second negative and third positive, the third negative and second positive, and the fourth negative and first positive quadrants likewise have functions with the same algebraic signs. 23 2~~~3 ~TRIGONOMETRY. 69. From the results above obtained is formed the following TABLE. Sie.Coin.iagetive't atNegative Quadant. Sine Cosne. anget. Ctan' Secnt.Cosecant. Quadrant. First. + + + + + + Fourth. Second. + - - - + Third. Third. - - + + - - Second. Fourth. - + - - + - First. VALUES OF THE TRIGONOMETRIC FUNCTIONS OF CERTAIN ANGWLES. 70. The definitions of the trigonometric functions already given (Art. 46-50) apply directly only to angles not exceeding a right angle. But by means of the formulaw which have been deduced from them we may now extend the definitions, so as to render them applicable to angles of any magnitude. 71. To find the SINE, S5'c. of 300 and of 600 Let A BC be an equilateral triangle, then B each of its angles equals one third of two right angles, or 600. Draw BD perpendicular to A C, then the angle ABD is equal to j A BC, A D isequaltoDC,and AD==j.BC==kAB. Therefore, A D C sin ABD= AD ~JAB_ AB AB or, since 300 and 600 are complements the one of the other, sin 300= cos 60o=' (21) whence by (10) cos 30' = sin 600= VI - = V3. (22) Then, by (13) and (5), tan 300 =cot 600" __ 1 (23) BOOK 11. 27 cot 30%=tan 600= V3:V3, (24) 12 cosec 30' sec 600= ~ 2. (26) 72. To find the SINE, ~5c. of 450* Since 450 is the complement of 450, sin 450 -- cos 450. Then making A = 450 in (8), we have sin-' 450 + cos2 450 2 sin-' 450 - 2 cos2 450 19 or sin,2 450 = cos2 450 J sin 450 cos 450 = Vj V2. (27) Hence by (13) and (5), tan 450 z=cot 45,2 1 (28) sec 450 = cosec 450 = V2. (29) 73. To find the SINE, c~c. of 00 and of 900. Since 00 and 9Q0 are complements the one of the other, sin 00 -- cos 90g. Then making a = b in (19) and (20), we have sin 00 =cos 900 =sin a cos a- cos a sin a=0, (30) Cos 00 sin 90' =Cos a Cos a ~ sin a sin a, or by (8), cos 00 = sin 900 =cos2 a ~ sin2 a -. (81) Hence by (13) and (5), tan 0 = cot 90' =.0, (32) cot 0 = tan 90' =~ ~ (33) sec 0= cosec 90' = +1, (34) Cosec 00 = sec 900 = =.() 28 28 ~~~~TRIGONOMETRYf. 74. To find the SINE, ~c. of 1800 Let a = b = 900 in (17) and (18); then, by means of (30) and (3 1), we have sin180 1 XO0+ 0X 1=O0, (36) cosl180' 0OXO0-1IX 1= -1. (37) Hence, by (13) and (5), 01 tan 80 -1 01 cot 1800 =-=E (38) sec 180' J = — 1, cosec 1800 = =. (39) 75. Tofinkd the SINE,!yc. of 270'. Let a = 18O0 and b =900 in (17) and (18), and we have sin 2700 = O X 0~+ (- 1) X 1 = - 1, (40) cos 270' = (- 1) X 0 -O0X 1 =0. (41) Hence, by (13) and (5), 0 -1 __0 tan 270 = 0=, cot 2700- 09 (42) sec 2700 =, cosec 270' - -1. (43) 76. To find the SINE, 4'C. of 3 600. Let a = b = 1800 in (17) and (18), and we have sin 3600 = O X (- 1) +(- 1) X 0 =0, (44) cos 3600 = (- 1) X (- 1) -O0X 0 1. (45) But these values for the sine and cosine of 3600 are the same as those for the sine and cosine of 0'. Hence, All the trigonometric functions' of 3 600 are the same as those (Or 00. 77. To find the SINE, 4Jc. of the supplement of an angle. Let a = 1800 in (19) and (20); then, by means of (36) and (37), we have sin (1800 -6b) = sin b, cos (1 800 - b) = cos b (46) BOOK H. 29 whence, by (13) and (5), sin b tan (180~ - ) = s b -tan, (7) -cos b (47) cot (180~ - ) -t b = -cot, (48) sec (180~ - b) = -- -.sec, (49) cosec (180~- b) = ib - = cosec b; (50) that is, the sine and cosecant of the supplement of an angle are the same as those of the angle itself; and the cosine, tangent, cotangent, and secant are the negatives of those of the angle. 78. It follows from the preceding article, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant are negative, as has before been shown geometrically (Art. 68, 69). 79. Tofind the SINE, Ryc. of a negative angle. Let a = 0~ in (19) and (20); then, by means of (30), (31), (13), and (5), we have sin (- b) = -sin b, cos (-b)- cosb, (51) tan (- b) - -tan b, cot (- b) = -cot b, (52) sec (- b) = sec b, cosec (- b) -cosec b; (53) that is, the cosine and secant of the negative of an angle are the same as those of the angle itself; and the sine, tangent, cotangent, and cosecant of the negative of an angle are the negatives of those of the angle. These results correspond with those obtained geometrically (Art. 68). 80. Tofind the SINE, Rc. of an angle which exceeds 180~. Let a = 180~ in (17) and (18); then, by means of (36) and (37), we have sin (180~+ b) = -sin b, cos (180~+ b) - -cos b, (54) tan (180~ + b) = tan b, cot (180~+ b) = cot 6, (55) sec (180~+ b) = -see b, cosec (180~+ b) = -cosec b; (56) TRIGONOMETRY. that is, the tangent and cotangent of an angle which exceeds 180C are equal to those of its excess above 180~; and the sine, cosine, secant, and cosecant of this angle are the negatives of those of its excess. 81. It follows from the preceding article, that the tangent and cotangent of an angle which exceeds 180~ and is less than 270' are positive; while its sine, cosine, secant, and cosecant are negative. Also, by considering b greater than 90~ (Art. 78), that the cosine and secant of an angle which exceeds 270~ and is less than 360~ are positive; while its sine, tangent, cotangent, and cosecant are negative. (See Art. 68, 69.) 82. To find the SINE, Sc. of an angle which exceeds 360~. Let a = 360~ in (17) and (18); then, by means of (44) and (45), we have sin (360~ + b) = sin b, cos (360~ + b) = cos b; (57) that is, all the trigonometric functions of an angle which exceeds 360~ are the same as those of the excess above 360~, so that 360~ may be suppressed as often as it can be, so far as the function of the angle is concerned. 83. The trigonometric functions of any angle whatever can now be readily expressed in those of an angle not exceeding 90~. Thus, 1. The trigonometric functions of any negative angle can be made to depend upon those of the corresponding positive angle (Art. 79). 2. Any angle exceeding 360~, as far as the trigonometric functions are concerned, may be replaced by an angle less than 860~ (Art. 82). 3. Any angle exceeding 180~ can in like manner be replaced by an angle less than 180~ (Art. 80). 4. The trigonometric functions of any angle exceeding 90~ may be made to depend upon those of an angle less than 90~ (Art 77, 78). For example, BOOK IL. 31 sin 600~- sin (3600 +240~) - sin 240~ = sin (180~ + 60~) -- sin 60~, tan (-1000~) = -tan 1000~ -tan (720~+ 280~) = -tan 280~ = -tan (180~ + 100~) = -tan 100~ = -tan (180~ - 80~) = tan 80~. 84. It will be seen from the preceding articles that the sine and cosine may have any value between -1 and +1; the tangent and cotangent, any value between -oo and +-~; the secant and cosecant, any value between -oo and -1 and between + 1 and +oo. It might also be shown that the versed sine, coversed sine, and suversed sine may have any value between 0 and 2. It will also be found that no trigonometric function changes its sign except when it passes through the value zero or the value infinity. 85. The values of the functions of 0~, 90~, 180~, 270', and 360~ can be readily recalled, by being represented geometrically. according to the definitions of Art. 54. sin 0~- 0, cos 0~ =OA = R 1, tan 0~- 0, and sec 0= OA= 1; sin 90~ - O' = 1, cos 90~ 0; sin 180~= 0, cos 180~- OA" - 1; sin 270~- OA' --- -1, cos 270~= 0. The tangent for either 90~ or 270~ would be a line drawn through A parallel to the secant, which would be A'A"' prolonged, and as they are Thus, A A 0 AJ' each to be limited by their mutual intersection, they must both be infinite. The cotangent of either 0~ or 180~ would be an infinite line drawn through A', and parallel to the cosecant, which would be AA" infinitely prolonged. vers 0~ = 0, vers 90~ = vers 270~ = A 0 - 1, vers 180~ = AA" = 2; covers 0~ = covers 180~ - Al'0 = 1, covers 90~ = 0, covers 270~ = A'A"' - 2; suvers ~ -= A"A = 2, suvers 90~ = suvers 270~ = A" -- I, suvers 1800 = 0. 82 TRIGONOMETRY. TABLE. Degrees. Sin'e. Cosine. Tangent. Cotangent. Secant. Cosecant-. 0 0 +1 0 Go +1 so 90 +1 0 so 0 so +1 180 0 -1 0 Go -1 so 270 -1 0 Go 0 co -1 360 0 +1 0 so +1 00 GENERAL FORMULA~. 86. From the four fundamental formulae (Art. 64), a large number of other formulae of general utility may be deduced. 87. To.find expressions for the products of -sines and cosines, and for their sums and differences. The sum and difference of -equations (17) and (19) are Sin (a +b) ~ sin (ab -2 sixka cos b, (58) sin (a -~b) - sin (a -b)=2 cos a~n b; (59) and the sum and difference of ( 18) and. (20) are cos (a +b) +cos (a -b) = 2cos acosb (60) cos (a +b) -cos (a - b)=-2sitva-ehvb. (61) Now, if in the formulas 'we let a+b=A, and a-Ib=B, that is, a =,3 (A +B), and b=~ (A-B), we shall haive, sin A,~ sin B= 2 sin j (A +B) cos (A -B), (62) sin A- sin B= 2cos j (A +B) sink (A -B), (63) cosA ~ cosB= 2cos k (A~B) cos (A-B), (64) cos B- cos'A = 2sin j (A +B) sin~ (A -B), (65) in which A and B represent any two angles, and consequently admit of every possible value. These formulte are of frequent BOOK II. 88 application, especially in calculations effected by logarithms; (58), (59), (60), and (61) serve to transform a product to a sum or difference, and (62), (63), (64), and (65) serve to transform a sum or difference to a product. 88. Dividing formula (62) by (63), we have sin A + sin B -sin j(A + B) cos j (A - B) sin A - sin B - cos (A + B) sin (A - B)' which, by means of (13) and (5), becomes sinA --- in = tan I (A ~ B) cot. (A-_B), sin A -sin Bn2 or, sin A + sin B tan (A +B) 66 sin A - sin B -tanI (A - B)'(6 and by (14), sn ~ i~ oiA B sin A - sinaB -cotjI (A + B)'(7 that is, The sum of the sines of two angles is to their difference as ~the tangent of half the sum of the angles is to the tangent of half their diff'erence, or as the cotangent of half their diffe~rence is to the cotangent of half their sum. 89. By means of (62), (63), (64), and (13), we obtain, sin A + sin B 2 sin4(A +B3) cos (A = B) cos A + cos, B 2 cosj (A + B) cos j(A -B)=tn(+B,68 sin A - sin B 2 cosi(A +B).sini(A -B) csA + cos B 2 cos~ (A +B) cosj (A - B)-tn(A B,69 that is, The sum of the sines of two angles divided by the sum of their cosines is equal to. the tangent of half the sum Qf.-the angle.; and The diffrence of the sines of two angles divided by the sum of their cosines is equal to the tangent of half the differenice of-the angles. 90. To find the tangent and cotangent of the sum of two angles sn eans of their tangents. Let A and B be the two angles; then, by, (1 3), tan A + ) =sin (A + B) cos (A + B)' 4 34 34 ~~~TRIGONOMETRY. or, substituting for sin (A + B), cos (A + B), their values from (1 7) and (1 8), sin A cos B + cos A sin B tan (A + B) = cos A cos B - sin A sinB Dividing all the terms of the numerator and denominator by cos A cos B, we have sin A sin B cos A X cos B tan (A +B)- aA tn s-tn AtsinB' (0 and, by, (13), cot (A +B) 1- -tan Atan B 1 tanA+tanB (1 91. To find the tangent and cotangent of the deference of two angles by means of their tangents. By (1 3), tan ( - B) sin (A - B) tan A - ) ~ cos (A - B) ' then, in like manner as in Art. 90, we have, tan (A -B)~ tan A - tan B 2 1 + tan A tanB'(2 and cot (A-B)I=1 tan A tan B tan A- tanB 92. To find the SINE, 4~c. of double an angle, by means of the functions of the angle itself. In the expression for sin (A ~ B) and cos (A~ B),~ let B= A, then sin 2A =sin A cos A +sin A cos A, or, sin 2 A = 2sin A cos A; (74) and cos2A=cosA cosA-sinA sin A, or,. cos 2 A — 7 COS2 A - sin2 A 1 (75) BOOK II. 35 Substituting in the latter, first the value of cos' A and then the value of sin2 A, from (9) and (1 1), we have cos2A = 1 2 2sin'A, (76) cos 2A = 2 cos2A-1(77) By means of (70) and (71), we have tan2A - 12 tan A tan2 A - 1- tan' A" (78) __1 - tan' A cot 2 A -.nA (79) 93. To find the SINE, 4rc. of half an angle by means of the sine or cosine of the angqle itself. Let j A = A in (76) and (77), and transpose; then 2 sin2 A = 1-cos A, (80) 2 COS'AA =1I+cos A; (81) whence 11- CosA Co A 11+ cos A sin kA 2 ' cok V 2, (82) takA sin A I_ /-COSA tanj A cos A V + cosA' (83) By multiplying both numerator and denominator by VI -cos A, or by VI + cos A, we obtain (12), tan.jA - cos A __ sin A (4 tan*A= sin A + cos A' (4 NATURAL SINES AND COSINES. 94. NATURAL SINES, COSINES, &c. are the values, of the sines, cosines, &c. expressed in natural numbers. 95. A table containing these values is called a table of nat-. ural sines and cosines. 96. The semi-circumference of a circle whose radius is 1 is equal to 3.1415926 nearly (Geom., Prop. XV. Sch. 2, Bk. VI.), 86 TRIGONOMETRY. and this divided by 10800, the number of minutes in 180~, w.l1 give.0002908882 for the arc of 1', which may be taken al-o for the sine of an angle of 1'. By means of formula (10), cos 1' = /1 - sin2 1' =.9999999577. Then by transposition of formulae (58) and (60), sin (a + b) = 2 sin a cos b - sin (a - b), cos (a + b) = 2 cos a cos b - cos (a - b), in which, making b equal to 1', and a, in succession, equal 1; 2', 3', &c., we obtain for the sines, sin 2' - 2 sin 1' cos 1- sin 0' =.0005817764, sin 3' = 2 sin 2' cos 1' - sin 1' =.0008726646, sin 4' = 2 sin 3' cos 1' -- sin 2' =.0011635526, &c., &c., and for the cosines, cos 2' = 2 cos 1' cos 1' - cos 0' =.9999998308, cos 31 = 2 cos 2' cos 1' - cos 1' =.9999996193, cos 4' = 2 cos 3' cos 1' - cos 2' =.9999993232, &c., &c., thus obtaining the sines and cosines up to 45~. The tangents may readily be found by dividing the sines by the cosines (13); and the secants, cotangents, and cosecants by dividing 1 by the cosines, tangents, and sines, respectively (Art. 51). 97. The sines, tangents, and secants of angles greater than 45~ are respectively the cosines, cotangents, and cosecants of their complements, which are less than 45~; and, by their definitions, cosines, cotangents, and cosecants are the sines, tangents, and secants of complements (Art. 50). Thus, sin 46~ = cos (90~ - 46~) = cos 44~, tan 51~ = cot 39~, cos 50 -= sin 40, cot 88~ = tan 20. BOOK IL 37 Tables, therefore, do not go beyond 45~; or, rather, are so arranged that each number answers as a function of both an angle less than 45~ and its complement greater than 45~. TABLE OF LOGARITHMIC SINES, COSINES, &o. 98. A TABLE of LOGARITHMIC SINES, COSINES, &C. contairnv the logarithms of the numbers expressing the natural sine-, cosines, &c. 99. Since the sines and cosines are never greater than 1, and tangents likewise, when under 45~, their logarithms properly have negative characteristics. But to avoid the inconvenience of these, the characteristics are, by common consent, increased by 10. Thus the characteristic 9 is used in the place of -1, 8 in place of. -2, &c. The radius, therefore, of the logarithmic sines, cosines, &c. is, as arbitrarily assumed, 1010, or 10,000,000,000. 100. In the accompanying table the degrees are given at the top and bottom of the page, and the minutes in the columns at the sides designated by M. The column headed D contains the increase or decrease for 1 second. This is obtained by taking one sixtieth of the difference between the logarithmic sine, cosine, &c. of an angle or arc, and that next exceeding it by 1 minute. The result is placed against the lesser angle or arc. TO FIND THE LOGARITHMIC SINE, &C. OF ANY ANGLE OR ARC. 101. If the angle or arc is less than 45~, look for the degrees at the top of the table, and for the minutes on the left; then, opposite to the minutes, on the same horizontal line, and in the column headed Sine, will be found the logarithmic sine; in the column headed Cosine will' be found the logarithmic cosine, &c Thus, the logarithmic sine of 190..23' -is 9.520990, -... ( " " cosine of 31~ 47/ " 9.929442, (" "< tangent of 43~ 5 " 9.970922. 4. 38 TRIGONOMETRY. 102. If the angle or arc is between 45~ and 90~, look for the degrees at the bottom of the table, and for the minutes on the right; then, opposite to the minutes, and in the column designated at the bottom Sine, will be found the logarithmic sine; in the column designated at the bottom Cosine will be found the logarithmic cosine, &c. Thus, the logarithmic sine of 800 11' is 9.993594, " " cosine of 65~ 59' " 9.609597, i" " cotangent of 73~ 35' " 9.469280. 103. If the angle or arc is between 90~ and 180~, subtract it from 180~, and take the logarithmic sine, &c. of the remainder. Thus, the logarithmic sine of 112~ is the logarithmic sine of 68~. " "' tangent of 98~ " " tangent of 82~. 104. If the angle or arc is expressed in degrees, minutes, and seconds, find the logarithmic sine, &c. of the degrees and minutes as before; then multiply the number opposite, in the column headed D, by the seconds, and add the product to the number first found, for sines and tangents, but subtract it for cosines and cotangents. Thus, if the logarithmic sine of 30~ 25' 42" is required, The logarithmic sine of 30~ 25' is 9.704395 Tabular difference, 3.59 Number of seconds, 42 Product, 150.78 150.78 Logarithmic sine of 30~ 25' 42" is 9.704546 It is customary to omit the decimal figures at the right, but to increase the last figure retained, by 1, when the figure at the left of those omitted is 5 or greater than 5. 105. The secants and cosecants are not included in the table, since they may be readily derived from the cosines and sines. By (5), sec A cos A = 1, and log sec A - log cos A = 0; but as log sec and log cos are each increased by 10 (Art. 99), the second member of the equation must be increased by 20, that is, BOOK II. 39 logarithmic secant = 20 - logarithmic cosine. (n like manner, logarithmic cosecant = 20 - logarithmic sine. Hence, to find the logarithmic secant, subtract the logarithmic cosine from 20; and to find the logarithmic cosecant, subtract the logarithmic sine from 20. Thus, The logarithmic secant of 65~ 59' is 10.390403 i" " cosecant of 30~ 25' 42 " 10.295454. To FIND THE ANGLE OR ARC CORRESPONDING TO ANY LOGARITHMIC SINE, &C. 106. Look in the column designated by the same name with the given logarithm for the sine, &c. which is nearest to the given one, and if the name be at the head of the column, take the degrees at the top of the table, and the minutes on the left; but if the name be at the foot of the column, take the degrees at the bottom, and the minutes on the right. Thus, The angle or arc corresponding to the logarithmic sine 9.681443 is 28~ 42'. The angle or arc corresponding to the logarithmic tan 9.984079 is 43~ 57'. The angle or arc corresponding to the logarithmic cos 9.731603 is 570 23'. 107. If the given logarithmic sine, &c. is not found exactly, or very nearly, then, to find the seconds, subtract from the given logarithm that next less in the table, to the remainder annex two ciphers, divide the result by the number in the column headed D, and the quotient will be the number of seconds to be added to the degrees and minutes of the lesser logarithm for sines and tangents, or to be subtracted for cosines and cotangents. Thus, to find the angle or arc corresponding to the logarithmic sine 9.938070. Given log sine, 9.938070 Next less, 9.938040 corresponding angle, 60~ 7 Diff. from column D, 1.21)30.00 25" The log sine 9.938070 has for its cor. angle or arc, 600 71 25' 40 TRIGONOMETRT. The angle or arc corresponding - to the logarithmic tangent 9.497200 is 170 26' 33"1. The angle or arc correspondin-g to thfe logarithmic cosine 9.792477 is 510 40' 30"1. EXAMPLES. 1. Required the logarithmic sine of 280 42'. Ans. 9.681443. 2. Required the logarithmic cos ine of 590 331 47/i* Ans. 9.704657. 3.. Required the logarithmic cotangent of 1270 2'. Ans. 9.877640. 4. Required the logarithmic -sine of 810 20'. Ans. 9.995013. 6. Required the logarithmic secant of 510 40' 30". Ans. 10.207523. 6. Required the logarithmic tangent of 740 21' 20". Ans. 10.552778. 7. Required the logarithmic cosecant-of 10'20 24' 41". Ans. 10.010270. 8. Required the logarithmic tangent of 10 59, 5 1".8. Ans. 8.5425-87. 9. Required the angle o'f the logarithmic'sin'e 9.999969. Ans. 890 19'. 10. Reqiir&1- the arc of the- logarithmic tangent 9.645270. Ans. 230 50' 17". 11. Required the angle of the 'logarithmic cosine 9.598075'. Ans. 660 39'. 12. Required the angle of the logarithmic cotangent 10.301470. Ans. 260 32' 31". 13S. Required the arc- of the logarithmic sine 9.893410. Ans. 5 10 28' 40". 14. Required the angle -of the logarithmic cosine 9.421157. Ans. 1050 17' 29". 15. Required the arc of the logarithmic tangent 9.692 125. Ans. 260 12' 20"... 16. Required the angle of the. logarithmic cotangent 9.421901. Ans. 750 12' 6"1. BOOK III. SOLUTION OF PLANE TRIANGLES. 108. THE SOLUTION OF TRIANGLES is the process by which, when the values of a sufficient number of their elements are given, the values of the remaining elements are computed. The elements of every triangle are the three sides and the three angles. Three of these elements must be given, one of which must be a side, in order to solve a plane triangle. The solution of plane triangles depends upon the following FUNDAMENTAL PROPOSITIONS. 109. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the hypothenuse into the sine of the -angle; and the side adjacent to an acute angle is equal to the product of the hypothenuse into the cosine of the angle. Let A B C be a triangle having a right angle at C; then, by (1), b * sin A=-, sin B - —; A-b Co h Xh' A b. C' therefore p h sin A, b== h sin B. (85) Again, by (4), cos J-, cos B=; - therefore b=h cos A, p= h cos B. (86) 110. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the other side into the tangent of the angle; and the side adjacent to an acute angle is equal to the product of the other side into the cotangent of the angle. 42 42 ~~~~TRIGONOMETRY. For, by (2), tan A - - tan B- = I b -p tan B. therefore p, = b tan A, (87) _ _ b Again, by (4), cot -A — therefore b =p cot A, b7)p = bcot B. (88) sides are proportional to the 111. In an~y plane triangle, the sines of th~e opposite angles. B Let A B C be any triangle, in which the sides opposite the angles A, B, C, respectively, are denoted by a, b, and c. c From one of the angles, as B, draw BD perpendicular to the opposite side A C, and denote the line B D by p. Then the A D b right-angled triangles, CIBD, A B D, give, by (85), p ==-asinC0, p==csin A; whence, a sin C = c sin A, which gives the proportion a c::sin A sin C. In like manner it may be proved that a:b::sin A: sin B, c:b::sin C: sin B, aid these three proportions give a: 1': c::sin A: sin B: sin C, which may also be written a b C sin A - sin B - sin C' (89) (90) (91) (92) (93) The angle C was acute, but had it been obtuse, or a right angle, the results would have. been the same. The proposition, therefore, applies in every case. . BOOK M. 48 1 12. In 'any plane triangle, the sum of any two sides is to their d~jference as the tangent of half the sum of the opposite angles is to the tangent. of half their difference. For, by (90), a:b::sin A:sin B; whence (Geom., Prop. X1I. Bk. II.), a + b: a - b: sin A ~ sin B: sin A sin B which may also be written, a + b -sin A + sin B a -b sin A - sin B But, by formula (66), sin A 4- sin B __tan j(A 4-B). sin A - sin B3- tan j(A - B)' thereforea ~ b -tan 4(A +B) a -. b -tan 4(A - B)' (4 or, as it may be written, 113. lIn any triangle, the square of any side is equal to the sum of the squares of the two other sides, diminished by twice the rectangle of these sides multiplied by the cosine of the included angle. Let A B C be any plane triangle, in B which the sides opposite the angles A, B, C' respectively, are denoted by a, 11 c. Draw BD from one of the angles, B, e perpendicular to the opposite side, AC.2 Then, if A is acute, we have by Geom -_____ etr~y (Prop. XII. Bk. IV.), A D b C a'-2~c2+ - 2b A D; but from the right-angled triangle AB BD, by (86), we have AD =c cosA; therefore, a2= b2 + c" - 2 bc cosd. (96) 44 TRIGONOMETRY. When, the -angle A. is obtuse; the point D will fall on the other side of A, and. we have by Geometry (Prop. XIII. Bk. IV.), a2 =l 0 +2 b AD. But since BA D is now the supplement of D A b 13 A (7, by'' -Art.- 7 7 we have, AD = c cos BAD =-ccos BAC/= -c Cos A. Substituting this value of AD, we have as before, a2== 1,2k ( 2 bc cos A. When A is a right angle and a the hypothenuse, cos A is zero (30), and (96) becomes a 19 + C2' and thus the formula (96) is true,- whatever the angle A may be'. In like manner we have b2- a'~ 2- 2ac cos B, (97) c( =a 2+ b2- 2ab cos C. (98) 114. The~ cosine of any angl of a' plane triangle is equal to the fr-action whose numerator is the sum of the squares of the containing' sides, dmnse&y thesur of the opposite side, and' whose denominator is twice the product of the containing s8ides. For, by (9 6), a2= 9~(2- 2 bc cos A, whence, ` csA bt- + ( - a2 Cos A.2 bc (99) Simiirly, from (97) and (98), we have CosB. 2c CosC 2b (1 00) 115. By these formulvathe angles of -a triangle can be found when the sides are given, but they cannot be conveniently applied ia' computation,by logarithms. We thea subtract both members of formula (99) from 1, and Ol04Aif 1.BOOK MI.'.45 and. substituting, for 1 -cos A its value, 2 sin.2.Aby (80), we have 2sin'j - b'+2c'-a' a'- (b-c)' 1- bc 2 bc (a +b -c) (a-b +c). - ~~2 bc whence, si2 1 A =(a 4b-c) (a-b+ c)* 11 4 bc 01 Let now 2 s = a +4 b ~ c, so that s is half the sum of the' sides of the triangle; then a+b - c =2 (,s —c), a-b c = 2 (s -b). Substituting these values in the preceding equation, and reducing, we have sin A = (S-b)~ c) (102) In like manner we may obtain sin B = y~ )(-)(103) sin ~ ~ ~ (s - a) bs - b)(14 That is, The sine of half of an~y angle in a plane triangle is equal to the square root of half the sum of the three sides less one of the adjacent sides, into half the sum less the other adjacent-. side, divided by, the rectangle of the two adjacent sides. 116. If 1 be added to both sides of (99), then, substituting for 1 + cos A its value, 2 cos2 A, by (81), we have b' ~ c' - a' (b +c)' -a" - (b + c ~a(b ~c - a) c (b~c+a) (b+c-a) whence, COO' A=4c(05 5~~~~~ ~ 46 TRIGONOMETRY. Let now s = half the sum of the sides of the triangle, as in Art. 115; then, b c + a = 2 s, b+c -a=2 (s-a). Substituting these values in the preceding equation, we have cos 3 A b -Va). (106) Similarly, cosB- = (S-b), ' (107) ac cos C (S - C). (108) ab That is, The cosine of half of any angle of a plane triangle is equal to the square root of half the sum of the three sides, into half the sum less the side opposite the angle, divided by the rectangle of the two adjacent sides. 117. Dividing (102) by (106), (103) by (107), and (104) by (108), we have, by (13), tan A _ - b (( - b) (s- c); (109) s (s —a) tan I B (s- a)(s c); (110) tan J C ( -a) (s- b)(111) V (8 — c) ' That is, The tangent of half of any angle of a plane triangle is equal to the square root of half the sum of the three sides, less one of the adjacent sides, into half the sum less the other adjacent side, divided by half the sum, into half the sum less the side opposite the angle. SOLUTION OF RIGHT-ANGLED TRIANGLES. 118. In a right-angled triangle, the side opposite to the right angle is called the hypothenuse; that adjacent to the right angle, and upon which the triangle is supposed to stand, is called the BOOK II. 47 base; and the other side adjacent to the right angle, the perpendicular. The base and perpendicular have been termed the sides about the right angle. Of the acute angles, that adjacent to the base has been termed the angle at the base, and the other the angle at the perpendicular. Thus, let A B C be any right-angled triangle, with the right angle at C, then h represents the hypothe- B nuse, b the base, p the perpendicular, A the acute angle at the base, and B the acute h angle at the perpendicular. 119. In order to solve the triangle, two A b C elements other than the right angle must be given, one of them being a side. Hence there will be four cases in which there may be given, respectively, I. The hypothenuse and an acute angle. II. A side about the right angle and an acute angle. III. The hypothenuse and a side about the right angle. IV. The two sides about the right angle. CASE I. 120. Given the hypothenuse and an acute angle. Let there be given, in the right-angled triangle A B C, the hypothenuse h and the acute angle A; to find the angle B, the perpendicular p, and the base b. z To find B. The angle B is the comple- A ment of A (Art. 44); hence, B — 90~ -A. To findp and b. By (85) and (86) we have p = h sin A = h cos B, b =- h cos A = h sin B; or, by logarithms, B h P CA-. b 4-C logp = log h +- log sin A = log h + log cos B, (112) og b = log h + log cos A = log h + log sin B. (118) 48 TRIGONOMETRY. That is, The logarithm of either side about the right angle is equal to the logarithm of the hypothenuse, plus the logarithmic sine of the opposite angle, or plus the logarithmic cosine of the adjacent angle. NOTE 1. As the logarithmic sine and cosine are increased by 10 (Art. 99), the resulting logarithm will be so much too great, and must be diminished by 10. This increase by 10 will affect the work wherever the logarithms of trigouometric functions are used. NOTE 2. The last figure of an answer may occasionally be found to differ from the one given in this work, when it has been obtained by the use of different formulae or tables. The results are not, however, generally carried so far as to admit of such a difference. When two methods of solving give different results, that is inserted which is most accurate, whether obtained by the isual method or not. EXAMPLES. 1. Given the hypothenuse of a right-angled triangle equal to 1785.395 feet, and the angle at the base equal to 59~ 37' 42"; to solve the triangle. Solution. The angle at the perpendicular =90~ - 59~ 37 42" =30~ 22' 18". Let, now, h = 1785.395 feet and A = 59~ 37' 42", and we have, by (112) and (113), h =1785.395 log 3.251734 log 3.251734 4 59~ 37' 42" log sin 9.935892 log cos 9.703813 p 1540.37 log 3.187626 b = 902.708 log 2.955547 Ans. Angle at the perpendicular, 30~ 22' 18"; perpendicular, 1540.37 feet; base, 902.708 feet. 2. Given the hypothenuse of a right-angled triangle equal to 25 yards, and one of the acute angles equal to 54~ 30'; to solve the triangle. 3. Given the hypothenuse of a right-angled triangle equal to 173.2 feet, and one of the acute angles equal to 370 2' 43"; required the other parts. Ans. Angle, 52~ 57' 17"; sides, 104.34 feet and 138.24 feet. BOOK IlL 49 CASE II. 121. Given a side about the right angle, and an acute angle. Let there be given (Fig. Art. 120) the side b and the angle A; to solve the triangle. To find B. The angle B is the complement of A; hence, B = 90~ - A. To find p. By (87) and (88), we have p = b tan A - b cot B; or, by logarithms, log p log b + log tan A = log b - log cot B. (114) To find h. By means of (113), we obtain log h = log b - log cos A = log b - log sin B. (115) Given the side p and the angle A; to solve the triangle. To find B. We have, as before, the angle B, the complement of A, or B- 90 -- A. To find b. By (87) and (88), we have b = p cot A =p tan B; or, by logarithms, log b = logp + log cot A = logp + log tan B. (116) To find h. By means of (112), we obtain log h = log p - log sin A = log - log cos B. (117) That is, The logarithm of either side about the right angle is equal to the logarithm of the other, plus the logarithmic tangent of the angle opposite, or plus the logarithmic cotangent of the angle adjacent to the former. The logarithm of the hypothenuse is equal to the logarithm of either side about the right angle, minus the logarithmic sine of the angle opposite, or minus the logarithmic cosine of the angle adjacent to the side. 5 * 50 50 ~~~~TRIGONOMETRY. EXAMPLES. 1. Given the side b of a right-angled triangle equal to 902.708 feet, and the acute angle A equal to 590 371 42"; to solve the triangle. Solution. The angle B = 900 - 590 37' 42" = 300 22' 18". By (114) and (115), we have b=-902.708 log 2.955547 log 2.955547 A =59037142/I log tanl10.232078 ar. co. log cos 0.296187 p =1540.37 log 3.187625 h=1785.395 log 3.251734 Ans. Angle B, 300 22' 18"; perpendicular, 1540.37 feet; hypothenuse, 1785.395 feet. 2. Given one of the sides about the right angle of a rightangled triangle equal to 14.52 rods, and the opposite angle equal to 35o 30'; to solve the triangle. 3. Given the perpendicular of a right-angled triangle equal to 3555.4 yards, and the angle at the perpendicular equal to 330 30' 47"; to solve the triangle. Ans. Angle at the base, 560 29' 13"1; base, 2354.4 yards; hypothenuse, 4264.3 yards. CASE III. 122. Given the hypothenuse and a szde about the tight angle. Let there be given (Fig. Art. 120) the hypothenuse It and the side p; to solve the triangle. To find A and B. By (1) and (4), 'we have sin A = cos B= = A' or, by logarithms, log sin A= log cos B =log p- log h. (118) To find b. By (85) and (86), we have b=h cosA=h sin B; or, by logarithms, log1 = log h +log cos A =log h +log sin B. (119) BOOK III. 51 Also, by Geometry (Prop. XI. Bk. IV.), we have h2 = p22+ 8b2 (120). whence, b2=h2 _ p2 =(h+ p) (h -p), b = /(h + p) (h -p); or, by logarithms, log b l= log (h + p) + ' log (h -p). (121) That is, The logarithmic sine of one of the acute angles, or the logarithmic cosine of the other, is equal to the logarithm of the side opposite the former angle, minus the logarithm of the hypothenuse. The logarithm of either side about the right angle is equal to the logarithm of the hypothenuse, plus the logarithmic cosine of the angle adjacent, or plus the logarithmic sine of the angle opposite. EXAMPLES. 1. Given the hypothenuse of a right-angled triangle equal to 1785.395 feet, and the perpendicular equal to 1540.37; to find the other parts. Solution. By (118) and (119), we have p= 1540.37 log 3.187626 h = 1785.395 ar. co. log 6.748266 log 3.251734 A- 59~ 37' 42" log sin 9.935892 log cos A 9.703813 B= 30~ 22' 18" log cos b= 902.708 log 2.955547 Ans. Base, 902.708 feet; angle at the base, 59~ 37' 42"; angle at the perpendicular, 30~ 22' 18". 2. Given the hypothenuse of a right-angled triangle equal to 73 feet, and one of the sides equal to 55 feet; to solve the triangle. 3. Given the hypothenuse of a right-angled triangle equal to 643.7 rods, and the base equal to 473.8; to find the perpendicular and the two acute angles. Ar3. Perpendicular, 435.73 rods; acute angles, 420 36' 12" 47~ 23' 48". 62 52 ~~~~TRIGONOMETRY. CASE IV. 123. Given the two sides about the right angle. Let there be given (Fig. Art. 120) the sides p and b; to solve the triangle. To find A and B. By (2) and (4), we have tan A =cot B = -! or by logarithms, log tan A = log cot B== log p. log b. k122) To find h. By (1), we have sinA= p' wvhence h ~?4;(123) or, by logarithms, log h =log p- log sin A; (124) Also, by (120), h2==p+b2,whence hzz '+(125) That is, The logarithmic tangent Of one of the acute angles, or the logarithmic cotangent of the other, is equal to the logarithm of the side apposite the former angle, mninus the logarithm of the side adjacent. The logarithm of the hypothenuse is equal to the logarithm of either side, minus the logarithmic sine of the angle opposite the side. EXAMPLES. 1. Given of a right-angled triangle the side p equal to 1540.37 feet, and the side b equal to 902.708 feet; to solve the triangle. ISolution. By (122) and (124), we have p = 1540.37 log 3.187626 log 3.187626 b = 902.708 ar. co. log 7.044453 A - 50 37 42"log ~n - ar. co. log sin A 0.064108 B- 00 2' 8" og oti10*32h:9 1785.395 log 3.251734 Ana. Hypothenuse, 1785.395 feet; acute angles, 590 37# 4~2U, 300 22' 18".* BOOK IIL 58 2. Given the perpendicular of a right-angled triangle equal to 65 feet, and the base equal to 72 feet; to find the hypothenuse and the two acute angles. 3. Given the perpendicular of a right-angled triangle equal to 2.269 rods, and the base equal to 126.9 rods; required the hypothenuse and the two acute angles. Ans. Hypothenuse, 126.92 rods; acute angles, 1' 1' 28", 88~ 58' 32". SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 124. Since there must be given three elements, one of which is a side (Art. 108), there will be four cases, the data in them being, respectively, I. One side and any two angles. II. Two sides and an angle opposite one of them. III. Two sides and the included angle. IV. The three sides. CASE I. 125. Given one side and two angles. Let there be given in the triangle A B C, B the side a, and the two angles A and B; to solve the triangle. To find C. Since the sum of the three / c angles must be 180~, we have C= 1800 - (A + B). A b C To find b and c. By (90) and (89), we have a: b: sin A: sin B, a: c: sin A: sin C; a sin B a sin C whence, b- c (126 sin A sinA1 or, by logarithms, log = log a + log sin B- log sin A, (127) log c = log a + log sin C- log sin A. (128) 54 TRIGONOMETRY. That is, The logarithm of the required side is equal to the logarithm of the given side, plus the logarithmic sine of the angle opposite the required side, minus the logarithmic sine of the angle opposite the given side. EXAMPLES. 1. Given of a triangle the side a equal to 9459.31 feet, the angle A equal to 71~ 3' 34", and the angle B equal to 53~ 26'; to find the sides b, c, and the angle C. Solution. C= 180~ - (71~ 3' 34" + 53~ 26') = 55~ 30' 26". Then, by (127) and (128), we have a- 9459.31 log 3.975859 log 3.975859 A 71~8 334" ar.co.log sin 0.024176 ar. co. log sin 0.024176 B- 53~ 26' log sin 9.904804 C- 55~ 30' 26" log sin 9.916032 b —8032.28 log 3.904839 c=8242.64 log 3.916067 Ans. Angle C, 55~ 30' 26"; side b, 8032.28 feet; side c, 8242.64 feet. 2. Given one side of a triangle equal to 110 rods, the opposite angle equal to 50~ 5', and an adjacent angle equal to 33~ 55'; to solve the triangle. 3. Given one side of a triangle equal to 654 feet, one of the adjacent angles equal to 41~ 0' 39", and the other adjacent angle equal to 55~ 34' 8"; to find the other parts. Ans. Angle, 83~ 25' 13"; sides, 432 feet, 543 feet. CASE II. 126. Given two sides and an angle opposite one of them. Let there be given in any triangle, C AB C, the two sides a, b, and the angle A opposite to one of them; to solve the b \a triangle. To find B. By (90), we have A c B' B a: b: sin A: sin B, b sin A whence, sin B =, (129) a1 BOOK 111. 55 or, by logarithms, log sin B= log b + log sin A - log a. (130) That is, The logarithmic sine of a required angle whose opposite side is given, is equal to the logarithm of that side, plus the logarithmic sine of the given angle, minus the logarithm of its opposite side. To find C. We have C= 180~ - (A 4- B). To find c. By (128), after Cis found, we have log c - log a - log sin C - log sin A. 127. Whenever the given angle is acute, and the side opposite to it is less than the side adjacent to it, there may be formed, as shown in Geometry (Prob. XI. Bk. V.), two triangles, each satisfying the given conditions, and, therefore, there will be two solutions. Thus (Fig. Art. 126) with two given sides a, b, equal respectively to tY B and A C, and the given acute angle A opposite the less side C B, there may always be formed two triangles, A B C, A ]B C, which have A, a, b in common, and the angles A B C, A B C supplements of each other. In one of them, therefore, the required angle is acute, and in the other it is obtuse. When the given angle is obtuse, the required angle must of necessity be acute, since a triangle can have but one obtuse angle. When the given angle is acute, and its opposite side is greater than the side opposite to the required angle, that must also be acute, since the greater angle must be opposite the greater side. When the side opposite the given angle is exactly a perpendicular let fall from C on A B, the required angle is a right angle. If the side opposite the given angle be less than the perpendicular, the solution is impossible, since there will be no triangle with the given parts. When two values are admissible for B, in case of ambiguity, two corresponding values will exist for C and c. 66 56 ~~~~TRIGONOMETRY. EXAMPLES. 1. Given of any triangle A B U, the side b equal to 216 yards, the side a equal to 117 yards, and the angle opposite the side a equal to 220 37'; to solve the triangle. Solution. By (130), we have a 117 ar. co. log sin 7.931814 b==216 log 2.334454 A 220 371 log sin 9.584968 B = 450 13' 55"1 or 1340 46' 5" log sin 9.851236 A + B 670 50 55"1 or 1570 23' 5" C 7=1800 67' 50' 55/11 -1120 9'S", or U = 180' - 1570 23' 5" = 220 36' 55". Then, by (128), we have a=117 log 2.068186 log 2.068186 C=~~1120 9' 5"1 log sin 9.966700 or=22036'55"logrsin9.584943 4-A22037'ar co.logsin0.415032 ar. co. log sin 0.415032 e =281.785' log 2.449918 or = 116.99 log 2.068161 Axis. Angle B, 450 13' 55"1, or 1340 46'5S"; angle (7, 1120 9'5", or 22' 36' 55"; side c, 281.785 yd., or 116.99 yd. 2. Given two sides of a triangle equal to 9459.31 feet and B032.28 feet, and the angle opposite the first side equal to 710 31 34"; to-find the other parts. Solution. a=9459.31 ar. co. log 6.024141 log 3.975859 1b = 8032.28 log 3.904839 A- 710 31 34"1 log sin 9.975824 ar. co. log sin 0.024176 B== 530 26' log sin 9.904804 CY= 180o - 1240.29' 34"1- 550 30' 26"1 log sin- 9.916032 = 8242.64 log 3.916067 Ans. Side, 8242.64 feet; angles, 530 26', '55"30,'26's 3. Given two sides of a triangle equal to 80 rods and 142.6 IBOOK IlL '57 rods,- and the angle opposite the second side equal to 96' to solve the triangle. 4. Given in a triangle A B C, the side a. equal to 32.1098 rods, the side b equal to 125.701 rods, and the angle A equal to 140 48'; to solve the triangle. Ans. Angle B, 900; angle C, 750 12'; side c, 121.531 rods. 5. Given two. sides of a triangle equal to 1540.37 feet and 760.9 feet, and the angle opposite the second equal to 300 22' 8"; to find the other side and angles. Ans. Impossible. CASE III. 128. Given two sides and the included angle. Let there be given in the triangle A B CB the sides a and b and the included angle C, to solve the triangle. To find A and B, we have a and A ~ B= 1800 -C A b C j (A + B) = 90 0 - C = complement of j C; whence, tan I. (A ~ B) == cot j. C. (131) Then, the half diffierence of A and B is found by means of (95), which gives a-f-b:a-b. tan j (A +B) tan I (A -B); whence, tan I (A -B)=a+ tan,X (A +B) ~ a- cot i. C, (132) or, by logarithms, logp tan ~.(A-B) =log, (a-b) -log (a~b) + log tan (A+B) =locg (a-b) - log (a +b) + log cot4 C'..(183) That is,. The logarithmic tangent of half the difference of the two re.. q'uired angles is equal to the logarithm of the difference of Oae 6 TRIGONOMETRY. given sides, minus the logarithm of their sum, plus the logarithmic tangent of half the sum of the required angles, or plus the logarithmic cotangent of half the given angle. Since e (A + B) is known, when ~ (A - B) is found, we have A = (A +- B) + ~ (A - B), B = (A + B) - (A- B). That is, The GREATER of the two required angles is equal to half their sum, plus half their difference; and the SMALLER angle is equal to half their sum, minus half their difference. To find c. By (128), we have log c = log a + log sin C- log sin A. EXAMPLES. 1. Given of any triangle A B, the side a equal to 9459.31 feet, the side b equal to 8032.28 feet, and the included angle C equal to 55~ 3) 26"; to find the side c and the angles A and B. Solution. A - B = 180~ - C = 124~ 29' 34", and 3 (A + B) = 62~ 14' 47". Then, by (133), a + b = 17491.59 a - b = 1427.03. (A + B) = 62~ 14'47 (A- B) - 8~48'47" B — 53 26' A = 71~ 3134// C - 55~ 30' 26" a - 9459.31 c - 8242.64 ar. co. log 5.757170 log 3.154433 log tan 10.278844 log tan 9.190447 ar. co. log sin 0.024176 log sin 9.916032 log 3.975859 log 3.916067 Ans. Side c, 8242.64 ft.; angle A, 7103/34"; angle B, 53~ 26'. 2. Given two sides of a triangle equal to 142.6 feet and 110 feet, and the included angle equal to 33~ 55'; to solve the triangle. BOOK III. 59 3. Given the two sides of a triangle equal to 153 rods and 137 rods, and the included angle equal to 400 33' 12"; to find the other parts. Ans. Side, 101.615 feet; angles, 78~ 13' 1" and 61~ 13' 47". CASE IV. 129. Given the three sides. Let there be given (Fig. Art. 128) the three sides a, b, and c; to solve the triangle. To flna A, B, and C. By (102), (103), and (104), we have (s - b) (s - c) sin i A e sin B= /(- a) (s- c) ac sin C= /(s — a) (s — b) or, by logarithms, log sin A log (- b) + log (s -c) - log b - log c logosin PA 2, (134) log sin 4 B= log (s-a) + log (s -c) - log a- log c ( log sin B B " 2, (135) log sin log (s - a) + log (s - b) - log a - log b(136) log sin. C 2 (136) That is, The logarithmic sine of half of any angle of a triangle is equal to the logarithm of the difference between half the sum of the sides and one of the adjacent sides, plus the logarithm of the difference between half the sum and the other adjacent side, minus the logarithms of those two sides, divided by 2. 130. A, B, and C can also be determined by formulae (106), (107), and (108) for the cosine of half an angle, and by formulae (109), (110), and (111) for the tangent of half an angle. When the half angle is less than 45~, the table will determine it from its sine with greater precision than from the cosine, and tice versa when the half angle is greater than 45~. 60 60 ~~~TRIGONOMETRY The: method by the tangent of half the angle is precise, and requires the use of but four logarithms. NOTE. This case may also be solved by drawing a perpendicular from the 'vertex to the base of the triangle, thus dividing it into two right-angled triaingles, of which the hypothenuses are known, and the sum of whose bases is the base of the original triangle. Let s and s' represent CD and D A (Fig. Art. 113), then (Geomn., Prop. XI. Bk-. IV.), p2 = —C2 - 8d ' a2_ 2 or, #.a2C1 yvhence, (s + a'.) (a -as) =(a + c) (a -c). Substituting bforas+a'f, b =~~C ac a form to which logarithms can be readily applied. Knowing s5+ a' and a - a', s and a' can at once be found, and thence the angles A, C, and B, by Art. 122. EXAMPLES. 1. Given of any triangle A BC, the side a equal to 216 yards, the side S equal to 217 yards, and the side c equal to 235' yards; to find the angles A, B, and U. Solution. By (134), (135), and (136) we have a= 216 ar~co.log7.665546 anreo.log7.6650546 b =21 7 ar. co. log 7.6 63 540 ar. co. log 7.663540 c =235 ar. co. log 7.628932 ar.co.log7.628932 s.-a=118 log 2.071882 log 2.071882 s-b=117 log2.068186 log 2.068186 a-c= 99 log 1.995635 log, 1.995635___ 2 )19.356293 2) 19.361995 2) 19.469154 log sines 9.678147 9.680998' 9.7-34577 j.A =280 27'/ 47" B= 280 40' 4"A4 C =- 3 20 52' 8"1.6. Ans. A 5660 65 34/1; B=.570 20' 8".8; C - 650 441 17"1.2. 2. Given the three sides of a triangle equal to.432, 543, and 654; to solve the triangle by means of the cosine. 3.,.. Given the three sides of -a triangle equal to 96.12, 162.34, and 98;, to solve the triangle by means of the tangent. Ans The angles, 320 14' 53"; 1140 24' 9"; 833 20' 58". BOOK IV. PRACTICAL APPLICATIONS. DETERMINATION OF HEIGHTS AND DISTANCES. 131. A HORIZONTAL PLANE is one which is parallel to the horizon. A VERTICAL PLANE is one which is perpendicular to a horizontal plane. A HORIZONTAL LINE is one which is parallel to the horizon. A VERTICAL LINE is one which is perpendicular to a horizontal plane. 132. A HORIZONTAL ANGLE is one the plane of whose sides is horizontal. A VERTICAL ANGLE is one the plane of whose sides is vertical. An ANGLE OF ELEVATION is a verti- D - B - cal angle having one side horizontal and the inclined side above it; as the angle CA B. An ANGLE OF DEPRESSION is a vertical angle having one side horizontal and the inclined side under it; as the angle A C D, IB A. 133. To determine the height of a vertical object standing on a itizontal plane. Let B be the top of the object, and, let it be required to find its height B BC. 6* 62 TRIGONOMETRY. Measure from the foot of the object, in the horizontal plane, any convenient distance, as A C, as a base line, and at A observe the angle of elevation CA B. Then, in the right-angled triangle A B C, we have known the side A C and the acute angle A; therefore we can determine the height B C by Art. 121. B EXAMPLES. 1. Standing on the edge of a moat 40 feet wide, I observe that the wall of a fort upon the opposite brink subtends an angle at the point of observation of 36~ 52' 12"; required the height of the wall. Ans. 30 feet. 2. The angle of elevation of the top of a flag-staff, measured on a horizontal plane, at a distance of 89 feet from the foot of the staff, is 41~ 29'; what is the height of the staff? 134. To find the distance of a vertical objec given, c^ ' a Let B C be the object whose heighis given, and let it be required to find the - distance A C. Measure the angle of elevation C A B, or the angle of depression D B A, which is equal to C A B. Then, in the right- / angled triangle A B C, we have known A the side B C and the angles; therefore we can find the distance A C by Art. 121. x,. t, its height being EXAMPLES, 1. A tree 91 feet in height stands on the same horizontal plane with a dial, at which the angle of elevation subtended by the tree is 832~ 22'; required the distance of the dial from the foot of the tree. Ans. 143.6 feet 2. From the top of a house whose height is 30 feet, I observe that; the angle of depression of an object standing on the same horizontal plane with the house is 86~ 52' 12"; required the BOOK IV. 63 distance of the qbject from the base of the house, and also the length of the line that will just connect the object with the top of the house. 135. To find the distance of an inaccessible point on a horizontal plane. Ct: A'..' ' ' C. Let C be the point inaccessible from A and B, and let it be required to find its distance from each of those points. Measure as a horizontal base line the distance between A and B, and observe the horizontal an- - -- gles CAB and CBA. Then, in the triangle A B C, there will be known the side A B and the angles; therefore the sides A C and B C can be found by Art. 125. EXAMPLES. 1. Wanting to know the distances of two objects from a tree, inaccessible by reason of an intervening river, I measured the distance in a straight line between the two objects, and found it to be 772.45 feet; I also found the horizontal angles formed by the extremities of the straight line with the tree to be 80~ 58' 4" and 43~ 33' 44". Required the distances of the objects from the tree. Ans. The one, 926.01 feet; the other, 646.16 feet. 2. Two ships are engaged in cannonading a fort by the seaside; the ships are 131.89 rods apart, and the two angles at the ends of the straight line connecting the ships, formed by that line and lines drawn to the fort, are 180 52' 13" and 152~ 11' 42". Required the distance of each ship from the fort. 136. To find the height of an inaccessible object above a horizontalplane. C: - ":. C- C.'C. - (3 -First Method. Let B be the top of the object, and let it be required to find the height B C. Measure a horizontal base line, A C,, of any convenient length, directly toward the object, and observe the angles of elevation, at A and C0. Then, in the triangle ABO *, since 64 TRIGONOMETRY. B OA is the supplement of C'C'B, we have known the side A CT and all the angles; therefore we can find the side AB by Art. 125. Then, in the B B right-angled triangle ABC, we have known the hypothenuse A B and the angles; there- / fore we can find the height BC 0 by Art. 120. EXAMPLES. 1. Required the altitude of a hill whose angle of elevation, taken at the foot of it, was 55~ 54', and 300 feet back, on the same horizontal plane with the foot, the angle was 33~ 20'. Ans. 355.71 feet. 2. Two observers at sea, 800 yards apart, noticed at the same instant a meteor bearing due east from each; to the one its angle of elevation was 57~, and to the other the same angle was 31~ 28'. Required the altitude of the meteor above the horizontal plane of the ships. Second Method. Let B be the I C. 3'.'' top of the object, and let it be re- A{ quired to find the height BC. Now, s suppose it is not convenient to measre a horizontal base line directly B -. -.. toward the object, and we measure ' it if any direction, A B2, also measuring the angles CA B' and CB'A. Then, in the horizontal triangle A B 0, we know the side A Bf and all the angles; therefore the side A ( can be found by Art. 125. Then, also, by observing the angle of elevation CAL B, we shall, in the right-angled triangle AB C, know the side A C and all the angles; therefore the height B 0 can be found by Art. 121.: "... EXAMPLE, 1. A person on one side of a river observed an eagle's nest on an inaccessible mountain-crag on the opposite side; and being desiroths of fascertainirrg its height above the 'level of the river, he measured along the;ahore a Btraight line 110 yards in length, and BOOK IV. 65 found the horizontal angles of its extremities with the object to be 33~ 55' and 966, and also the angle of elevation at the latter to be 45'. Required the height of the nest above the water. Ans. 240 feet. 137. To find the distance between two objects separated by an impassable barrier. (,, 6+: - 'r) "Z. a L.- -(t 3./ ' L.c A ( — 7 Let A and B be two objects separated by an impassable barrier, and let it be required to find the distance, A B, between them. Take any point, C, from which A and B are / both visible and accessible. Measure CA and CB, and also note the angle A CB. C Then, since in the triangle A B C the two sides CA and GB, with their included angle, are known, the distance A B can be found by Art. 128. EXAMPLES. 1. Two bounds of a lot have between them an impassable morass, and, wishing to find their distance apart, I have taken their distances from a third point, which could be seen from each. These distances are 124.75 and 171.41 rods, and the angle at that point subtended by the bounds is 99~ 25'. How far are the bounds apart? Ans. 227.91 rods. 2. The distance between two trees cannot be directly measured, in consequence of an intervening obstacle, but within sight of each is a third tree, and their distances from this are known to be 274.65 and 396.11 yards, and the angle at that point subtended by the two trees is 8~ 56' 5". Required the distance between the two trees. 138. To find the distance between two inaccessible objects. Let C and D be the objects, and A and B two accessible points, from which both the objects are visible. Measure the base line A B, and observe the angles D A B, D B A, CA B, and C B. Then, in the A B 66 TRIGONOMETRY. triangle DAB, since we have the side A B and all the angles, we can find the side B D by Art. 125. In the triangle A B C we have the side A B and all the angles, hence we can find B C. Then, B D and B C being found, we have in the triangle B C D the sides BD and B C, with their included angle; therefore we can find the distance CD by Art. 128. EXAMPLE. 1. Wanting to ascertain the distance between a tree, D, and a flagstaff, C, on the opposite side of a river from me, I measured along the shore, on the horizontal plane with the objects, a base line, A B, of 110 yards. At A, the angle DA B equals 96~, and C A B equals 29~ 56'; at B, the angle D B A equals 33~ 55', and C B A equals 133~ 50'. Required the distance between the tree and the flagstaff. Ans. 261.81 yards. 139. To find the distances from a given point, of three objects whose distances from each other are known. Let it be required to find the distances from D, a given point, of three objects, A, B, and C, whose distances from each other are known. Observe the angles A D C and / BDC. Describe a circle about the tri- / angle ADB, and draw AE and EB; / then the angle A B E is equal to the / angle A D E, since both are measured. by half of the same arc A E (Geom.,............. Prop. XVIII. Bk. III.); also the angle B A E is equal to the angle B D E, for a like reason. Now, in the triangle A E B, the side A B and all the angles are known, hence the side A E may be found by Art. 125. Again, the sides of the triangle A B C being given, we may find the angle B A C by Art. 129; then, in the triangle A E C, there will be known the two sides A C, A E, and the included angle CA E, so that the angle A CE may be found by Art. 128. Then, in the triangle A C D, we shall know the side A C and the angles A CD and A D O; therefore we can find the distance A D BOOK IV. 67 by Art. 125, and thence the other two 'distances, C D and B D. EXAMPLE. 1. On approaching a harbor, at the point D, I observed three headlands, A, B, and C. Now it appeared from a chart that the distance from A to B was 800 yards, from A to C 600 yards, and from B to C 400 yards; the angle A D C I found by observation to be 33~ 45', and the angle B D C to be 22~ 30'. What was the distance of each of the headlands from me? Ans. A, 710.19 yards; B, 934.29 yards; C, 1042.52 yards. DETERMINATION OF AREAS. 140. The AREA of any figure, or its quantity of surface, is deermined by the number of times the given surface contains some other area, assumed as the unit of measure, as a square inch, a square foot, &c. The areas of parallelograms, triangles, trapezoids, &c. can be determined by direct application of the principles of Geometry; but sometimes it is convenient to determine areas, especially of triangles, by means of their lines and angles, which requires the aid of Trigonometry. 141. To find the area of a triangle by means of two of its side. and the included angle. Let A B C be any plane triangle, in B which are given the sides b and c and the included angle A, to find the area of the triangle. Draw the perpendicular, p, from C B to the opposite side, A C; then, since the area of the triangle is equal to half the product of its base by its altitude A D b (Geom., Prop VI. Bk. IV.), area of A B C=- b p. (137) But, by (85), p=c sin A; whence, area A B C=; b c sin A, (138) 68 TRIGONOMETRY. or, by logarithms, log area A B C7= log ~ b - log c - log sin A. (139) That is, The logarithm of the area of a triangle is equal to the logarithm of half of either side, plus the logarithm of either of the other sides, plus the logarithmic sine of their included angle. EXAMPLE. 1. Required the area of a triangle which has two of its sides equal to 105 and 85 feet, and the included angle equal to 28~ 5'. Ans. 2100 sq. ft. 142. In like manner, the area of any parallelogram may be found, when two of its adjacent sides and the included angle are known; for the diagonal divides a parallelogram into two equal trangles (Geom., Prop. XXXI. Cor. 1, Bk. I.). EXAMPLE. 1. What is the area of a piece of ground, in the form of a parallelogram, which has two adjacent sides equal, respectively, to 120 and 212 rods, and their included angle equal to 85~ 30'? 143. To find the area of a triangle by means of a side and the angles. In the triangle A B C (Fig. Art. 141), let the side c and the angles be given, to find the area of the triangle. By means of Art. 111 we have c sin B (140).id by (85) p =c sin A. Substituting these values in (137), we obtain c2 sin A sin B area of B C_= siA snB (141) or, by logarithms, log 2 areaABC7 21 ogc+log sin A+-log sin B-log sin C. (142) That is, The logarithm of double the area of a triangle is equal to twice the logarithm of either side, plus the logarithmic sines of its adja-' cent angles, minus the logarithmic sine of it opposite angle. BOOK IV. 69 EXAMPLES. 1. A triangular lot has a side equal to 45 rods, and the adjacent angles equal to 70~ and 69~ 40'; required the area of the lot. Ans. 1378.41 sq. rods. 2. Given of a triangular field A B C, the angle A equal to 31~ 27', the angle B equal to 101~ 31', and the included side A B equal to 30 rods; required the area of the field. 144. To find the area of a triangle by means of its three sides. Let A B C (Fig. Art. 141) be the given triangle. Then, by (138), area of A B C =- b c sin A; but, by taking twice the product of the values of sin j A and cos j A, in (102) and (106), we have, by (74), 2 sin A -V s (s-a) (s- b) (s -c), (148) in which s denotes half the sum of the sides of the triangle. Substituting in the preceding equation this value of sin A, we obtain area of A B C - b c X - s (s-a) (s -) (s-c); whence, area of A B C == s (s - a) (s - b) (s - c); (144) or, by logarithms, log~rea A ~_log 8 ~ log (s —a) - + log (s —b) +log (s —c) Ig reaABC- 2 (145) at is, The logarithm of the area of a triangle is equal to hdlf the sum of the logarithm of half the sum of the sides and the logarithms of the remainders obtained by taking each side separately from half the sum of the sides. EXAMPLES. 1. Given the three sides of a triangle equal to 30, 40, and 60 rods, respectively; required the area of the triangle. Ans. 533.27 sq. rods. 2. A certain fort is in therform of an equilateral: triangle, whose sides are each 600 feet; required the area occupied by the fort. 7'* 70 TRIGONOMETRY. MISCELLANEOUS PROBLEMS. 1. The angle of elevation of a vertical tower is observed to be 300, at the end of a horizontal base line of 100 yards, measured from its foot. Required the height of the tower. 2. A rope-dancer wishes to ascend a steeple 100 feet high, by means of a rope 196 feet long. If he can do so, find at what inclination he must be able to walk up the rope. 3. From the summit of a pier which rises 100 feet above the margin of a river, the angle of depression of the opposite margin was found to be 33~ 16'. Required the width of the river. 4. If the distance of the moon from the earth be taken at 238500 miles, and the angle subtended by the semidiameter of the moon be 15' 33".5 at that distance, what is the moon's diameter? Ans. 2158 miles. 5. A point of land was observed by a ship at sea to bear east by south, that is, 11~ 15' S. of E.; and after sailing northeast 12 miles, it was found to bear southeast by east, that is, 33~ 45/ S. of E. Required the distance of the headland from the ship at the last observation. Ans. 26.07 miles. 6. From the top of Mont Blanc, 3 miles high, the angle of depression of the remotest visible point of the earth's surface is 2~ 13' 27". Required the diameter of the earth, supposing it to be a perfect sphere; and, also, the utmost distance from which the mountain is visible. Ans. Diameter, 7958 miles; distance, 154.5 miles. 7. A side of the base of a square pyramid is 200 feet, and each edge is 150 feet; required the slope of each face. Ans. 26~ 34', nearly. 8. I have a meadow in the form of a parallelogram, whose two adjacent sides are 20 rods and 18 rods, including an angle of 78~ 9'; the same has been divided into two equal lots by a fence running diagonally. Required the area of each lot.. ~~- ~~Ans. 176.16 square rods. 9. A traveler wishing to know the distance and height of a mountain-top over which he had to pass, took the angle of its BOOK IV. 71 elevation at two stations, in a direct line towards it, the one 3 miles, or 5280 yards, nearer the mountain than the other, and found the angles to be 2~ 45' and 3~ 20'. Required the horizontal distance of the mountain-top from the nearer station, and its height. Ans. Distance, 24840 yards; height, 1447 yards. 10. From the top of a light-house the angle of depression of a ship at anchor was observed to be 4~ 52', from the bottom of the light-house the angle was 4~ 2'. Required the horizontal distance of the vessel, and the height of the hill on which the light-house is placed, the height of the light-house being 60 feet. Ans. Horizontal distance, 4100.4 feet; height, 289.12 feet. 11. When a tower 150 feet high throws a shadow 75 feet long upon the horizontal plane on which the tower stands, what is the sun's altitude (Art. 189)? Ans. 63~ 26' 6". 12. The sides of a triangle are equal to 3 and 12, respectively, and the included angle is 30~; find the hypothenuse of an equal right-angled isosceles triangle. Ans. 6. 13. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal to 40~; then from another window, 18 feet directly above the former, the like angle was 37~ 30'. Required the height and distance of the steeple. Ans. Height, 210.4 feet; distance, 250.8 feet. 14. Two pulleys, whose diameters are 6 inches and 4 feet 3 inches, respectively, are placed at a distance of 3 feet 6 inches from centre to centre. What must be the length of a belt which shall connect them, by passing arouad their circumferences, without crossing? Ans. 15 feet 5.9 inches. 15. A tower is surrounded by a circular moat. At noon on a certain day, the shadow of the top of the flag-staff is observed to project 45 feet beyond the edge of the moat. When the sun is due west, on the same day, the shadow projects 120 feet beyond the moat. The distance between the extremities of the shadows is 375 feet. The angle of elevation of the top of the flag-staff from any point of the edge of the moat is 60~. Find the height of the tower, and the altitude of the sun at noon. Ans. 311.77 feet; 54~ 10' 57". BOOK V. SPHERICAL TRIGONOMETRY. DEFINITIONS. 145. SPHERICAL TRIGONOMETRY treats of methods of computing spherical triangles. 146. A SPHERICAL TRIANGLE is a portion of the surface of a sphere bounded by three arcs of a great circle, each of which is less than a semi-circumference. The three planes in which the arcs lie form a polyedral angle at the centre of the sphere. The ANGLES Of a spherical triangle are the diedral angles made by the plane faces which form the polyedral angle. 147. The sides and angles of spherical triangles are usually both expressed in degrees, minutes, &c. The circumference, however, is sometimes supposed to be divided into 24 equal parts, called hours; each hour into 60 equal parts, called minutes of time; each minute into 60 equal parts, called seconds of time. Then a side is expressed by the number of hours, minutes, seconds, and decimal parts of a second, which it contains. Hours, minutes, and seconds are denoted by h., m., and s. Thus, 3h. 35m. 5.8s. RELATIONS BETWEEN THE SIDES AND ANGLES OF SPHERICAL TRIANGLES. 148. In any spherical triangle, the sines of the sides are proa portional to the sines of the opposite angles. 1100K V. Let A BC be any.,pherical B triangle; A, B, and C the angles B opposite to its sides a, 6, and c,a respectively; and 0 the centre of c the sphere. 0 Take any point B' in O B,A D b fand~ draw B' D perpendicular to A the plane A 0 C; from D draw DA', D C', perpendicular to 0OA, 0 C, respectively; join B'A', B' C'. B' C' 0 is a right angle (Geom., Prop. VI. Bk. VII.); therefore, B' C' = OB' sin B' 0 C' = OB' sin a, and B D=B' C' sin B' C' D =B' C' sin C == OB' sin a sin C. In like manner, B' D = 0 B' sin c sin A; and, by, the two preceding equations, O B' sin a sin C OB' sin c sin A, sin a sin A whence, si i '(146) or, in the form of a proportion, sin a: sin c::sin A:sin G. (147) In a similar way it may be proved that sin a sinb sin A sin B, (148) sin c sin b::sin C: sin B.. (149) The figure supposes a, c, B, C, &c. each less than 9001, but -the relation stated may be shown to hold when the figure is modified to meet any case whatever. For instance, if C alone is greater than 900, the point D will fall beyond 0 C, instead of between O C and 0 A; then, B'C'D will be the supplement of C, and thus, since the sine of an angle and its supplement are the -same, the sine of B' CID is still equal to the sine of C. 7* 74 74 ~~~~TRIGONOMETRY. 149. 'In any spherical triangle, the cosine of any side is equal. to the product of the cosines of the other two sides, plus the product of the sines of those two sides into the cosine of their included angle. Let A B C be any spherical tri-B angle, 0 the centre of the sphere. Draw the plane B' Al C' perpendicular to 0 A. Then the an —C gle B'A' a is equal to the angle0 A, the angle B' 0 a measures A the side a, and in the triangles AA' B' a', 0 B' a we have, by Art. 113, 2 2 B' C'= A' B'~A' a- 2A' B'X Al a1cos A, B' U=B'~O0 a-2 O B'X O C'cos a. Subtracting the first eq uation from the second, observing, that 2 2 ~ 22 o B' -A' B' and o a -A' C' are each equal to 0 A', since the triangles 0 A' B', 0 A' a are right-angled at A', we have O0=2 OA ~2A'B'XA' acosA-2 OB'X 0 alcosa; therefore, cos a + ~ Co A Suibstituting the functions derived from the triangles 0 A' B', o A' a1, we have cos a= cos bcos c+sinabsin ccos A. (150) In lie manner may be deduced Cos b==cos c cos a +sin c sin a cos B,..(151) cos c cos acos b+sin asin b cos, C'. (1,52) The preceding. construction. s upposes -the sides 1, and c, which contain the angle A, to be both less than 900, but the formula obtaied may be shown to be applicable, in all cases. BOOK V. 7 5i - 150. In any spherical triangle, the cosine of any angle is equal to the product of the sines of the other two angles into the cosine of their included side, minus B the product of the cosines of those two angles.B Let A'B'R C' be the pola, r triangle C of A B C'; denote its angles by A', a B', and C", and its sides by a', b/, A and c'. Then (Geom., Prop. IX._A Bk. IX.), A'==180'-a, B'=1800-b, tY1800-c; al=180'-A, b'==180'-B, c'==1800-C'. Applying (150) to A'B'C', we have cos a' Cos b/ cos c' + sin b/ sin c' cos A'; or, by (46), - cos A = cos B cos C- sin B sin C'cos a; whence, cos A == sin B sin C cos a -cos B cos C'. (153) In like manner may be deduced cos B — sin C sin Acos b- cos C cos A, (154) cos C== — sin A sin B cos c -cos A cos B. (155) 151. In' any spherical triangle, the cotangent Of one side into the sine of another side is equal to the cotangent of the angle &pposite thefirst side into the sine of the included angle, plus the cosine of the second side into the cosine of the included angle. By (150) and (152) we have cos a cos bcos c +sin bsin ccos A, cos c = cos a cos b + sin a sin b cos C; and by means of (147), sin C sin c ~sin a Substituting these values of cos c and sin c in the first equation~ we obtain 76 TRIGONOME 11~Y cos a=(cos acosb+ sin a sinbcosC) cosb-jsin- sinV)csAsi or, cosa =cosa cos~b-j- sina sinb cosb cosC~sina sinb cotA sin a. Therefore, transposing cos a cos2 b, and observing that, by (1 1), cos a - cos a cos 2b cos a sin 2 b we have cos a sin' b = sin a sin b cot A sin 0C + sin a sin b cos b cos 0', and dividing the whole by sin a sin b, we obtain cot a sin b= cot A sin C+j cos bcos C. (156) 152. By interchanging the letters in (156), we obtain cot a sin c cot A sin B +cos c cos B, (157) cot b sin a =cot B sin C-i- cos a cos C, (158) cot b sin c= cot Bsin A ~ cos c cos A, (159) cot c sin a =cot C sin B + cos a cosll, (160) cot csin b=cot Csin A +cos bcos A. (161) 153. The formulm developed in the preceding articles are general, and apply to every case of spherical triangles, but require some transformations to render them more convenient for logarithmic computations. The formulae (150), (151), and (152) of Art. 149 are considered the fundamental frmule of spherical trigonometry, since from them all its other formulke may be deduced. 1054. To express the sine, cosine, and tangent of hal an angle of a triangle as functions of the sides. By means of (150) we have cos a - cos b -cos c cos A -= ibsn (162) but this formula is not suited to logarithmic computation. We then subtract each member of the equation from 1, and obtain (Art. 63), 1 -cosA= 1-cosea -cos bcos c cos (b -c) - cos a I - cosA = I sin bsin c I- sin bsia c BOOK V. 77 Substituting for 1 - cos A it's value, 2 sin2 I A (80), we obtain 2 sin2 f A -= cos (b -c) - cos a sin bsin c Now if, in (65), we make A = a, and B =- b-c, k1 (A + B) (a + b-c), (A.B) = (a-b +c), then, cos (b -c) -cos a = 2 sin,t (a + b-c) sin (a-b +c), which, substituted in the preceding equation, gives sin2A A= sini(a-b+c) sinj(a-+-b-c) (163) sin b sin C Let, now, s =half the sum of the sides of the triangle; then, a +b-c =2 (s -c), a -b f-c = 2 (s -b). Substituting these values in the last equation, and reducing, we have sin AA= /i (-) i (sc) (164) n (sinbcsinc( -a.Similarly, V sncinsin B = /s in c) sin~s a), (165) sin C /in= s-a si~ b) (166) Adding each member of equation (162) to 1, and observing that 1 + cos A = 2 cos2 I A (81), by means of (65), we have cos jA.sin I(a+b+c) sinjI(b~c-a) COS ~~~~~sin b sin c Introducing s (a + b ~ c), and reducing, we have /in s sin (s -a) Cos fA= V sin bsin c (167) Similarly, co si i sissn sb (168) Cos snssi s-)(169) 78 TRIGONOMETRY. Again,.dividing (164), (165), and (166) by (167), (168), and (169), respectively, we obtain /in (s -b) sin (s -c) ta V sin ssin (s —a) (170) ~ Sili(s -c) sin (s- (171 /sin (s -a) sin (s in (s b) a V sin s sin (s -c)(12 155. To express the sine, cosine, and tangent of half a side of a triangle as functions of the angles. By means of (153) we have Cos a =cos A +cos Bcos C (173) Whence, ~~~~~csA o o C cos A +cos (B +C)sin Bsin C sin Bsin C and srn'~a- cos I(A + B +Ccos I(B + C- A), (174) Maing k(A + B + C) =S, substituting, and reducing, we have 8rn a = - scsS-A) sin.b = V os sSB) (175) sinfrc = Q/cscsSO In lie manner, we obtain joB '- ~.7J~7S7LC) Os (SA) 1 ( cos a __ - ~ uin~uinB JS A BOOK V. '(A7 Hence, 1a=~/ -cos~co(S-AN tan a / - ~ cos(-B) co(S-C =a/ - cos S cos (S - B)'17 ~1os(- C) cos (S -A)' tan c. =7 _C0s S Cos (S -C) co(SA ) cos (S -B) Since S is always greater than 900 and -less than 2700 (Geom., Prop. X. Bk. IX.), cos S is always negative, and therefore. - cos S in the numerators of the first and third of the above sets of formula3 is essentially positive. 15 6. To prove Napier's Analogies. sin A sin B Let m= ~~~~sin a sin b then, sin A = m sin a, sin B = m sin b, sin A + sin B= m (sin a ~ sin ), (178) sin A - sin B= m (sin a -sin b). (179) By (1053) and (154) we have cos, A + cos Becos C= sin Bsin Ceos a rn sin CYcos a sin s, cos, B + cos A cos C =zsin A sin C cos b m sin C sin a cos56. Adding these equations, factoring, and reducing by (17), (cos A +cos B) (1 +cosC0)=msin-C sin (a +b). (180) Dividing (178) by (180), and multiplying by sin C, smnA+-sin B sin C sin a +siu b cos A +cos B 1H-CosC sin (a+b) 11 Now,. by means of (62), (63), and (74), we obtain sin a +sin b Cosj(a,.- (182j uin(a~b cosi(a +b 80 80 ~~~~TRIGONOMETRY. isin a -sin b -sin -4(a -b) and2 13 sin (a+b-f-b s-inj (a + b)(13 Substituting in (181) the value of each expression, from (68), (84), and (182), tan I (A + B) x tan j C =cos (a -b) cos (a + b)' or, cosi(a+ b) cot jC Cos (a -b) tan j(A +B)(14 In like manner, from (179) and (180), sin A -sin B sinGC sin a- sin b cos A + cos B 1+cos C sin (a+b)' whence, by (69), (84), and (183), tan.1 (A -B) x ta C=sn (a-b sinj (a + b)' or, sin I(a +b) cot IC sin j (a - b) -tanj (A -B)(15 Formulam (184) and (185) may be thus expressed: cos j (a + b): cos,X (a -b):: t -f, C: tan.1 (A+ B), (18 6) sin I (a + b): sin ~f(a-b)::cot.1 C: tan j (A -B). (1 87) That is, V, The cosine of half the sum of two sides of a spherical triangle is to the cosine of half their dfference as the cotangent of half the included angle is to the tangent of halfthe sum of the other two angles. The sine of half the sum of two sides of a spherical triangle is to the sine of half their difl'erence as the cotangent of half the included angle is to the tangent of half the defference of the other two angles. By applying (186) and (187) to. the polar triangle, Art. 150, we obtan BOOK V. 81 cos (A B co- 1 ( -BOO taV. a J( ) (81) sin (A+B,):sin I (A-B): tan c: tan J (a-b). (189)._ That is, The cosine of halfthe sum of two angles of-a spherical triangle is to the cosine of half their diff'erence as the tangent of half the included side is to the tangent of half the sum of the other two sides. The sine of half the sum of two angles of a spherical triangle is to the sine of half their diff~erence as the tangent of half the included side is to the tangent of half the difference of the other two sides. The above four proportions are called, from their inventor, Napier's Analogies. RELATIONS BETWEEN THE SIDES AND ANGLES OF RIGHT-ANGLED SPHERICAL TRIANGLE8. 157. The -sine of either oblique angle is equal to the sine of the opposite side, divided by the sine of the h~jothenuse. Let A B C be any spherical trian- B gle, right-angled at C. Ily means of (146) we have sin A - sin sin C; A -sin h but, as C- 900, sin C ==1, and; C sinnAA' In like manner, sin b *sin B= sin (191) 158. The cosine of either oblique angle is equal to t.&e tangea of the adjacent side, divided by the tangent of the hypothenuse. By means of (1 61) we have cot h sin b - cot C sin A + cos Cos. but, if C 900, then -cot C =0, and 8 82 TR1uGONOIN~I'TRr. cot h sin b = cos b cos A, or, cos A= cothsinb -cot Atan; tan b(12 whence, cos A = tan ~92 Also, by, means of (160), tanp cos B = ai (193) 159. The tangent of either oblique angle Its equal to, the tangent of the opposite side, divided by the sine of the adjacent side. *By means of (156) we have *cot psinb = cot Asin C cosb cos C; but, by making C = 90', sin C-= 1, cos C-= 0, and cot A == cotp sin b = i Sbl tan p' or, tan A -= (194) Also, by means of.(158), tan b tan B== (195) 160. The sine of either oblique angle is equal to the cosine of the other, divided by the cosine of its opposite -side. By means of (154) we have cos B= sin A sin C cos b -cosA cos CO, which, by making C = 900, becomes Cos B== cos b sin A, cos B whence, sin A (196) 1n- like manner, by means of (153), cos A sinB= co.(197) 161. The osine of the hypothenuse is equal to the product of the cosines of the other two sides. BOOK V. 8 By means of (152) we have cos h = cos p cos b ~ sin p sin 6 cos 0, which, by making C = 900, becomes cos Ih = cos p cos b. (198) 162. The cosine of the hypothenuse is equal to the product of the cotangents of the two oblique angles. Bymeans of (155) we have cos C= sin A sin B cos hs - cos A cos B which, by making C 900, becomes sin A sin B cos As = cos A cos B, cs=cos A cos or, Cs sin A sin B ctAot(199) 163. The preceding formulae may readily be remembered from their similarity to the corresponding ones for plane triangles; and, for convenience of reference, they are brought together in the fol-. lowing TABLE. 1. sin nAA 2. sin B=h 3. cos A tan b 4. cos B tan p tan h tan b 5. tan A= ~6. tanb.. sinb tan in p 7. sin A==cos B 8.Cos A 7. sn A cos V8. in B co 9. cos h =cos pcos b. 10. cos h =cot Acot B. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 164.. The solution of spherical triangles is the process by which, when the- values of a sufficient number of their six eleme nts are given, we calculate the values of the remaining elements. 84 TRIGONOMETRY. In order to solve a right-angled spherical triangle, two of its elements, other than the right angle, must be given. 165. The formulae requisite for the solution of right-angled spherical triangles are readily furnished by means of the relations demonstrated in the foregoing articles. Thus, sin sin p sin PA-sin h gives sin b sin b sin B —n sin h A tan b tan h cos B ta p tan A tan b tan B- tan sin p tan A p b sin b.sin B cos A sin BR -- " cos p. cos B sin A = -c " cos b sin p = sin A sin h, sin b = sin B sin A, cos A = cot h tan b, (200) (202) cos B = cot h tan p, sin p = cot B tan b, sin b = cot A tan p, cos A = sin B cos p, cos B = sin A cos b, which, with equations (198) and (199), cos h = cosp cos b, cos h = cot A cot B, enable us to determine every case of right-angled spherical triangles. For every one of these ten equations is a distinct combination, involving three of the five quantities, p, b, h, A, B; and five quantities, taken three at a time, can be combined only in ten different ways. NAPIER'S CIRCULAR PARTS. 166. If, in any right-angled spherical triangle, the right angle be left out of consideration, the two sides adjacent to the right angle, and the complements of the hypothenuse and of the two other angles, are called the five circular parts of the triangle. BOOK V. 85 Thus, in the spherical triangle B A B C, right-angled at C, the circular parts are p, b, and the com- plements of h, A, and B. p A 167. When any one of the five b parts is taken for the middle part, the two adjacent to it, one on either side, are called the adjacent parts, and the other two parts are called the opposite parts. Then, whatever be the middle part, we have as THE RULES OF NAPIER. I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 168. Napier's rules may be proved by showing that they agree with the results already established, Art. 165. Thus, 1. Let b be taken for the middle part; then p and the complement of A will be the adjacent parts, and the complements of B and h will be the opposite parts, and by the rules we have sin b = tan (com. A) tan p, sin b. cos (coin. B) cos (corn. A); whence, by Art. 50, sin b = cot A tan p, sin b = sin B sin A which agree with (205) and (201). In like manner, if p be taken as the middle part, sin p = tan (com. B) tan b, sin p = cos (con. A) cos (com. h); whence, sin p = cot B tan b, sin p = sin A sin A, which agree with (204) and (200). 2. Let the complement of h be taken as the middle part; 8* 86 TRIGONOMETRY. then the complements of A and B will be the adjacent parts, p and b the opposite parts, and we have sin (com. h) = tan (corn. A) tan (comr. B), sin (com. h) cos p cos b; whence, cos h = cot A cot B, cos h - cos p cos, which agree with (199) and (198). 3. Let the complement of A be taken as the middle part; then b and the complement of h will be the adjacent parts, p and the complement of B the opposite parts, and we have sin (com. A) = tan (com. h) tan b, sin (com. A) == cos (com. B) cos p; whence, cos A = cot h tan b, cos A = sin B cos p, which agree with (202) and (206). In like manner, sin (com. B) = tan (cor. h) tan p, sin (cor. B) = cos (cor. A) cos b; whence, cos B - cot h tanp, cos B = sin A cos b, which agree with (203) and (207). 169. Any element of a spherical triangle is less than 180~ (Geom., Art. 505, 539). Two parts are said to be of the same species when they are in the same quadrant, that is, when they are both less, or both greater, than 90~; and of diferent species when one terminates in the first and the other in the second quadrant. 170. In order to determine whether a part sought is less or greater than 90~, the algebraic signs of the terms should be observed, according to Art. 68 or 78. When, however, the part sought is determined by its sine, since the sines in both the first and second quadrants are positive, there will be two solutions, BOOK V. unless the ambiguity be removed by one of the following rules - 1. In any right-angled spherical triangle, an oblique angle and its opposite side are always of the same species. For, by (205), sin b = cot A tan p, in which, since sin b is always positive, cot A and tan p must always have the same sign, that is, A and p must be of the same species. 2. When the two sides about the right angle are of the same species, the hypothenuse is less than 90~, but when they are of different species, the hypothenuse is greater than 90~. For, by (198), cos h = cos p cos b, in which, if cos p and cos b have the same signs, cos h will be positive, but if they have unlike signs, cos h will be negative. 171. In the solution of right-angled spherical triangles, there will be six cases to consider, in which there may be given, respectively, I. The hypothenuse and an oblique angle. II. The hypothenuse and one side. III. One side and its adjacent oblique angle. IV. One side and its opposite oblique angle. V. The two sides about the right angle. VI. The two oblique angles. CASE I. 172. Given the hypothenuse and an oblique angle. Let there be given in the right- B angled spherical triangle A BC, the hypothenuse h and the oblique angle,/ A; to solve the triangle. To find p. Make p the middle A C part, and we have, by Napier's rules, b or by (200), sin p = sin A sin h, 88 TRIGONOMETRY. or, by logarithms, log sin p = log sin A ~ log sin &i (208) TO find b. Make the complement of A the middle part, an(d we have, by Napier's rules, or by (202), cos A= cotkh tan b whence, tan b = tan Ii cos A, (209) or, by logarithms, log tan b = log tan h + log cos A. (210) TO find B. Make the complement of h the middle part, and we have, by Napier's rules, or by (199), Cos h=cot A cot B; whence, cot B== cos h tan A, (211) or, by logarithms, log cot B = log cos h ~ log tan A. (212) Thus, b and B, by observing the algebraic signs, are determined without ambiguity; and p, though determined by its sine, is 'not ambiguous, since 'it must be of the same species as A (Art. 170). EXAMPLES. 1. Given in a right-angled spherical triangle A B C', rightangled at C, the hypothenuse h equal to 1050 34', and the angle A equal to 800 40'; to solve the triangle. Solution. By (208), By (210), By (212), h, log,-sin+9.983770 log tan-10.555053 log cos- 9.428717 A, log sin+9.994212 log cos+ 9.209992 log tan+1O.784220 p,logsin+9.977982 b, log tan- 9.765045 B~ogcot-10.212937 H1ence, p = 710 541 3311, b - 149' 47' 37"1, B-= 1480 30' 54". 2. Given in the spherical triangle A B (1, right-angled at (1, BOOK V. 89 the hypothenuse h equal to 70~ 23' 42", and the angle A equal to 66~ 20' 40"; to find the other parts. Ans. p, 59~ 38' 26"; b, 48~ 24' 15"; B, 52~ 32' 55". CASE II. 173. Given the hypothenuse and one side. Let there be given (Fig. Art. 172) the hypothenuse h and the side p; to solve the triangle. To find A. Make p the middle part, and we have, by Napier's rules, or by (200), sin p = sin A sin h; whence, sin A =sin ' (213) sin h' or, by logarithms, log sin A = log sin p - log sin h. (214) To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (203), cos B = cot h tan p, or, by logarithms, log cos B = log cot h + log tan p. (215) To find b. Make the complement of h the middle part, and we have, by Napier's rules, or by (198), cos h = cos p cos b; whence, cos b = os h(216) or, by logarithms, log cos b = log cos h - log cos p. (217) Here, as in the preceding article, b and B are determined without ambiguity, for there is only one angle less than 180' corresponding to a given cosine; and A must be of the same species asp. 90 TRIGONOMETRY. EXAMPLES. 1. Given in a right-angled spherical triangle A B C, the hvpothenuse h equal to 91~ 42', and the side p equal to 95~ 22' 30"; to solve the triangle. Ans. A, 95~ 6'; B, 71~ 36' 45"; b, 71~ 32' 12". 2. Given in a right-angled spherical triangle, the hypothenuse equal to 70~ 23' and a side equal to 48~ 24'; to solve the triangle. CASE III. 174. Given one side and its adjacent oblique angle. Let there be given (Fig. Art. 172) the side b and the angle A; to solve the triangle. To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (207), cos B = sin A cos b, or, by logarithms, log cos B= log sin A + log cos b. (218) To findp. Make b the middle part, and we have, by Napier's rules, or by (205), sin b -cot A tan p; whence, tanp - tan A sin b, (219) or, by logarithms, log tan p = log tan A + log sin b. (220) To find h. Make the complement of A the middle part, and we have, by Napier's rules, or by (202), cos A - cot h tan b; whence, cot h = cos A cot b, (221) or, by logarithms, log cot h = log cos A + log cot b. (222). ..-BOOK -V. 91 - 91 EXAMPLES. 1;. Given in a spherical triangle A B C, right-angled at C, the side b equal to 290 46' 8", and —the angle A equal to 1370 2412' 2ii; to solve the triangle. Ans. B,- 540 14 16"/; p, 1550 27' 54"; h4, 1420 9' 13";2. Given in a spherical triangle A B C, -right-angled at C, the side p equal to 1490 47' 23", and the angle B equal to 800'40'; to find-the other parts. CASE IV. 175. Given one side and its opposite oblique angle. Let there be given in a spherical triangle A B U, right-angled at C', the side p and the opposite angle A; to solve the triangle. A To find h4. Make p the middle part, and we have, by Napier's rules, or by (200), \A'. (223) whence, or, b y logarithms, sin p =r sin A sin h; s in'p rsin Is in A'. log sin A = log sin p log sin -A. (224). To find b. Make b the middle part, and we have; by NapileA.' rules, or by (205), sin b = cot A tan p, or, by logarithms, log sin b log cot A + log tan p... (225) To find B. Make the complement of A the middle arto arnd we have, by Napier's rules, or by (206),- cos A = sin B cosp; ' whence, sin B=_ o Cos P or, by logarithms, log -sin B =-log cos A4- -log coo p.. (226) (227) 92 TRIGONOMETRY. Here, since all the unknown parts are determined by their sines, and since there are always two angles less than 180~ corresponding to a given sine, h, b, and B may be taken either acute or obtuse; hence there may be two solutions. For, produce A B and A C till they meet in A', then we have a second triangle, A' B C, which satisfies the given conditions, for it has a right angle at C, the given side p, and A' equal to A, the given angle. But h', b', and B, the other parts of the second triangle, are respectively the supplements of h, b, and B of the first triangle. When, however, p is given equal to A, we have A, b, and B, each equal to 90~, and the triangle A'B C is equal to the triangle A B C (Geom., Prop. XII. Bk. IX). When p and A are both equal to 90~, h is also equal to 90~, and b and B are equal, but indeterminate. EXAMPLES. 1. Given in a spherical triangle A B C, right-angled at C, the side p equal to 36~ 31', and the angle A equal to 37~ 25'; to solve the triangle. Ans. h, 78~ 20', or 101~ 40'; b, 75~ 26', or 104~ 34'; B, 81~ 12', or 98~ 48', when carried only to minutes. 2. Given in a spherical triangle A B C, right-angled at C, the side b equal to 79~ 30', and the angle B equal to 89~ 35'; to solve the triangle. CASE V. 176. Given the two sides about the right angle. Let there be given (Fig. Art. 172) the sides p and b; to solve the triangle. To find h. Make the complement of h the middle part, and we have, by Napier's rules, or by (198), cos h = cos p cos b, or, by logarithms, log cos h = log cos p + log cos b. (228) BOOK V. 93 To find A. Make b the middle part, and we have, by Napier's rules, or by (205), sin b cot A tanp; whence, cot A cot p sin b, (229) or, by logarithms, log cot A = log cot p + log sin b. (230) To find B. Make p the middle part, and we have, by Napier's rules, or by (204), sin p = cot B tan b; whence cot B = sin p cot b, (231) or, by logarithms, log cot B = log sin p + log cot b. (232) These formula determine h, A, and B without ambiguity. EXAMPLES. 1. In a spherical triangle A B C are given the sides about the right angle, p equal to 48~ 24' 15", and b equal to 59~ 38' 27"; to solve the triangle. Ans. h, 70~ 23' 42"; A, 52~ 32' 55"; B, 66~ 20' 40". 2. Given in a right-angled spherical triangle, the side p equal to 95022'301/, and the side b equal to 71 32' 14"; to find the other parts. CASE VI. 177. Given the two oblique angles. Let there be given (Fig. Art. 172) the angles A and B; to solve the triangle. To find A. Make the complement of h the middle part, and we have, by Napier's rules, or by (199), cos h = cot A cot B, or, by logarithms, log cos h = log cot A + log cot B. (233) To find p. Make the complement of A the middle part, and we have, by Napier's rules, or by (206), 9 .94 TRIGONOMETRY. cosA =sin B cosp; =cos A whence, Co si B (234) or, by logarithms, log, cos p log cos A -log sin B. (235) To find b. Make the complement of B the middle part, and we have, by Napier's rules, or by (207), cosB= sin A Cos b; whence, Cos b cos B 26 oby logarithms, log cos =b log cos B -log sin A. (237) Here h, p, and b are determined without ambiguity. EXAMPLES. 1. In a right-angled spherical triangle A B C are given the two, oblique angles, A equal to 440 50', and B equal to 650 49'5 3"; to solve the triangle.. Ans. h, 630 10' 4"1; p, 380 591 11"; b., 54' 30'. 2. Given, in a right-angled spherical triangle, the two oblique angles, A equal to 125' 30', and B equal to 800 40'; to find the other parts. QUADRANTAL TRIANGLES. 178. A QUADRANTAL TRIANGLE is a spherical triangle having one of its sides quadrantal, or equal to 90'. Quadrantal triangles may be B solved in the same manner as right-angled. spherical triangles, by means of the polar triangle. A' Let 4 BC be a quadrantal tri- AC% angle, and Al BV C denote a tri- A' b angle polar to it; then, by Art. C 160, we have A 61 AI1800.-..p, B'=180'- b, U'=1800 - h p'=180 - A, b = 1800 -B, h' =180' — Ci. BOOK V. 95 Now, if the side h be taken equal to 90~, its corresponding polar angle C' will also equal 90~; hence the polar triangle will be right-angled, and can be solved by application of the preceding formula for right-angled spherical triangles, and thus the required parts of the quadrantal triangle may be determined. A triangle, one of whose sides is a quadrant, may also be solved by laying off a quadrant on one of the other sides, prolonged if necessary, and connecting this last point with the other extremity of the original quadrant by the arc of a great circle, thus making the original quadrantal triangle either the difference or the sum of a bi-quadrantal and a right-angled spherical triangle. Solving the latter solves the original triangle. Thus, AC" measures B, CA C" - 90~ - A, C C" = 90~ - p, A C C" = 180~ - C, and solving the triangle A C C"1 also solves the triangle A B G. EXAMPLES. 1. Let there be given, in a quadrantal triangle ABC, the side h equal to 90~, the angle A equal to 54~ 43', and the angle B equal to 42~ 12'; to find the other parts. By taking the supplements of the given parts, we have in the polar triangle, p' = 125~ 17', b' = 137~ 48', whence A', B', and h' are determined as in Art. 176, and the supplements of these give the required parts of the quadrantal triangle. Ans. p, 64~ 34' 40"; b, 48~ 0' 16"; C, 115~ 20' 5". 2. Given two sides of a quadrantal triangle equal to 72~ 53' and 51~ 4', to find the angle opposite to the quadrantal side. Ans. 104~ 24' 21". SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 179. In the solution of oblique-angled spherical triangles, it -is sometimes found convenient, especially in removing an ambiguity, to refer to one or more of the following propositions, of which the first four have been demonstrated in Book IX. of the Geometry. 96 TRIGONOMETRY. I. Any side of a spherical triangle is less than the sum of the other two. II. The sum of the sides is less than 360~. III. The sum of the angles is greater than 180~. IV. The greater side is opposite the greater angle, and conversely. V. Any angle is greater than the difference between 180' and the sum of the other two angles. VI. A side which difers more from 90~ than another side, is of the same species as its opposite angle. For, by (150), we have cos a = cos b cos c + sin b sin c cos A; whence, cos A cos a - cos b cos c cos A.. sin b sin c in which the denominator is always positive. Then, if a differs more from 90~ than b or than c, cos a is numerically greater than cos b, or than cos c, and we have cos a > cos b cost, hence, the sign of the numerator, and consequently the sign of cos A, is the same as that of cos a, that is, A and a are in the same quadrant. VII. An angle which differs more from 90~ than another angle, is of the same species as its opposite side. For, by (153), we have cos A = sin B sin C cos a -cos B cos C; whence, cos A + cos B cos C cos a --- cos a sin B sin C in which, if A differs more from 90~ than B, or than C, cos A is numerically greater than cos B, or than cos C, and the sign of cos a is the same as that of cos A, that is, a and A are in the same quadrant. VIII. When the sum of two sides is greater than, equal to, or ess than 1800, the sum of the two opposite angles is the same. BOOK V. -97 For, by means of (188), tan 2 (a + b) cos (A + B) =tan ~ c cos I (A - B), in which the second member is always positive, since e c and I (A - B) are each less than 90~, so that the factors of the first member, tan I (a + b) and cos - (A + B) must have the same sign. Therefore,.- (a + b) and ~ (A + B) are of the same species. 180. In the solution of oblique-angled spherical triangles, there are six cases, the data in them being, respectively, I. Two sides and an angle opposite one of them. II. Two angles and a side opposite one of them. III. Two sides and the included angle. IV. Two angles and the included side. V. The three sides. VI. The three angles. CASE I. 181. Given two sides and an angle opposite one of them. Let there be given, in the oblique- c angled spherical triangle A B C, the A sides a and b, and the angle A; to solve the triangle. b\ To find B. We have, from (148), C sin b. sin B sin A, sin a or, by logarithms, log sin B log sin b - log sin a + log sin A. (288) To find C and c. We have, by Napier's analogies, (186) and (188), cot =cos (a + b) tan (A + B), cos (a - b) co (A + B) tan c cos (A-B) tan (a+ b); or, by logarithms, 9* 98 TRIGONOMETRY. log cot C = log cos 3 (a + b) - log cos, (a- a) + log tan (A - B), (239) log tan, c = log cos ' (A +B) - log cos ( (A - B) + log tan 3 (a + b), (240) which determine - C and c, and thence C and c. In this case, since B is found from its sine, it will sometimes admit of two values, the one supplementary to the other. When B has two values, C and c must each have two corresponding values. Whether both values of B are admissible must be determined by one of the propositions of Art. 179. Thus (Prop. VI.),.if b differs more from 90~ than a, B must be of the same species as b, and there can be but one solution; but if b differs less from 90~ than a, there may be two solutions. Or (Prop. VIII.), if only one of the supplementary values of B makes j (A + B) of the same species as j (a + b), there can be but one solution; but if both values of B fulfil that condition, there will be two solutions. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the side a equal to 63~ 50', the side b equal to 80~ 19', and the angle A equal to 510 30'; to solve the triangle. Solution. By (238) we have a = 63~ 50' ar. co. log sin 0.046958 b = 80~ 19' log sin 9.993768 A - 51~ 30' log sin 9.893544 B = 59~ 15' 57", or 120~ 44' 3" log sin 9.934270 As b differs less from 90~ than a, both values of B are admissible, and we have 3 (b-a) = 8~ 14' 30", 3 (a+b) = 72~ 4' 30", } (A+B) = 550 22'58" or 86~ 7'2", and 3 (B-A) - 3~ 52' 58 or 34 37'2". The cosines of 3 (b-a) and 3 (B-A) are the same as those of 3 (a-b) and I (A-B), respectively, by Art. 79. Hence, by (239) and (240), BOOK V. 9 99 (b-a) ar. co. log cos~ 0.0041508 ar. co. log cos.-f 0.004508 _4 (a-fb) log cos-f 9.488229 log cos-f 9.488229 kX (A-+B) log tan~10.160964 or log -tan+-1 1. 1683 14 jC log cot-f 9.6537-01 or~ log cot-f10.661051 - C= 650 441 53"1 or 120 18' 42",9 6C= 1310 29' 46'- or 240 371 24". j (B-A) ar. co. log cos-f 0.000998 or ar. co.lIogcos-f- 0.084618 ~-(A-fB) log cos~ 9.754418 or log cos-f 8.830687 (a-fb) log, tan+10.490161 log tan~10.490161 C ~~~log, tan-j10.245577 or log tan-+ 9.405466 c =60' 23' 57"1 or 140 16' 18", c -1200 47/ 54"1 or 280 32' 36". 2. Given, in an oblique-angled spherical triangle, two sides equal to 990 40' 48"1 and 640 23' 15"1, and an angle opposite to the first of' these equal to 950 38' 4"1; to find the other side and angles. Ans. Side, 1000 49' 30"; angles, 650 33' 10"1 and 970 26' 30"., CASE II. 182. Given two angles and a side opposite one of them. Let there be given, in the oblique-angled spherical triangle A B C (Fig. Art. 181), the angles A and B, and the side a; to solve the triangle. To fnd b. We have, fro'm (148), sin B sin b= sin sin a, or, by logarithms, log sin b = log sin B - log sin A ~ log sin a. (241) To find C and c. We use equations (239) and (240), a's in the last article. This case is exactly analogous to Case I., and gives rise to the same ambiguities, as may be shown by -passing to the polar triangle. 100 TRIGONOMETRY. If B differs more from 90~ than A, b must be of the same species as B, and there can be but one solution; but if B differs less from 90~ than A, there may be two solutions. (Prop. VII. Art. 179.) Or, if only one of the supplementary values of b makes i (a + b) of the same species as 1 (A + B), there can be but one solution; but if both values of b fulfil that condition, there will be two solutions. (Prop. VIII. Art. 179.) EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 135~, the angle B equal to 60~, and the side a equal to 155~; to find the other parts. Ans. C, 98~ 3' 4" or 16~ 57' 1"; b, 31~ 10'17" or 148~ 49/ 43"; c, 143~ 42' 57" or 10~ 2' 6". 2. Given, in an oblique-angled spherical triangle, two angles equal to 97~ 26' 30" and 65~ 33' 10", and the side opposite to the first equal to 100~ 49' 30"; to find the other parts. CASE III. 183. Given two sides and the included angle. Let there be given, in the oblique-angled spherical triangle A B C (Fig. Art. 181), the sides a and b, and the included angle 7; to solve the triangle. To.find A and B. By means of Napier's analogies (186) and (187), we have tan. (A + B)- (ab) cot cos j (a +- b) tan ~ (A - B) - cot " C; sin j (a - b) cot or, by logarithms, log tan (A+ B) = log cos z (a-b) - log cos 3 (a-+b) + log cot J C, (242) log tan ( (A-B) = log sin, (a-b) - log sin ~ (a+b) + log cot 7 C, (243) which determine, (A + B) and, (A - B). The sum of these values gives A, and the second subtracted from the first gives B. BOOK V. 101 To find c. We use equation (240), as in the first two eases. The value of c might also be obtained by (147) or (149); but as it is thus determined from its sine, it would be necessary to remove the ambiguity by means of the principles contained in Art. 179. As A, B, and c may all be found by means of tangents, there can be but one value for each. It will be observed that.1 (A+ B) must always be of the same species as If (a ~ b). (Prop. VILL. Art. 179.) EX AM-PL ElS. 1. Given, in an oblique-angled spherical triangle A B C, tlhe side a equal to 700, the side b equal to 38' 30', and the included angle (1 equal to 3lV 34' 26"; to solve the triangle. Solution. (a-f-b) = 540 15', J (a -b) = 150 45k, and I C 150 47' 13"; then, By (242), By (243), k (a+b) ar.co.logcos+ 0.233402 ar. co. log sin + 0.090672 -(a -b) log cos + 9.983381 log, sin+f- 9.433675 ~~ C log00 cot + 10.548635 log cot ~ 10.548635 (A + B9) log, tan + 10.765418 I (A-B) logtan+ 10.072982 J (A +B) =800 15' 41"1 J (A -B)-490 47' 30" A - 1300 3' 11" B- 300 28' 11"J By (240), (A - B) 490 47' 30"/ ar. co. log cos + 0.190058 (A + B) == 800 15' 41" log cos ~ 9.228282 ~.(a ~ b) = 540 15' log tan + 10.142730 c = 20' log tan + 9.561070 Ans. Angle A, 1300 3' 11"; angle B, 300 28' 11"; side c, 400. 2. Given, in an oblique-angled spherical triangle, an angle equal to 480 36', and the two adjacent sides equal to 1120 22' 58kJ" and 890 16' 53k"; to find the other parts. 102 102 ~~~TRIGONOME~TRY. CASE IV. 184. Given two angles and the included side. Let there be given, in the oblique-angled spherical triangle A B C (Fig. Art. 181), the angles A and B, and the included side c; to solve the triangle. To find a and b. By means of Napier's analogies (188) and (189), we have tan ~-(a ~ b) co2 AB tan j C cos ~(A + B) tan ~.(a-b) =tsin A-B sin. (A + B) tn~C or, by logarithms, log tan j (a ]-b) = log cos (A-B) - log, cos (A +B) + log tan'~ c, (244) logtan'j- (a-b) =log, sin ~-. (A-B) - log sin I (A~ B) ~ log tan l-c, (245) which determine.- (a + b) and I. (a - 6), and thence a and 6. To find 0. We use equation (239), as in the first two cases; but (147) or (149.) may be employed, as in the last case. This case is analogous to Case Ill., and gives rise to no ambiguity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle A B 6, the anggles A and B equal to 1190 15' and 700 39', and the side c equal to 520 39' 4"1; to solve the triangle. Ans. Sides a and b, 1120 22' 58k" and 890 16' 54".1; angle 6C, 480 3 6'. 2.. In an, oblique-angled spherical triangle, given two angles equal to 1300 3' 11"1 and 310 34' 26"1, and the included side equal to 880 30'; to find the other parts. CASE V..185. Given the three sides. Let there be given, in the oblique-angled spherical triangle A B C (Fig. Art. 181), the sides a, 6, and c; to -solve. the tri. I BOOK V..108 To0 jnd A, B, and C, we have, by (164), (165),- and (166), sin A - (s -b) sin (s-c) in - ~~~sin bsin c sin --- in(s c)sin (s -a) sin ~ B /sinsinc)si a sins c= ysin is - a) sin (s -b); or, by logarithms, logsin~.A log sin (s-b)+loog sin (s-c)-log sin b-log snc26 logsnjB= -log sin (s-c)+log, sin (s-a) -log sin c-log sin a 27 log in IV =log sin (s-a)+loog sin (s-b)-log sin a-log sin b (248) A, B, and C can also be determined by formulae (167), (168), and (169) for the cosine of' half an angle, and by formulae (170), (171), and (172) for the tangent of half an angle. Since the half-angles must be less than 900, there is no ambiguity in determining the angles by any of these formnulae. EXAMPLES. I. Given, in an oblique-angled spherical triangle, the side a equal to 700, the side b equal to 380, and the side c equal to 400 to find the angles. Solution. 8=74', s-a=40, s-b= 360, s-c=340. By (246), By (247), By (248), s -bI, logsin9.769219 log sin 9.769219 s-C, logsin9.7475e2 log sin 9.747562 a - a,. ~~~log sin 8.843585 log sin 8.843585 b, ar.co.log sin 0.210658 ar.co.log sin 0.210658 c, ar.co.logsinO0.191933 ar.co.logsinO0.191933 a, _____ar~co.log sin 0.027014 ar.co.logsin0.027014 2) 19.919372 2) 18.810094 2 ) 18.850476 A, log sin 9.959686 frB, log sin 9.405047 j. C, log sin 9.425238 -104 TRIGONOMETRY. A = 650 41' 33".7 1- B = 140 43' 18" 3 =0150 2 6'21"1.7 Ans. A, 1310 23' 7"; B, 290 26' 36"; (v, 3O0 52' 43". 2. Given, in an oblique-angled spherical triangle, the sides equal to 112' 22' 59", 890 16' 53"1, and 520 391 4"1; to solve the triangle. CASE VI. 186. Given the three angles. Let there be given, in the oblique-angled spherical triangle A BO (Fig. Art. 181), the angles A, B, and C; to solve the triangle. To find a, b, and c, we have, by (175), /cos S cos (S -A) sin 3-a = V sin B sin C /coS cos_(S -B) 3- V ~~sin Csin A /Cos S Cos (8 - C) sin c= sin Asin B or, by logarithms, log ~inj a log cos S+log,, cos (S-A)-log sin B-log sin C 29 log sin j = log cos S+log cos (S-B) ----og, sin C-log sin A (250) logsin~ =log Cos S+log cos (5-QG-1og sin A-log sin B(21 a, b, and c can also be determined by formulae (176) for the cosine of half an angle, and by formulae (177) for the tangent of half an angle. Here a, b, and c are determined without ambiguity. EXAMPLES. 1. Given, in an oblique-angled spherical triangle, the angle A equal to 1200 43' 37",y the angle B equal to 1090 55142"1, and the angle C equal to I1160 38' 33"; to find the sides. Ans. a, 1150 13' 26"; b, 980 21' 39"; c, 1090 50' 20". 2. Given the angles in an oblique-angled spherical triangle equal to 1310 23' 7"1, 290 26' 36"1, and 300 52' 43"; to 'solve the triangle. BOOK VI. APPLICATIONS OF SPHERICAL TRIGONOMETRY TO ASTRONOMY AND GEOGRAPHY. 187. The CELESTIAL SPHERE is the spherical concave surrounding the earth, in which all the heavenly bodies appear to be situated. 188. The ZENITH is that pole of the horizon which is directly overhead. 189. The ALTITUDE of a heavenly body is its distance above the horizon, measured on the arc of a great circle passing through that body and the zenith. 190. The DECLINATION of a heavenly body is its distance north or south of the celestial equator, measured on a meridian. 191. The altitude of the celestial pole is equal to the latitude of the place where the observer is located. For the distance from the zenith to the celestial equator is the latitude of the place, and the distance from the zenith to the pole is its complement; but the distance from the zenith to the pole is also the complement of the altitude of the pole; hence the latitude of the place and the altitude of the pole are equal 192. To find the time of the RIsING AND SETTING OF THE SUN at any place, the sun's declination and the latitude of the place being given. IS Let P represent the celestial north pole, E A Q the celestial equator, HA O the rational horizon, S the place of the sun's rising, S' the posi10 106 TRIGONOMETRY. ton of the sun at 6 o'clock, PE P' the meridian of the given place, PB P the meridian passing through S, and P AP the meridian 90~ distant from PE P', passing through SI. From the time of the sun's rising to 6 o'clock, it will pass over S S, the arc of a small circle, corresponding to B A, the arc of a great circle. The length of BA, expressed in time (Art. 147), will then give the amount to be taken from or added to 6 o'clock, to give the time of the sun's rising or setting. B S is the sun's declination, P 0 is the latitude of the place (Art. 191), and Q 0, which measures the angle BA S, is its complement; hence, in the right-angled spherical triangle A B S, there are known the side B S and the angle B A S, from which, by Art. 175, sin BA - tan BS cot BA S, or, log sin B A = log tan sun's decl. + log tan lat. of place. After reducing the arc B A to time, at the rate of 159 to an hour, or 4m. to a degree, it must be added to 6 o'clock for the time of the sun's setting, and subtracted for its rising, when the declination and latitude are both north or both south; but subtracted for its setting and added for its rising, when one is north and: the other south. The preceding reasoning rests upon the assumption that the sun's declination does not change between sunrise and sunset, which, although not strictly true, is accurate enough for our present purpose. The time obtained is apparent time, and a correction must be applied if we wish to find mean time, or that indicated by the clock. Another correction is necessary for refraction. Neither of these corrections has, however, been applied to the answers that follow. EXAMPLES. 1. Required the time of the sun's rising and setting in Edinburgh, latitude 55~ 57' N., when the sun's declination is 23~ 28' N. Ans. Rises, 3h. 20m. 7s.; sets, 8h. 39m. 53s. 2. What is the time of the sun's rising and setting in latitude 60~ 3' N., when the sun's declination is 23~ 28' S.? Ans. Rises, 9h. 15m. 33s.; sets, 2h. 44m. 27s. BOOK VAL. -107 3. Required the time of the sun's rising and setting in places whose latitude is 48~ S., when the sun's declination is 15~ S. 193. To find the IOUR OF THE DAY at anyplace, the latitude of the place and the sun's declination and altitude being given. Let Z represent the zenith, and z Z D an arc of a great circle drawn p through the zenith and the sun's \ S place, S; P B /P, a meridian drawn through the sun's place, &c., as in H the last article. As before, the arc A B, added to or subtracted from 6 o'clock, will Pa 9 give the time when the sun is at S; but it will be more convenient to use its complement, E B, which is the time before or after 12 o'clock. E Z is the latitude of the place, and P Z is its- complement; B S is the sun's declination, and P S is its complement; S D is the altitude of the sun, and Z S is its complement; hence, in the spherical triangle Z P S, the three sides are known, and the angle ZP S, or the arc E B, may be found by Art. 185. If either the sun's declination, or the latitude of the place, is south, it must be considered negative in taking its complement. unless the south pole is taken as a vertex of the triangle, when north will be negative. EXAMPLES. 1. Required the apparent time of day in the morning, at a place in latitude 390 54' N., the sun's declination being 17' 29 N., and its corrected altitude 15~ 54'. Ans. 6h. 25m. 30s. A, M. 2. In latitude 36~ 39' S., when the sun's declination was 9~ 27' N., its corrected altitude was observed, in the afternoon, to be 10~ 40'; what was the apparent solar time? Ans. 4h. 36n. 10s. P. M. 3. Required the apparent time of day in Boston, latitude 42~ 21' N., when the sun's declination is 20~ S., and its corrected altitude 15~ 15', the sun being east of the meridian. 108 TRIGONOMETRY. 194. To find the SHORTEST DISTANCE between two places on the earth's surface, and the BEARING of one from the other, their latitudes and longitudes being given. Let Z and S represent the two z points on the earth's surface, and P p the north pole of the earth. P Z and 7\ P S are the complements of the lati- B tudes of the two places, and the are A EB, or the angle ZP S, is the difference of their longitudes; hence, in the spherical triangle S P Z, the two sides P P Z and P S, and their included angle P, are known, from which the side ZS, and the angles ZSP and S Z P may be found by Art. 183. The distance Z S can easily be reduced to miles by allowing 69.16 statute miles, or 60 nautical miles, to a degree. The answers which follow are given in statute miles. If one place is south and the other north of the equator, the south latitude must be considered negative in taking its complement. EXAMPLES. 1. What is the distance and bearing of Jerusalem, lat. 310 47' N., long. 35~ 20' E., from London, lat. 51~ 30' N., long. 6' W.? Ans. Distance, 2248 miles; bearing, S. 66~ 31' E. 2. Required the distance and bearing of Cape Horn, lat. 550 58' S., long. 67~ 21' W., from London. Ans. Distance, 8363 miles; bearing, S. 360 59' W. 3. Required the distance and bearing of Quito, lat. 0~, long. 78~ 45' W., from San Francisco, lat. 37~ 49' N., long. 1220 14' W. A T A BLE OF LOGARITHMIC SINES, COSINES, TANGENTS, AND COTANGENTS, F03 EVER? DEGREE AND MINUTE OF THZ QUADRANT. 16 M. 1 Sine. 0 - OD 1 6.463726 2.764756 3.940847 4 7.065786 5.162696 6.241877 7.308824 8.366816 9.417968 10.463725 11 7.505118 12.542906 13.577668 14.609853 15.639816 16.667845 17.694173 18.718997 19.742477 20.764754 21 7.785943 22.806146 23.825451 24.843934 25.861662 26.878695 27.895085 28.919879 29.926119 30.940842 31 7.955082 82.968870 38.982233 34.995198 35 8.007787 36.020021 37.031919 38.043501 39.054781 40.065776 41 8.076500 42.086965 48.097183 44.107167 45.116926 46.126471 47.135810 48.144953 49.153907 50.162681 51 8.171280 52.179713 53.187985 54.196102 55.204070 66.211895 57.219581 68.2271-34 69.234557 60.24185 5 I Cosine. LOGARITHMI1C SINES, COSINES, 00 D. 5017.17 2934.85 2082.31 1615.17 1319.68 1115.78 966.53 852.54 762.63 689.88 629.81 579.37 536.41 499.38 467.14 438 81 413.42 391.35 371.27 353.15 336.7~2 321.75 308.05 295.47 283.88 273.17 263.23 253.99 245.38 237.33 229.80 222.73 216.08 209.81 203.90 198.31 193.02 1t88.01. 183.25 178.72 174.41 170.31 166 39 16~2.65 159.08 155.66 152.38 149.24 146.22 143.33 140.54 137.86 135.29 132.80 130.41 128.10 125.87 123.72 [21.64 "D.P Cosine. j D. I Tang. D. Cotang.I 10.0000O0 - 00 00 60.000000 1 6.463726 3.536274 59.000000:.764756 5017.17.235244 58.000000.940847 24.059153 57.000000 0 7.065786 2082.3 2.934214 56.000000.00.162696 1615.17.837304 55 9.999999 0.241878 1319.68.758122 54.999999.308825 1115 78.691175 53.01.3885 9.53.9999QCI..366817.633183 52.999999.ot.417970 852.54.582030 51.01 ~~~762.6.3.999998.463727 76 362.3 63.01 689-88 9.999998 7.505120 2.494880 49.99999.oi.542909.457091 48.9 9.577672 579.38.422328 47.s999996.01 5.36.42.999996.01.609857 536.42.390143 46.999996.01.639820 4.360180 45.ot.667849 4867.1.332151 44.999995.oi.69417 4.305821 43.01.7190a441 3-73.999994.oi.719004 41.73.280997 42.999993.742484.257516 41.01 785951 371128 9.999992.o1 7764761.235239 40 153~ 17 9.999992 oi 7 78595 336 73 2.214049 39.999991 01.806155 33673.193845 38.999990.01.825460.174540 37.999989 02.843944 308.06.156056 36.999988.861674 2)5.4;.138326 35.999988.02.878708 283.90.121292 34.999987.02.895099 273.18.104901 33.999986.02.910894 263.25.089106. 82.02.254.0 b106 ~31.999985.02.926134 254.01 073866 31.999983.02.940858 245.40 059142.30 237.35 9.999982.02 7.955100 229.81 2.044900 29.999981.02.968889 222-75.031111 28.999980.02.982253 222.75 017747 27.9999.9 02.995219 216 10 004781 26.999977.02 8.007809 209.83 1.992191 25.999976.02.020045 203.92.979955 24.999975.02.031945 198.33 968055 23.999978.02.043527 193.05 956473 22.999972.02.054809 188.03.945191 21.999971.02.065806 183.27 934194 20 'e.9999s s ~~178.74 ~O.999969 8.076531 174.44 1.923469 19.999968.02 -086997 170.34.913003 18.999966.02.097217 166.42.902783 17.999964.03.107202 15268.892797 16.999963.03.116963 15910.883037 15.999961.03.126510 15568.873490 14 -999959.03.135851 15241.864149 13.999958.03.144996 149.27.855004 12.999956.03.153952 146-27 "846048 11.999954.0.162727 6.837273 10 148-36 9.999952.03 8.171328 140-57 1.828672 9.999950.03.179763 13790.820237 8.999948.03.188036 13532.811964 7.999946.03.196156 13284.803844 6 9994.03.204126 13044.795874 5.999942.04.211953 12814.788047 4.999940.04.219641 12590.780359 3.999938.04.227195 12. 72805 2.0493 123.76.999936.04.234621.765379 1.999934'.241921 121.68.758079 0 Sine. D Cotang. P. Tang. IM. TANGENTS, AND COTANGENTS. lo 19 I I I I M. Sie0 b.241boo 1.249033 2.256094 3.263042 4.269881 6.276614 6.283243 7.289773 8.296207 9.302546 10.308794 11 8.314904 12.321027 13.327016 14.332924 15.338753 16.344504 17.360181 18.355783 19.361315 20.366777 21 8.372171 22.377499 23.382762 24.387962 26.393101 26.398179 27.403199 28.408161 29.413068 30.417919 31 8.422717 32.427462 33.432156 34.436800 86.441394 36.445941 37.450440 38.454893 39.469301 40.463665 41 8.467985 42.472263 43.476498 44.480693 45.484848 46.488963 47.493040 48.497078 49.601080 60.505045 61 8.608974 62.612867 63.616726 64.620551 65.624343 66.628102 67.631828 68.635523 69.639186 60.542819 Cosine. I I D). Cosine. D. 119.63 9999932.04.999932 117.68 '999929.04 115.80 999927.04 113.98 999925.04 112.21 999922.04 110.50 '999920 04 108.83 999918.04 107.21 '999916.04 105.65.04 104.13 '999910.04.999910 102.66.04 9.999907 1.1-2 999905 0 98.7 999902 ( 99.82:~.01 9 999899 97.14 '999897.0O 9586.05 94.60 '999891 05 93.38 '999888 ( 92.19 999885.05 91.03 '999882 05 89.90 998.05 9.999879 87-72.999876.Oo 84667.9987 999870 85.64 999867.05 84.64 '999864 83 66 1999861 82.71 '999868 '05 81.77.05 80.86.999851.05.999851 79.96 06. 9.999848 99984. 78.723 994.06 7740 9998~38 0 76.57 999834.06 7 999831 06 74-22 '999827.06 7 999823 73.43 999820.06 2 999816 72.00 06. 9.999812 71-2 —.06.999809 6310.999805.06.999801 69.24 06 9.999797 6859.07 67940.07.999793 67.34.999790 07 6 999786 6.60 *999782.0 6 999778.07 65.48.07. 9.999774 64.89.07.999769 64.31.07.999765 63.19 9.07.99976 0 63211 9.07.99978. 62.0.07 9.999753 62.11).07.999748 61.58.07.0999744 610:6.999740.07 60-5 99.07.999735 8.241921.249102.256165.263115.269956.276691.283323.289856.296292.302634.308884 8.315046.321122.327114.333026.838866.344610.350289.355895.361430.366895 8.372292.377622.382889.388092.393234.398315.403338.408304.413213.418068 8.422869.427618 432316.436962.441660.446110.450613.465070.469481.463849 8.468172.472464.476693.480892.486050.4891710.493250.497293.601298.606267 8.609200.513098.616961.620790.624586.628849.632080.536779.539447.543034 ITang. D. Cotang. 1.7580796 119.67.750898 69 117.72.743835 68 115.84.736885 67 114.02.730044 66' 112.25.723309 65 110.57.716677 64 108.87.710144 53 107.26.703708 62 104.70.697366 61.691116 60 102.70 1.684964 49 101.26.678878 48 93.87.672886 47 98.51.666976 46 97.19 5.661144 46.655390 44 94.65.649711 43.644106 42 91.08.638670 41 89.95.633105 40 1.627708 89 88.85.622878 88 87.77.617111 37 86.72.611908 36 85.70.606766 36 84.70.601686 34 83.71.696662 83 82.76.691696 82 81.82.686787 31.5.1 81982 80 80.021 1.677131 29 79.14.6728832 28 78.30.667685 27.668038 26 76.63.668440 26 75.83.663890 24 76.05.649387 23 74.28.644980 22 73.62.640619 21 72.79.686151 20 72-06 72.06 1.631828 19 71.65.627646 18 70.66.628307 17 69.98.619108 16 66.31.614960 16 68 65.610830 14 68.01.606760 10 67.38.602707 12 66.76.498702 11 66.16.494783 10 65-55 65.66 1.490800 9 64.96.486902 8 64.39.488089 7 637.82.479210 6 63.26.475414 6 62.72.471651 4 62.18..467920 61.66.464221 2 61.13.4063 A 6.2.456916 0 I i D I- Sine. I D. I Cotang. I D. I Tang. - I I 20 LOGARITHMIC SINES, COSINES, 20 I M I — sine. I --- I Coiine- I D. I Tang. I D. I Cotang. I.1 0 1 2 3 4 6 6 7 8 9 10 11 12 13 14 16 16 17 19 20 21 22 23 24 25 26 27 28 29 80 81 32 83 34 36 36 37 38 39 40 41 42 43 44 46 46 47 48 49 60 61 62 63 64 66 66 67 68 69 60 b.b4z61v.646422.649995.663539.657054.660540.663999.667431.670836.674214.6775666 8.680892.684193.587469.690721.693948.697162.600332.603489.606623.609734 8.612823.616891.618937.621962.624965.627948.630911.633864.636776.639680 8.642663.646428.648274.661102.663911.666702.669476.662230.664968.667689 8.670393.673080.676761.678406.681043.683666.686272.688863.691438.693998 8.696643.699073.701689.704090.706577.709049.711607.713962.716383.718800 60.04 69.55 59.06 68.58 58.11 57.65 57.19 66.74 66.30 65587 55.44 55.02 64.60 64.19 63.79 53.39 63.00 52.61 62.23 61.86 61.49 61.12 60.76 60.41 60.06 49.72 49.38 49 04 48.71 48.39 48.06 47.75 47.43 47.12 46.8~2 46.52 46.22 46.92 46.63 46.35 46.06 44.79 44.61 44.24 43. 97 43.70 43.44 43.18 42.92 42.67 42.42 42 17 41. 92 41.68 41.44 41.21 40.97 40.74 40. &1 40.29.99973 1.999726.999722.999717.999713.9997108.999704.999699.999694.999689 9.999686.999680.999676.999670.999665.999660.999655.999650.999645.999640 9.999635.999629.999624.9996 19.999614.999608.999603.999597.999592.999586 9.999581.999575.999570.999564.999558.999553.999547.999541.999535.999529 9.999524.999518.999512.999506.999500.999493.999487.999481.999476.999469 9.999463.999466.999450.999443.999437.999431.999424.999418.999411.999404.07.07.07.08.08.08.08.08.08.08.08.08.08.08.08.08.08.08 I 08:09.09.09.09.09.09.09.09.09.09.09.09 0,99.09.09.09.10.10.10.10.10.10:10.10.10.10.10.10.10.10.10?5.D4iul54.646691.650268.653817.557336.660828.564291.567727.5711;37.674520.577b77 8.581208.584514.687795.691051.6942E3.697492.6006747.603839.606978.610094 8.613189.616262.619313.622343.625352.628340.631308.634256.637 184.640093 8 642982.645853.648704.651637.664352 -657149.659928.662689.665433.668160 8.670870 -673563.676239.678900.681644.684172.686784.689381.691963.694529 8 697081.699617.702139.704646.707140.709618.712083.714634.716972 719396 60.12 69.62 59.14 58.66 58.19 67.73 67.27 66.82 566.38 66.96 55.62 65.10 64.68 64.27 63.87 63.47 63.08 52.70 62.32 61.94 61.58 61.21 50.86 60.50 50.15 49.81 49.47 49.13 48.80 48.48 48.16 47.84 47.53 47.22 46.91 46.61 46.31 46.02 45.73 45.44 46.16 44.88 44.61 44.34 44.07 43.80 43.54 43.28 43.03 42.77 42.52 42.28 42.03 41.79 41.55 41.32 41 08 40.85 40.62 40.40 1.40ti618I6 60.453309 60.449732 58.416 183 57.442661 I56.439172. 55.435709 6a4.432273 53.4288639 62.425480 5 1.4221123 60 1.418792 4 9.415486 4 8.412205 47.408949 46.405717 45.402508 44.399323 43.396161 42.393-022 41.389906 40 1.386811 39.383738 38.380687 37.377657 36.374648 35.371660 34.36S692 33.366744 32.362816 31.359907 30 1.357018 29.354147 28.351296 27.348463 26.345648 25.342861 24.340072 23.337311 2~2.334567 21.331840 20 1 329130 19.326437 18.323761 17.321100 16.818456 16.815828 14.313216 13.310619 12.308037 It.305471 10 1.302919 9.800383 8.297861 7.295364 6.292860 6.290382 4.287917 3.285465 2.283028 1.280604 0 T ngno. I m.I Co.ine. D. I nine.j I D. JCno~ali D). I 'i7 TANSGENTS, AND COTANGENTS. 30 21. | Sine. 0 8.71bb00 1.721204 2.723595 3.725972 4.728337 5.730688 6.733027 7.735354 8.737667 9.739969 10.742259 11 i 8.744536 12.746802 13.749055 14.751297 15.753528 16.755747 17.757955 18.760151 19.762337 20.764511 21 8.766675 22.768828 23.770970 24.773101 25.775223 26.777333 27.779434 28.781524 29.783605 30.785675 31 8.787736 32.79787 33.791828 34.793859 35.795881 36.797894 37.799897 38.801892 39.803876 40.806852 41 8.807819 42.809777 43.511726 44.813667 45.815599 46.817522 47.819436 48.821343 49.823240 60 825130 51 8.827011 62.828884 58.830749 54.832607 55.834456 66.836297 57 888130 58.839956 69.841774 60.F4351 I 1). Cosine. 9.999404 40.06 99'398 39.84.999391 39.62.999384 3-4.999384 391.4 l.999378 39. 19.999371 38.98.999364 38.77.9993 38.57.099357 38.57 638:J.999350 8.999343 0 <99>336 37.'93 3 9.b99329 37.76.99322 37.56 99315 ^37.:7 (.999308 39.999301 3.98.999294 36.79 999286 38.61.999279 36.42.999272 36 21.9992765 36.0. 35.8 9950 3.70.999242 35.53.999235 35.35.999227 35.18.999220 35.01.999212 34.84.999205 34.67.999197 34.51.999189 31.34 3418 9.999181 34.02.999174 33.86.999166 33.86.999158 33.70 4.999150 33.53.999142 33-3.23.999126 3:.08.999118 32.93 999110 32.78.999110 9.999102 36.999094 32.49 999086 32.34.999077 32.9.999069 32.05.999061 31.91.999053 31.77.999044 31 63 999036 31.49 999027 31.35 9999019 31.22 9.999019.999010 31.08.999002 30.95.998993 30.82.998984 30.69.998976 30.56.998967 30.43. 99 958 30.30 9989950 30.17.998954.99M>4l I I). I Tang. I D. I Cotang. I -. '; ' f L 0ws{iJ{. jX.11.11.11.11.11.11.12.12.12.12.12.12.12.12.12.12.13.13.12.12.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.14.13.14.13.14.14.14.14.14.14.14.14.14.14.14.15.15.15 b.7 Itdu.721806.724204.7265S8.728959.731317.733663.735996.738317.740626.742922 8.745-U7.747479.749740.7 51'J'3.754227.7564'53.75o668.760872.76306.5.76-CA46 8.767417.769578.771727.773C66.775995.778114.780222.7S2320.784408.786486 8.7t554.790613 792662.794701.796731.798752.800763.802765.804758.806742 8.808717.810683.812641.814589.816529.818461.820384.822298.824205.826103 8.827992.829874.831748.883613.835471.887321.839163.840998.842825.F44(;44 1.280604 60 409.17.278194 59.275796 58.273412 57 3J.52 |.271041 56 y3.30) ^.268663 55 38.)).266337 51 318 8'.264004 1;3 1 38 1.(2616t3 -!.33 13.259374 1 38 2i | 3 ~.di.257078 |, ) 387 1 -21.2547 3 4. 37.81.252521 48 3768.250260 47 37.2').248011 43 37.2) 37 810.245773 45 37.92 1.243547 44. 1 241332 43 f3 l7:.239128 43.3;i ).236935 41 l3 3.:8.234754 40 3d. 18 1.232583 39 3j 03.230422 38 35. 8.228273 37 35.05.226134 36 35.48.224005 35 35 31.221186 34 35.1 l.219778 33 3 97 l.217680 32 34-80.215592 31 34.64.213514 30 i. 1.211446 29 34 -31.209387 28 34 14.207338 27 3:3 81,.205299 26 331 8.203269 25 3:16.03.201248 24 31.5'2.199237 23 3 1 2~.197235 22 31-21.195242 21.2 193258 20 77 1.191283 19 32.77.189317 18 32.62.187359 17 32 48.185411 16 32 33.183471 15 32 13.181539 14 32.05.179616 13 3191 1.177702 12 3 l.77.175795 11 3L.635.173897 10 33 1.172008 9 313;.170126 8 31.23.168252 7 31.10.166887 6 30.96.164629 6 30.83.162679 4 30.70.160837 8 80 57.159002 2 30.45.157175 1 30.32.155356 0,.,,, I I i ill i ii i ~ ~ ~ ~ I Cosine ID. I Sine. I D. Cotang. I D. I Ting. I M J 86' 22 LOGARITIHMIC SINiS, COSINES, 40 12 13 14 6 7 10 11 12 13 14 15 16 17 18~ 19 20 21 22 23 24 25 16 27 21.8 29 130 31 '32 33 34 35.36 87 38 409 40 43 44 4~. 48 I49.1. 56 130 I Sine. I 1) I Co.-1 I1). I I~g 1). I_ Cotangr. I.1 I. ".., -. 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I 1) 10 11 12 13 14 16 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 38 87 88 89 40 41 42 43 44 46 46 47 48 49 50 61 62 53 64 65 66 67 68 69 60 9.1l4350 i149.144453 14.96.145349 I14.93:146243 149 14 1 6 14.87.148036 14.84.148915 14.81.148982 14.78.149862 14.75.151569 14.72.152451 146 14.66.1552083 14.67.155957 1.5 1597 14.54.156830 14.51.157700 14 48.158569 14.45.159435 142.160301 14.42.161164 14.39 -9.162025 14.36.1628805 14.33.163743 14.30.164600 14.27.16545.4 14.22.166307 14 19.167159 14.10.168008 14.13.168856 14.10.1619702.14.07 -9.170547 14.05.171389 14.0~2.172230 13.99.173070 13.96.173908 13.94.174744 13.91.175578 13.88.176411 13.86.177242 13.83.178072.13.80 9.178900 13.77.179726 137.180551 13.72.181374 13.69.182196 13.66.183016 13.64.183834 13.61.184651 39.185466 13.59.186280 13.56 9.187092.13.53.187903 13.51.188712 13.46).189519 1.3.43.190325 13 41.191130 13.38.191933 1:1.3f0.192734 13.33.193534.3.1943"32 -Cosine. D.T Cosine. jD. T 'ang. 9.995753 93 U147b03.3073.148718.996717.30.149632.995699.30.150544.995681.0 151454.995664.0 152363.954 30 153269.995628 31 154174.995610.30 155017.995591.80 155978.99 073 1,568771.80 9.995555 9.157775.9557 30.158671.995519.30 159565.995501.160457.995482.31.16 1347.995464 3.162236.995446 3.163123.995427.1164008.995409 31.164892.995390.3 165774 9.995372.31 665.3 1 9 1 6 5.995353.1675032.953 0.168409.995316.31 169284.995297.31 117.995278.31 171029.995260 31 171899.995241 32.172767 *995222 3 173634.995203.32 ri4499 9.995184 9.132. 6.3 2 ~ 1 5 6.995165.176224.995146.32 177084 9917 32 174.995127 32.1787942.99510 8 32. 78 9.995079 32.179655.9950 0 32 1 80508.9950.31 32.181360.995032' 32.182211.995013. 3 183059 9.99t 32 9.183907 329974 184752 9995 32.185597 9995 32.186439.994916 32 187280.994896 33 1 88120.994877.18 95.994857.189794.994838.3190629.994818:33.191462 9.994798 3. 9 192294.994779.3193 124.994759.193953.994739 33 194780.994719 33 19-5606.994700 33 196430.994680 33.19753.994660:33.198074.994640 33 1968F04.994620 33.199713 Sine. ID.I Cotang. i I1). 15.26 15.23 15.2() 15.17 15.14 15.11 15.08 15.05 15.02 14.99 14.96 14.93 14 90 14.87 14.84 14.81 14.79 14.76 14.73 14.70 14.67 14.64 14.61 14.58 14.55 14.53 14.50 14.47 14.44 14.42 14.39 14.36 14.33 14.31 14 28 14.25 14.23 14.20 14.17 14.15 14.12 14.09 14 07 14 04 14.02 13.99 13.96 13393 13.91 13.89 13.86 13.84 13 81 13.79 13.76 13. 74 13.71 13.69 13.66 13.64 I Cotansr. I 0.bb2197.85 1282.850368.849456.848,546.847637.846731.845826.844923.844022.843123 0.842225.841329.840435.839543.838653.837764.836877.835992.835 108.834226 0 833346'.832468.83 1591.830716.829843.8289711.828101.827233.826'366.825501 0.824638.823776.822916.822058.821201.820345.8 19492.81S640.81 7789.816941 0.816-093.E15248.814403.813j561.8 2720.81188.81 1042.810206.809371.808538 0.807706.806876.800047.805220.804394.803570.602747.E01926.801106 8C00287 60 69 68 67 66 65 54 53 62 61 49 48 47 46 45 44 43 42 41a 40 39 38 37 36 35 34 33 32 81 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15' 14 13 I11 10 9 8 7 6 6 4 3 2 1 I wwmI i 1). I Tang. I 0 5%1. I tgo.. TANGENTS, AND COTANGENTS. 9~ i I - '.. Sine. D. Cosine. I). 0 I;.lJ4;lJJGjJ'; I'-"~i i 1.1,5129 13.28.994500.33 2.1959.5 13.26.99450.33 3.196719 13.23.994560 33 4.197511 13.21.994540.34 5.198302 13.18.994519 4 6.199091 13.6.994499 34.199879 13.13.994479.34 8.200666 13.11.994459.34 9.201451 13.08.994438 34 10.202234 13.06.994418.34 13.04,.34 11 9.203017 1304 9.947 34 9. 12.203797 13.01.994377 *.34 13.204577 12.99.1,47 '3A. 14.205354 12.96.9943 '3. 15.206131 12.94.994316 34 16.206906 12.92.994295.3 17.207679 12.83.994274.34 18.208452 12.87.994254.3 19.209222 12.85.994233 1.35 20.209992 12.82.994212. 12.80 35 21 9.210760; 9994191.. 9. 22.211526 12.78.9941 1 23.212291 12.75 9941'0*,0 24.213055 12.73 ~ (41 9* 25.213818 12.71.994108'35 26.214579 12.68 AL 1994087.3. 27.215338 12.66 ).994066 35 28.216097 12 64.994045~.5 29.216854 12.61 994024 1.35. 30.217609 12.59.994003.3. 31 9.218363 19.993 ' 9.5 32.219116 12.55.993fL60 35 83.219868 12'50.993939*~ i 34.220618 12.50.99318-5. 35.221367 12 48.993896'.3 36.222115 12.46.99375.36.2 87.222861 12.44.993C54.36.2 38.223606 12.42.93~3 6.2 39.224349 12.39.993811.36. 40.225092 12.37 99379 '36.2 41 9.225833. 12.35. '6.36.2 9. 993768' 9.2 42.226573 12.33.993746.36.2 43.227311 12.31.993725 36. 2 44.228048 12.28.993703.36.2 45.228784 12.26.99361.36.2 46.229518 12.24.993660.36.2 47.230252 12.22.993638.6.2 48.230984 12.20.993616.36.2 49.231714 12.18.993594.36.2 60.232444 12.16 993572.87.2 51 9.233172 12.14 9.993550. 9.2 62.233899 12.12.993528 37 2 53.234625 12.09.993506.37.2 54.235349 12.07.993484 3.2 55.236073 12.05.993462.37.2 56.236795 12.03.993440 37.2 57.237515 12.01.993418 37.2 58.238235 11.99.99 9.,36 37 69.238953 | 11.97.993374 37 | 2 60.239670 95.993351.2 ICosine DI). jSine. ID. I Co ' l.l' ]..1.11.3.200529.201345.202159.202971 2037t2.204592 205400.206207 207013 207817 208619 209420 210220 211018 211815 212611 213405 214198 2149f9 2157L0 216568 217356 218142 218926 219710 220492 221272 222052 222630 223606 224382 225156 225929 226700 127471 Z28239 Z29007 229773 130539 131302:32065!32626 233556;34345 235103;35859 '36614 237368 368120 38872 39622 40371 41118 41865 42610 43354 44097 44L39 45579 46319 i D. Cotang. 0O.b002b7 60 13.61.799471 69 13.59.798655 68 13.56.797841 57 13.54.797029 56 13.52.796218 55 13.49.795408 54 13.47.794600 53 13.45.793793 52 13.42.792987 61 13.40.792183 60 13.38 0.791381 49 13.35.790580 48 13.33.789780 47 13.31.788982 46 13.28.788185 45 13.26.787389 44 13.24.786595 43 13.21.785802 42 13.19.785011 41 13.17.784220 40 13.15 0.783432 39 13.12.782644 38 13.10.781858 37 13.08.781074 36 13.05.780290 35 13.03.779508 34 13.01.778728 33 12.99.777948 32 12.97.777170 31 12.94.776394 30 12.90 0.775618 29 12.98.774844 28 12.88.774071 27 12 86.773300 26 12.84.772529 25 12.81.771761 24 12.79.770993 23 12.77.770227 22 12.73.769461 21 12.71.7G6698 20 l12-69 0.767935 19 12.69.767174 18 12.67.766414 17 12-65.765655 16 12.62.764897 15 12 60.764141 14 12.58.763386 13 12.56.762632 12 13.54.761880 11 12.52.761128 10 12.50 0.760378 9 12.48.759629 8 12.46.758882 7 12.44.758135 6 12.42.757390 6 12.40.756646 4 12.38.755003 3 12.36.755161 2 12.32.754421 1 12.32.753681 0,tanT. ID. I Tang. I M...trtn ~~... I 238 LOGARITHMIC SINES, COSINES, 100 I I IMl I Sine. I D. I1 Cosine.I D. V 1 2 3 4 5 6 8 9 10 I11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 33 36 37 38 39 40 41 42 43 44 4.5 43 47 48 43 6 3 5 1 52 53 54 55 56 57 58 59 60 U.J OI,.2403F6.241101.241S14.242526.243 237.243947.244656.245363.246069.246775 9.247478.248181.248883.249583.250282.250980.251677.252373.253067.253761 9.254453.255144.255834.256523.257211.257898.258583.259268.259951.260633 9.261314.261994.262673.263351.2640~27.264703.265377.266051.266723.26 7395 9 268065.268734.269402.2.970069.270735.271400.272064.272726.273388.274049 9.274708.275367.276024.276681.277337.277991.278644.279297.279948.280599 11.93 11.91 11.8.0 11.87 11.85 11.83 11.81 11.79 11.77 11.75 11.73 11.71 11.69 11.67 11.65 11.63 11.61 11.59 11.58 11.56 11.54 11.52 11.50 11.48 11.46 11.44 11.42 11.41 11.39 11.37 11.35 11.33 11.31 11.30 11.24 11.22 11.20 11.19 11.17 11.15 11.13 11.11 1 1.10 11.08 11.06 11.05 11.03 11.01 10 99 10.98 10.96 10.94 10.92 10.91 fo.89 10.87 10.86 10.84 9.9913351 *9933299 3.993307 3 *9932E5 3.993262 3.993240.37 *993217.37.993195 3.993172.38.993149.38.9931-27.38 9.9 93"1 04. 38.993081 3.38.993059 3.993036.9930 13 3.992990 3.992967.38 *992944.38.992921 38.992~-98.38.38 9.992875 3.992F52 38.992829 3 -992F06 3.9927~,3.9927,59 3.992736:j.992713.992690 3.992666 3 3 9 9.992564.992572 3.992549 3.992525 3.992501 3.992478 40.992454 4 40 9.992406.992382.40.999359.40 *992335 ~40.992311.40.992287.40.992263.40.992239.40.992214 4.992190.40.40: 9.992166 4.-992142 4.992117 41'.992093 *41.992069 *4.992044 *41.992020 *41.991996 41.991971 *41.991947 4 Sine. D1. -iang. D. Cotang. 1.2 44 1 9 12.30 0.75s16b 0.247057 12.28.752943 6 9.247794 12-2'.752206 5 8.248530 12.24.751470 5 7:249264 12 22.750736 5 6.249998 12.20 750009 55.250730 118 74927 54.251461 12-17.748539 53.252191 1~215.7478-09 562.252920 12 13.747080 51.253648.1:1.7463529 5 0 9.254374 12.09 0.745626 49.255 100 12.07.744900 48.255S24 12.05.744176 47.256547 12.03.743453 46.257269 12.01.742731 45.257990 12.00.742010 44.2-58710 11.98.741290 43.259429 11.96.740571 42.260146 11-94.7398.54 41.260L63 1.2.739137 40 9.261578 11.90 0.738422 39.262292 11-89.737708 388.263005 11-87.736995 37.2637 17 11.85.736283 36.264428 11.83.735572 35.265138 11-81.734862 34.265847 11-79.734153 83.266555 11-78.733445 82.267261 176.732739 31.267967.174.732033 30 9 268671 112 0.731329 29.269375 11-70.7-30 6 25 28.270077 117 729923 27.270779 169.72922 1 26.2714 799 16.728521 25.2772178 1..727822 2 4.272876 11-62.727124 23 2753 11.60.726427 2 2.. 74269 11580.725731 21.274964 11:57.725036 20 9.275658 15 0 724342 19.276351 H-53.72.3649 18.277043 f.722957 17.277734 115.722266 16.278424 11-48.7215 76 15.279113 11-47.720887 14.279801 11.45.720199 13.280488 11.43.719512 12.281174 11.41.718826 1 1.281858 114 718142 10 9.282542 11.38 0.71 7458 9.283225 11.36.7167,75 8.283907 11.35.716093 7.284588 11.33.715412 8.285268 11:3l.714732 6.285947 11 30.714053 4.286624 11 28.7133 76 3.287301 11 2'3.712699 2.287977 12.712023 1 -28865'2 ~ ' 711348 0 Cotatng. I D. ITang. MI I _I Cos',ine. D. I 7`20 .TANGENTS, AN D.CO.TANGENTS.11~ I M. - 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 82 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 I 63 54 55 56 67 58 59 60 _ I. I Sine. I 9.2i-0599.281248.28U197.282544.283190.283036.284480.285124.285766.286408.287048 9.2VE7687.288326.288964.289600.290236.290870,291504.292137.292768.293399 9.294029.294658.2952V6.295913.296539.297164.297788.298412.299034.299655 9.3002 76.300895.301514.302132.302748.303364.303979.304593.805207.305819 9.806430.307041.307650.308259.808867.309474.310080.310685.311289.311893 9.812495:313097.813698.314297.814897.315495.816092 A316689.317284.317879 Cosine. D. Cosine. 10.82 10.81.99192.991897' 10.79. 9917 10.77.991F48 10.76 10.74.991348 -.991[37 10.2.91 77 1 0.7.991749 10.69.991724 10.67 99169 10.66. 10.64 9.991674 1063.991649 10.63 ocr9 10.61.991624 10.59.991599 ]0.58.991574 10-56.991549 10.56 qor9 ]0.54.991524 10.53.991498 ]0M.991473 10.50!.991448 10.48 9.91422 10-4.991397 10.46.991372 10.45 v0.43.991346 10.42.991321 -4.991295 ]0.40.991270 10.39.991244 10.37.991218 10 36 12 10.34!.9*91193 10.32 9.991167 10.321 991141 ]0.29.991115 }.991115 IO*V8.991090 10.28 10 26.991064 10.25.991038 10,23.991012.9909t6 10.22.99096 10.20 *.990960 10.20.990934 10.19 990 10.17 9.990908 10-16.99062 10.14,990855 ]0.13.990829 ]0.1.990E03.990803 100. 990750 ]0.08.990724 10 07.990672 10.05 o09 10.04.9006 71 10.03 9.990644 1.990618 10.01.990591 9100.990565 9.98.990538.9-7.990511 9.94 00485 9.93.990458:9,91.990431.990404 D. | Sine. 7 m i! I D. I Tanz. -.41.41.41.41.41.41 41.42.42.42.42 1.42.42.42.42.42.42.42.42.42.42.42.42.43.43.43.43.43.43 43.43.43.43.43.43.43.43,43.43,43.44.44.44.44.44.44.44.44.44.44 A44.44.44.44.44.44.46:45.46.45 3D. I.2fbt652.2E9326.289999.290671.291342.292013.292682.293350.294017.2946G4.295349 9 296013.296G77.297339.298001.298662.2)99322.299980.300638.301295.301951 9.3026107.303`61.303914 304567.305218.305869.306519.307168.307 15.30R463 9 309109.809754.310398.811042.311685.312327.812967.313608.314247.314885 9.315523.316159.316795.817430.316064.31E697.319329.319961.320592.821222 9.821851.822479.823106.823733.824358.824983.825607.826231.8326853.827475 I 11.23 U.71148 1.710674 11.22.710001 11.20.709329 11.18 708658 11 17.707987 1 1.5.707318 1I I..706650 11.12.705983 1.011.705316 1107.704651 0.703987 11.064.703323 11.03.702661 -0.701999 11 oo0 *.701338 11.008.700678 ]0986.700020 0.96.699362 ]0.95.698705 10-92 *698049 0.697893 10.90.696739 10-87.696086 10 87.695433 10 86.694782 10.83.694131 ]0.81.693481 0-80 1.692832 10.80.692185 10.78.91537 10.75 0.690891 10.74.690246 10173.689602 10.71.688958 0-70 1.688315 10.70.687673 1068 |.687033 10.65.686392 10.64.68753 10.62 *686115 10.61 0.684477 10.610 1.688841 10-0.6068205 p10 67.682570 10.57.681936 10.55.681303 10.54.680671 10.53.680039 10-51..679408 10.48.678778 1048; 0.678149 10.47.677521 10.45 676894 10.44.676267 10.43 676642 10.41.67b017 10.40.74393 16.89.678769 10.37 -ai47 10.36.7295 n I d"tn.n. I -0 60 59 58 57 56 55 54 53 52 51 60 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 83 32 31 30 29 28 -27 26 25 24 28 22 21 20 19 18 17 161 16 14 18 12 11 89,8 7 '.6 6.4 -I,0 I t * * Cottn.g. I D. T-l/is 1, '30 LOGARITHMIC SINES, COSINES, 120 W T Sine. 1___ D ICosine. ID. I ang. a 4 6 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 80 31 82 83 84 85 37 38 39 40 41 42 43 44 46 46 47 48 49 60 61l 62 63 64 66 66 67 68 69 60 v3. d 1 N.318473.319066.819658.320249.320840.321430.322019.322607.323194.323780 9.324306.324950.325534.326117.326700.327281.327862.328442.329021.329599 9.3301`76.330753.331329.8331903.332478.333051.333624.334195.334766.835337 9.335906.336475.337043.337610.338176.338742.339306.339871.340434.340996 9.341558.342119.342679.3432~9.343797.344365.344912.845469.346024.346579 9.347134.347687.348240.348792.349343.349893.360443.360992.35 1640.862088 9.90 9.88 9.87 9.86 19.84 9.80 9.77 9.76 9.75 9.73 9.72 9.70 9.69 9.68 9.66 9.64 9.62 9.61 9.60 9.58 9.57 9.56 9.54 9.53 9.52 9.50 9.49 9.48 9.46' 9.45 9.44 9.43 9.41 9.40 9.39 9.37 9.36 9.35 9.34 9.32 9.31 9.30 *9.29 9.27 9.26 9.2.5 9.24 9.2~2 9.21 9.20 9.19 9.17 9.16 9.15 9.14 9.13 9.991)04U 4.990378 4.990351.990324 4.990297 4.990270 4.990243 5.990215 4.990188 4.990161 4.990134 4 9.990107 4.990079.46.990052.46.990025.46.989997.989970.46.989942 4.989915.46.989887 4.989860.46.46 9.989832.989804.46.989777.46.989749 4.989721 4.989693 4.989665 4.989637 4.989609 4.989582 4.47 9.989553 4.989525 T.989497.989469 4.989441 4.989413 T.989384 4.989356 4.989328 4.989300 4 9.989271 4.989243 4.989214.989186 4.989157 4.989128 4.989100.4.989071 4.989042.48.989014 4 9.988985. 48.988956.48.988927.4.988898.48.988869.48.988840.48.988811.48.988782 4.988753 4.988724 49__ Sine. ID. U.32.7474.3218095.328715.329334.329953.330570.331187.331803.3324 18.333033.333646 9.334259.33487 1.335482.336093.336702.337311.3379 19.338527.339133.339 739 9.340344.340948.34 1552.342155.342757.343358.343958.344558.345157.345755 9 346353.346949.347545.348141.34 8 735.349329.349922.350514.351106.351697 9.352287.352876.353465.354053.354640.355227.35.5813.356398.356982.357566 9 358149.358731.859313.359893.360474.361053.361632.362210.362787.363364 1). jCot-ang. 0.65o-7:2,526 10.35 6190 10.33.67 1215 10.32.670666 10.23 6704 10)28.669430.668'113 10.26.619 10.25.667 582 10.24.666967 10.21..6663154 0.665741 10.209.665129 10.19.664518 101.17.663907 10.16.663298 10. 15.662689 10.12.662081 10.12.661473 10.10.660867 10.4)8..660261 107 0.659656 10.07.659052 10.06.658448 10.04j.657845 10.1)3.657243 10.4)2.656642 10.00.656042 9.8.655442 9.98.654843 9 6 427 9.96 0.653424 9.4.653051 9.3.652455 9.92.651859 9 91.65 1265 9.9.65067 1 9.88.650078 9.87.649416 9.86.641194 9.85.648303 9.83.0671 9.82 0.6477124 9 81.646535 9 80.645947 9.79 654 9.7.645360 9.76.644773 9.5.644187 9.4.643602 9.3.643018 9.71..642434 9.0 0.641851 9.70.641269 9.69.640687 9.68.640107 9.67.639526 9.66.638947 9.65.638368 9.63.637790 9.62.637213 9.1.636636 P. ITang. 60 59 58 57 56 55 54 5 1 5 0 4 8 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 82 31 30 29 28 27 26 25 24 23 2 2 2 1 20 1 9 1 8 1 7 1 6 1 5 1 4 1 3 12 1 1 10 9 8 7 6 6 4 3 2 1 0 I - -. -- - I Cosine. I D. — I iCottian. I M. TANGENTS, AND COTANGENTS. 130 3,1 Ill I Sine. 0 9.352088 1.352635 2.353181 3.353726 4.354271 6.354815 6.355358 7.355901 8.356443 9.356984 10.357.524 1 1 9.358064 12.358603 13.359141 14.359678 15.3602 15 16.360752 17.361267 18.361822 19.362356 20.362889 21 9.363422 22.363954 23.364485 24.365016 25.365546 26.366075 27.366604 28.367131 29.3637659 30.36818-5 31' 9.368711 32.369236 33.3697161 34.370285 35.870808 86.371330 87.371852 38.372373 39.372894 40.373414 41 9333 42.374452 43.374970 44.375487 45.376003 46.376519 47.377035 48.3771549 49.878063 50.378577 51' 9.379089 52.379601 53.380113 54.380624 65.381134 56.381643 57.882152 58.382661 59.383168 60.383675 CosineT ID. i 9.11 9.10 9.09 9.08 9.07 9.05 9.04 9.03 9.02 9.01 8.99 8.98 8.97 8.96 8 95 8.93 8.-92 8.91 8.90 8 89 8 88 8 87 8.85 8.84 8.83 8.82 8.81 8.80 8.79 8.77 8.76 8.75 8.74 8.73 8. 72 8.71 8.70 8.69 8.67 8.66 8.65 8.64 8.63 8.62 8.61 8.6 00 8.59 8 58 8.57 8.56 8.54 8.53 8.52 8.51 8.50 8 49 8.48 8.47 8.46 8 45 ICosine. I D. 9.988724.988695 4.988666 4.988636 4.988607 4.988578 4.988548 4.9E8519 4.988489 4.988460 4.988430 4 9.988401.4.988371 4.988342 4.988312 4.988282.50.988252.50.988223 5.988193 5.988163 5.98 8133 5 9.988103 5.988073 'so0.988043.988013 5.987983 5.987953 5.987922.987892).987862 5.987E.32 50 9.987801.9877140 5.9877110 5.987679 5.987649 5.9871618 51.987688 5 9.98 57496.987465.51.987434.51..987403.51.987372.52.987341.52.987310.52.987279.52.987248.52.9871217.52 9.987186..2.987155 52.987124.52.987092.52.987061.52.987030.52.986998 52.986967 5.986936 5.986904 5 Sine. ID.I I Tang. U.31i3364.36,3940.3645 15.365090.865664.366237.366f 10.367382.367 953.368524.369094 9.369663.370232.3707b9.371367.371933.372499.373064.373629.374183.374756 9.3753 19'.37588~ 1.376442.37 7003 ".o7563.378 122 MM78 1.379239.379797.380354 9 380910.381466.382M00.3825715.3833129.383682.3842-34.38471.86.386.337.3W858 9.386438.386987.3871536.388084.38F631.38978.389724.390270.390815..391360 9.391903.392447.392989.393531.394073.3946 14.895154.395694.396233.396771 Cotang. D. Cotang. 0.636636 6 0.9.6t;0.636060 59 9..635485 5 8.9.58.634910 67 9 5 634336 566 9.5.633 763 655 9.4.633190 54 9.3.632618 53 9.52.632047 52 9.51.631476 5a1 9.50.6.130906 50 9.49 9.8 0.630337 49 9.6.629768 48 9.46.629201 47 9.4.628633 46 9.3 628067 45 9.2.627501 44 9.42.626936 43 9.41.626371 42 9.40.625807 41 9.8.625244 40 9.387 0.624681 3 9.5.624119 38 94.623558 837 933.622997 86 9.2.622437 3.5 9.32.621878 84 9.31.621319 83 9.0.620761 32 9.29.620203 3 1 9.28.619646 80 9.276 0.619090 29 9.26.618534 2 8 9.25.617980 27 9.24.617425 26 9.23.616871 25 9.22.616318 24 9.21.615766 23 9.20.615214 22 9.19.614663 2 1 9.8.614112 20 9.17 9.15 0.613562 1 9 9.4.613013 1 8 9.14.612464 17 9 13.611916 16 9.12.611369 15 9.10.610822 '14 9.109.610276 18 9.09.609730 12 9.08.609185 1 1 9.07.608640 10 9.5 0.608097 9 9.5.607563 8 9.04.607011 7 9.03.606469 6 9.0-2 606927 6 9 01.058 9.00.6053846 4 8.99 I.604846 8 8.98 8.7.603229 0 I I D. I D. I Tang.M I IV 32 LOGARITHMIC SINES, COSINES, 140 II M.I I Sine. I I) I Cosine. I D. I Tanz. I 0 1 2 3 4 S 6 7 8 9 10 I11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 80 81 82 83 34 35 86 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Si. 52 53 54 55 56 57 58 59 60 9.383675.384182.384687.385192.385697.386201.386704.387-207.3877'09.388210.3887 11 9.389211.3897 11.390210.390708.39 1206.391703.392199.392695.393191.39368.5 9.394179.394673.395166.395658.396150.396641.397132.397621.3198111.898600 9.399088.399575.400062.400549.401035.401520.402489.402005.402972.403455 9.403938.404420.404901.405382.40,5862.406341.406820.407299.407777.408254 9.408731.409207.409682.410157.410632.411106.411579.412052.412524.412996 8.44 8.43 8.42 8.41 8.40 8:39 8.1'8 8 37 8.36 8.35 8.34 8:33 8 32 8 31 8 30 8 28 8.27 8.26 8 25 8 24 8 23 8 22 8 21 8 20 8 19 8.18 8.17 8.17 8.16 8.15 8.14 8.13 8.12 8.11 8.10 8 09 8 08 8.07 8 06 8 05 8.04 8.03 8.02 8.01 8.00 7.99 7.98 7.97 '7.96 7.95 7..44 7.94 7.93 7.92 7.91 7.90 7.89 7.88 7.87 7.86 9.986V8i4 9967.986873.52.397309.986841 ' 397846.986809.5.398383.986778.5.398919.986746.5.399455.986714 53.399990.986683 3.400524.986651 53 401058.986619 -5 401591.986587 I.53.402124 9.986555 9425.986523.5:403187.986491.403718.986459 404249.986127.5:1 404778.986395 5 405308.986363.13 405836.986331 54 406364.986299 54 406892.9862.66 54:407419 9.986234 54 9.407945.986202 54 408471.986169 54 408997.986137 54 409521.986104 54 410045.986072 5 410569.986039 54 411092.986007 54 411615.985974 54 412137.985942 54.412658 9.985909 54 9 4h173,9.985876 55 413699.085843 55 414219.985811.5 414738.98,5778 55 415257.985745 55 415775.9857,12.ss:416293.985679 5.416810.985646 417326.985613.s:417842 9.985580 9485.985547, 418873.985514 5 419387.985480.5 419901.985447.ss.420115.985414.5 420927.985380.56.421440.985347.56.421952.985314.56 422463.985210.6 4229714 9.985247 9 4.56 8.985213 56.423993.985180 56.424503.985146 56.425011.985113 56.425519.985079 56.426027.985045 56.426534.985011 56.427041.984978 56.427547.984944 56.428052 Sii~e. f P. Cotsg. ). Cotangm. 8.96 O632 8 196.602615 8.95.601617 8.94.601081 8.93.600545 8.93.600010 8.92.599476 8890.598942 8 89.598409 8 88.597876 8 87. 0.597344 8 835.596813 8.85.596282 8 84.595751 8 83.595222 8.82.594692 8 81.594164 8 80.593636 8 78.593108 8 77..592581 8 6 0 592055 8 76.591529 8.7 591003 8.74.590479 8 74.589955 8 73.589431 8.72.588908 8 71.588385 8 70.587863 8.68..587342 8 I6 3.586821 8.t67.586301 8.66.585781 8.65.585262 8.64.584743 8.3.584225 8.63.583707 8.62.583190 8.60.582674 8.59..82158 8.8 0.581642 8.8.581127 8.57.580613 8 56.680099 8.55.679585 8.55.579073 8.54.678560 8.53.584 8.52 784 8.51 673 8.50.577026 8.9 0.576516 8.49.676007 8.48 759 8.48.574989 8.47.574481 8.46 537 8.45.573466 8.44.5702959 8.43.572453 8 3 71948 T) j Ta ng. 59 58 57 56 55 54 53 52 51 50 49 48 47. 46~ 45. 44~ 43 42 41. 40 39 38 37, 36 35 34 33 32 31 30 29' 28 27, 26 25 24, 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 I Cosi;,c. 6. 1 D. I M TANGENTS, AND CUTANGrNTS 150 33 M. I Sine. 0 9.41296 1.41346 2.4139138 3.414408 4.414878 15.415347 6.415815 7.4162E3 8.416751 9.41721 7 10.417684 I1I 9.418150 12.418615 13'.4 19079 14.419544 15.420007 16.420470 17.4209.33 18.421395 19.421857 20.422318 21 9.422778 22.423238 23.423697 24.424156 25.424615 26.425073 27.425530 28.425987 29.426443 30.426899 31 9.427354 82.427809 33.428263 34.428717 35.429170 36.429623 37 '.430075 38.430527 39.430978 40.43-1429 41 9.431879 42.432329 43.432778 44.433226 45.433675 46.434122 47.434669 48.435016 49.435462 60.435908 61 9.436353 62 4367198 63.437242 64.437686 65.438129 66.4386572 67.439014 68.439456 69.439897 60.440338 Co~lne D. Ij 7.85 7.84 7.831 7.83 7.82 7.81i 7.83 7.79 7.78 7.77 7.76 7.75 7.74 7.73 7-71 7.72 7.71 7.70 7.6) 7.68 7.67 7.6 7 7.66 7.65 7.64 7.63 7.62 7.61 7.60 7.60 7.59 7.58 7.57 7.56 7.155 7.54 7.531 7.52 7.52 7.51 7.60 7.49 7.49 7.48 7.47 7.46 7.45 7.44 7.44 7.43 7.4~2 7.41 7.40 7.40 7.39 7.38 7.37 7.36 7.36 7.35 Cosine. D. I Tang. 9.984944- 9.428052.984910 57 428557.984876 57:429062.984842 57 429566.984808 57 430070.984774 57 430573.984740 57 431075.9847106 57 431577.984672 57 432079.984637 57 432580.984603 57 4303080.57. 9.984569 I 9.433580.984535 57 434080.984500 57 4345769.984466 57 435078.98.443' 57 4355796.984397.58 4360'03.984363.58 436570.984328 8.437067.984294 58.437563.9842.259.58.43b059 9.984224 9.438554.984190 58.439048.984155 58 354.984120 58.440036.984085 58.440529.984050 8.441022.984015 8.4415 14.983981 8.442006.98346.8 442497 J.083911.58:442988.58. 9.983875 58 9 437.983840.443968.C8E3'05 59 444458.98377i0 59 444947.983735 59 445435.983700 59 445923.983664.69 446411.983629.5 446898.983594 59.447384.983558 59.447870 9.983523 9.448356.983487.9448841.983452 59 449.326.983416:59.449810.983381.9450294.9833405.450777.983309 59 451260.983273 9.451743.983238. 60.452225.983202.60:452706 9. 8 1 6. 6 0. 9 4 3 8 9.816 60 9438.983130 60.453668.983094 *60.454148.963058 60.454628.98302.2 60.455107.982986 60.4666586.982950 60.466064.982914 '60.46642.982878.60.457019.982842.457496 Sine. I D. Cotanlg. 1). ICotting. 8 2.67lu4b 60 8.2.571443 5 8.41.670938 58 -8 43).570434 5 7 8.39.569930 56 8.38.569427 65 8.38.568925 54 8.36.668423 53 8.35.667921 62 8.34.567420 51I 8.33..666920 G0 8.2 0.566420 49 8.2.565920 4 8 8.32.565421 47 8.31.664922 4 6 8.30 564424 45 8 2).663927 44 8.28.663430 43 8.28.562933 42 8.26 -.562437 41' 8.25..561941 40 8.4 0.561446 31) 8.24.660952 38 8.23.660457 371 -8.23.559L.64 3 8.22.65947 1 35 8.21.65 8 97i8 34 8 20.658486 33 8.19.657994 321 8.19.557503 31 8.18.557012 30 8.7 0.656521 2~9 8 16.556032 28 8.16.5556,42 27 8 15.555063 26 8.14.654565 25 8.13.654077 24 8 12.653589 23 8.1~2.653102 22 8.11.652616 _Z 8. 10.662130 20 8.09. 0.561644 1 8.09.561159 18 8.08.650674 1 7 8.07.650190 1 6 8.06.649706 1 5 8.06.649223 1 4 8.05.648740 13 8.04.648257 12 8.03.647776 11.8.02.6,47294 10 8.2.0.546813 9 8.01.6 6 3.646372 6 7.98.644414 4 7.6.543936. 8 7.96.543-458 2 7.96.642981 1.642504 0 i I D. _I 0 -1). I T'r S I 3' Li 14 740 io 4 ~LOGARITHMIC SINES, COSINES, 160 'I.Sine. I 0 1 2 3 4 5 6 7. 8 9 10 11 12 13 14 16 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41. 42 43 44 45 46 47 48 49 50 51' 52 63 64 55 66 57 68 69 60 9.4403-38.440778.441218.441658.442096.442535.442973.443410.443847.444284.444720 9.445155.445590.446025.446459.446893.447326.447759.448191.448623.449054 9.449485.44991.5.450345.450775.4512,04.451632.452060.452488.452915.453342 9.453768.454194.454619.455044.455469.4'55893.456316.456739.457162.457584 9.458006.458427.458848.459268.459688.460108.460527.460946.461364.461782 9.462199.462616.463032.493448.463864.464279.464694.465108.465522.465935 D. Cosine. D. 7.4 9938-2842.60.982805 7.3.98'769.60 7..3~2.982)7,33.61 7.1.982696.6 7.31.982660.61 7.30.982624.61 7.29.9825b7 6 7.28.982551..61 7.27.982514.61 7.27.98 61 7.26 9.982441 6 7.25:982404.6 7.24.982367.61 7.3 982331.61 7.23.982294.61 7.22.982257.61 7.21.9.82220.61 7.20.982183.62 7.20.982146.62 7.19 919.62 7.18 -.62 7.7 9.982072.6:.7 982035 6 7.16.981998.62 7.16.981961.62 7.15.981924.62 7.14.981886.62 7.13.981849.62 7.13.981812.62 7.12.9817.62 7.11:981737. 62 7.0-9 981699 7.0 981662 6 70 981625.63 7.07.981587.63 7.07 i.981549:63 7.06.981)12 '63 70 981474 '63 7.05.981436 63 7.04.981399:63 7.04.981361 ~.6 7.03. 63 7.02.918 '63 7.01.981247.63 7.01.981209 '63 7.00 9811 63 6.99.981133 64 6.98.98109.5 64 6.98 9807 64 6.7 981019 6.96.980981.64 6.95...64. 6.5 9 980942 6 6.95.98090.64 6.94.98046.64 6.93.980866.64 6.92.980789.64 6.91.980750:64 6.90:980712.64 6.90 9861.64 68 980635.64 6.9 980596 -D sdne.ID IT ang(. I D. U.45i490.4571973.458449.458925.459400.459875.460349.460823.461297.461770.4622142 9.4627 14.46316.463658.464 129.464599.465069.465539.466008.466476.466945 9.467413.467880.4 6 8347.468814I'.469280.469746.470211.470676.4 7 Ili i.471605 9 472068.472532.472995.473457.473919.4 74 3c8:1.474842.47 5303.475763.47162,23 9.476688.4771~42.477'601.478059.478517.4 78 975.479432.479889.480345.480801 9 481257.4817 12.482 167.482621.483075.483529.483982.484435.484887.485339 7.94 7.93 7.93 73)2 7.1 7.t90 7.1)0 7.89 7.88 7.87 7.87 7.86 7.85 7.85 7.84 7.83 7.83 7.82 7.81 7.8)) 7.80 7.79 7.78 7i.77 7.76 7.7 7.74 7.73 7.73 7.72 7.71 7.71 7.70 7.69 7.69 7.68 7.67 7.67 7.66 7.65 7.65 7.64 7.63 7.63 7.62 7.61 7.61 7.60 7.59 7.58 71.57 7.57 756 7.55 7.55 7.54 7.53 Cotang. 0-54:~304 60.542027 I5.540600 56.540125 b55.539651 r 4.539177 63.538703 62.538230 5 1.5637758 5 0 0.5372Lc'6 419.536L'-14 48.536342 47.5353871 46.535401 45.534931 44.534461 43.533992 42.533524 41.533055 40 0.532587 39.532120 38.531653 37.531186 36.5030 720 35.5302,54.34.529789 33.529324 32.528859 31.628395 30 0.527932 29.52-7468 28.527005 27.526543 26.526081 25.5256 19 24.525158 23.524697 22.524237 21.523777 20 0.523317 19.522858 18.522399 17.521941 16.621483 15.521025 14.520568 13.520111 12.519655 1 1.6 19.1909 10 0.518743 9.518288 8.517833 7.617379 0.6516925 6.5164711 4.516018 8.615565 2.515113 1.514661 0 Tng. Cosine. ICotang..I D. TANGENTS, AND COTANGENTS. 170 1.I Sinle. 09.465935 1.466348 2.466761 3.467173 4.467585 5.467996 6.468407 7.468817 8.469227 9.469637 10.470046 1 1 9.470455 12.470863 13.4712711 1 4.471679 15.472086 16.472492 17.472898 18.473304 1 9.473 710 20.474115 2 1 9.474519 22.474923 23.475327 24.475730 25.476133 26.476536 27.476938 28.477340 29.4777141 30.478142 81 9.478542 32.478942 33.479342 34.479741 36.480140 36.480539 37.4809371 38.481334 39.481731 40.482128 41 9.482525 42.482921 43.483316 44.483712 45.484107 46.484501 47.4848,95 48.485289 49.485682 50.48607-5 51 9.486467 52.486860 53.487251 54.487643 55.488034 66.488424 57.488814 68.489204 69.489593 -60.489982 ID. ICosine. ID. I Tan-cr. I I I - - i 6.88 6.88 6.87 6.86 6.85 6.85 6.84 6.83 6.83 6.82 6.81 6.80 6.80 6.79 6.78 6.78 6.77 6.76 6.76 6.75 6.74 6 74 6.73 6.72 6. 72 6.71 6.70 6. G9 6.69 6.68 6.67 667 6.66 6.65 6.65 6.65 6.63 6.63 6.62 6.61 6.61 6.60 6.59 6.59 6i.58 6.57 6.57 6.56 6. 5f 6.55 6.54 6.53 6.5.3 6.52 '6.51 6.51 6.50 6.50 6.49 6.48 9.9805!96.980558.9805 19.980480.980442.980403.980364.980325.980286.980247.980208 9.9F,0169.960130.980091.980052.98,0012.9199713.97,i9934.979895.979855.979816 9.979776.97.9737.979697.979658.979618.97,9579.979539.979499.979459.979420 9.979380.97340.979300.979260.979220.978o180 (7.87140.871C9100.4W79059,9780,0 1 9 9.867897'9.97893.978898.978858.978817.978777.978736.978696.978655.9718615 9.978574.978533.978493.978452.978411.978370.978329.978288.978247.978206.64 9.485339:64.4857.4.65.486242.65 -486683:6 4871143:65.487593.65.488043 488IS492.65.488941.65.:1.65. 4b89838.65 94(1.65.65.4911L10.65.492073.65.492519.65 -429(js9.66 431.66.493E4 1.66.493294:66 9.494299.66 458.66.495163.6 4956:30.66 461.66.496073.66.496515.6.4967957.66 482.66.497841.66 9.498282.66.499603.67.500042.67.500481.67.500920.67.501359.67.501,177:67.502235.67.5026712:6 503109 679.503546.67.503U82.67.5044 18.67.504854.67.505289.67.505724.67.506159 68.506593:68.507027.68.5071460 6(8 9.507883.68.508326.68.5087159.68.509191.68.509622.68.5 10054.68.510485.68.510916:6 511346 5117 Cotang. I D. ICotang. 7.3 0.514661 60 7.2.514209 59 7.52.513758 58 7.51.513307 57 7.1 512857 56 7.8.511508 53 7.48.511059 62 7.7.510610 6 1. 510162 50 7.46 7.5.509267 4 8 7.4.508820 47 7.4.608373 46 7.3.507927 45' 7.4-017481 44 7.2.507035 43 7.42.506590 42 7.4.506146 41 7.40..05701 40 0.505257 39 7.40.504814 38 7.9.504370 37 7.38 '.503927 36 7.7.603485 35 7.36.503043 34 7.36.502601 3 7.5.602159 82 7.4.601718 8 1 7.4.501278 30 7.3 0.500837 29 7.3.500397 2 8 7.32.499958 2 7 7.1.499519 26 7.31.499080 25 7.30.498641 24 7.30.4 98203 23 7.20.497765 22 7 -28.497328 2 1 7.28.496891 20 7 27 0.496454 9 7 27.496018 1 8 7.2.495582 1 7 7.25.495146 16 7.25.494711 1 5 7.24.4-94276 1 4 7.24.493841 13 7.23.493407 12 7.22.492973 1 1 7.22..492540 10' 7.21 0.492107 9 7.1.491674 8 7.20.491241 7 7.19.490809 6 7.19.490378 6 7.18.489946 4 7.18.489515 B. 7.17 489084 2 7.16:488054 1.488224 0 I m LFoin. I Sinle. I i I). ___ITang. I M ' I 7211 36 LOG A-RITIISIIC SINES, COSINES, Igci IM S ine. j 1) 0*5 9.489982 6.48 1.490371 64 2.490759 6.47 3.491147 6.46 4 -.491535 6.46 5.491922 6.45 6.492308 6.44 7.492695 6.44 8.493081 6.43 9.493466 6.42 10.493851 6.42 11 9.494236 6.41 12.494621 6.41 13.495005 6.40 14.4953S8 6.39 15.495772 6.39 16.496154 6.38 17.496537 6.37 18.496919 6.37 19.497301 6.36 20.497682 6.3.6 21 9.498064 6.35 22.498444 6.34 23.498825 6.34 24.49920;4 6.33 25.499584 6.32 26.499963 6.32 27.500,3-11 6.31 28.500721 6.31 29.601099 6.30 30.501476 62 81 9.501854 6.29 32-.5 02 2 31 6.29 33.602607 6.28 34.50.2984 6.27 3.5.503360 6.26 86.50373.5.6.26 37.604110 6.25 38.594485 6.25 39.504860 6.24 40.605234 62 41 9.50.5608 6.23 42.65081 6 22 43.6063364 6.2~2 44.606727 6821 45.5071099 6.20 46.5074741 6.20 47.507843 6.19 48.5082 14 6.19 49.608585 6.18 60.608956 6.18 51 9.509320 6.17 6.)2.609690 6.16 63.610066 6.16 54.61@434 6.1.5 55.610803 6.15 56.6111172 6.14 67.6115440 61 68.611907 6.13 60.612642 81 ICosine..978165.978124.978083.97t042.978001.977i59.977918.977877.977835.977794 9.977752.9777 11.977669.977628.977586.977544.977503.977461.9774 19.977377 9.977335.977293.977251.977209.9717167.977125.977083.97704 1.976999.976957 9.976914.976872.976830.976787.976745.976702.976660.976617.976574.976532 9.976489.976446.976404.97It6.361.9763,18.97,6275.976232.976189.976146.976103 9.976060.976017.975974.975930.975887.975844.975E00.971575.7.975714.975670 i h Tang. 9.511776 68 512206.68 5612635 -68 513064 69 513493.69.513921.69.514349.69.514777.69.515204.69.515631.69.51605 7.69 9 5164814.69 616910.69.5173:35.69 5617761.69.51s1s5.69.518610.70.619034.70.519458.70.519S82.70.520305.70 9.520728.7 521151.70 5621573.70 521995.70.522417.70 56228,38.70.5232.59.70.523680.70 '524100.70 5624520.70 9 5241;39.71 '526198.71 5626615.71 562 7033....71 5627451.71.527E68.71.528285.7 628702.71.71 9.529119 57.29535.71.529950.71.530366.71 5630781.71.531196.71 '531611.72 '5320215.72 '632439.7 32853.72.72 9.533266.72.533679.72.634092.72.634504 72.634916.72.635328.72.635739.72 536150.7 53656 1.72.5-3697 D-ICotang. I I D. 7.16 7.16 7.15 7.14 7.14 7.13 7.13 7.T12 7.12 7.11 7.10 7.10 7.09 7.09 7.08 7.08 7.07 7.06 7.06 7.05 7.05 7.04 7.03 7.03 7.03 7.02 7.02 7.01 7.01 7.00 6.99 6.99 6.98.6.98 -6.97 6.97 6.96.6.96 6.95 6.986 6.184 6.93 -6.93 6.93 6.92 6.91 6.91 6.90 0..90 6.89 6.89 6.88 6.87 6.87 6.86 6.86 6 85 6 85 6.84 D. I.I I Cot.-II'. I U.460Z:44 bO.487794 159.487365 58.486936 57.486507 56.486079 55.485651 54.485223 5 3.484796 62.484369 61l.483943 5 0 0.483516 49.483090 48.482665 47.482239 46.481815 45.481,390 44.480966 43.480542 42.480118 41.479695 40 0.479272 39.478849 38.478427 87.478005 36.477683 35.477162 34.476741 33.476320 32.475900O 31.475480 80 0.475061 29.474641 28.4741219 27.4738U3 26.473385 25.472)967 24.472549 23 A4721;32 22.471715 21.4 7129hJ8 20 0.470881 19.470465 18.470050 17.469634 16.4692 L9 15.468804 14.468389 13.467975 12.467561 11.467147 10 0.466 734.9 A66,321.8.465908 7.465496.6.465084 tZ.464672.4.464261 Z.463850 2.463439..46530128.0 TanIIg. I ~ I-0. I 7:.0 .TANGENTS, A&ND COTANGENTS 140 M.ISine. L - 09.512642 1.513009 2.513375 3.513741 4.514107 5.514472 6.914837 7.5152,02 8.515566 9.515930 l0.5162ul 11 9.516657.51702.0 1.517382 14.51T745 1,5.518107 6.518468 17.518829 18.519190 19.519551 20.519911 21 9.5 2102)7 1 22.520631 23.520990 24.521349 25.521707 26.522066 27.522424 28.522781 29.523138 30.523415 31 9.523852 32,.524208 33.524564 34..524920 35.525275 36.525630 Si7.525984 38.526339 39.526693 40. 527046 41 9.5-27400 4. 1.527753 41.52E105 41.52.84-58.8.529!64 1).30565 52 531265 63.6.3114M 54.6.319G3 55 I.632312 5,6.5.3 2.0 21 57.533009 58.5 333 3 7 59.533 iC 4 60.534053 jCosine 10. G.1i2 6.11 6.11 6.10 6.09 6.09 6.08 6.08 6.07 6.07 6.06 6.05 6.05 6.04 6.04 6.03 6.03 6.02 6.01 6.00 6.00 6.00 5.99 5.99 5.98 5.98 5.97 5.96 5.965 5.9,5 5.95 5.94 5.94 5.9:1 5.92 5.91 5.91 5.90 5.90 5.89 5.89 5.88 5.88 5.87 5.87 5.86 5.86 5.85 5. 8 3 5 84 5.84.5.831 5.82 5.82 5.81 5 81 5.8( 5.81 5.71 CJosine. I D Tang 9.956710 9.536972.976627.7.537382.975583:7.637792.975539.73.538202.975496 73.538611.975452 J3.539020.975408 73.539429.975365 73~.539837.975321 73 540245.97527,7 73:540653.975233.7 5641061 9.975189 9.541468.97 5 1415, 7.541875.975101 73.542281.9750 57 7.542688.97o013.543094.9.4969 7..543499.974925.543905.974880 74.544310.974836.74.544715.974792.7.545119 -9.974748 9.545524.974703 5.45928.974659 6.46331.974614 ~ 546735.9.4570 6.47138.9.4525.7 547540.974481 ~ 547943.974436 ~ 548345 9,4391.7.548747.974347.649149 9.974302 9.549550.9794257 5.49951.974212 ~ 550352.974167.550752.974122 ~ 551152.974077 75.551552.9704032 75.551952.97,3987.7.552351.973942 '.552750.973897 75.553149 9.973852 j~9.653548.973807.7.553946.973761 ~ 554344.9737116 76.547141.97367 76.55139.9743625 7.555536.97350.76 65l I1.034.76.556329.973449.76.5572 9939.:76 -.5712 1 9.T38 76 9.557517 7673.557913.973307 *76.580 973261 76.6s8702:973215 *76.559097 936 *76.559491 97124 *6.6559815 973078 76.507 17303 ~ 56.1673.972916.5606 6.84 6.83 6.83 6.82 6.82 6.81 6.81 6.80 6.80 6.79 6.79 6.78 6.78 6.77 6.76 6.76 6.75 6.75 6.74 6.74 6.73 6.73 6.72 6.72 6.71 6.71 6.70 6.70 6.63 6.69 6.68 6.68 6.67 6.67 6.66 6.66 6.65 6.65 6.65 6.64 6.64 6.63 6.663 6.62 6.62 6.61 6.61 6.60 6.59 -4.59 6.59 6.58 6.58 6.57 6.57 6.56 6.156 6.55 10. CoGtang. 0.468028.60.462618 59.462208 68.461798 67 461389 566.460980 65.460571 654.460163 53.459`755 52.459347 61.468939 50 0.46F632 I49.458125 48.467719 47.4571312 46 456906 45 466501 44.466095 43:466690.42.4662E5 41.464881 40 0.454476 3.4640712 38.468669 87.463265 36.462862 86.462460 84.462057 83.461655 82.461253 SI.456081 30 0.450450 29.460049 28.44948 27.449248 26.448848 26.448448 24.448,048 23.447649 22.447250 21.446861 20 0.446462 19.446054 IS.446656 17.446259 16.444861 15.444464 14.444067 13.443671 12.448276 11.442879 10 0.442483.9.442087..441692 7.441298.6.440903 6.440509 4.440115 8.439721 2 A43932 1.4 3 C34 0 I D. D0. ITring.- I la. 7JJ 38 LOGAIbTHMlIC SINELS, COSINES, 200 I MI Sine. I D. Cosine. I D. I Tang. I D. Cotang. I I I U 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 88 39 40 41 42 43 44 45 46 47 48 49 50 61 62 63 54 55 56 57 58 59 60 9.534052.534399.534745.635092.535438.635783.536129.536474.536818.537163.537507 9.537851.538194.538538.538880.639223.539565.639907.540249.640590.540931 9.541272.641613.541953.642293.642632.542971.543310.543649.543987.544325 9.544663.545000.545338.645674.546011.646347.546683.547019.547354.547689 9.648024.648359.548693.549027.649360.649693.650026.550359.550692.551024 9.651356.551687.6552018.552349.552680 -6526S0.553 010.553341.553670.554000 564329 Cosine. ] 5.78 5.77 5.77 5.77 5.76 5.76 5.75 5.74 5.74 5.73 5.73 5.72 5.72 5.71 5.71 5.70 5.70 5.69 5.69 5.68 5 68 5.67 5.67 5.66 5.66 5.65 5.65 5.64 5.64 5.63 5.63 5.62 5.62 5.61 5.61 5.60 5.60 5.59 5.59 5 58 5.58 5.57 5.57 5.56 5.56 5.55 5.55 5.54 5.54 5.53 5.53 5.52 5.52 5.52 5.51' 5.51 5.50 5.50 5.49 5.49 9 9729b6.972940.77.972894.77.972848.77.972802.77.972755.77.972709.77.972663 77.972617 77.972570 7.972524 *77 9.972478 77.972431 77.972385 *78 78.972338 78.972291.78.972245.78.972198.78.972151.78.972105i.'8.972038 78.78 9.972011.971964 78.971917.7.971870 78.971823.78.971776 78.971729 78.971682.971635 79.971588 7.79 9.971540.971493 '9.971446..971398 79.971351 79.971303 7.971256.7.971208 79.971161 79.971113 79.79 9.971066.971018.80.970970.80.970922.8.970874 80.970827.0.970779 80.970731 80.970683 80.970635 80.80 9.970586 80.970538 80.970490 80.970442.80.970394.80.970345.81.970297.81.970249.81.970200.81.970152 Sine. I D. V.00 LUDD V. 4;j 6 VU 4tj u.661459 6.55.438541 69.561851 6.54.438149 68.562244 6.54.437756 57.662636 6.53.437364 56.663028 6.53 -486972 55.663419 6.53.436581 64.563811 6.52.436189 63.664202 6.52 -435798 52.664592 6.51.435408 51 6.51 1.564983.435017 50 9 66,5373 6.50 0.434627 49.56-5763 6.50.434237 48.666153 6.49.433847 47.566542 6.49.433458 46.566932 6.49.433068 45.567320 6.48.432680 44.667709 6.48.432291 43.668098 6.47.431902 42.668486 6.47.431514 41.568873 6.46 431127 40 6.46 9.569261 6.45 0.430739 39.569648 6.45.430352 38.570035 6.45.429965 37 -6710422 6.44.429578 36.670809 6.44.429191 35.671195 6.43.428805 34.6711-581 6 43.428419 33.571967 6.42.428033 32.572352 6.42.427648 31.572738 6.42.427262 30 9.573123 6.41 0.426877 29.6 '13507 6.41.426493 28.573892 6.40.426108 27.674276 6.40.425724 26.574660 6.39.425340 25.575044 6.39.424956 24.676427 6.39.424573 23.576810 6.38.424190 22.576193 6.38.423807 2 1.576576 6.37.423424 20 9.676958 6.37 0.423041 19.677341 6.30.422659 1 8.577723 6.36.422277 17.578104 6.36.421896 16.678486 6.35.421514 15.678867 6.35.421133 14.679248 6.34.420752 13.579629 6.34.420371 12.680009 6.34.419991 1 1.580389.419611 10 6.33 9.580769 6.33 0.419231 9.581149 6.32.418851 8.681528 6.32.418472 7.581907 6.32.418093 6.582286 6.31.417 '114 6.582665 6.31.417335 4.583043 6.30.416957 3.583422 6.30.416578 2 583800 6.29.416200 1,6641 77 41-5823 - 0. 1. -1 —.1 - -- 9.561066.561459 6.55.561851 6.54.562244 6.54.562636 6.53.563028 6.53.563419 6.53.563811 6.52.564202 6.52.564592 6.51.564983 6.51 9565373 6.50.565763 6.50.566153 (.49.566542 6.49.566932 6.49.567320 6.48.567709 6.48.568098 6.47.568486 6.47.568873 6.46 9.569261 6.569648 6.45.570035 6.45.570422 6.45.570809 6.44.571195 6.44.571581 6.43.571967 43 6.42.572352 642 6.42.572738 6.42 9.573123 6.4.573507 6.41.573892 6.41.674276 6.40.674660 6.40.675044 6.39.575427 6.39.575810 6.39.676193 6.38.576576 6.37 6.37 9.576958 67.577341 637.677723 6.3.578104 636.578486 6.3.578867 635.579248 6.3.579629 634.680009 634.580389 633 9.680769 i 6.33.681149 632.681528 632.581907 6.32.582286 6.31.582665 6.31.583043 6.30.683422 6*30 583800 6.29 6584177 Coang. { nD. i t V.4asy;e4 ou,438541 59.438149 58.437756 57.437364 56.436972 55.436581 54.436189 53.435798 52.435408 51.435017 50 0.434627 49.434237 48.433847 47.433458 46.433068 45.432680 44.432291 43.431902 42.431514 41.431127 40 0.430739 39.430352 38.429965 37.429578 36.429191 35.428805 34.428419 33.428033 32.427648 31.427262 30 0.426877 29.426493 28.426108 27.425724 26.425340 25.424956 24.424573 23.424190 22.423807 21.423424 20 0.423041 19.422659 18.422277 17.421896 16.421514 15.421133 14.420752 13.420371 12.419991 11.419611 10 0.419231 9.418851 8.418472 7.418093 6.417714 6.417335 4.416957 3.416578 2.416200 1 413523 - 0 Tang. I M......... - I D. I Colstnz- I D. I Tang. I I TANGENTS, AN D COTANGEN' 210,M1 Sine. 0 U.554329 1.654658 2.554987 3.555315 4.555643 5.555971 6.556299 7.556626 8.556953 9.557280 10.6557606 11 9.557932 12.558258 13.558583 14.558909 15.559234 16.559558 17.559883 18.560207 19.560531 20.560855 21 9.561178 22.561501 23.561824 24.562146 25.562468 26.562790 27.563112 28.563433 29.563755 30.564075 31 9.564396 32.5647 16 33.565036 34.565356 85.5656716 36.565995 37.5663 14 38.566632 39.566951 40.567269 41 9.567587 42.567904 43.568222 44.568539 45.568856 46.569172 47.569488 48.569804 49.570120 50.570435 51 9.570751 52 5 71066 63.571380 5a4.571695 55-.572009 56.572323.57.5712636 58.572950 59.573263 60.5 7 357i5 ICosine D. I Co~zine. I D).I Tang. I 5.48 5.48 5.47 5.47 5.46 5.40) 5.45 5.45 5.44 5.44 5.43 5.43 5.43 5.42 5.42 5.41 5.41 5.40 5.40 5.39 5.39 5.38 5.38 5.37 5.37 5.36 5.33 5.36 5.35 5.35 5.34 5.34 5.33 5.33 5.32 5.32 5.31 5.31 5.31 5.30 5.30 5.23) 5.2.) 5.28 5.28 5.23 5.27 5.27 5 26 5-2 2 5 2i3 5.2,5 5.24 5 24 5-21 5 23 5 21 5 22 5.22 5.21.970103.970055.970006.969957.969909.969860.9698 11.969762.9697 14.969665 9.969616.969567.9695 18.969469.969420.969370.969321.969272.969223.969173 9.969124.969075.969025.968976.968926.968877.968827.96877 7.968728.96E678 9.96. 62.8.b6878.968528.968479.968429.968379.968329.968278.968228.968178 9.968128.968078.968027.967977.967927.967876.967826.967775.967725.967674 9.967624.967573.967522.967471'.967421.9673790.967319.967268.967217.967166. 81'.81.81L.81.81'.81.81.81.81.81L.82.82.82.82.82.82.82.82.82.82.82.82.82.82.8.1.81J.83.8:1.33. 83.83.83.83.83.83.81i.83.83.84.81.84.84.84.84.84.84.84.84.81.84.84.84 -85.8.5.8.5.85.584555.584932.585309.585686.586062.586439.586815.587 190.587566.687941 9.588 36.586H91.589066.589440.589814.590188.59562.5909435.591308.59168.1 9.592054.592426.592798.593170.593542.593914.594285.594656.595027.595398 9.595768.596138.596508.596878.597247.597616.597986.598354.598722.599091 9.599459.599827.600194.600562.600929.601296.601662.602029.602395.602761 9.603 127.603493.603858.604223.604588.604953.605317.6u56F2.606046.60641( D. ICotang. 6.9 0.415823 60 6.29.415445 59 6.28.415068 58 6.28.414691 57 6.28.414314 56 6.27.413938 55 6.27.413561 54 6.26.413 185 53 6.26.412810 52 6.26.412434 5 1 6.25.412059 5a0 6.25 0 411684 49 6.24.411309 I48 6.24 A.41(934 47 6.23.410560 46 6.23.4 101 F6 45 6.23.409812 44 6.22.409438 43 6.22.409065 42 6.2~2.408692 4 1 6.21.408.319 40 6.1 0.407946 39 6.21.4075`74 388 6.20.407202 37 6.29.40629 36 6.19.406458 35 6.18).406O86 34 6.18.40571.5 33 6.18.405344 32 6.18.4049713 31 6.17..404602 30 6.7 0.404232 29 6.17.403862 28 6.16.403492. 27 6.16.403122 26., 6. 16-.402753 25. 6.15.4 02 38F4 24' 6.15.402015 23 6.15 4166 2 6.14.401646 22 6.131.400909 20, 6.3 0.400541 19 6.13.400173 18 6.13.3998,06 17 6.12.399.438 16 6.12.399o 7,1 15 — 6.11 ME3It 04 14 6.11.398338 1.3 6.10.397971 I I 6.10.897605 11 6.10.8.97239 10 6 09.0.896873 9 6.9.M 507 8 6.09 1.3 14 7 6.8.895777 6 6.08.895412 5 6.07.895047 4 6.7.394683 3 6 07.894318.2 6.07.893954 1 6.06.8935900 0 I). 'ran g. i fM D I Sine. I1).cia I~ Co!-nngI.3 4J LOGrAIITH311C SINES,7 COSINES, 220 I I I ISine.I.L.I1U Cosinle. jU. I T.-it g. I 1). I Cotanig. I_ 3 4 5 6 7 8 9,10 14 15 16 18 ~19.20 21 22.23 24 25 26 27 28 29!31 * 22 84 35 86 38 37 40 42 43 44 46 46 47 48 49.60 62.64 65 -,56 657 68:59 60.578888.574200.574512.674824.676136.575447.5757'58.676069.676379.576689 9.576999.677309.6776 18.677927.578236.578545.578853.679162.679470 9.680085.580392.680699.681005.681312).581618.681924.582229.68253.5.582840 9.583145.683449.683754.584058.684361.684665.684968.686272.686674.685877 9 686179.586482.6867183.587085.587886.687688.587989.688289.688690.688890.9.689190.689489.689789.590088.690387.690686.690984.691282.691580 691878 I I j I 5.21t 5.~2J 5.20 5.19 5.19 5. 19 5.18 5.18 5.17 5.17 5.16 -5.16 5 16 5.1.5 5.15.5.14.5.14 6.13 5.13 5.13 5.12.5.12 5.11 5.11 5.11 5.10 5.10 5.03 5.09 5.09 5.08 5.08 5.07 5.07 5.06 5.06 5.06 5.05 5.05 5.04 5.04 6.03 5.03 5.03 5.02 5.02 5.01 5.01 5.01L 5.00 5.08) 4.99 4.99 4.99 4.98 4.98 4.97 4.97 4.97 4.96 9.94166.967115.85.967064.85.96701.3.85.966961.85.966910.85.966859.8.5.966,808.85.966756,.85.966705.86;.966653.86 9.96G602.86.966550.86.966499.86.966447.86.9663"95.86.966344.86.966292.86 96240.86.9GG188.8.9661-36.86 9.966085. 87.9660313.87.965981.87.965J28.87.965876.8.9650 24.87.9657742 87.9670.87.965768.87.995615.87 9653.87-.965611 8.965458.87.965406.87.965353.965301.8.965248.88.965195.88.965143.88.965090.88 9.965037..8.964984.88.964931:88.964879.88.964826.88.964773.88.964719.88.964666.8.964613.89.89 9.964507.89.964454.89.984400.89.964347.89.964294 8.9.964240.89.9641 87.8.964133.8.964080.8.96402,6 Sine. ID. 9.606410.6067,73.607137.607500.607863.608225.608588.608950.609312.609674.6 1 0036 9.610397.610759.611120.611480.611841.612201.612561.6 12 9 2.1.61321081.6 1364 1 9.614000.614359.614718.615077.615435.615793.61615 1.6 16509.616EG7.61722-4 9.61 75F2.617939.61829.618652.619008.619364.619721.620076.620432.620767 9.621142..621497.621852.622207.622561.622915.628269.623623.928976..624330 9.624683.625036.6253888.625741.626093.626445.626797.627149.827601.627852 6.06 6.06; 6.05 6.05 6.04 6.04 6.04 6.03 6.03 6 03 6.02 6.02 6 02 6.01 6.01 6.01 6.00 6.00 6.00 5.99 5.99 5.98 5.98 5.98 5.97 5.97 5.97 5.96 5.96.5.96 5.95 5. 95 5.95.5.94 594.94 593 5.93 5. 93 5.92 5.92 5.92 5.91 5.91 5.90 5.90 5.90 5.89 5.89 5.89 5.88 5.88.5.88 6 87 5 87.5.87 5.8(1 5.86 5.816 5.85 0.I- I..393.2217.392863.392500.392137.3917475.391412 -.391050.390688.390326.389964 0.389603.389241.388880.388520.388159.387799.3 87 4 2"9 ~.387079.386719.38G359 0.386000.385641.385282.384923.384565.384207.383S49.383 49(',1.3843 133.382776 0.382418.381'48.380606.880279.379924.3795G8.379213 0.378868.378503.378148.377793.877439,.377085.3767131.87677.376024.376670 0.875317.374964.874612.374259.373907.873555.373203.372851.872499 372148 59 58 57 56 55 64 53 52 51:50 49 48 47.46.45.44 '43 42 -41 40 39.38 37.36 35.34.33.32).31 60 29 28 27 26 23 24 23 22, ~21 20 19 18 17 16 15 14 13 12 11 10 1.9 ~8 4 2 0 I1 I Cosine. I 1). I -,Cotang. I D. I Tang. _I M.; f~7D TANGJENTS9, AND COTANGENTS. 230 41 - x.I Sine. -0 9.591878 1.592176 2-.592473 3.592770 4.593067 5.593363 6.593659 7.593955 8.594251 9.594547 10.594842 11 9.595137 12.595432 13.595727 -14.596021 '15.501 16.590609 17.596903 18.597100:19.597490 20.5977E3 21 9.69E075 22.5983068 23.5OEOGO 24.598952 25.5992144,26'.59536 27.599827 '28.600118 ~29.600409 -30.600700 31 9.6,0099,0 '32.601280!33.601570 34.601860 #3,5.002150 30:.6024309 7.602728 8,.603017 9. 6 03IS0 5 40.603594 I41: 9.603882 1ti2.:604170 j43.604457 44.604745 645.605032 ~46- 605319 147.605606 '48..605892;4.606179 #60.606465 I 9.606751 2 607036.6073822 4 I 607607 65.607892 6s.608177 P.608461 8].608745 ~9.609029 60.609313 C!osine I D. I Cosine. I D. 'lang. 4i.96 4.95 4.95 4.95 4.94 4.94 4.93 4.93 4.93 4.92 4.92 4.91 4.91 4.91 4.90 4.90 4. 89 4.89 4.89 4.88 4.88 4.87 4.87 4.87 4.86 4.86 4.85.4.85 4.85 4.84 4.84 4.84 4.83 4.83 4.82 4.82 4.82 4.81 4.81 4.81 4.80 4.80 4.79 4.79 4.79 4.78 4.78 4.78 4.77 4.77 4.76 4.76 4.76 4.75 4.75 4.74 4.74 4.74 4.73 4.73 UtJUf4tzf.963972.963919.963865.963811.963757.963704.903650.963596.963542.963488 9.963434.963379.963325.963271.963217.96310,13.963108.963034.962999.902045 9.962E890'.962836.902781.902727.902072.962017.962562.902508.902453.9202398 9.962343.9622E88.962233.962178.962123.962067.962012.961957.961902.961846 9.961791.961735.961680.961624.961569.961513.961458.961402.961346.961290 9.961235.961179.961123.961067.96101.1.960955.960899.960843.960786.960730 I.89.89.-89.90.90.90.90.90.90.90.90.90.90.90.90.90.91.91.91.91.91.91.91.91.91.91.91.91.91.92.92.92.92.92..92.92.92.92.92..92..92,.92.92.93.93.93.93.93.98.93.93.93 ~.93.93.93.94.941.628203.628554.628905.629255.629606.629956.630306.630656.631005.631355 9.6317'04.632053.632401.632750.633098.633447.633795.634143.634490.634838 9.635185.635532.6358719.6 36212 6.6365 72.630919.637265.637611.6379566.638302 9.638647.63C992.639682.640027.640371.640716.641060.641404.641747 9.642091.642434.642777.643120.643463.643806.644148.644490.644832.645174 -9.645516.645867.646199.646540.646881.647222.647562.647908.648248.648688 D. Cotang. _1 5.5 0.372148 60 5.5.371797 59 5.85.371446 5F 5.85.371095 57 5.84 870745 5 6 5.84.370394 55 5 83.370044 54 5.8.3.369694 53 58.83.369344 52 S 82J.368995 51 5.82.36t645 50 58.3682S6 4 O 82.367947 48 81.867599 47. 5.81.367260 46 5 81.366902 46 5.80.366553 44 5.80.366205 48 5.80.365857 42 5.9.365510 41: 5.9.866162 I40 5.9 0.364815 8 5.78.364468 38 5 78.364121 37 5.78.363774 36. 5.7.363428 3 5 5.7.363081 34 5.7.362735 33' 5.6.362389.82 5.76.362044 31 5.76..361698 30 5.5 0.361838 29..361008 28. 5.5.360663.27.~ 5.5.360318 26: 5.4.359973 25.5.4.359629 24., 5174.8984 2 5.3.368940 22> 5.73 5.72.8.68253 20i& 5.2 0.357909 19 5.2.857566 18. 5.72.357223 17~ 5.72.356880 16. 5.1.366537 16' 5.71.582 13. 5.70.8658105 2 5.70.856516810 5.70.854182 10 -5.69.686 l 0.854484, 9; 5.69.844 8 5.69 844 5.69.868601 7 5.68.868460 6 5.8.868119 V 5.68 f 862778 4'~ 5.67.862438 a. 56.7.352097:. b.67.817.344160 I J i I I i I I I I i I i i I -. I I D. ___I! ___e ___________________ 1) ( Sine '. L I Coan.[ P. __ Tang._, - ________I____ awl 42 LOGARITHMIC SINES, COSINES, 240 I I Al I Sine.I D 0 I 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 833 34 35 37 38 89 40 41' 42. 43 44 45 46 47 48.49 60.51. 52 53 64 55 566 57 58 60 9.6o9313.609597.609880.610164.610447.610729.611012.611294.611576.611858.612140 9.612421.612702.612983.613264.613545.613825.614105.614385.614665.614944 9.615223.615502.6 15781.616060.616338.616616.616894.6 17172.617450.617727 9.618004.618281.618558.618834.619110.619386.619662.619938.620213.620488 9.620763.621038.621313.621587.621861.622135.622409.622682.622956.623229 9.623502.623774.624047.624319.624591.624863.625135.625406.625677.625494 I I I 4.73 4.72 4.72 4.72 4.71 4.71 4.70 4.70 4.70 4.69.4.69 4.69 4.68 4.68 4.67 4.67 4.67 4.66 4.66 4.66 4.65 4.65 4.65 4.64 4.64 4.64 4.63 4.63 4.62 4.62 4.62 4.61 4.61 4.61 4.60 4.60 4.60 4.59 4.59 4.59 4.58 4.58 4.57 4.57 4.57 4.56 4.56 4.56 4.55 4.55 4.55 4.54 4.54 4.54 4.53 4.53 4.53 4.52 4.52 4.52 )Cosine. D. 9.0ij730.960674 9.960618 9.960561.960505 4.960448 9.960392 9.960335 9.960279 9-1.960222.960165 9.94. 9.960109 9.960052 9.959995.9.959938.959882 9.959825) 9.959768' 9.959711 9.959654 9 _99 96 9.959539.5.959482 9.959425 9.959368 9.959310.96.959253.96.959195.96.959138.96.959081.96.959023.96 9.958965 9.958908.96.958850.96.958792.96.958734.96.958677.96.958619.96.958561.96.958503 9.958445.9 9.958387 9.958329 9.958271 9.958213 9.958154:97.958096 9.958038 9.957979.97.957921 9.957 863.:97. 9.957804.957746.98.957687.98.957628.98.957570 -98.957511:98.957452.98 953.98.973.98.957276 ___ Sine. D.; Tang. I)D. _ICotang. I 9.64.583.6.648923 5.66.649263 5.66.649602 b6.649942 5.66.650281 56.650620 5.6.6509-59 5.6.651297.5.64.651636 5.64 ~6519 74 5.64 5.63 9 652312 5.63.652650 5.63.652988 5.63.653326 5.62.653663 5.62.654000 5.62.654337 5.61.654674 5.61.655011 56.655348. 9.655684 5 60.656020 56.656356 5.60.656692 55.657028 55.657364 55.657699 55.658034 5.58.658369 5.58.658704.55 9.659039 15 58.659373 55.659708 55.660042 55.660376 55.660710 5.56.661043 5.56.661377 5.56.661710 55.662043.5 9.662376 5.5.662709 55.663042 5.5.663375 5154.663707 5.5.664039 55.664371 55.664703 55 9.665697 5 52.666029 5.52.6663360 55.666691 55.667021 55.667352 55.667682 55.668013 5.50.6683413 5..668672 Cotang. I D. 0.351417.35 1077.350737.350398.350058.3497119.349380.349041.348703.348364.348026 0.347688.347 012.346674.346337.346000.345663.345326.3449E9.344652 0.344316.343910.343644.343308.34 2 9 72.342636.312301.341966.341631.341296 0.340961.340627.340292.339958.339624.339290.3.18957.3038623.308290.337957 0.337624.337291.336958.336625.336293.335961.335629.335297.334965 ~834634 0.3343013.833971.333640.333309.332979.332648.332318.331987.331657.331328 Tang. >60T I59 58 57 156 55 54 53 52 151 501 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 I -- - -- I - - - - - - -. fC.Osine. jI) M. j TANGENTS, AND COTANGENT&. 2-50 43 - Cotani. 1 -.....-.- I I '-'. I 181. I Sine. 0S '.625948 1 *626219 2.626490 3.6267160 4.627030 6.627300 6.627570 7.627840 8.628109 9.628378 10.628647 11 9.628916 12.629185 13.629453 14.629721 15.629989 16.630257 17.630524 18.630792 19.631059 20.631326 21 9.631593 22.63159 23.632125 24.632392 25.632658 26.632923 27.633189 28.633454 29.633719 30.633984 301 9.634249 32.634514 33.634778 34.6350492 35.635306 36.635570 37.635834 38.63 C 0 C7 39.636360 40. 6 36G2 3 41 9.6,36- 6 42.637 148 43.637411 44.63'7673 45.637935 46.638197 47.638458 48.638720 49.63F981 50.639242 51 9.68,9503 52.639764 53.640024 54.640284 55.640544 56.640804 57.641064 58.641324 59.641584 60.641842 Cosine D. I Cosine. I 1). I ianI cotam Z. I.1 9). 6I6Ibng7 1 Ll.I 4.51 4.51 4.51 4 50 4.,50 4.50 4.49 4.49 4.49 4.48 4.48 4.47 4.47 4.47 4.46 4.46 4.46 4.46 4.45 4.45 4.45 4 44 4.44 4.44 4.43 4.43 4.43 4.4~2 4.42 4.42 4.41 4.41 4.40 4.40 4.40 4.39 4.39 4.39 4.38 4.31 4 31 4'.3 4.31 4.31 4 3, 4.3~ 4.3, 4.3 4.3 4.3 4.3 4.3.957217.957 158-.957099.957040.956981.956921.956862.956803.956 744.956684 9.956625.956566.956506.956447.956387.956327.956268.956208.956148.956089 9.956029.955969.955909.955849.955789.955729.955669.955609.955548.955488 9.955428.955368.955307.955247.955126.955186.955065.955005.954883 9.954823 * 954 762.954701.954640.95457.954518.954457.954396.954335 9.954213 4.954152 4.954090.954029 3.953968 3.953906 2.953845 2.953783 12.953722.953660 -J Sine..98.9.8.98.98.98.98..99.99.99.99.99.99.99.99.99.99.99.99 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1 00 1.00 1.00 1.00 1.00 1.00 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.01 1.02 1.02 1.02 1.02 11.02 1.02 1.02 1.02 1.02 1.02 1.02 1.02 1.02 1.03.669002.669332.669661.669991.670320.670649.670977'.671306.671634.6711963 9.672291.67 2619.672947.673274.673602.673929.674257.674584.674910.675237 9.675564.6 7589 1,0.676216.676543.676869.677194.677520.677846.67817 1.678496 9 678821.679146.679471.6797 95.68012-0.680444.680768.681092.681416.681740 9.682063.682387.682710.683033.683356.68367 9.684001.684324.684646.684968 9.685290.685612.685934.686255.686577.686898.687219.687540.687861.68F182 5.50 5.49 5.49 5.49 5.48 5.48 5.48 5.48 5.47 5.,47 5.47 5.47 5.46 5.46 5.46 5.46 5.45 5.45 5.45 5.44 5.44 5.44 5.44 5.43 5.43 5.43 5.43 5.42 5.42 5.42 5.42 5.41 5.41 5.41 5.40 5.40 5.40 5.40 5.39 5.39 5.39 5.39 5.38 5.38 5.38 5.38 5.87 5.37.5.37 5.37 5.36 5.36 5.36 5.36 5.35 5.35 5.35 5.35 5.34 -U~3-3327 V.330998 51.330668 54.330339 51.830009 54.829660 5 1.329351 51.329023 5 q.328684 5,.828366 5.3203W 5 0.327709 4'.327381 4.327053 4'.326726 4.326398 4.326071 4.325743 4.325416 4.325090 4.324763 4 0-.324436 3.324110 3.3237E4 3.323457 3.323131 3.322806.322400.322154.321829.321504 0.321179.320854.320529.320205.319880.319556.819232.318908.818584.818260 0.317937.317613.317290.316967.316644.316321.315999.316676.315354.315032 0.314710.814388.814066.318745.313428.813102.312781.812460.812139.311818 I 9 r 7 5 4 8 2 1 0.9 8 6 14 13 12 11 10 29 28 26 I14 23 22 21 20 19 18 17. 16 13 12 11 20 18 7 6 6 4 8 2 1 0 t D.ICotang. I D.I Tang. I 183 640 4?i LOGARITHMIC SINES, COSINES, 260 I | Sine. D Cosine. 1. | 'Iang. | o. (:o. ___,_ 0 9.641b42 10o, o - "4 4.31 4 03 1.642101. 953599 40.6188502. 64'.31148 5 2.642360 1.0537.683bb2.3 5. 1 5 4 31.311177 3.642618 431.953475 1.689143 5..31057 57 4.642877.953413 3.69463 5..310537 56 5.643135 430.9 53352.689783 1017 5 6.643393 4.30.95320 1'03.69010330 7 54 7.643650 4.953228 '.690423.30957 53 8.643908 4.29.953166 1 03.690742.30258 9.644165 J.953104.691062 2 308938 51 10.644423 4.29 953042.691381 2 30S619 50 4.28 1.03 ) 5.3 11 9.644680. 9.954980 9691700. 000300 49 12.644936.952918 1 04.692019.307981 48 13.645193.95855 04.692338.3662 47 14.645450.952793 04.692656.30734 4 4.27 1.04 5.31.307344 46 15.645706 7.952731.69297.5.307025 45 16.645962.27.952669 0.693293 5.306707 44 17.646218 4.26.952606 104.693612 O.306358 43 18.646474 4.26.952544 04.693930.36070 5.04 19.646729 4.26.952481 04.694248 530.306070 4 20.646984 4.25 104 6924 5.30.305752 4 20.646984 4.25.952419 104.694566 5.29.305434 40 21 9.647240 9.952356 1 9.6941 3 0.305117 39 22.647494 424.952294 104.69501.304799 38 23.647749 4.24.952231 104.69 1 5.29 3044 37 2.680 4.24 1.0168 5.29.30442 37 24.648004 4.24.952168.695 33 29.304164 36 25.648258 424.952106.696153 5.29 303847 3. 26.648512 4.24.952043 4.696470 5.28.303530 34 28.649020 4.23 51910 6.28. 27.648766 4.23.95190.69677 5.28.303213 33 4.2 14 1 6913 62.302997 32 28.649020 4.23.951917.697103.302597 32 29.649274 423.951854.697420 528.302580 31 30.649527 4.22.951791.697736.302264 30 31 9.649781 9.951728 1.0 9.6903 0.301947 29 32.650034 422.951665 1.0.693369.301631 28 51602 1.05 5.27:83.650287.951GO2.695.30131 27 514.21 1.0 699001 5,2.30115 27,84.650539 4.21.951539.699001 5..300999 26 85.650792 4.21.951476.69931 5.26.300614 25 86 -.651044 421.951412.699632.300368 24.651297 420 951349 106 699947 2.300053 23 38.651549 4.2.951286 O.700263 5.2.299737 22 4 20 1.06 5.25 89.651800 419.951222.70o078.299422 21 40.652062 4.19.951159.2.70093 525.299107 20 il 9.652804 4,. 9.951096 i o 9.701208 24 0.291792 19 42.652555 48.951032 1.06.701523 524.2 477 18 43.652806.950968 ].06.701837 524.29163 17 44.653057 418.950905 106.702152 524.2748 16 45.653308 4.18.950841.7024606 5.24.297534 15 46.653558 4.18.950778 o.702780.297220 14 47.653808 4.950714 106.703095.29605 13 48s.654059 ~~ 4.17 1:06.09 5.23.29605 13 48.64059 417.950650 06 703409 523.296591 12 49.654309.950586.703723.296277 11 50.654558 4.16 1.06 5.23 60.654558 4.950522.704036 ^ 295064 10 4.16 1.07. 704036.22 1.654808 46 9.950458 07 9.704350 22 0.295650 9 4,16.950394 62.65058.950394 1.07.704663 22.295337 8.655807 416.950330 10.704977 22.295023 7 64.665556.950266 1.07 705290 522.294710 6 65.6558065 415.950202.705603 521.294397 5 4. 0 16 1.07 5.21 56.656054 4.950138.705916 21.294084 67;.656303 414.950074 107 706228 21.293772 8 68.666551 4:14.950010 1.07.706541 5.293459 2 69.656799..949945 1.07.706854 5.293146 1 60 657047.949881.707166 29321C,3i' 0 Cogine. I). Sine. I D. Cotang. I). |Tang. I M. TANGENTS, AND COTANGENTS. 4 3 270 Si;:e. I). Cosine. I D. Tan. I. Cotanrg. 0 b.65i047.13 94 l 9. i 0 5. 0292[34 60 I.6725 4.13 94916 1.07 707478 5.20.292522 59 2.67542 4..1777,207.707710 52.29210 58 3.65775o0 4.12 1949 07.70103 5.20 3.65.,20. 349068 W102.2or 92 2,, 6 7 4.6580037 4. 12 108 064 5.20 15 6 66 5.658254 412.949558 1.08 708726.1.291274 55.658531 4.1 19494 0894 5.19 29 3 7.658778 4.11.9494 108.709349 5.19.205 3 8.659025 4 1.949361 108 7090.GO 9.290:40? 9.659271 1.949300 1.708 097' 5..2900-9 t 10.659517 54192305 1.10o 5.18.281718 60 4.10 108 5.18.10 1.08 11 9.659763 0 9.9 410 9.7 1503, 0.269407 12.660009.949 15.71004 5.18.2Ec0O6I 43 11-03 5.18 13.660255 *..949040.718 5.2518 47 14.660501 409.95 1892H75.1 4 15.660746.4948910 1 711836.285164 45 4 03D 1.08 5.17 16.660991 403.94845 10.71146.2751744 17.6G133G.94c~r 60..12846.267_544 17.661236 4.08.94780 1.0.712456 517.27544 1.0614 1 408.948715 1.7127G6 517.287"34 42 19.66176 407.48I650 1.09 713076 516.20924 41 20.6G 1' 70.071C0594.7133,6.266014 40 4.07 1.09 5.16 21 9.u6 1i:21t.b4~519 9.7*1S,656 0.2tG004 39 22.662459 4.07.948454 10.71400 5.16.2855 38 23.66 2,70:1 4.07 10.714314.9E25556 387 4.06 1.09 5.16 24.662946.4. 432 13 09.71464 5.15.285376 836 25.663150 4.94257..71430 3.285067 35 26.663433 4.06.945192 109 154 15.284758 34 27.663677 4. 9 48126 1551A 5..284449 33 28.663920 4.0.94060 1.09 5.71 14.284140 32 29.664163 4.05 94795 1.09.71GI3 5.14.26333 81 30.664406.947929.71G-77.283523 30 -9648 4.04 1.10 5.14 31 9.664648 9.947663 9 'i G )0.283215 29 32.664I1 4.04 947797 1.10.717033.282907 28 33.665133 4.04 947731 1 I.717401 5.282599 27 34.665375 4..947665 110.7i17709.1.282291 26 35.665617 4.947600 1.0.71017.281083 25 36.665659 4.947533 10 71325 5 13.21670 24 37.666100 4.02 947467 1.10 71633 5.13.21367 23 38.666342.947401 1 l.718940 5.1.261060 23 89.666583 4.947335 1 719248.20752 21 40.666524.9472690 1 719555 5. 0445.719555.250445 20 4.01.1.10 5.12 41 {.667065 9.947203 9.71902 0.260138 13 42, 667305.947136 110.720169...270831 18 43.667546 4.0.947070 1..720476 5.27~,524 17 44.667756 4.947004 1. 7207f3.279217 16 45.668027.946937 1. I.71059.27511 13 46.606267.946871 1.72136 ',6.276G04 1 t 47.666506.94604 1 1.721702 11.27298 13 48.665746.946738.722009 5.10.277991 12 49.665986.946671 7 '2.7-2313.277685 11 50.609225.946604 1.1.722G21.277379 10 51 9.6094G4;9.946538 9.722127 0 0.277073 9 52.669703 3.98 946471 111.72323 5.10.27678 8 53.660942.98.946404 1 '.723538 5.09.276462 7 54.67011 3.9.940337 1.723344.27106 6 55.670419 3.97.946270 1. 724149.275851 5 56.670658 97.9403 112 724454 5.09.275546 4 7.670CL6.9.946136 12.74759.276241 09 58.671134 94G 1.12 725065 08.274935 569.671C72 3.96 9400 1.12 5.08.274631 1 60.671C09. 3.96.94535 1 12 C74.08.274326 0 Co!nCe D. | Sine. 1D. ICo an, 1). Tang. JM Ir He~~~~~~~~~~~~~~~~~~~~~~~~~~~~,. * - VI LOGARITHMIC SINES, COSINES, 280 M1 I Sine. I D 1 2 4 5 6 7 8 9 10 I11 12 13 14 15 16 17 18 19 20 21 23 24 25 26 27 23 29 30 31 31I 33 34 3-5 36 37 33 "3 4) 413 41 43 51 53 6)8 59 69 9.671609.67 1847.672084.672321.672558.672795.673032.673268.673505.673741.673977 9.674213.674448.674684.674919.675155.675390.675624.675859.676094.676328 9.676562.676796.677030.677264.677498.677731.677964.678197.678430.678663 9.678895.679128.679360.679592..679824.680056.680288.680519.680750.680982 9.681213.681443-.681674.681905.682135.682595.682825.683055.683284 9.683514.683743.683972.684201.684430-.684658.684887.685343.685571I 3.96 3.95 3.95 3.95:3.94 3.94 3 94 3.94 3.93 3.93 M3. 3. 0,2 3.0b2 3.0)2 3.02 3.91 3.01 3.91 3.91 3.90 3.90 3.90 3.00o' 3.89 3.89 3.89 3.88 3.88 3.88 3.88 3.87 3.87 3.87 3.87 3.86 3.86 3.86 3.85 3.85 3.85 3.858 3.84 3.84 3.84 3.84 3.83 3.83 3.83 3.83 3.82 3.82 3.82 3.82 3.81 3.80 3.80 3.80 Cosiae D. T) ang. 9.945935.1 97 ~256 74.945868 1 2 725979.945800 1.12.726284 9473 1.12.726588.945666 1.12 '726892.945598 1.12.727197.945531 1,12.727501.94,5464 "12.727805.945396 1 13.728109.945328 1.13 728412.945261 1 3 728716 1.13 9.945 193 9 729020.945125 1 729323 1 13.945058.729626.944990 1 13 729929 *944922 1.13 703 1.13 303.944854 1.13.730535.944786.730838.944718 i1. 13 731141.944650 1113 3144.944582 14 731746 9.944514 1.14 9724.941446 1.14 '732351 9437 1.14.732653.944309 1.14 '732955.944241 1.14.733257.9 44 1 7 ' 1.14.733558.944104 1.14 733860.9440.36 1.4 734162.943967 1.14 734463.943899 1.14.734764 9.943830 1.14 9.735066.943761 1.14 735367.943693 1.lo 735668.943624 1. 15 735969.9435,5.5 736269.943486 1.15 736570.943417 1.15 736871.943348 II" 737171.943279 1.5 737471.943210 1~737771 lb1 9.943141 1.15 9.738071.943072 1.15 738371.943003 1.15 738671.942934 1.15 738971.942864 1.Ic i39271.942795% 1 739570.9 42)7 2 6 I6 70u,9870.942656 116.740169.9421587 16 740468.942517 1 6 706 9.942 448 1.16 9. 741066.942378 1 Jr.741365.942308.741664.942239..6 741962.942169 1.16.742261.942099 1.16.742559.942029..6 742858.941959 1.16 '743156.941819 1.7 743454.941819 1.7.7437152 Siune. D. Cotang. 5.8 274021 5.08:2771 5.07.273412 5.07.2734108 5.07.272103 5.07.272499 5.07.27-2195 5.0(6.2721891 5.06 219 5.06.271588 5.06.271284 5.06 0.270980 5.05.270677 5.0.270374 5 J5.270071 5.05.269767 5.05.269465 5.05.269162 5.04.268859 5.04.268556 5.04 5.4 0.267952 5.4 267649 5.03.267347 5.03.267045 5.03.266743 5.03.266442 5.03.266140 5.02.265838 5.02.265537 5.02. 265236 5.2 0.264934 5.02..264633 5.02..264332 5.01..264'031 5.01.263731 5.01.2653430 5.01.263129 5.01.262129. 5.0.26 2529 5.00 5W 0.2)619219 5.00.2616'29 5.00.261329 4. D ).261029 4.9.260729 4.9.260430 4.9.260130 4 90.25953i 4.98.259533 4.98.293 0.258934 4.98.258635 4.98.258336 4.98.258038 4.7.257739r 4 97.257441 4.7.2571414 4.7.256844 4.7.256546 flf Tangj, I D. I Cotang-. 0.243 - li) 59 58 57 56 55. 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 87 36 35 34 33 32 31 30 29 28 27 26 23 24 23 22 21 20 19 18 17 16, 15 14 13 12 11 10 91 8' 6 4 3 2 1. i '1 CsMn& I - 1.-, 1VW* I CIO TANGENTS,, AND COTANGENTS. 29~ 47. 1 Sine. 0 9.685571 1.685799 2.686027 3.686254 4.686482 5.686709 6.686936 7.687163 8.687389 9.687616 10.687843 11 9.688069 12.688295 13.688521 14.688747 15.688972 16.689198 17.689423 13.689648 19.689673 20.690098 21 9.690323 22.690548 23.690772 24.690996 25.691220 26.691444 27.691668 28.691892 29.692115 80.692339 31 9.692562 33.6927:5 33.693008 34.693-:31 35.693453 33:.693676 37.693 {98 38.694120 33.694342 40.694564:41 9.6947L6 42.695007 43.695229 44.695450 45.695671 46.695C92 47.696113 48.696334 49.696554 50.696775 51 9.696995 52.697215 53.697435 64.697654 55.697874 56.698094 5.7.696313 58.696532 9.696751 60!.;9e,7o0 [,Cosine. I I 1). I Cosine. I D. I Ta-zig. I D. I Cotnng. I- I.. I.-.-.. -1 3.80 v.v4lbLu 1.17 V-i4;JV-5U' 4.96 W 3.79.941749 1.17 744050 4.!jo.255950 69 3.1-9.941679 1.17 744348 4.1..:G.255652 58 3.79.941609 1.17 '744645 4 G.255355 rd.941539 i.255057 66 '3-79 1.17 "44943 4-OG.941469 74,5240.254,760 6.5 -U-78.941398 1.17 4. Of) 3.78 1.17 '745538 4.9b.254462 54.941328 -2 3.78 1.17 '745835 4. 5.254165 6 3.78.941258 1.17 -746132 4 5.253F68 5-3 3.77.941187 1.17.746429 4.253571 6 1.941117.746726 -253274 50 -3.77 1.17 4 3.77 9.941046 1.18 D. 747023 4!,4 0. 215 2 9 74 7 4 9.940975.747319.2626EI: 4 8 3.77.94090' 1.18.747C10 4.!'4.2523E4 4 7 3.76 4 1.18 4.!;4 3. 7 6.940834 1.18 7 4 710113 4 t 4.252087 46 -3.7f;.940763 j.,R 4 `20 3 4.(4.231791 45-.940693 4 o 1).2514D5 44 -3.7G.94062-1 1. is 4 G, 1 4.93 -92511VIY9 43 4 3.75 1 18 93 3.75.940551 1.18 4 O.! 7 4.93 250'903 40 3.75.9404EO 1.18.7493"3 4!A:250607 4 T.940409.749689 -250'311 40 3.75 1.18 4.93 9.940838 9.749VF5 0.250010' 8 9 3.74.940267 1.18.10502EI 4 93.249719 3 8 3.74.940196 1.18.1,505,16 4.4.2.24,9424 87 -3.74 1.18 -I 4.92.940125 50'72.24912" 30. 3.74 1.19 4.92 3.73.940054 I 19 51161, 4 f.,2.24SE33 35-.93991-1 76 5 1,16 2.246538 81 3.73 ' ' 1.11) 4.92 3.73.9319911 1.19.7151757 4.92.248243 337 3.1-3.939F40 1.19 "j, 5 2 521 4.9,.247948 n 13.72.93976S 19.75 24 7 4.9.247653 8 1,.939607.7,52642,.247338 2.0, 3.72 i.'Io 4.9 3.72 9.93962.5 1-1.9 9,75:,"037 4.91 0.247063 20 3.71 -M554 1.19. 17 5.1j, -1 3 1 4.91 '24670 2& 3.71.939482 1.19.733526 4-91.24G474 27 3.71.939410 1.19.750"120 4.90.246IIE0 26'.9303"9.731115.245M 25. 3.71 1. 19 34409 4.90 3.70.939267 1.20 4.90.2 4 3 5 0, 1 zt.930195 3.70 1 20 '754 703 4.90.245"ri 23, 3.70 1.20 '754001, 4.90.245003 22: 3.70.939052 1 20 '7531,91.244709 21. - 4.90 1938980 So Z44415 3.09 1.20 4.89 3.69 9.93SU08 L20 9.755878 4.63, 0.244122, 19..91,8836.243C 3.GV 1.20 1756172 4.89 3 28' 18 169.93kc" 763 1.20 756405 4.89.243535 rr 3.68.93C4391 L20 '7361,59 4.80.243"41 10.03EC19 "37052.2 4'2 9 4 8 D 3.68 - 9 %')' C 5 4 71.20 7i,70'45 4.89.242655 14 3.68.9384'15 1.20 4 4.88.24"362 137 3.68 1.20 "57G"s 4.88 -938402. P.5703"I.24i.069 12 -3 67.0UIE3UO 1.21 1 4.88 3.67 1.21 '750224 241776 It.9 "E258 4.88:241483 Iff. 0 '75C.514 8.67 1.21 4.88 1 ). I Cosinc. 3.80 994181.941749 [ 9~.941679 3 '79.941609. 37.941539. 38.941469 3U78.941398 3.78.941328 3. 8.941258 377.941187 377.941117 37 9.941046 3.77.940975 3.76.940905 3.7 ',..940834 3.76 i.940763 3 o,.940693 '3.7.940621 3.75'.940551 3.7 5.940480 1375.940409 3.75 _ 9.940338 3.74.940267 3..4.940196 3 -74.940125 3.73 3'!3.9399211 3.73 3. 3.939f40 3 72.939768 3:72.939697 3.72 9.939625 3.72 2.939554 371.939482 3.71.939410 3.71.939339 3.71.939267.70.939195 3.70.939123 3.70.939052 3.09.938980 3.69 9.938U08 3.6.938836 369.936763 3.8.936#91 3.68.936619 3.68 9354.68.938475 3 67.938402 3 67.93E330 3.67.93C258 3.67 9.361 3 66.938113 3.66 3 *.93. 0409 3.o5.937967 3. 5.937C22 3.65 ^365.937749 *6.937531 D. | Sine. I I. I tnlg. | 1.17, 9.743752 1 117 ' 744050 > '17.744348 1~,,.744645 1.17 1 7. 44943 1.17.745240 1.17 745538 1 17 745835 1 17.746126 8 17 746429 1.178 118 9.747023 1. I.747319.747613.7198 3'7 3 1 18.496389 I.749C1 1' 19 7497013 1.18 1. }171.7 493703 1.18 749689 1 18 9.749985 750261 1 18 1.50576 1.18;'21 1 9 751727 1.19 G5114 1.19. 1u19.352954 1 19.72347 119.52G432 1 9.752937 19.753526 1.1.075520 1.19 19. 754409 1.20.754703 1 20 120 9.755878 1,20.756172 1.20 757052 1.20 1.20.757345 1.20.757638 1.21 57931 1.21. 75S224 1.21.756517 1. 21 1.21 9.758810' 1.21.759102 1 21 i21 i.759395 1 21.7596~7 11.759979 121.760272 1 21.760504.7C1145 1.21 g D.7C1d39 D. "Cotan. \ rv I II lr~ I f ~.,~-.l I X.,utung. I...w........, I I, I ~tHHIg. | 96 256248 660 4.6.255950 59 4.96.255652 58 4. 0.255355 17 496.255057 66 4.90.254760 55 4.91r.254462 54,4.95.254165 53 4 5.253868 52 4 65.253571 61 4:5.253274 50 -4 J4 0.252977 i 49 4 - |4.2526f1 48 4.!,4.252364 4 47 49 4.252087 40 4 4 ~.251791 43 4.(4.251495 44 4.93.251199 43.3 250903 42 4 9.3.250607 41 493.250311] 40 493 4 0.250015 39 4 2.249719 38 4.62.249424 87 4.92.249128 3 4 92.248633 35 4 -2.246538 34 4.92.248243 33 4.92.247948 32 4.91.247653 31 4.91.247358 3.0 4.91 0.247063 29 4.91.240670C 2& 4.91.246474 27 4.90.24G1E0 26' 4.245.85 25. 4.90 |.245591 28 4.90.245297 23. 4.90.245003 22 4.90.244709 21 4.89. 244415 20 4.83i 0.244122 19. 4.89.243628' 18 4.89.243535 tr 4.89.243241 16 4.89.242948 15 4.242655 14 4.88.242362 13 4.88.242069 12 4.88.241776 1 4.88. 4.88!.241483 t1 4.88 0241190. 9. 4.87 i.240898 8 4.87! 4.87.2240605 1 4.87.240313 6' 4.87.240021 5 4.87.239728 4 4.87 1.239436 21 4.8.230144 2 4.86.238=t t.2385F11 0 - I P t D. I, im,, D4 I Sinai I D. I Cotang. I - Uu D- I LTarg. 1-' I 48 M I Sine. 0 9.6989710 1.699189 2.699407 3.699626 4.699844 6.700062 6.700280 7.700498 8.700716 9.700933 10.701151 11 9.701368 12.701585 13.701802 14.702019 15.702236 16.702452 17.702669 18.702885 19.703101 20.703317 21 9.703533 -22.703749 23.703964 24.704179 25.704395 26.704610 27.704825 28.705040 29'.705254 30.705469 31 9.705683 32.705898 8.3.706112 34.706326 35.706539 36.706753 387.706967 88.707180 389.707393 40.707606.41j9.707819 42.708032 43.708245 44.708458 45.708670 46.708882 '47.709094 48.70930 '49.709618 60.709730 61 9.709941 62.710163 53.710364 '54.710575 6i 5.710786.66.710997 ~67.711208 768.711419.69.711629 ~60.711839 ICosine. LOGARITHMIC SINES, COSINES, IWO D. 3.6~ 3.6 3.6k 3.64 3 6~ 3.63 3. 6 3.63 3.62 3.62 3.62 3.61 3.61 3.61 3.61t 3.60 3.60 3.60 3.60 3.59 3.59 3.59 3.59 3.59 3.58 3.58 3.58 3.58 3.57 3.57 3.57 3.57 3.56 3.56 3.56 3.56 3.55.355 -3.55 3.55 3.54 3.54 3.54 3.54 3.54 3.53 3.53 3.53 3.53 3.52 3.52 3.52 3.52 3 51 3.51 3.51 3.61 D.0 I Cosine. I D. I Tang,. I I). 9.937531 9.761439.937458 1.21.7617131.937385 1.22.76M2023.937312 1.22.762314.937238 1.22.762606.937165 1.22.762897.937092 1.22.763188.937019 1.22.763479.936946 1 22.737.936872 1 2.764061.93 6 7 99 1.22.7643152 1.22. 9.936725 9.764643.936652 1.22.764933.936578.765224.936505 1.23.765514.936431 1:23.765805.936357 1.23.766095.936284 1.23.7466385.936210 1.23.766675.936136 1.23 695.936062 1.3 676695 1.23-.675 9.3988 9.767545.935914 1.23 767834.935840 1.23 '76h124.935766 1.~23 '768413.93.5692 1.24 768703.935618 1.24 '768992 9353 1.24.769281.935469 1.24.7o9570 1.24395 769860.935320 1.24.770148 1.24 9.935246 12 9.70437.9305 171 2.770726.935097 12.771015.935022 1:24' 771303.934948 1 24.771592.934E73 1.24.771880 4.934798 1.24.77#2168 4.934723 1.2 772457 4.934649 1 25 772745.934574 105 '733 4 1.25 4703 9.934499 1 25 19773321 A.934424 7736084.914 1.25 739 4 9819 1.25.773896.9341274 1.2.'47498 4 1.25.4447 4..934048 1.25.775046 4,.933973.775333 4.933898 1.25 775621.933822 1.26.775908 4 993771.26. '. 4..9336717 1.26 9.776195 4.933567 1.26.776482 4.93359 1.26.776769 4.933520 1.26.777056 4 9345 1.26.777342 4.933369 1.26.777628 4.933293 1.26.777915 4.933217 1.26.778201 4..933141 12.778487 4..933066.778774 Sine.ID. Cotang. I 4.8( 4.8! 4.8( 4.8C 4.85 4.85 4.85 4.85 4.8.5 4.84 4.84 4 84 4.84 4 84 4.84 4.81 4.8.1 4.83 4. 8 3) 4.831 4.83 4.82 4.82 4.82 4.82 4.82 4.82 I.8 I 1.81 L81 [.81 181 [.81 -80.-80.80.80.79.79.79,79 '79,79 798 7 8 78 78 78 7 8 78 7 7 77 7 7 ICotanm. 0.-K )db5b-6'-.238269.237D77 5.23768o6 57.237394 56.23 '1 03 5.5.236812 5 4.236521 53.236230 5 2.235939 5 1.235648 5 0 0.235357 4 9.2.3506 7 48.234776 47.234486 46.234195 45.233905 44.233615 43.233325 42.233035 41.232745 40 0.232455 3 232166 38.231876 37.231587 36.231297 35.231008 34.230719 33.230430 32.230140 31.229852 30 0.229563 29.229274 28.228985 27.22E697 26.228408 25.228120 24.227832 2.3.227 543 22.227255 21.226967 20 0.226679 1 9.22639291 8.226104 1 7.225816 1:1.220552 9 1).225241 1!.224954 11.224667 11.224379 1 1.224092 10 0.223805 9.223518 8.223231 7.22294.5 6.222658 5.222372 4.222085 3.221799 2.221512 1 221226 0 Tnng...~ I 59 0 TANGENTS, AND COTANGENTS. _ 310 Sine. D. Cosine. D. Tan,.711839s;j9.933066~77t77 0.2"21226- 6'O0 0 1839 3.50 1.26 9779060 77.220940 59 1.712050.50 939990 1.27.779060.220654 58 2.712260 3.50 9323 1.27?7 4.76.220368 57 3.712469 939 1.27.779963 4.76 22002 4.712679 39.9327 1.27.918.76.219287 65 5.712889 349 932685 1.27 7.032 4.76.219611 54 6.713098 3.932609 1.27.7I489 476.219225 543 7.713308 349.939533 1.27.780775 4.76.218922 62.713517 3.48.93 7 1.27.781060 4.76.218 60 4"- *.x 93245.781346 7.13726 3.48 1.27.218654 10.71 3.48 93 1.27 1 4.5.2183690 1 9 9714144 3.48 9.932228 1.27 9.781916 4 0.218084 49 12.714352 3.47 9315 127.782201 475.217799 48 13.714352 3.47 932075 1:28 78246 47.217514 47 13.714561 3.47 1'28.714769 3.4 931998 1.28 782771 4.75.217229 46 21.714978 3.47.931921 1.28.783056 475.216944 45 [ 936845 1..28 4. 1 16.715186 347 931845 1.783341.216659 44 17.715394 3.46 931768 1.28.783626 474.216374 43 18.715602 3.46.931691 1.28.78310 74.216090 42 19.715809 3.46 931614 1.28.78195 474.215805 47 20.7160 31537 784479.7.215521 40 19.?lg0 9.31614. 0.215236 210 9.71622409 1.28.78474 4.74 22.716413.3.46 9313 12 7848 0. 215236 38 716 3 3.41.28.214952 22.71643 3.4 931229 128.78501 4.2146684 7 4 24 3.71 3 t5.933 4.73 23.716639 3.45.9311 1.289 785 473.214384 8 24 716846 3.45 1.29.785616.140 85 25:71053 3456.931052 1.29.78560 473.213816 84 26.7172059 3.4.93 1.29 7618 4.73.213532 3 27.717466 3.44 9309198 1.29.78675 473.213248 82 28.717673 344 930921 1.29.787036 4.73.212964 81 29.717879 3.44 930784 1.29.78703.2126815 4 30.718085 3.43.90566 1..32121 4. 31 9.718291 9.93068 1.29 9.787603 4.72 0.212397 29 54.7 22994 3. 33 *^ S89? 1.31 ^ ^ ^ 4 6 ^052 1 7229 32.718497 3.3.930611 1.29 4.72 212114 2 33.718703 3 93033 1.29.78810 4.72 3 25 34.18909 33.930456 9 788453 4.72. 211547 26: _1.218694 35.919114 93038 1.28.8 4.2.216 2 86.19320 3.42 930300 1.30 789019 4.72.21098 2 37.719525.93023 1.30.789302 4.71 2 3. 930145 1.30 789585 4.71 21013 2 38.719730 3.42 930067 1.30.789868 4.71 210984 20 39.719935 3.41 929989 1.30.790151 4.71 3.4 1.307 40.720140 3.341 *9^499 ^ 4.7 83 2 41 9.720345 9.929911 1.30 9.790433 4.71 0.209567 19 64.720549 3.41.92833 1.30 11.209284 18.720549 3.40 9275.79016 4.27090 17 13.720754 34 92975 130.79099 4.71.2083719 16 44.720958 3.40 6 1.30.791689 4.71.208437 15 45.721162 3.40 929599 13. 79156 4.70.208154 14.721576 3.40.929521 1.30.791846 4.70.207872 13 46.72156 3.40 929442 792128 4.70 4.721 3.40.929.792410 4.70 207308 11 48.721774.929364 1.31 792692 4.70.20730 1 49.721978.92928 1.31.79294207026 10 -.7221r81 I.92920~7 1.31 7994 4.70 6 0 - 0 ---- ^ --- -- ^ W.7 9 8 2 5 6 ).?O32 0 6414. 51 9.722385 9.929129.1 9.793256 4.70 0206746 8 52.722588.929Q50 1.31 53.7938 4.69 201 7 3.722791 8.38.9289 131 791 469 205899 54.722994 3.38.92 1.31.794101 4 205617 55.723197 3.38.928815 1.31 7 4.69 56.723400 3.38. 6 1.31.79464 4.69 2050558 58.72305.928578 4131.79 4.69 87.928499 11.795508 4.68.204492 1 59.724007 3:37.928420 1. 79789.204211 60.724210 18.73 Cosine D. Sine. I. Cotang. D. 3,4.4 5 o 50 MN Sine. 0 b.Y144.41 0 1.7244 1 2 2.724614 8.724816 4.725017 5.725219 6.725420 7.725622 8 -725823 9.726024 10.726225 11 9.726426 12.726626 13.726-827 14.727027 iS.727228 16.727428 '17.727628 1 8.727828 19.728027 20.728227 21 9.728427 -22.728626 23.728825 24.729024 25.729223 26.729422 27.729621 28.729820 29.780018 80..730216 81 -9.730415 32.730613 83.730811 84.731009 35.731206 87.73160?, 88.7,31799 39.731996 40.732193 -41 -9.732390 42.732587.43.782784 '44.732980.45.733177 -46.733373 47.7385o9 -48.733765 49.733961 50.734157 S1 9.734353 62.734549 63.784744 64.784939 65.735135 66.735330 67.736525 68.73-5719 59.735914,60.73610o ICosine. r LOG C 1::1T RMIC -SIX ES, COSINES, 39.0 L.37 3.37 3.37 3.36 3.36 3.36 3.36 3.35 3.35 3.35 3.35 3.35 3.34 3.34 3.34 3.34 3.34 3.33 3.33 3.33 3.33 3.33 3.32 3.32 3.32 3.31 3 31 3.31 3 31 3.30 3.30 3.30 3.30) 3.30 3.29 3.29 3.29 3.29 3.29 3.28 3.28 3.28 3:28 3.28 3.27 3. 27 3.'27 8.27 8.-27 3.26 8:26 3:,26 8.26 3:25 3.25 3.215 8325 '3.25 -S.24 2.24 IICosine. I D. 9448420 13.928342 1.32.928263 1.32..928183 13.928104 1.32.928025- 1.32.927946 1.32.927867 1.32.92-77.87 1.32.927708 1.32 1.32 9.927549 113.927 4 70 13.92"7390 1.33.927310 133 M9223 1 33.927151 13.9270711 1.33.926991 1.33~.926911 1.33.926831 1,33 9.926751 1.3.926671 1.33.926591 1.33.926511 1.34.926431 1.34.926351 1.34.92627.0 1.34.926190 1.34.-926110 1.34.926029 A.34. *9.925949 13..9.,868 134:'92578813.92.5i07 13.925626 1.34.925545 13.925465 13.92-5384 1 as5.925303 1.35 -.925222 13 9- 9250141 1.35.:92 5060 13.924979 1.35.924897 1.35.924816 1.35.924735 18.92465.4 '1.86 '.924572 1.36.924491 1.36.9244091.1:6. 0.924828 1.861.'924246 1.36 -..924164 1 36.924083 1386.924001 1.36.923919 1:36.923837 1.36.923755 3.923673 1287.923591 Sine.iaD 4 570_I.796351:796632.796913.797194.797475.797755.798036.798316.7,98596 9.798877.79,9157.799437.799717.799997.800277:,800557.800836.801116.801396 9.801675.80 1955.802234.802513.802792.803072.803351.803630.803908.804187 9.804466.804745.805 023.805302.805580.805859-.806137.806415.806693.806971 9.807249.807527.807805 A.80,8083 A.808361.808638.808916 A.809193 4 A8094 71 4 809748 4 L.810025 4.8103029.810580.81-0857.811134.811410.,811687.811964.812241 -.812517 Dotang~ D. 4418 4.68 4.68 -4 (;8 4.68 -4.68 4.68 4.68 4.67 4.67 4.67 4.67 -4.67 4.67 4.67 4.66 4.66 4-466 4.66 4.66 4.66 4.66 4.66 -4.65 -4.65 4.65 4.65 4.65 4.65 4.65 4.65 4.64 4.64 4.64 4.64 4.64 4.64 4.64 4.63 4.63 4.63 1.63 1.63 1.63 1.63 1.63 L.62 L.62 L.62 k62 L.62 L62.62.62.-62.6l.61.6l.6l.61 Cotang..203930 Oy.203649 58.203368.57.203087 5 6.2018-06 55..202525 54.202245 53.201964 52.201684 51.201404 6-0 0.201123 49.200843 48.200563 47.200283 46.200003 45.199723.44.199443 43.199164 42.198884 41.198604 40 0.198325 39.198045 88.197766 37.197487 86.197208 85.196928 34.196649 83.1963710 -82.196092 81.195813 S0 0.1595534 29.195255 28.194997 27.194698 26.194420 25.194141 24.193863 23.1931585 22.193307 21.193029 20 0.192751 19.192473.18.192195 17.1-91917 16.191639 15.19t362 14.191,084 13.190807 12.190529 11.190252.10 0.189975 9.189698 8.189420 7.189143 6.188866 6.188590 4.188313 3.188036 2.187759.1.1287483 0 Tang. - L I ON...6 D. t I TANGENTS, AND COTANGENTS. 51 330,Sine D.I Cosine. D. I Tang. D. Cotang. 0 — l.7'., ~ 3 91iu JoJ. 4.6160 33 3.24 3 1.812794 4.61.187206 9.92.3509or ~.Y 1.37.hlais4 4.61.186930 58 7364 _ 3.24.923427 1.37.813070 4 61i.186653 3.736692 3.23?923:35137.813~23 4.60.1861014 45 1.716886 3.23.923181 137.81399 4.60.18514815 3 3.734080 3.23.923098 1.37.81415 4.60.18654 53 6.737274 3.2.923016 137.14 4.60.186529 52 7.37463 3.23 *922133 1.37.81478 4.60.184916 5 8.743661 3.2.92251 1.37.815.18721 9.737855 3.2. 9227 1.34.196 10.738048 93.22 6 1.38 8.15279 4.603 12.738434 3.2 922603 138.15831 459.183893 47 13.738627 3.22.923520 1.816107 4..183618 46 14.738820 3.21.922438 13 8.816382 4.59.183342 45 33.7623 3.21.922355 1.38.816658 459.183067 46 1.73920 3.2.92 138.81.182791 48 16.739206 3.21L.922189 1.38.817209 4.59.182516 42 17.739398 3.21 922106 138.817484 4s59.182241 41 1.739590 3.20.922023 1.38.817759 4.59.181965 40 21.739783 3.20.921940 1.38.818035 4.58 0.739975 3.20 921857 0.181690 39 91.740167.20139 9.818310..181415 38 21:.740187 30 921774 1 39.8185E5 4sg.181140 37 2.45 3.20 921691 139.818860 4.58.18065 36 24.740742 3.19 921524 139.819410 4.58.180 34 25.740934 3.19 9214 139.819684 4.58.180316 3 26.741125 3.19.9213 139.819959 4.58.179766 32 27.741316 3.19.921254 1.39.820234 4.58.179492 31 28.741508 3.18.92110 1.39.820508 4.5.179217 30 29.7415699 3.18 921190 139.820 4.57 30.741889.3.18 9.20078 4.57.178 2 9.742080 31 9.921023 1.39.821057 4.57.178694 2 31 0 - n3.18.920939 1.40.82 332 4.57.178344 27 32.742271i 3.18.922 086 140.821606 457 33.742462 3.17.920772 1.40 821880 457.178120 26 34.742652 3.17.920688 140.8221534 457.177835 24 36.743033 3.17 920604 1 40.822429 4.57.177297 23 37.743323 920520 140.822703 457.177023 22 38.743413 3.17.920436 1:40.822977 4.56.176750 21 39.743602 3.16.920352 1.40.823524 4.56 39.74360 3 9.16:920184 920.176202 19 4 9.7439.16 9.82398 4.5.17928 4 7 9.743982 3.920099 1:40.824072 4.5.1659 42.744171 3.16:920015 140.824345 4.56.175655 17 43.744361 3.15.919931 1.40.824619 4.56.175381 16 44.744550 3.15 919846 141 824893 4.56.17484 1 45.7404739 4.59.1783610 1 46.744928.15 9196 141.825166 4.56.174661 13 3.15.919677 141.825439 4.55 174287 12 48.745306 8.14.919293 1.41.825213 4.55.174014 11 50.74563 3.14.919424 1.41.826259 4.55 51.459 3.14 9919339 1.4 826532.173468 95 5.746280 3.14.19169 141. 55 817264962 52.734601 3.1423 *[.8276351 8. 172376 7 54.746248 3.13.919085 1 41 64.746436 3.13.919000 1.41.827624 4.55.2103 4 56.746624 3.13:918915 142.827897 4.5.171803 8 56.7469812 3.13 12.828442 54 1128 1 58.747187 3.12.918745 142.828715 4.b5.171013, 159 3.12.918674 42.828967 60.7347562.91857. 8289.9264 52 LOGrARITH-MIC SINES, COSINES, 340 i i I M1 Sine. 0 9.741562 1.747749 2.747936 3.748123 4.748310 5.748497 6.748683 7.748870 8.749056 9.749243 10.749429 11 9.749615 12.749801 13.749987 14.750172 15.750358 16.750543 17.750729 18.7509 14 19.751099 20.751284 21 9.751469 22.751654 23.751839 24.752023 25.752208 26.752392 27.752576 28.752760 29.752944 80.753128 31 9.753312 32.753495 33.753679 84.753862 35.754046 36.754229 37.754412 38.754595 39.7547718 40.754960 41 9.755143 42.755326 43.755508 44.755690 45.755872 4.3.756054 47.756236 48.756418 43.756600 50.756782 51 9.756963 52.757144 53.757326 54.757507 55.757688 66 757869 57.758050 58.758230 59.758411 60.758591 Cosine. I ) I Cosine. 1D). I lang. I JD. I Cotantig. I I V v16574 V b2b987 3.12 9 1.42 3.12 I916489 1.42 '829.60 3i. 1 2..1d404 1.42 '829532 3.12 911318 1.42:829805.918233 1.30077 3.11 14 3 911147.830149 3.11 1.42L 3.11.918062 14.830621 3.11.917976 1 830893.917891 1 831165 1 917805.81.43 831437.917719 1:43:831709 3.10 9.917634 1.43 9.831981 3.10:917548 1.43.832253 3.09.917462 1.43 832525 3 917376 1.832796 3.09 1.43 3.09.917290 143.833068 3 917204.833339 3 917118.833611 3.09 1.44 3.08.91703 144.833882 3 916946.834154 3.08 1.44 3 916859.834425 3.08 1.44tl 9.916773 9.834696 3.08 ~ ~~~.44 3.08.916687 144.834967 3.08.916600 144.835238 3.08.916 514 144.835309 3.07.916427 1:44.835780 3.07.916311 144.836051 3.07 916234 144.836322 3.07.916167 145.836593 3.07.916081 145.836864:915994.837134 3.06 1.45 3.06 9.15907 145.837405 3.06.915733 1.45 3.06.915646.837946 3.05.915645 145.838216 3.05 9159 145 838487 3.05 915472 145.838757 3.05.91538I 145.8390.7 3.05.915297 1:45.839297 3.04 1.45.839568.915123 1 839138 3.04 1.46 3.04 9.915035 1.46 9.840108 3.04.914948.46.340378 30 91486.4 840647 3.04 9 O 1.46.8,91427 403917 3.04 1.46..91461 413587 3.03.85 1.46 8 3.03.914598 146.841437 3.03.914510 1.46 841726 3.03.914422 1.46.841996 3.03.914334 14 42266 3.03 > 1.46.8 3.02 912 1.47 842535 3.02 9.914158 1.47 9.842805.914070 -843074 3.02 1.47 3.02.913982 147.843343 3.02.913894 147.843612 3.2 91368,43882 3.01 06 1.47 8 0 913718 147.844151:913630 147.844420 3.01 913545 1:47.844689 8.01:913436 1.47.844958.91_336.845227 D. I Sine. IP. I Cotang., 4.54 4.54 4.54 4. 534 4.34 4.533 4.53 4.53 4.5:3 4 53 4.53 4.53 4.53 4.53 4.53 4.52 4.52 4.52 4.52 4.52 4.52 4.52 4.52 4.52 4.51 4.51 4.51 4 51 4.51 4.51 4 51 4.51 4.51 4.51 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.49 4.49 4.49 4.49 4.49 4.49 4.49 4.49 4.49 4.49 4.49 4.48 4.48 4.48 4.48 4.48 I v.1 1U19 60.170740 59.170468 58.170195 57.169923 56.169651 55.169379 54.169107 53.168135 52.168563 51:168291 50 0.168019 49.167747 48.167475 47.167204 46.166932 45.166661 44.166389 43.166118 42.165646 41.165575 40 0.165304 39 165033 38.164762 37.164491 36.164220 35.163949 34.163678 33.163407 32.163136 31.162866 30 0.162595 29.162325 28.162054 27.161784 26.161513 25.161243 24.160973 23.160703 22.1604-32 21.160162 20 0.159192 19.159622 18.159333 17.159083 16.158813 15.158543 14.158274 13.158004 12.157734 11.157465 10 0.157195 9.156926 8.156657 7.156388 6.156118 6.155849 4.155580 8.155311 2.155042 1.1547s73 0 Tang. A. I D.I TANGENTS, AND COTANGENTS. 350 V I C -, I I). ICosine. ID. I Tang.I D._ I 2 a 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 81 I82 33 84 386 87 88 89 40 41 42 43 44 46 46 47 48 49 50 61 62 63 54 66 56 67 68 69 60 I 9.758591.758772.758952.759132.759312.759492.769672.759852.760031.760211.76039 9.7(50569.760748.760927.761106.761285.761464.761642.761821.761999.76217-7 9.7623"56.762534.762712.762889.763067.763245.7634229.763600.763777.763954 9.764131.764308.764485.764662.764838.765015.765191.765367.765544.765720 9.76589,6.766072.766247.766423.766598.7667 74.766949.767124.767300.767475 9.767649.767824.767999.768173.768348.768622.768697.768871.769045 3.01 3.00 3.00 3.00 3 00 3.00 2 99 2.99 2.99 2.99 2.99 2.98 2.98 2 98 2 98 2.98 2.98 2.9 7 2.97 2 97 2.97 2.97 2.96 2.96 2.96 2.96 2.96 2.96 2.95 2.95 2.95 2.95 2.95 2.94 2.94 2.94 2.94 2.94 2.94 2.93 2.93 2.93 2.93 2.93 2.93 2.92 2.92 2 '32 2.92 2.92 2 91 2.91 2.91 2.91 2.191 2.90 2.90 2.90 2.90 2.90.913276.913187.913099.913010.912922.912833.912744.912655.912566.912477 9.912388'.912299.912210.91212 1.912031.911942.911853.911763.911674.911584 9.911495.911405.911315.911226.911136.911046.910956.910866.9 107 76.910686 9.910596.910506.910415.910325.910235.910144.910054.909963.90987 3.9091782 9.909691.909601.909510.909419.909328.9(G92371.909146.909055.908964.908873 9.908781.908690.908599.908507.908416.908324.908233.908141.908"049.9Q7858I 1.47 1.47 1.48 1.48 1 48 1.48 1.48 1.48 1.48 1.48 1.48 1.48 1.49 1.49 1.49 1.49 1.49 1.49 1.49 1.49 1.49 1.49 1.49 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.51 1.51 1.51 1.51 1.51 1.51 1.51 1.51 1.51 1.51 1.51 1.52 1.52 1.52 1.52 1.52 1.52 1.52 1.52 1.52 1.52 1.53 1.53 1.53 1.58 1.53.845496.845764.846033.846302.8465 70.846839.847107.847376.847644.84790 13 9. F4~,18.848449.848 7 1 7.848986.849254.849522.849790.850058.850320i5.850593 9.8508,61.851129.8513296.851664.851931.852466.8527033.853001.853268 9.853535'.853802.854069.854336.854603.854870.855137.855404.855671.855938 9.856204.85647 1.8 5 673'7.857004.857270. E57"I537 O.857 803.858069.858336.858602 9.858868.859134.859400.859666.859932.860198.860464.8607 30.860995 X81261c 4.48 4.48 4.48 4.48 4.48 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.47 4.46 4.46 4.46 4.46 4.46 4.46 4.46 4.46 4 46 4.46 4.46 4.45 4.45 4.45 4.45 4.45 4.45 4.45 4.45 4.45 4.45 4.45 4.44 4.44 4.44 4.44 4.44 4.44 4.44 4.44 4.44 4.44 4.44 4.43 4.43 4.43 4.43 4.43 4.43 4.43 4.43 4.43 4.43 Cotang. 0.154773 6.154504 59.154236 68.153967 6 7.153698 66.153430 65.153161 54.152893 6 3.152624 52.152356 5 1.152087 50 0.151819 4 9.151551 4 8.151283 4 7.151014 46.150746 46.150478 44.150210 43.149942 42.149675 4 1.149407 40 0.149139 39.148871 3 8.148604 37.148336 3 6.148069 36.147801 34.147534 33.147267 32.146999 3 1.146732 30 0.146465 29.146198 2 8.145931 27.145664 26.145397 26.145130 24.144863 23.144596 22.144329 21.144062 20 0.143796 19.143529 1 8.143263 17.142996 16.142730 16.142463 14.142197 18.141.931 12.141664 1 1.141398 1 0 0.141132 9 M14OM 8.140600 7.140334 6.140068 6.139802 4.139536 8.139270 2.139006 1.138739 0 I I I I- CO:,re!PC I D. I.,_ne. ID I) Cotang. I D. I T n..1. 4'1 i 54 l lI Sine. 0 9.762l1V 1.769393 2.769566 3.769740 4.769913 5.770087 6.770260 7.770433 8.770606 9.770779 10.770952 11 9.771125 12.771298 13.771470 14.771643 15.771815 16.771987 17.772159 18.772331 19.772503 20.772675 21 9.772847 22.773018 23.773190 24.773361 25.773533 26.773704 27.773875 28.774046 29.774217 30.774388 31 9.774558 32.774729 33.774899 34.775070 35.775240 36.775410 37.775580 38.775750 39.775920 40.776090 41 9.776259 42.776429 43.776598 44.776768 45.776937 46.777106 47.777275 48.777444 49.777613 50.777761 51 9.777950 52.778119 53.778287 64.778455 55.778624 56.778792 57.778960 58.779128 59.779295 60.779463 Cosine. LOGARITHMIC SINES, COSINES, 386 D I Cosine. I D. 'l Tang. | 1). _ 2.90 2.89 2.8 -2.89 2 89 2.89 2 88 2.88 2.88 2.88 2.88 2 88 2.87 2.87 2.87 2.87 2.87 2.87 2.86 2.86 2.83 2.86 2 86 2.86 2 85 2.85 2.85 2.85 2.85 2.85 2.84 2.84 2.84 2.84 2.84 2 84 2.83 2.83 2.83 2.83 2.83 2.83 2.82 2.82 2.82 282 282 281 2.81 281 2.81 281 281 2 80 2.80 2.80 2.80 2.80 2.80 2.79 9.9U795.53.907866 15.907774 1.53.907682 1.53.90750 1.53.907498 1.53.907406 1.53.907314 1.53.907222 1.54.907129 1.54.907037 1.54 9.906945 1,.54.906852 1.54.906760 1.54.906667 1.54.906575 1.54.906482 1.54.906389 15.906296 11.55.906204 1.55.906111 1.55 9.906018..905925 155.905832 155.905739 155.905645 1.5.905552 1.55.905459.55.905366 1i.5.905272 16.905179 1.56 1.56 9.905085 1.904992 156.904898 1 5.904504 1.56.904711.56.904617 1b6.904523 16.904429 1.5.904335 1.57.904241 7 9.904147.904053 157.903959.57.903864 1.57.903770 1.57.903676 157.903581.57.903487 157.903392.5 1.58.903298 1.58 9.903203 158.903108 158.903014 158.902919 158.902824 15.902729 158.902634 158.902539 1.5.902444 159.902349 Sine. I D. -- v. O Ul Z O I 4.43 U-ldbdv b 0.86 152 7 4.43.138473 69.861792 4 42.138208 58.862058 4.42 - I "a 4' 9 4 2ra 7.8623 '3 4.42.1376717 56.862,589 4.42.137411 515.862S54 4.42.137146 54.863119 4.42.136881 63.86338a' 4.42.136615 62.863650 4 42.136350 5 1.863915 '136085 50 4.42 9.864180 4 42 0.135820 i 49.864445 4.42.135555 48.864710 4.42.135290 47.864975 4.41.135025 46.865240 4.41.134760 45.865505.134495 44 4.41.865770 4.41.134230 43.866035 4.41.133965 42.866300 4.41 -1337100 41.866564.133436 40 4.41 9.866829 4.41 0.133171 39.867094 4.41.132906 38.867358 4.41.132642 37.8676213 4.41.132377 36.E67887 4.41.132113 35.868152 4.40.131848 34.86841G.4-40.131584 83.868GEO 4.40.131320 32 J6, "945 4.40. 1 31 I 0 5 53 1.869 1)9 '130794 30 4 40 9.E6947.3 4.40 0.130527 29.869737 4.40.130263 28.870001 4.40.129999 27.8190M5 4.40.129135 26.870529 4.40.129471 25.870793 4.40.129207 24.871057 4.40.128943 23.871321 4.40.128679 22.8715F5 4.40.128415 21.871849 4.39 '128151 20 9.872112 4.39 0.127888 19.8723716 4.39.127624 18.872640 4.39.127360 17.872903 4.3.1).127097 16.873167 4 1).126833 15.8734-30.126570 14.8 '13694 -4.3'.126306 13 4 - 3.).873957 4-33.126043 12.874220 4. 49.125TEO I 1 874484 I...".125516 10 y.b61261 -.861527 4.43.861792 443.862058 4 42.862323 4.42.862589 442.862S54 4 42.863119 442.863385 442.863650 4.4.863915 442 9.864180 4.42.864445 4.42.864710 4.4.864975 441.865240 441.865505 441 4.41.865770 4.41.866035 4.4.866300 441.866564 9.866829 4 41.867094 441.867358 4.4.867623 441.867887 4.41.868152 440.868416.44.868GC0 4.40.6tf945 440.8G692~9 4 40 9.E69473 4.40.869737 440.870001 440.870265 4.40.870529 4.40.870793 440.871057 440.871321 440.871585 4.40.871849 4:39 9.872112 439.872376 4.3.872640 4.3.872903 43.873167 4.873430 '3.873694 4.3,4.873957 43..874220 4..$ 874484 49 9.874747.39.875010 4.39.875273 438.875536 438.875800 438.876063 438.876326 438.876589 438.876851 438.877114 Cotang. D. Cotang. 0.13b739 60.138473 59.138208 568.137942 57.137677 56.137411 55.137146 54.136881 63.136615 62.136350 51.136085 50 0.135820 49.135555 48.135290 47.135025 46.134760 45.134495 44.134230 43.133965 42.133700 41.133436 40 0.133171 39.132906 38.132642 37.132377 36.132113 35.131848 34.131584 33.131320 32.131055 31.130794 30 0.130527 29.130263 28.129999 27.129735 26.129471 26.129207 24.128943 23.128679 22.128415 21.128151 20 0.127888 19.127624 18.127360 17.127097 16.126833 15.126570 14.126306 13.126043 12.125780 11.125516 10 0.125253 9.124990 8.124727 7.124464 6.124200 6.123987 4.123674 8.123411 2.123149 1 122886 0 Tang. / M. I I D. { Cotang. I D. I Tang. I ALJ TANGENTS, AND COTANGCENTS. 370 I Sine.~ D. Co.ine. iD. I Tang. D. otang...79 9.902349 15 9.877114 38.1226 60.77963.902253.877377.122623 59 2.79.902158 438.122360 58.76 2.79 902158..8777640 438 122097 5 2.779798 2.79.063 1.59.877903 38 121835 6 3.779966 2.79.90211215 5 4.'ir9798 9.901967 1.59.8778165 4:38 4.780133 2.79.90 6 59.878428.780300 2.78 9 2.59. 84728 4.38.121309 4 6.780467 2.78.901776 159.878691 438.121047 53 6.780463 2.78.901681 1.59.878953 4.37.120784 62.901585 1.59.879216 7.120522 61 8.780801 2.78.901490 79478 4.37.120259 50 9.780968 2.8.034 1.59:741.43 0.119997 49 2.78.9019.89741 64.37 10.781134.7 0 1.60 o.1199975 4 90.781013 2.7.9012.88000 4.37 119354 1 9.781301 2.77 1-60.880265 4-37 47 11 94.2 47 11 8.181501 2~71.901202 1.609 12.781468 2.77.901106 160.880528 4-37 46 13.781634 2.77.901010 160.880790:119210 4 14.781800.7.900914 160.881052 4.3.118948 45 is.781966 2.77.900818 1:60.881314 4.37.118686 44 16.782132 2.77.900722 1 60.8817 4.36.118424 43 16 2.77 902 8817.7.111 4 17.782916046;9 17.782298 2.76.900626 i:;o.88183 4.37.118161 42 18.782464 2.76.900529 1"60.882101 4.37.1176 4 19.782630 2.76.900433.882363 4.36. 4 20.782796 2.76 9.8 262 -11;,37 3' 9.900337 161.8826 4.36.11353 389 2 9.782961 2.76:900240 166.88288 436.1176913 8,-~~~~~~~~~~~~~6 4.3.165 37 22.783129 2.7.900144 61.883148 436.1162 8 23~ 7839161 8840 1639:8 24.783458 2.75.900047.883410 436.116908 85 2.7836238 2.75.899951 161.883672 436.11638 85 26 -.783788 2.75.899854 1.61.888934 4.36.116806 84 26.7839 2.75.899757 161.884196 4.36.1154 83 28.783953 2.75.899660 1:61.884457 4.36.11543 82 28.784118 2.75.899564 1.61.884719 436.115281 81 8018444 74 89467.612.884 3.11020 80 29.784282 2.74 9899467 162 98 436 0.11758 29 s.8 4.114749 28 80.784447 2?r74.1202 156219.885242 2.74 ~897 4 ~ N.780189923~ 1562.114235 27 32.7841 2.74.899 162 885765 4.36.113974 26 83.784941 2.74 899078 1.62.886026 4.36.113712 25 84.78105 2.74 898981 162.88288 436.11341 24 35.785269 2.73:8988 1-62.886549 4.35.113190 28 36.785433 2.73.89878 1-62.886810.112928 22 87.785597 2.73 898789 1162.887072 4:35.112667 21 88.785761.73.898689 1.62.887333 4.35.112406 20 89.785925 2.73.89892 1.6 887594 4. 40.7868 2.73.98494. 163 4 35 0.112145 19 9.898397 1 963.887 855 435.111884 18 41 9.786252 2.72.898299 1.63.888116 4.35.111234 17 42.786416 2.72 89820 1.63.88816.1116.786579 2.72.898104 1-63.88837.35.111361 16 4.7896 2.72.8979084 6.888639 4835.mo 44.786742 2.72.898006 163.888900.35.11080 14 45.786906 2}72.897908 1:63 889160 4.35.11008408 46.787069 2.72.897810 1.63.889421 4-35.110318 12 47.787232 2.7:1 897712 1.63.889682 4.35.110057 12 48.787395 2.71.897614 1'63.889943.3. 1109 96 1 49.787557 2.71.897616 1.3 8.109 50.787720 1 63.890 0.109585 - 2.71 9.897418 1 964.890465 4.34.109275 51 987883.897320 164.890725 4 194 52.788205 2.71.897222 1.64.890986.3.108753 6 53 78808 2.71.897123 164.891247. 34.108493 54.788370 2.70.897025 164.891507 43.108282 4 65.788582 2.70.897025 1.64.891768 434.107972 66.788694 2.70.896828 1.64.892028 4.34 10797211.... ~~~~0066,,, 57.788856 2.70.896729 1.64.892289 44.107411 2 58.789018 2.70.89663291.64.6925 410 ~ '1 181ro a,?o.896631 42 60.789342.896532.8280.17 3 '.Sine0 D. Cotat 6 Cosine ~ ~ ~ ~ ~ ~ ~ ~. 3 1631 4 56 M Sine. W i 9.789342 1 I.789504 2 I.789665 8.789827 4 1.789988 a.790310 I.790471 8.790632 9.790793 10.790954 12.791275 13.791436 14.791596 I1&.791757 18.791917 17.792077 18.792237 19.792397 b.792557 21 9.792716 22.792876 23.793035 24.793195 25.793354 28.793514 27.79,3673 28.793832 29.793991 t0.794150 81 -.794308 -ft.794467 33.794626 34.794784 35.794942 3,.795101 37.795259 38.795417 8* 757 40.795733 41 9.795891 -42.796049 43.796206 44.796384 45.796521 48.796679 47.796886 48.796993 49.797150 fi0.797307 5.1 9.797484 52.797821 53.797777 '54.797934s 55.798091 88.798247 57.798403.58.J98560 89.798716 80.7872 Cosine. j LOGARITHMIC SINES, COS1NE~, 380 — I D. 2.69 2.69 2.69 2.69 2.69 2.69 2.68 2.68 2.68 2.68 2.68 2.68 2.67 2.67 2.67 2.67 2.67 2.67 2.66 2.66 2.66 2.66 2.66 2.66 2.65 2.65 2.65 2.65 2.65 2.65 2.64 2.64 2.64 2.64 2.84 2.64 2.64 2.83 2.63 2.63 2.63 2.63 2.63 2.63 2.82 2.82 2.82 2.82 2.82 2.61 2.81 2.81 2.81 2.81 2.81 2.61 2.61 t.60 1.80 2.60 Coie. D'.I T gng. ID. jCotang. 9.896532 9.b9261 0T170 7T900.896433 1.64.893070 4.94 169 0 5.896335 1 1.65 8331 4.34.106669 58.896236 1.65.893591 4.34.106409 57.896137 1.65.893851 4.34.106149 56.896038 1.65.894111 4.34.105889 55.895939 1.65.894371 4.34.105629 54.895840 1.65 '894632 4.34.105368 53.895741 1.65.894892 4.33.105108 52.895641 1.65:895152 4.33.104848 51.895542 1.65.895412 4.33.104588 60 9.895443.1.65 9867 0.104328'i 49.854.66 89593 4.33 -.104068 4 8.895244 1.6 896192 4.3.103808 47.895145 1.66 89-6452 4.33.103548 46.895045 1.66 '896712 4.33.103288 45.894945 1.66 '896971 4.33.103029 44.894846 1.66 '897231 4.33.102769 43.894746 1.66 '897491 4.33.102509 42.894646 11.66 897751 4.33.102249 41.894546 1.66 '898010 4.33.101990 40 9.894446 -1.66 9887- 4.33 0- 070'8.894346 1.67:898530 4.3.101470 38.894246 1.7 898789 43.101211 37.894146 1.67 '899049 4.33.100951 36.894046 1.67 '899308 4.32.100692 36.893946 1.67 '899568 4.32.100432 34.893846 1.67 '899827 4.32.1001730 33.893 745 1.67 '900086 4.32.099914 82.893645 1.67 '900346 4 32.099654 81.893544 1.67 '900605 4.32.099395 s0.1.67. 4.32. 9.893444' 9.900864 0.0-99136 29.893343 1.68 901124 4.32.098876 28.893243 1.68 '901383 4.32.098617 27.893142 1.6 901642.3.098358 26.893041 1.68.901901 4.32.098099 25.892940 1.68.902160 4.32.097840 24.892839 1.68.902419 4.32.097581 23.892739 1.68.902679 4.32.007321 22.892638 1.68.902938 4.32.097062 21.892536 1.68.903197 4.32.096803 20 9.945 1.68. 4.31 -1923.69 9.903455 4.1 0.096545 - 19.892334 1.9.903714 4.31 026 18.892233 1:69.903973 4.1.096027 1 7.892132 1.9.0232 4.31 058 16.892030 1.9.904491 4.31.959 1.891929 1.69.904760 4.31.095260 1 4.891827 1.9.905008 4.31 042 13.891726 1.9.905267 4.31 -973 1.891624 1.69.905526 4.31.094474 11.891523 1.0-.905784. -.094216 10 9.891421 1.70 9.906043 4.1 0.093967.891319 1.70.906302 4.31.093698 8.891217..0 906560 4.31 034.891115..0 906819 4.81 038.891013.907077 4.81 23 1.70 4.1.0929234.890911 1.0.907836 4.814.890809 1.0.064 48.092406 8.890707 10.907852 4.81.092148 29.890605 '70 1.6914.81.965 1.70.908111 4.80.0918891.890503 ___.908369.961 0 Sine. ID. jCotang. I. Tang. IM D.i TANGENTS, AND CUTANGENTS. 390 MI. Sine. D. Cosine. ID. I -f D. Cot I I 0 9.972..6 9.890503 9.9Ob369 00916 6.7 2.60.890400 17 908628 30.09132 59 2.799184 2.60.890298 1.71. 6 430.09866 3 9.89951 2.59.889990 171.909660 4.30.090302 6 5.799651 2.59.889888 1.71.909918 430.090823 53 6.799806 2.59 889785 1.71.910177 4.30.0956 52 7.79996.7 43 0898 4 ~~99495 2'59 1.71.910693 4.30.06b307 1.80017 2.59.889679 171.917035 4.30 O 9 11 9 800582 2 58 9889271 1 72 9.911469 4.30 0.087296 47 1 **I C I C Ci:72 9S0 1 724 43 12 809737 2.58.889181 1.732.91724 4.30.088018 47 18.801165 257.888651 1.72 91 4-2.081641 4 319.801819 2.57.888548 1.72 91 7 4 0648213 43 20.801973 2.57.888444.3.913529 4.29.08 1 40 21 9.80128 2.5 9 888341 173 9.913787 4.23.086213 39 2 9~ 802128 2 5T 1 74|914 4 4.29 085956 31 25.888237 1.73 9104 4 08566 22.802282 2.56.888134 1.73.914302 4.29.085409 35 23.802436 2.56.888030 1.73.914561 4.29.0840 24.802589 2.56.887926 1.73.94i817 423.085608 3 25.802743 2.56.887892 1.73.91 3071 4.2 1.0846103 3 26.802897 2.56.88718 1.7|3.08492 5 31.803664 2,8871 1.7 91 4.803 2..887 1.7.91 4.23.083123 27 33.803917 2.55:8869 1.7.915147 4.29.082460 25 948.42'.916 0.082869 29 3 9803664 2.55 8865 1.74.917391 4.29.082352 24.804876 2.54 886780 1.74.917648 1.2 082095 23 33 F03970.9.54 886676 1.74.91679 4.2.081837 2 34.04581 2.54.886571 1.74 4.28.081580 21 38.804734 2.54 8864668 1.74.918420 428 082 2 36.80448 2.54 88630 1.74.914677 4.28.0 2 40.800509 2.54.88657 1.75 9 3081066 19 48.04731 92.5 9.91834 4.28:08080 21 2.54.886127 1.75.919191 4.28 08052 17 40 E.805343 2.54.886072 1.75 919448 4.28 080295 16 43.805343 2.53.885942 1.75.919705 4.28 0800380 15 43.05495 2.53 85837 1.75.919962 4.328.07971 1 44.805647 253.885732 1 5.90219 4.28 078954 13 4.805951 2.53 38562 117.920476 4.28 079206 12 47.806103 2.53.85522 1.75. 4.28.079010 11 4.806254 2 53.885416 1.75 920990 4.28.0750 10 49.806406 2.52 0661 1 9 0 2 078497 4 r0.806557 2 885311 1.921247 4.28 51 '9.06709 288850 16 9.921503 4.8 78240 8 52 806860 22.885100 17".921270 4.28.0677983 532.52.884994. 1.76.922017 4.28.077726 6. 5.807113 5.884889 17;.9 7 4.28.077470 5 54.807163 2 52.884783 1.76.922530 4.28.07723 4 55.807314 2.5. 884677 11.922787 4 0765213 4 56.807465 2.51.884572 176.923044 4.28.0756002 67.807615 2 51 766 51 884466 176.923300 4.28 59.80797 2'51.884360 1.76.92355 0 59.807917 176 7 4.29 7IS7 60.808067.5 884254.923813 C~osine D I ~i Cncn1..D12 I 19.08647] 40 68 LOGAEITHMIC" SINIES, COSINES, i i I I I 1 6 7 8 9 10 11 12 13 14 16 16 17 18 19 20 21 22 23 24 25 26 27 i28 29 30 Si 32 33 34 345 36 37 33 39 40 41 '42 43 44 45 46 471.48.49 560 I.51I 6 2 63 6 4.68 671 59 60 Sine._ 9.808067.808218.808368.808519.808669.808819.808969.809119.809269.809419.809569 9.809718.809868.810017.810167.810316.810465.810614.810763.810912.811061 9.811210.811358.811507.811655.811804.811952.812100.812248.812396.812544 9.812692.812840.812988.813135.813283.813430.813578.81372.5.81387-2.814019 9.814166.814313.814460.814607.814753.814900.815046.816193.815339.815485 9.815631.8157i78.815924.'816069.816215.'816361.816507 A316943 -I ID.I Cosine. I D.I Tang,. I D. 0 2.51.884148 2.51.884042 2.51.883936 2.50.883829 -2.5(0 883723 2.50.883617 2.150.883510 2.50.883404 250.839 2.49.883191 2.49 9.883084 2.49.882977 2.49.882871 2.49.882764 2.48.882657 2.48.882550 2.48.882443 2.48.882336 2.48.882229 2:48.882121 2 48 9.882014 2.47.881907 2.47.8817199 2.47.881692 2.47.881584 2.47.881477 2.47.881369 2.47.881261 2.46.881153 2:46. 881046 -2 46 9.880938 2.46.880830 2.46.8807229 2.40.880613 '2-46.88050.5 2.45.880397 2.45.880289 2.45.880180 2.45.880072 2:45. 879963 -2.4.5 9.879855 '2 45.879746 2.44.879637 '244.879529 2 44.879420 2 44.879311 2 44.879202 '244.879093 2-44,.878984 23.878875 2.43 9.878766 2.43.878656 2.43.878547 2.43.878438 2:43.878328 2 43.8 7821 I9 2.42.878109 2:42.877,999 2.42.877S90 ~.877780 -D. sine. I 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.77 1.-78 I1.78 1.78 1.78 1.78 1.78 1.78 1.78 1.79 1.79 1.79 1. 79 1.79 1.79 1.79 1.79 1.79 1.80 1.780 1.80 1.80 1.80 1.80 1.80 1.80 1.80 1.81 1.81 1.81 1.81 1.81 1.81 1.81 1.81 1.81 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.82 1.83 1.83 1.83 1.83 9.923813.924070.924327.924583.924840.923096.925352.925699.925865.926122.926378 9.926634.926890.927147.927403.9276-59.927915.92817 1.928427.92'6813.926U40 9.9291L6.9 I9 4.52.9297081.929964.920220.930475.930731.930987.93 1243.931499 9.9,311755.9320 10.932266.932522.932778.933033.933289.933545.933800.934056 9.934311.934567.934823.935078.935333.935589.935844.936 100.936355.936610 9.936866.937121.937376.937632.937887.938142.938398.938653.938908.939163 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4.27 4 27 4 27 4.27 4.27 4.27 4.27 4 27 4:27 4.27 4.27 4.27 4:26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.26 4.25 4.25 4.25 4.25 4.25 4.25 4.25 4.25 4.25 I i Cotanig. I U.VigO1b 60..075930 59.0756 73 58.075417 57.075160 66.074904 55.074648 54..074391 53.074135 52.073878 51.073622 50 0.073366 49.073'110 4 8.072853 47.072597 46.072341 45.072085 44.071829 43.07 1573 42.071317 41.071060 140 0.070804 3.070548 38.07 0292 37.070036 36.069780 35.069525 34.069269 33.069013 32.068757 31.068501 30 0.068245 29.067990 28.067734 27.067478 26.067222 25.066967' 24.066711 23.066455 22.066200 21.065944 20 0.065689 19.065433 18.065177 17.064922 16.064667 15.064411 14.064156 13.0603900 12.063645 11.063390 10 0.068134 -9.062879 8.062624 7.062368 6.062113 5.061858.4.061602 3.061347 2.061092 1 06O938317 0 Tang. j Ml. I -- Css,;Ine.I D. I Cotang. I D.I 4;,O TANGENTS, IND COTANGENTS.Sv 410 M. Sine. D. Cosine. I D. I Tang. D. Cotang. 0 9.b16943. 9b7'7h0 9939163 0.060837 60 7.817088 1.877670.939418 4.25.06052 59.80 S 2.42 1.87 834.25 2.817i233 24.87060 1.83 939673.060327 58 3.817379 2.42.877450.939928 4.25.060072 57 2.42 1.83 3.817324 2.42.877340 1.83:940183 425.059817 6 5.817668 2.41 877230 184 940438 4.2 059562 5 6 81112.41 1.84 694 4.25.059306 64 6.817813 2.41 877120 1 940949 4.25.059051 3.818103 2.41 6 184 4.25.058796 52 9.818247 2.41 876789 1.84 4.25.0582 6 11 9.818536 9.876568 9.941968 0.058032 49 2.40 8 1.84 94 3 425 057777 48 12.818681 2.4 8760 457 184 13.818825 2.40 8 1.84 94 425.057522 47 2.40 1.88 42 14.818969 240 876236 184.942733 42.057267 46 15.819113 240.87625 185.942914588 4.057012 451 10.81925 2.40.876014 85 943243 4.25.058286757 4 2.43 1.88 4.25 17.819401 240.875804 185.943498 425.056502 43 18.819545 240.875793 1.85.943753 l5.056248 42 19.8196887568 185.94400 2 055993 41 20:819832 2.39 875571..944262 25.055738 40 2.39 1.85 4.25 21 9.819976.875459 85 9.944517 25 0.05483 39 22.820120 2.39 875348 185.944771 42.05529 8 23.820263 2.39 875237 1 85.945026 424.054974 *24.820406 239 E75126 186 945281.054719 6 28.820979 874680 1.86 424 5 25.820550 238.875014.945535 4.054465 85 26.820693 238.74903 186.945790 424.054210 84 27.820836 238. 74791 186.946045 424.053955 83 28.820979 2.38.874680 r186.946299 44.053701 82 29.821122.874568 186.946554 42.053446 31 30.821265.874456.946808.053192 80 2.38 1.86 4.25 31 9.821407 238 9.874344 1.86 9.947063 424 0.052937 29 82.821550 2.38.874232 1.87.947318 424.052682 28 33.81693 2.37.874121 18.947572.052428 27 84 821835 237.874009 187.947826 4.052174 26 35.821977.873896 87.948081 424.051919 25 2.377 378 1.87 983 4.2.051664 36.822120 2.3787.948336 424 051664 24 37.822262 237.873672 187.948590 424.051410 23 39.822546 237.873448 187.949099 424.050901 21 40.822688.873335 1 949353.0506478O 2.36 a..1.874.244 41 9.822830 236 9.873223 187 9.949607 424 0.050303 10 42.822972 236.873110 188.949862 424.050138 18 2.40 1.88 4.24 43.82314 2.36.72998 188.950116 424.0498E4 17 45.823397.872772 88.950626.04935 15 2..36 67 659 1.88 5 2.24 46.823539 2.872659 88.950879 424.049121 14 47.823680 235.872547 188.951133 424.048867 13 48.823821 235.872434.951388 4.048612 1 49.823963 35.87221 88.951642 424.048358 11 50.824104 2.35 87208 88..951896 424.48104 10 51 9.824245 235 9.872096 189 9.952150 44 0.047850 9 52.824386 35 871981.89 952405 424 04795 3.824527 35.871868.952659 424.047341 54.824668.871755 89.952913 047087 2.34 1.8 5 4-~23 55.824808.871641 8 9.953167.046833 5 56.824949 234.871528 19.953421 423.046579 4 2.34 1.89 5 4 8 57.825090.871414 89.53675.046325 8 20^2.34 1 89: 7 4.23 8.825230.81301.953929.040071 2 2.34 1.89 4.23 59.825371. 871187.95413 43.045617 1 4234.04794 -60 M.82511.71073.954437.04563 0 Cosine. 11)D Sii. I 1).I Cootn.~ 1).I.Tang. 5.6 48 2 60 LOGARITHMIC SINES, COSINES, a2 I I I 31'T I Sine. I D. I Cosinle. I1). I Tang. I) 1 4 S 6 7 8 9 10 11 12 13 14 15 16 17 -18 19 20 21' 22 23 24 25 26 27 28 29 30 31 32 33 84 35 36 37 38 39 40 41' 42 43 44 45 46 47 48 49 60 51 62 63 54 65 66 67 68 60.825651 2.33.82.15791 23.825931 2.33.826071 2.33.826211 2.i33.826351 2.33.826491 2 33 - 826631 23.826770 2.33.826910 2.3~2 2.32.827189 2.32.827328 2.32.827467 2.32.827606 2.32~.827745 2.32.827884 2.31.828023 2.31.828162 2.31.828301 23 -9.8Q8439.2.31.828578 2.31.828716 2.31t.828855 2.30.828993 2.30.829131 2.30.829269 2 30.829407 2.30.829545 2.3().829683 2.30 9.829821 2.29.829959 2.29.830097 2.29.830234 2.29.830372 2.29.830509 2.29.830646 2.29.830784 2.29.830921 22.831058 22 9.831195. 2.28.831332 2.28.831469 2.28.83 1606 2.28.831742 2.28.831879 2.28.832015 2.27.832152 2.27.832288 2.27.832425 -2.27 9.832561 2.27.832697 2.27.832833 2.~27.832969 2.26.833105 2.26.833241 2 26.8333777 2.26.833512 2.26.833648 22 833783 22 cos'ine. I I). I I 9.b7 1073.870960 1.90.870846 1. 90.870732 1.90.87 0618 1.90.870504 1.90.870390 1.90. 879027'6 1.90.870161 1.90.870047 1.90.869933 1.91 9.869818 1.91.869704 1.91.869589 1.91.869474 1.91.869360 1.91.869245 1.91.869130 1.91.869015 1.91.868900 1.92.868785 1.92 9.868670 1.92.868555 19.868440 19.868324 1.9 2~.868209 1.92.868093 1.92.8679708 1.92.867862 1.93.867747 1.93.867631 1.93 9.867515 1.93.867399 1.93.867 283 1.93.867167 19.867 051 1.93.866935 1.93.866819 1.94.866703 1.94.866586 1.94.866470 1.94 9.866353.1.94.866237 1.94.866120 1.94.866004 1.94.865887 1.95.865770 1.95.865653 1.95.865536 1.95.865419 1.95.865302 1.95 9.865185.1.95.865068 1.95.864950 1.95.864833 1.95.864716 1.96.864598 1.96.864481 1.96.864363 1.96.864245 1.96.864127 19 Sine. I D. I 1U.954437.954691.934945.953200.955454.955 707.955961.956215.956469.956723.956977 9.957231.957485.957739.957993.958246.958500.958754.959008.959262.959516 9.959769.960023.960277.960331.960784.961038.961291.961545.961799.962052 9.962306.962560.962813.963067.9063320.963574.963827.964081.964335.964588 9.964842.965095.965349.965602.965855.966105.966362.966616.966869.967123 9 967376.967629.967883.968136.968389.968643.968896.969149.969403.969656 4.2:3 4.23 4.2.3 4.23 4.2.3 4.23 4.2 3 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.2:3 4.2:3 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.21 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.23 4.22 4.212 4.22 4.22 4.22 4.22 4.22 4.2~2 4.2~2 4.22 4.22 4.2~2 4.22 4.2~2 4.22 4.22 4.22 4.22 4.22 4.22 j Cotang.I.1 U.U4bb63 60.045309 59.045055 58.044E00 57.044546 566.044293 55.0440039 54.043785 53.043531 52.043277 51.043023 50 0.042769 4.042515 48.042261 47.042007 46.041754 45.041500 44.041246 43.040992 42.040738 41.040484 40 0.0402031 3 039977 38.039723 37.039469 36.039216 35.038c962 34.038455 32.038409 33.038201 31.037948 30 0.037694 29.037440 28.037187 27.036933 26.036680 25.036426 24.036173 23.035919 22.035665 21.035412 20 0.035158 19.034t.'05 18.034651 17.034398 16.034145 15.033891 14.033638 13.033384 12.033131 11.032877 10 0.032624 9.032371 8.032117 7.031864 6.031611 6.031357 4.031104 3.030851 2.030597 1.030344 0 Tag-IM. I N. I I I IN 470 61 TANGENTS, AND COTANGENT 430 S. M.ISine. 0.3137 1.833919 2.834054 3.834189 4.834325 6.834460 6.834595 7.834730 8.834865 9.834999 10.835134 11 9.835269 12.835403 13.835538 14.835672 15.835807 16.835941 1?.836075 18.836209 19.836343 20.8364 77 21 9.836611 22.836745 23.836878 24.837012 25.837146 26.837279 27.837412 2 8.837546 29.837679 30.837812 3 1 9.8379405 32.838078 33.838211 34.838344 35.838477 36.838610 37.838742 3U8.838875 39.839007 40.839140 41 9.839272 -42.839404 43.839536 44.839668 46.839800 46.839932 47.840064 48.840196 49.840328 50.840459 61 9.840591 62. 8407229 63 I.840854 64 -840985 65.841116 66.841247 67.841378 68.841509 59.841640 60.841771 Caz~inet D.I Cosine. I z I 2.26 2.25 2.25 2.25 2.25 2 2-5 2.25 2.25 2.25 2 24 2 24 2.24 2.24 2.24 2.24 2.24 2.24 2.23 2.23 2.23 2.~23 2.23 2.23 *2 23 2.22 2.22 2 22 2.22 2.22 '2.22 2.~22 ~2.22 2.~21 2.21t 2.'21 2.21 2.21 2.21 2.21L 2.21 2 20 2.20) 2.20 2.20 2.2(0 2.20 2.20 2.19 2.19 2.19 2.19 2.1.9 2.11. 2.1It 2. IV 2.11 2.11 D..864010.863892.863774.863656.863538.863419.863301.863183.863(164 862946 9.862827.962709.8662190.962471.8.62353.862234.862115.861996.861877.861758 9.861638.861519.861400.861280.861161.861041.860922.860802.860682.860-5,62 j9.860442.860322.860202.860082.869962.859842.859721.859601.859480.859360 9. 8592 39.859119.858998 858877.858756.859635.8585 14.868393.858272.868151 9.858029.857908.857786.857666.867643.857422.857300.857178.857056.866934 -jSine..96.96.97.97.97.97.97.97..97 -97..98 ~.98 L.98 L.98 1.98 1.98 [.98 L.98 1.98 1.98 1.99 1.99 1.99 1.99 1.99 1.99 1.99 1.99 1.99 2.00 ~2.00 ~2.00 ~2.00 2.00 2.00 2.00 2.01 2.11 2.01 2.01 2 01 20W 2.0' 2.0 2.0: 2.0: 2.0' 206 2 0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 im Tang. -9.-969656.969909.970162.970416.971o669.970922.97 17175.971429.97 1692.P71935.9721L.8 9.972441.972694.972948.973201.97 3454.9737 07.973960.974213.974466.97 4719 9.974973.975226.975479.975732.975985 976238.976491.976744.976997.977250 9 977503.977766.979009.978262.978515.978768.979527.9797E0 19.980033. 990296. 980538 1.980791. 981044 2.981297 2.981550 2 913 2.982036 2.9823056 2 9..982309 92826 2.982814 12.983067 2:.983320 13.983573 13.983826 0.984079.3984331.3984584 984837 Cotang. D. ICotalng. 42.030091 59 4.22.0238 58 4. 22.029594 67 4.22 02b331 56 4.22.02907 8 55 4.22.028825 54 4.2~2.028571 53 4.22.028318 52 4.22.028065 61 4.22.027812 50 4.~22.0.027559 4 4.22.027306 48 4.22.027062 47 4.22.026799 46 4.22 026546 45 4.22 0'62,93 ~44 4.22.026040 43 4.22.025787 42 4.22.025534 41 4.2~2.02528 1 40 4.22 0.025027 89 4.22.024774 38 4.22.024521 37 4.2~2.024268 86 4.22.024015 36 4 22.023762 34 4.2~2.023509 83 4.22.023256 82 4.22.023003 31 4.22.270 8 4.2 0.022497 29 4.22.022244 28 4.22.021991 27 4.22.021738 26 4.22.021485 26 4.22.021232 24 4.22.020979 23 4.2~2.020726 22 4.212.020473 21 4.2~2.020220O 20 4.22 0.019967 1 9 4.22.019714 18 4.22.019462 1 7 4.21.019,209 163 4.21 01956 15 018 OI703 1 4 4.21.018450 13 4.21.018197 1 2 4.21.017944 1 1:017691 10 4.2 t~~~ 4.21 0.017438 0 4.21.017186 8 4.21.016933 4.21.016680 16 4.21.016427 6 4.1.016174 4 4.21.016921 8 4.21.015669 2 4.21.016416 1.015163 0 1 D. Tang. M I...b. 460 v." LOGARITHM-IC, SINES, COSJNES~' &C. 4401 -Sine. 0 9.b41771 1.841902 2.842033 8.842163 4.842294 5.842424 6.842.555 7.842685 8.842815 9.842946 10.843076 11 9.843206 12.843336 13.843466 14..843595 15.-843725 16.84,3865 17.843984 18.844114 19.844243 ~20.844372 21 9.844502.22.844631 2.3.844760:24.844889 25.845018 26.845147.27.845276 28..845405.29.845533':830.845662 '81 9.845790!82.845919 33.846047 ~34..846175, 3~5.846304 386.846432 37.846560 38.,.846688 P8 -.846816 40.846944 41 9.847071 42.847199 48.847327 44.847454 45~.847582 46.847709 47.847836 48.841964 49.848091 60.848218 51 9.84834& 62.848472. 68.848599 64..848726 i56.848862 156.848979 '57.849106 168.849232 ~9.849859 A0.849485 I D i I 2.18 2.18 2.18 2.17 2.17 2.17 2.17 2.17 2.17 2,17 2.17 2.16 2.16 2.16 2.16 2.16 2.16 2.16 2.15 2.15 2.15 2.15 2.15 2.15 2.15 2.15 2.15 2.14 2.14 2.14 2.14 2.14 2.14 2.14 2.14 2.14 2.13 2.13 2.13 2.18 2.13 2.13 2.13 2.13 2.12 2.12 2.1.2 2.12 2,12 2. 12 2.12 2.12 2.11 2.11 2.11 2.11u Cos~iie. D. I Tang., 9.856934 9.0984-837'.856812 20.986090.856690 2.03.985343.856568 2.04.985596.856446 2.04.985648.856323 2.04.986101.856201 2.04.986354.8660798 2.04.986607.855956 2.04.986860.855833 24.987112.855711 2.0.987365 9.558.2.05 -9.558 2 05 9.987018.855465 205.98~816 I.855342 2.05.98823.855219 2.05.988E376.855096 2 05.98F629.854973 2.05.98882.854850 25.989134.854727 12_00.9E37.854603 206..98l9640.8544~0 ) 2.1).E9899 9.8.54356 2 06 9.910145.85423.3 2.06.990398.854109 2.06.990651.853986 2.06.990903.853862 2.06.991156.8537038 2.06.991409.8-53614 2.07.991662.853490 2.07.991914.853366 2.07.992167.853242 2:7-.92420 9.853118 2.07 9 992672.852994 2.07.992925.852869 2.07.993178.824 2:7.993430.852620 2.07.993683.852496 2 08.993936.852371 2.8.994189.852247 2.08.994441.8 52112 2 2.08.994694.851997.2.081-.9994 9.851872 2 09& 9.995199.851747 2.08.995452.851622 20..995705.851497 2.9995957.851372 ~ 996210.8512,46 2.09.996463.85112 2.09.996715.850J96 2.09.996968.850E870 2.09.997221.850745 2. 9977 9.850619 2.09 9 997726.850368 2.10.998231.850242 2.10.998484.850116 2.10.998737.849990 2.10.998989.849864 2.10.999242.849738 2.10.999.849611 2.1o.999748.849485 a0000000 Sine, 1 17. _ICotalg.. 1)D. ICotango. I I 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21 4.21t 4.21 4.21 4.21 4.21 4.21 4.21 4.21 K4.21 4.21 4.21 4.21 4.21.4.21 4.21 4.21 4.21 4.21 I UVIVL4it)6.014910.014657.014404.014152.013899.013646.013393.013140.012888.012635 0.012382.012129.011877.011624.011371.011118.010866.010613.010360.010 107 0.0098655.009602.009349.009097.008844.008691.008338.008086.007833.007580 0.007328.007075.006822.00667 0.006317.006064.0058 11.005559.005306.005053 0.004.01,.004548.004295 004043.003790o.003537.003285.008032.002779..002527 0.002274).002021.001769.001516.001263.001011.000758.000505.000253 0 000000 59 56 55 54 53 52 51 50 49. 48 47, 4,6 45 44. 43 42 41. 40 319. 38 37 36 33 32 31 30 29 28 276 26 24' 212 21 20r 19 18 17 16 16 14 13a 12 11 10 I I L2U I I II). 11, Tang. I A TA B LEi CONTAINING THE LOGARITHMS OF NUMBERS FROM 1 TO 10,000. No. Log. No.,Log. No. Log. No. Lo~g. No. Log. I 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.30103t~ 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 23 1.361728 43 1.633468 63 1.799341 83 1.919078 4 0.602060 24 1.380211 44 1.643453 64 1.806180 84 1.924279 6 0.698970 25 1.397940 45 1.653213 65 1.812913 -85 1.929419 6 0.778151 26 1.414973 46 1.662758 66 1.819544 86 1.934498 7 0.845098 27 1.431364 47 1.672098 67 1.826075 87 1.939519 8 0.903090 28 1.447158 48 1.681241 68 1.832509 88 1.944483 9 0.954243 29 1.462398 49 1.690196 69 1.838849 89 1.949390 10 1.000000 30 1.477121 50 1.698970 70 1.845098 90 1.954243 11 1.041393 31 1.491362 51 1.707570 711.851258 91 1.959041 12 1.079181 32 1.505150 62 1.716003 72 1.857332 92 1.963788 13 1.113943 33 1.618514 53 1.724276 73 1.863323 93 1.968483 14 1.146128 34 1.531479 54 1.732394 74 1.869232 94 1.973128 15 1.176091 35 1.544068 55 1.740363 75 1.875061 95 1.977724 16, 1.204120 361.556303 56 1.748188 76 1.880814 96 1.98'2271 17 1.230449 37 1.568202 57 1.755875 77 1.886491 97 1.986772 18 1.255273 38 1.579784 58 1.763428 78 1.892095 98 1.991226 19 11.278754 39 1.591065 59 1.770852 79 1.897627 99 1.995635 20 11.301030 40 1.602060 60 1.778151 180 1.9030901100 2.000000 2 LOGARITIMS I N.- o I 1 [ 2 1 3 4 f 5 6 7 8 9 ID. 1000 0 000000434 00868 001301 001734 002166 002598 003029i003461 003891 432 1 4321 4751 6181 6609 6038 6466 6894 73211 7748 8174 428 2 8600 9026 9451 98761010300 010724!011147 011570,011993 012415 424 31012837 013259013680014100 4521 4940 6360 57791 6197 6616 42C 4 7033 7451 7868 8284 87CO 9116 9532 9947 020361 020775 41C 6 021189 021603 022016 022428 022841 023252 023664 024075 4486 4896 412 6 5306 5715 6125 6533 6942 7350 7757 8164 8571 8978 408 7 9384 9789 030195 030600 031004 031408 031812 032216 032619 033021 404 810334241033826 4227 4628 6029 6430 6830 6230 6629 70281400 9 7426 7825 8223 8620 9017 9414 9811 040207 040602 040998 397 110 041393 041787 042182 042576 042969 043362 043755 044148 044540 0449321393 1 6323 6714 6105 6495 6885 7275 7664 8053 8442 8830 390 2 9218 9606 9993 050380 0507c6 C 51153 061538 051924 052309 052694 38C 310530781053463053846 4230 4613 4996 5378 5760 6142 6524 383 4 6905 7286 7666 8046 8426 8805 9185 9563 9942 060320 373 60606098 061075 061452 061829 062206 062582 062958 063333 063709 4083 376 6 4458 4832 5206 6580 5953 6326 6699 7071 7443 78151373 7 8186 8557 8928 9298 96681070038 070407 0707761071145071514370 81071882 0722501072617072985 073352 3718 4085 4451 4816 61821366 91 66471 5912 6276 66401 70041 7368 7731 80941 8457 8819|363 120'079181 079543 G79904 080266 080626 080987 081347 081707 082067 082426 360 1!08278510831441083503 3861 4219 4576 4934 5291 6647 6004]357 2 6360 6716 7071 7426 7781 8136 8490 8845f 9198 9552]355 3 9905 090258 090611 090963T091315 091667 092018 092370 092721 093071 352 4 093422 3772 4122 4471 4820 5169 5518 58661 6215 65621349 6 6910 7257 7604 7951 8298 8644 8990 9335 9681 1000261346 6100371 100715 101059 101403 101747 102091 102434 102777 103119 3462 343 7 3804 4146 4487 48281 6169 6510 6851 6191 6531 6871]341 8 7210 7549 7888 8227 8565 8903 9241 9579 9916 110253 338 9 110590 110926 111263 111599 111934 112270 112605 112940]113275 3609 335 13011139431114277 1146111114944 115278 115611 115943 116276 116608 116940;333 1 7271 7603 7934 8265 8595 8926 9256 9586 9915 120245 330 2 120574 120903 121231 121560 121888 122216 122544 122871 123198 35251328 3 3852 4178 4504 4830 5156 6481 6806 6131 6456 67811325 4 7105 7429 7753 8076 8399 8722 9045 9368 9690[130012[323 6 130334 130655 130977 131298 131619 131939 132260 132580 132900 3219 321 6 3539 3858 4177 4496 4814 5133 5451 6769 6086 64031318 7 6721 7037 7354 7671 7987 8303 8618 8934 9249 95641316 8 9879 140194 140508 140822 141136 141450 141763 142076 142389 142702 314 911430151 3327 3639 3951 4263 4574 4885 5196 6507 68181311 140 146128 146438 14674811470581147367 147676 147985 1482941148603 148911 309 11 9219 9527 98351150i42150449 '- 0756 151063 151370 151676 151982 307 2162288115259411529001 3205 3510 3815 4120 4424 4728 50321305 3 6336 6640 6943 6246 6649 3852 7154 7457 77591 0611303 4 8362 8664 8965 9266 9567 9868 16018116046911607691161068!3011 6 1613681161667 161967 162266 162564 162863 3161 3460 3758 40551299 6 4353 4650 4947 6244 5541 6838 6134 6430 6726 702212971 7 7317 7613 7908.8203 8497 8792 9086 9380 9674 9968 295 81170262 170555 170848 171141 171434 171726 172019 172311 172603 172895 293 91 3186 3478 3769 4060 4351 4641 4932 6222 6512 68021291 150 176091 1763811766701776959 177248 17753611778251178113 178401 178689 289 1 8977 9264 9552 9839 180126 180413 180699 180986 181272 181558i287 2 181844!182129 182415182700 2985 3270 3555 3839 4123 4407:285 3 46911 4975 6259[ 5542 6825 6108 6391 6674 6956 72391283 4 75211 7803 80841 8366 8647 8928 9209 9490 9771 190051 281 6 1903321190612 190892 191171 191451 191730 192010 192289 192567 2,846 279 6 3125 3403 3681 3959 4237 4514 4792 6069 6346 8623 278 7 6900 6176 6453 6729 7005 7281 7556 7832 8107 83821276 81 8657 8932 9206 94811 9755 200029 200303 2005771200850 201124 274 9 201397 201670 201943 202216 202488 2761 3033 3305 3577 3848 272 I I I V.f U I 1 I1 2 | 3 1 4 11 5 6 1 7 1 8 1 9 1 D J OF NUMBERS. 8 0 rN.) 0 I 1 1 2 1 3 1 4 11 5 6 1 7 1 8 1 9 i D. _., 'InlrI l )I 11 IUl 2 3 4 5 7 8 9 170 1 2 3 4 S 6 7 8 9 1 2 3 4 6 6 7 8 9 6826 9515 212188 4844 7484 220108 2716 5309 7887 230449 2996 5528 8046 240549 3038 5513 7973 250420 2853 255273 7679 260071 2451 4818 7172 9513 271842 4158 6462 0404391 204663 204934 7096 7365 7634 9783 210051 210319 212454 2720 2986 5109 6373 5638 7747 8010 8273 220370 220631 220892 2976 3236 3496 5568 6826 6084 8144 8400 8657 230704 230960 231215 3250 35041 3757 5781 6033 6285 8297 8548 8799 240799 241048 241297 3286 3534 3782 5759 6006 6252 8219 8464 8709 250664 2509081251151 3096 3338! 3580 2555141255755 255996 7918 8158 8398 260310 260548 260787 2688 2925 3162 6054 5290 5525 7406 7641 7875 9746 9980 270213 272074 272306 2538 4389 4620 4850 6692 6921 7151 I 205204 205475 7904 8173 210586 210853 3252 3518 5902 6166 8536 8798 221153 221414 3755 4015 6342 6600 8913 9170 231470 231724 4011 4264 6537 6789 9049 9299 241546 241795 4030 4277 6499 6745 8954 9198 251395 251638 3822 4064 256237 256477 86371 8877 261025 1261263 3399 3636 57611 5996 8110 8344 270440 270679 27701 3001 60811 6311 7380] 7609 8441 211121 3783 6430 9060 221675 4274 6858 9426 231979 45i7 7041 9550 242044 4525 6991 9443 251881 4306 256718 9116 261501 3873 6232 8578 270912 3233 6542 7838! vUv AU &vuolv 87101 8979 211388 211654 4049 4314 6694 6957 9323 9585 221936 222196 4533 4792 7115 7372 9682 9938 232234'232488 4770 6023 7292 7544 9800 240050 242293 2541 4772 6019 7237 7482 9687 9932 252125 252368 4548 4790 256958 257198 93:55 9594 261'M9261976 41091 4346! 6467 6702 8812 9046 271144 271377 3464 3696 5772 6002 8067 82961 92471269 211921 26q 4579 26( 7221 264 9846 262 222456 261 5051 259 7630 25E 230193 25( 232742 251 5276 253 77951 25M 2403001 25C 27901 24b 62661248 7728 24( 250176 245 26101243 60311242 2574391241 9833 239 262214 238 4582 237 6937 235 9279 234 271609 233 3927 232 6232 230 8525 229 I}nfi'l/1A! Ii) ilr'l tI O)li)0 l ofL.f.v- I i)7 1 190 2787541278982 2792111279439 279667 279895!2801231280351 2805781280806!228 1 2810331281261 281488 281715 2819421282169 2396 2622 2849 30751227 2 3301 3527 3753 3979 4205 4431 4656 4882 6107 5332 226 3 5557 6782 6007 6232 6456 6681 6905 7130 7354 7578 225 4 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 5 290035 2902571290480 290702 290925 291147 291369 291591 291813 292034 222 6 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 221 7 4466 4687 4907 6127 6347 5567 6787 6007 6226 6446 220 8 6665 6884 7104 7323 75421 7761 7979 8198 8416 8635 219 9 88531 9071 9289 9507 9725 9943 300161 300378,300595 300813 218 200 301030 301247 301464 301681301898 3021 14 302331 3025471302764 3029801 217 1 3196 3412 3628 3844 4059 4275 4491 4706 4921 51361216 2' 5351 5566 5781 5996 6211 6425 6639 6854 7068 7282 215 3 7496 7710 7924 8137 8351 8564 8778 8991 9204 94171213 41 9630 9843 310056 3102681310481 310693 31090613111181311330 311542:212 51311754311966 2177 2389 2600 2812 3023 3234 3445 36561211 6 3867 4078 4289 4499 4710 4920 6130 5340 5551 6760i210 7 56970 6180 6390 6599 6809 7018 7227 7436 7646 7854 209 8 8063 8272 8481 8689 8898 9106 9314 9522 9730 99381208 9320146 320354 3205621320769 320977 321184 321391 321598 321806 322012i207 210 3222191322426 322633 322839'323046 323252 323458 323665 323871 324077 206 1 4282 4488 4694 4899 6105 6310 65161 6721 5926 61311205 2 6336 6541 67451 6950 7155 7359 7563 7767 7972 8176 204 3 8380 8583 8787 8991 9194 9398 9601 9805'330008 330211 203 4 330414 330617 330819 3,~1022 331225 331427'331630 331832 2034 2236 202 6 2438 2640 2842 3044 324611 3447 3649 3860 4051 4253 202 6 4454 4655 4856 5057 6257, 5458 5658 6859 6059 62607201 7 6460 6660 6860 7060C 7260] 7459 7659 7858 8058 82571200 8 8456 8656 8855 90541 92531i 9451 9650 9849,340047;3402461199 9,340444 340642 340841 3410391341237 1341435 34163213418301 2028 22251138 __1.. r I 1.. I 1.. I r\?-~ll~R~ i I I L I I I i I I I 0 1 1 0 i I I I r t I I I I 0 N.. 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370883 185 5 371068 3712533 3 37437371622 371806 1991 2175 236;0 2544 2728 184 6 2912 3096 3280 3464 3647 3831 4015 4198 4382 45;5 184 7 4748 4932 6115 5298 5481 5664 5846 6029 6212 6394 183 8 657 7 6759 6942 7124 7306 7488 7670 7852 8034 8216 182 9 8398 8580 8761 8943 912 4 9306 9487 966;8 9849 380030181 240 380211 380392138057 33807541380934 381115 3812963814763816i56381837181 1 2017 2197 2377 2557 2737 2917 3097 3277 3456 3636 180 2 3815 3995 4174 4353 4533 4712 4891 5070 52491 5428 179 3 5606 6785 5964 6142 6321 6499 6677 6856; 7034 7212 178 4 7390 7568 7746 7923 8101 8279 8456 8634 8811 8989 178 5 9166 9343 9520 9698 9875 390051!390228390405 390582390759 177 6390935 391112 3912883914i64391641 1817 1993 2169 2345 2521|176 7 26;97 2873 3048 3224 3400 3575 3751 3926 4101 4277 176 8 4452 4627 4802 4977 5152 6326 5501 5676 5850 6025 175 91 6199 6374 6548 6722 6896 7071 7245 7419 7592 7766174 250 397940"398114 39828713984l61 398634 '398808'39898139915413993281399501 173 1 9674 9847 4000201400192i400365 400538 400711 400883 401056 401228 173 2401401401573 1745 1917 2089 22!61 2433 2605 2777 2949 172 3 3121 3292 3464 3635 3807 3978 4149 4320 4492 4663 171 4 4834 56005 5176 5346 5517 5688 5858 6029 6199 6370 171 5 6540 6710 6881 7051 72'1 7391 7561 7731 7901 8070 170 6 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 169 7 9933 410102 410271 410440 410(;09 410777 410946411114411283 411451 169 8411620 1788 1956 2124 22'193 2461 2629 2796 2964 31321168 9 3300 3467 3635 3803 3970 4137 4305 4472 4639 4806 167 260 414973 415140 415307 415474 415641 415808 415974:416141 416308 416474 167 1 6641 6807 6973 7139 7306 7472 7638 7804 7970 8135 166 2 8301 8467 8633 8798 8964 9129 9295 9460 9625 9791 165 3 9956!420121420286 420451 420(;16 420781 420945 421110 421275 421439 165 4 421604 1768 1933 2097 2261 2426 2590 2754 2918 3082 164 6 3246 3410m 3574 3737 3901 4065 4228 4392 4555~ 4718 164 6 4882 5045 5208 5371 5534 5697 5860 6023 6186 63491163 7 6511 6674 6836 6999 7161 7324 7486 7648 7811 7973 162 8 8135 8297 8459 8621 8783 8944 9106 9268 9,29 9591 162 9 9752 9914 430075430236 430398 430559 4307201430881,431042 431203,11; 270"43136414315265431685 431846 432007l432167 4323228 432488 432649 432809 1 61 1 2969 31301 3290 3450 3610 3770 3930 4090 4249 4409160 2 4569 47291 4888 5048 5207 5367 5526 5685 5844 6004 159 8 6163 6322 6481 6640 6799 6957 7116 7275 7433 7592 159 4 7751 7909 8067 8226 834 8542 87011 8859 9017 9175 158 6 9333 9491 9648 9806 9964 4401224402794404371440594440752 158 6 440909 441066 41224!441381 441538 1695 1852 2009 2166 2323157 7 24Mf 2637 2793 2950 310(;6 3263 3419 3576 3732 38891157 8 4045 4201 4357 4513 46691 4825 4981 61371 5293 5449115" 9 5604 57601 5915 60711 622611 6382 6537 66921 6848 70031155 "m o 1 _"1 T..4 11 5 "7. 9_ OF NUMBERS. 5 N. j I 1 I 2 3 4 II 6 i 7 8 a.4 IL).I ZSU 4441 bb 1 8706 2 450249 3 1786 4 3318 5 4845 6 6366 7 7882 91460898 2901462398 1 3893 2 5383 3 6868 4 8347 5 9822 6 471292 7 2756 8 4216 9 5671 8861 450403 1940 3471 4997 6518 8033 9543 461048 4042 7016 8495 9969 471438 2903 4362 r~ki A 4474 68144 6263 44t'-1 i 9015 91701 93524 450557 4507111450861r 2093 2247 2406 3624 3777 3936 5150 6302 5454 667-0 6821j 6973 8184 8336w 8487 9694 9845 99919 4611981461348 461491 462697 4628471462991 4191 4340' 4496 5680 5829 5977 7164 7312 74(;C 8643 8790 8938 470116 470263 470416 1585 1732 1878 3049 3195 3341 4508 4653 4799 5962 6107 6252 144 UJ J3 9478 451018 2553 4082 7125 8638 460146 1649 46314u 4639 6126 7608 9085 470557 2025 3487 4944 6397 44ZS08 448242 9633 9787 451172 451326 `2 706 2851 4235 4387 5758 5916 7276 7428 8789 8946 460296 460447 1799j 1948 44b6u1 9941 451479 3012 4546 6061 7571 9091 460597 2 0 9 44855245Io 450095 154 1633 154 3165 153 4692 153 6214 152 7731 152 9242 151 460748 151 2248 150 463744 150 5234i149 67191149 8200114b 9675 148 471145 147 2610 146 4071 146 55261146 6976 145 41i;9t6 4'788 6274 77563 9233 4710704 2171 3U33 5096 6541 463441 4936 6423 7904 9386 470851 2318 3771 5231 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O 1 1 2 iL 3 1 4 11 6 'I.I OF NUMBERS. 7 0 1 2 2 3 1 4 _11 5 1 6 i 7 1 8 9 00 602060 2169 602277 602386 602494 602603 60271 602819 60t22928 603036 10& 1 3144 3253 3361 3469 3577 3686i 3794 3902 4010 4118 10 2 4226 4334 4442 4550 4658 4766 4874 4982 5089 6197 10 3 5305 5413 5521 5628 5736 5844 5951 6059 6166 6274 108 4 6381 6489 6596 6704 6811 6919 7026 7133 7241 7348 107 6 7455 7562 7669 7777 7884 7991 8098 8205 8312 8419{107 6 8526 8633 8740 8847 8954 9061 9167 9274 9381 9488 107 7 9594 9701 9808 9914 610021 610128 610234 610341 610447i6105541107 9 1723 1829 1936 2042 2148 2254 2360 2466 2572 26781106 410 6127841612890 612996 613102 613207 613313 613419 613525 613630 613736 10 1 3842 3947 4053 4159 4264 4370 4475 4581 4686 4792 106 2 4897 6003 5108 5213 6319 6424 5529 6634 6740 5845 105 3 6950 6055 6160 6265 6370 6476 6581 6686 6790 6895F105 4 7000 7105 7210 7315 7420 7525 7629 7734 7839 794 I10O 5 8048 8153 8257 8362 8466 8571 8676 8780 8884 89891105 6 9093 9198 9302 9406 9511 9615 9719 9824 9928 6200321104 7 620136 620240 620344 620448 620552 620656 620760 620864 620968 1072 104 8 1176 1280 1384 14881 1592 1695 1799 1903 2007 21101104 9 2214 2318 2421 2525 262811 2732 2835 2939 3042 31461104 t2016232491623353 623456 623559 623663 623766 623869 623973 624076 624179 103 i 4282 4385 4488 4591 4695 4798 4901 5004 65107 5210103 2 5312 6415 5518 6621 5724 6827 5929 6032 6135 62381103 3 6340 6443 6546 6648 6751 6853 6956 7058 7161 72631103 4 7366 7468 7571 7673 7775 7878 7980 8082 8185 8287 102 5 8389 8491 8593 8695 8797 8900 9002 9104 9206 9308 102 6 9410 9512 9613 9715 9817 99196300211630123 630224630326102 716304281630530 630631 630733 630835 630936 1038 1139 1241 1342 102 8 1444 1545 1647 1748 1849 1951 2052 2153 2255 2356 101 9 2457 2559 2660 2761 2862 2963 30641 3165 3266 3367 101 430 633468l633569 63367016337711 633872 63397316340746341 75634276 634376 101 1 4477 45781 -679 47791 4880 4981 5081 5182 6283 6383'101 21 6484 5584 5685 57851 5886 5986 6087 6187 6287 6388100 3 6488 6588, 6688 6789' 6889 6989 7089 7189 7290 7390 100 4 7490 7590 7690 7790 7890 7990 8090 8190 8290 8389 100 6 8489 8589 8689 8789 8888 8988 9088 9188 9287 9387 100 6 9486 9586 9686 9785 9885 9984640084640183640283l640382 9 7 640481 640581 640680640779 640879 640978 1077 1177 1276 1375 99 8 1474 1573 1672 1771 1871 19701 2069 2168 2267 2366 99 9 2465 2563 2662 2761 2860 2959 30581 3156 3255 3354 99 440 643453 643551 643650 643749 643847 643946 644044 6441431644242 644340 98 1 4439 4537 4636 4734 4832 4931 6029 6127 6226 6324 9 2 6422 5521 6619 6717 6815 6913 6011 6110 6208 6306 98 3 6404 6502 6600 6698 6796 6894 6992 7089 7187 7285 98 4 7383 7481 7579 7676 7774 7872 7969 8067 8165 8262 98 6 8360 8458 85655 8653 8750 8848 8945 9043i 9140 9237 97 6 9335 9432 9530 9627 9724 9821 99196500166650113 650210 97 7650308 650405 650502 6505996506966507931650890 0987 1084 1181 97 8 1278 1375 1472 1569 1666 '17621 2859 1956 2053 2160 97 9i 2246 2343 2440 2536 2633 27301 2826 2923 8019 3116 97 450 663213663309 653405 653502653598 6536951653791 653888 53984 s5480( 1 4177 4273 4369 4465 4562 4658 4754 4850 4946 6042 9 2 5138 6235 6331 5427 6523 6619 6716 6810 5906 60021 9 3 6098 6194 6290 63861 6482 6577 66s'3 6769 6864 6901 9 4 7056 7152 7247 73431 7438 7534 7629 7725 7820 7916 96 6 8011 8107 8202 8298 8393 8488 8584 8679 8774 8870 95 6 8965 9060 9155 9250 9346 9441 9536 9631 9726.9821 95 7 9 16 660011 660106 660201 660296 660391 660486 660581660676660771 96 8660865 09601 1055 11501 1245 1339 1434 1529 1623 1718 95 9 1813 1907 20021 2096, 2191 2286 2380 2476 2569 2663 95. 0 1 2 1 3 4 l 5 6 I 7! 8 I -9 _D 8 LOGARITHMS WN.7i 0 1 2 1 3 j 4 II 5 1 6 1 7 1 8 9 ID. 60 662758 6628521662947 6ti3041 663135 6632 3066332)4 663418 663512 6636071 94 1 3701 3795 3889 3983 4078 4172 42661 43601 4454 45648 94 2 4642 4736 4830 4924 5018 5112 5206 299 63593 5487 94 3 6581 5675 5769 5862 5956 605') 6145 6237f 6331 6424 94 4 6518 6612 6 70,-) 6799 6892 6986 7079 71731 7266 7360 94 6 7453 7546 7640 7733 7826 7920 8013 81061 8199 8293 93 6 8386 8479 8572 8665 8759 8852 8945 90381 9131 9224 93 7 9317 9410 9503 9596 96891 9782 9875 9967'670060 670153' 93 8'670246 670339 670431 670524 6706171 670710 670802 670895 0988 10801 93 91 1173 1265 1358 1451 1543 16361 17281 1821 1913 20051_93 47067209S 672190!672283 672375 6724671 672560 672652 672744 672836 672929 92 1 3021 3113 3 205 3297 3390i 3482 3574 3666 37 58 3850 92 2 3942 4034 4126 4218 4310 4402 4494 4586 4677 47169 92 3 4861 4953 5045 6137 52281 5320 5412 5503 5595 5687 92 4 5778 5870 5962 6053 6145 6236 63 28 6419 6511 6602 92 5 6694 6785 6876 6968 7059 7151 7242 7333 7424 7516 91 6 7607 7698 7789 7881 7972 8063 8154 8245 8336! 8427 91 7 8518 8609 8700 8791 8882i 8973 9064 91551 9246: 9337 91 8i 0428 9 5 19 9610 9700) 97911 9882 9973~6800631680154:680245 91 91680336 680426 680517 680607 6806981 680789 6808791 0970j 10601 11511 91 480 681241 681332 681422Z681513 681603 681693 6817184 681874 681964 682055 90 1 21451 22,35 23261 2416 2506 2596 2686 27,77 2867 2957 90 2 3047, 31:37 3227' 3317 3407 3497 3587 3677 3767 3857 90 3 3947 4037 4127 4217 4307 4396 4486 4 7 6 4666 4756 90) 4 4845 4935 5025 5114 5204 5294 5383 547-3 5563 5652 90 5W 5742 5831 5921 6010 6100 6189 6 27 9 6368 6458i 61547 8 9 6j 6636 6726 6815 6904 6994 7083 7172 72(61 7351 7440 8 9 7 7529 7618 7707 7796 7886 7975 8064, 8153 824 2 8331 89 8 8420 8509 8598 8687 8776 8865 89531 9042 9131! 9220- 89 9 9309 9398 9486 96751. 9664 9753 98411 99301690019 6901071 89 490 690 196 690285 690373 690462 690550 690639 690728 690816 690905 6909193 89 1 1081 1170 1258 1347 1435 1524 1612 1700 1789 1877 88 2 1965 2053 2142 2230 2318 2406 2494 2583 2671 2759 88 3 2847 2935 3023 3111 3199 3287 3375 3463 3551 3639 88 4 3727 3815 3903 3991 4078 4166 4254 4342. 4430 4517 88 6 4605 4693 4781 4868 4956 5044 5131 6219 5307 53.94 88 6 5482 5569 5657 5744 5832 5919 6007 6094 6182 6269 87 7 6356 6444 6531 6618 6706 6793 6880 6968 7055 7142 87 8 7229 7317 7404 7491 75781 7665!77521 7839 7926 8014 8 7 91 8101 8188 8275 8362: 84491 8535, 86221 87091 87961 8883 8 7 500 O9897G 6990571699144 699231 699317 699404 699491:699578 6996641699751 87 ii 9838 99241700011 700098 700184 700271 7100358 700444 700531 700617 87 2:700704 700790 0877 0963 1050 1136 1222 1309 1395 1482 86 3! 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'140757 740836'740915 '40994 741073i 79 1 I152' 1230 1309 1388' 146-7 1546 1624 1703 1782' 1860 79 1,1439 2018- 2096 21751 2254 2:3321 2411 2489 25681 26471 79 3 27 251 2804 2882 2961 3039 3118 3196 3275 3353' 3431 78 4 3510' 3588 3667 3745' 3823 3902 3980 4058 4136 42151 78 5 4293 4371 4449 4528 4606! 4684 4762 4840 4919 4997 78 65075i 6153 5231 5309 5387' 5465 5543 6621 6699 5777 78 7 5855 5933 6011 60891 6167 62451 6323 6401 6479 65561 78 8 6634 6712 6790 68681 6945! 7023 7101 7179 7256 73341 78 9 7412 7489 7567 7645; 7722 7800 7878 7955' 8033 81101 78 560 748188 748266 748343 748421l748498 748576 74865317487131 748808 748885 77 1 8963 9040 9118 9195 9272 9350 9427 9504: 9582 9659 77 2 9736 9814 9891 9968 750045 750123 750200 750277,750354 750131 77 3 750508 750586 750663;750740 0817 0894 0971 1048' 1125 1202 77 4 1279 1356 14331 1510 1587 1664 1741 1818t 1895 1972 77 j 20481 2125 2 2021 2279 2356 2433 2509 2586, 2663 240' 77 6 2816,-i 2893 2970 3047 3123 3200 3277 3313 3430 3506 777 7 3583' 30660 37361 3813 3889 3966 4042! 4119 4195 4272' 77 8 43481 4425' 4501 45781 4654 4730 4807 4883 4960 5036 76 9 5112; 5189 5265 5341; 5417 5494 5570 6646 5722 5799 76 570j755875 755951 1 756027 756103.756180 756236 7563,32 16408 756484 756560 76 j 6636; 6712 6788 6864 6940 7016 7092 7168 7244 7320 76 2 7396 7472 7548 7624 7700 7775 7851 7927 8003 8079 76 3 8155 8230' 8306 8382 8458 8533 8609 8685 8761 8836 76 4 8912' 8988 9(63 9139 9214 9290 9366 9441i 9517 9592 76 5 9668 9743, 9819 9894; 9970 760045 7601211760196 760272 760347 76 6-1760422 760498 760573 760649.760724 0799 087 5 0950 1025 1101 76 71 11761 1251; 1326 14)2 1477 1552 1627 1702 1778 1853 76 8 1928' 2003! 2078 2163, 2228 2303 2378 2453i 2529' 2604 75 9 2679 2754 2829 2904 2978 3053' 3128' 3203' 3278 3353' 75 WFI. I h 1 2 3 4 5 11 6 6 7 8 9J. 11 10 LOGARITHMS I i I I t | I | |.1 i t i. I I I.1I u I I I z! 3 I 4 11 b I 6 i 7 1 I j U. 80 763428763503 763578 763653 763727 763802 763877 763952 764027 764101 75 1 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 75 2 4923 4998 5072 6147 6221 5296 6370 6445 5520 6594 75 3 5669 6743 6818 6892 6966 6041 6115 6190 6264 6338 74 4 6413 6487 6562 6636 6710 6785 6859 6933 7007 7082 74 5 7156 7230 7304 7379 7453 7527 7601 7676 7749 7823 74 6 7898 7972 8046 8120 8194 8268 8342 8416 8490 8564 74 7 8638 8712 8786 8860 8934 9008 9082 9156 9230 9303 74 8 9377 9451 9525 9599 9673 9746 9820 9894 9968 770042 74 9 770115 770189 770263 770336 770410 770484 770557 770631 770705 0778 74 590 770852 770926 770999 7710731771146 771220 771293 771367 771440 71514 74 1 1587 1661 1734 1808 1881 1965 2028 2102 2175 2248 73 2 2322 2395 2468 2542 2615 2688 2762 2835 2908 2981 73 3 3055 3128 3201 3274 3348 3421 3494 3567 3640 3713 73 4 3786 3860 3933 4006 4079 4152 4225 4298 4371 4444 73 5 4017 4590 4663 4736 4809 4882 4955 5028 5100 6173 73 6 5246 6319 5392 6465 6538 6610 5683 5756 5829 5902 73 7 6974 6047 6120 6193 6265 6338 6411 6483 6556 6629 73 8 6701 6774 6846 6919 6992 7064 7137 7209 7282 7354 73 9 7427 7499 7672 7644 7717 7789 7862 7934 8006 8079 72 600 7781651 778224 778296 778368 778441 778513 778585 778658 778730 778802 72 1 8874 8947 9019 9091 9163 9236 9308 9380 9452 9524 72 2 9596 9669 9741 9813 9885 9957 780029 780101 780173 780245 72 3 780317 780389 780461 780533 780605 780677 0749 0821 0693 0965 72 4 1037 1109 1181 1253 1324 1396 14418 1540 1612 1684 72 5 1755 1827 1899 1971 2042 2114 2186 2258 2329 2401 72 6 2473 2544 2616 2688 2759 2831 2902 2974 3046 3117 72 7 3189 3260 3332 3403 3476 3546 3618 3689 3761 3832 71 8 3904 397'5 4046 4118 4189 4261 4332 4403 4475 4546 71 9 4617 4689 4760 4831 4902 4974 56045 5116 5187 5259 71 610 785330 778401 785472 785543 786616 785686 785757 785828i785899 785970 71 1 60411 6112 6183 6254 6325 6396 6467 6538 6609 6680 71 2 6751 6822 6893 6964 7035 7106 7177 7248 7319 7390 71 3 7460 7531 7602 7673 7744 7815 7885 7956 8027 8098 71 4 8168 8239 8310 8381 8451 8522 8593 8663 8734 8804 71 5 8875 8946 9016 9087 9167 9228 9299 9369 9440 9510 71 6 9581 9651 '722 9792 9863 9933 790004 790074 790144 790215 70 7 790285 790356 790426 790496 790567 790637 0707 0778 0848 0918 70 8 0988 1059 1129 1199 1269 1340 1410 1480 1550 1620 70 9 1691 1761 1831 1901 1971 2041 2111 2181 2252 2322 70 620 1792392 792462 792532 792602 792672 792742 792812 792882 792952 793022 70 1 3092 3162 3231 3301 3371 3441 3511 3581 3651 3721 70 2 3790 3860 3930 4000 4070 4139 4209 4279 4349 4418 70 3 4488 4558 4627 4697 4767 4836 4906 4976 6045 6115 70 4 5185 6254 5324 6393 6463 5532 6602 6672 6741 6811 70 5 5880 6949 6019 6088 6158 6227 6297 6366 6436 6505 69 6 6574 6644 6713 6782 6852 6921 6990 7060 7129 7198 69 7 7268 7337 7406 7475 7646 7614 7683 7752 7821 7890 69 8 7960 8029 8098 8167 8236 8305 8374 8443 8513 8582 69 9 8661 8720 8789 8858 8927 8996 9065 9134 9203 9272 69 130 799341 799409 799478 799547 799616 799685 799754 799823 799892 7999611 69 1 800029 800098 800167 800236 800305 800373 800442 800511 800580 800648 69 2 0717 0786 0854 0923 0992 1061 1129 1198 1266 1335 69 3 1404 1472 1541 1609 1678 1747 1815 1884 1952 2021 69 4 2089 2158 2226 2295 2363 2432 2500 2568 2637 2705 68 6 2774 2842 2910 2979 3047 3116 3184 3252 3321 3389 68 6 3467 3625 3594 3662 3730 3798 3867 3936 4003 4071 68 7 4139 4208 4276 4344 4412 4480 4648 4616 4685 4753 68 8 4821 4889 4967 5025 5093 6161 5229 6297 5365 6433 68 9 5501 5569 5637 5705 5773 5841 5908 5976 6044 6112 68 -N )0-o I_ 1 2 3 4 115 1 6 7 "8 9 ID. 0 OF NUMBERS. 11 N. 0 1 2 3 4 6 6 7 640 80618 8046248 806316 806384 806451 01806519 87 806655 806723 806790 68 1 6858 6926 6994 7061 7129 7197 7264 7332 7400 7467 68 2 7535 7603 7670 7738 7806 7873 7941 8008 8076 8143 68 3 8211 8279 8346 8414 8481 8549 8616 8684 8751 8818 67 4 8886 8953 9021 9088 9156 9223 9290 9358 9425 9492 67 6 9560 9627 9694 9762 9829 9896 9964 810031 810098 810165 67 6 810233 810300 810367 810434810501 810569 810636 0703 0770 0837 67 7 0904 0971 1039 1106 1173 1240 1307 1374 1441 1508 67 8 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 67 9 2245 2312 2379 2445 2512 2579 2646 2713 2780 28471 67 650 812913 812980 813047 813114 813181 813247 813314 813381 813448 813514 67 1 3581 3648 3714 3781 3848 3914 3981 4048 4114 4181 67 2 4248 4314 4381 4447 4514 4581 4647 4714 4780 4847 67 3 4913 4980 5046 6113 5179 6246 6312 56378 6445 6511 66 4 6678 6644 6711 6777 6843 6910 6976 6042 6109 6175 66 6 6241 6308 6374 6440 6506 6573 6639 6705 6771 6838 66 6 6904 6970 7036 7102 7169 7235 7301 7367 7433 7499 66 7 7566 7631 7698 7764 7830 7896 7962 8028 8094 8160 66 8 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 66 9 8886 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 660 8195441819610 819676 819741 819807 819873 819939 8200041820070 820136 66 11820201 820267 820333 820399 820464 820530 820595 0661 0727 0792 66 2 0858 0924 0989 1055 1120 1186 1251 1317 1382 1448 66 3 1514 1579 1645 1710 1775 1841 1906 1972 2037 2103 66' 4 2168 2233 2299 2364 2430 2495 2560 2626 26911 2756 65 5 2822 2887 2952 3018 3083 3148 3213 3279 3344 3409 66 6 3474 3539 3605 3670 3735 3800 3865 3930 3996 4061 65 7 4126 4191 4256 4321 43861 4451 4516 4581 4646 4711 66 81 4776 4841 49061 4971 50361 5101 6166 5231 6296 6361 65 9 542' 6491 6556 6621 668611 6751! 6815 5880 6945 6010 65 670 826075 82614082620826 26334262082 33482399 826464 826528 826593 8266586 65 1 6723 6787 6852 6917 6981 7046 7111 71765 7240 7305 65 2 7369 7434 7499 7563 7628 7692 7757 7821 7886 7951 65 31 8015 8080 8144 8209 8273 8338 8402 8467 8531 8595 64 41 8660 8724 8789 8853 8918 8982 9046 9111 9175 9239 64 6 9304 9368 9432 9497 9561[ 9625 9690 9754 9818 9882 64 6 9947 830011183007518301391830204 830268 830332 830396 830460 830525 64 7 830589 0653 0717 0781 0845 0909 09',3 1037 1102 1166 64 8 1230 1294 1358 1422 1486 1650 1614 1678 1742 1806 64 9 18701 1934 1998 2062 2126 2189 2253 23171 2381 2445 64 680 832509 832573 832637 832700 832764 832828 832892 832956 833020 833083 64 1 3147 3211 3275 3338 3402 3466 3530 3593 3657 3721 64 2 3784 3848 3912 3975 4039 4103 4166 4230 4294 4357 64 3 4421 4484 4548 4611 4675 4739 4802 4866 4929 4-993 64 4 6056 6120 6183 6247 6310 6373 6437 6600 6564 6627 63 6 6691 6754 6817 6881 6944 6007 6071 6134 6197 6261 63 6 6324 6387 6451 6514 6577 6641 6704 6767 6830 6894 63 7 6957 70201 7083 7146 7210 7273 7336 7399 7462 76256 63 8 7588 7662 7716 7 715 7778 7841 7904 7967 8030 8093 8156 63 9 82191 82821 8345 8408 8471 8534 8597 86601 8723 8786 63 690 838849 838912 838975 839038 839101 839164 839227 839289 839352 839416 1 9478 9541 9604 96671 9729 9792 9856 9918 99811840043 631 2 840106 840169 840232 840294840357 840420 840482 840545 840608 0671 6 3 07331 0796 0859 0921I 6984 1046 1109 1172 1234 1297 63 4 1359 1422 1485 1547 1610 1672 1735 1797 1860 1922 63 I 6 1985 2047 2110 2172 2235, 2297 2360 2422 2484 2547 62 6 2609 2672 2734 2796 28591 2921 2983 3046 31081 3170 62 7 3233 3296 3357 3420 3482 35441 3606 3669i 3731 3793 62 8 385 3918 3980 4042 4104 4166 4229 4291 43531 44156 62 9 4477 4539 4601 4664 47261 4788 4850 4912 49741 5036 62 _ _,,,,,,, _,. _~~~~~~~~~_ [N I 0 1 I 2 1 3 I 4 1 6 6 I 7 I 8 I 9 I)D J LOGARITHMS N.i U I 1I 1 2 I 3 1 4 II 5 1 6 1 7 1 8 1 9 1.) 700 8450981845160 845222 845284 815146 845408 845470845 3'2 84559: 8456561 62 1 6718 6780 5842 6904 5966! 6028 6090 6151 6213 6275 62 2 6337 6399 6461 6523 6585 6646 67081 6770 6832 68941 62 3 6955 7017 7079 7141 7202 7264 7326 7388 7449 7511 62 4 7573 7634 7696 7758 7819 7881 7943 8004 80i66 8128 62 6 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 6 8805 8866 8928 8989 9051 9112 9174 9235 9297 9358 61 7 9419 9481 9542 9604 96651 9726 9788 9849 9911 9972 61. 8 850033 850095 850156 850217 850279 80350340 850401 850462 850524 850585, 61 9 0646 0707 0769 08;M3 08911 0952 1014 1075 1136 1197 61 10 85'1258 851320O851381 85144' 851503 851564 8511/25 851686 851747 851809 61 1 1870 1931 1992 2053 2114 2175 2236 2297 2358 2419 61 2 2480 2511 2602 2663 2724 2785 2846 2907 2968 3029 61 3 3090 3150 3211 3272 3333 3394 3455 3516 3577 3637 61 4 3;98 3759 3820 3881 3941 4002 4063 4124 4185 4245 61 6 4306 4367 4428 4488 4549 4610 4670 4731 4792 4852 61 6 4913 4974 6034 5095 5156 5216 5277 6337 5398 6459 61 7 5519 6580 6640 6701 6761 6822 6882 5943 6003 6064 61 8 6124 6185 6245 6306 6366 6427 6487 6548 66;08 6668 60 9 6729 6789 6850 6910 69701 7031 7091 7152 7212 7272 60 720'85733218573931857453 857513'857574 1857634 8576941857755 857815 857875 60 1 7935i 7995 8056 8116 8176 8236 8297 8357 8417 8477 60 2 8537 8597 8657 8718 8778] 8838 8898 8958 9018 9(078 60 3 9138 9198 9258 9318 9379] 9439 9499 9559 9619 9679 60 4 9739 9799 9859 9918 99781860038 860098 8601581860218 86027'8 60 6860338 860398 860458860518 860578' 0637 0697 0757 0817 0877 60 6 0937 0996 1056 1116 1176 1236 1295 1355 1415 1475 60 7 1534 1594 1654 1714 1773 1833 1893 19521 2012 2072 60 8 2131 2191 2251 2310 2370 24:30 2489 2549 2608 26681 60 9 2728 2787 2847 2906 29661 3025 3085 3144] 3204 326;31 60 730 863323 863382 863442 866350 1 863561 i863t6i2i863680 863739l863799 863858! 59 1 3917 3977 4036 4096 4155 4214 4274 4333 4392 44521 59 2 4511 4570 4630 4689 4748 4808 4807 4926 4985 5045 59 3 6104 6163 5222 6282 53411 5400 645469 6519 5578 5637 59 4 6696 6755 6814 6874 6933 5992 6051 0110 6169 6228 59 65 6287 6346 6405 64;65 6524 6583 6642 6701 6760 6819 59 6 6878 6937 6996 7055 7114 7173 72:)2 7291 7350 7409 59 7 7467 7526 7585 7644 7703 7762 7821 7880 7939 79981 59 8 8056 8115 8174 8233 8292 8350 8409 8468 8527 8586 59 9 86441 8703 8762 8821 8879 8938 8997 9056 9114 9173 59 740 869232 869290 869349 869408 869466 8569525 8695841869'6421869701 186,76 ) 59 1 9818 9877 9935 9994 870053 870111 8701701870228 870287 8703415 59 2 870404 870462 870521 870579 0638m 0696 0755 0813 0872 0930 58 3 0989 1047 1106 1164 1223 1281 1339 1398 1456 1515 58 4 1573 1631 1690 1748 1806 1865 1923 1981 2040 2).98 58 5 2156 2215 2273 2331 2389m 2448 2506 2564 2622 21,81 58 6 2739 2797 2855 2913 29721 3030 3088 3146 3204 3262 58 7 3321 3379 3437 3495 3553 3611 36G9 3727 3785 3844! 58 8 3902[ 3960 4018 4076 4134 4192 4250 4308 4366 4424 58 9! 44821 4540 4598 46561 47141 4772 4830 488z] 49451 5003 58 Zl'.n Qil AK l IoFTrI la' la,?' 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OF NUMBERS. 13 N.I 0 1 i 2 8 4 6 - 7 1 8 1 9 1D. 760 880814j880871 880928 880986 881042 881099 881156 881213 881271 881328 67 1 1385 1442 1499 1556 1613 16701 1727 1784 1841 1898 67 2 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 67 3 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 67 4 3093 3150 3207 3264 3321 3377 3434 3491 3548 *3605 57 5 3661 3718 3775 3832 3888 3945 4002 4359 4115 4172 67 6 4229 4285 4342 4399 4455 4512 4569 4625 4682 4739 67 7 4795 4892 4909 4965 56022 56078 56135 56192 6248 6305 67 8 5361 5418 56474 5531 5587 5644 5700 5757 5813 5870 67 9 6926 o083 6039 6096 6152 6209 6265 6321 6378 6434 66 770 886491 886547 886604 886660 886716 886773 886829i88688i 886942 886998 56 1 7054 7111 7167 7223 7280 7336 7392 7449 7505 7561 56 2 7617 7674 7730 7786 7842 7898 7955 8011 8067 8123 66 3 8179 8236 8292 8348 8404 8460 8516 8573 8629 8685 66 4 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 58 5 9302 9358 9414 9470 9526 9582 9638 969i 9750 9806 66 6 9862 9918 9974890030 890086 890141 890197 890253 890309 890365 66 7 890421 890477 890533 0589 0645 0700 0756 0812 0868 0924 66 8 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 66 9 1637 1593 1649 1705 1760 1816 1872 1928 1983 2039 56 780 892095892150 89220)892262 8923 17 892373 892429 892484 892540,892595 56 1 2651 2707 2762 2818 2873 29291 2985 3040 3096 3151 56 2 3207 3262 3318 3373 3429 3484 3540 3595 3651 3706 56 3 3762 3817 3873 3928 3984 4039 4094 4150 4205 4261 665 4 4316 4371 4427 4482 4538 4593 4648 4704 4759 4814 55 5 4870 4925 4980 5036 56091 56146 6201 56257 56312 6367 66 6 6423 56478 5533 56588 5644 5699 5754 56809 5864 6920 55 7 6975 6030 6085 6140 6195 6251 6306 6361 6416 6471 55 8 6526 6581 6636 6692 6747 6802 6857 6912 6967 7022 55 9 7077 7132 7187 7242 7297 7352 7407 7462 7517 7572 65 790 897627 897682 897737 8977921897847 897902 897959 898012 898067 898122[ 65 1 8176 8231 8286 8341 8396 8451 8506 8561 8615 8670 55 2 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 55 3 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 55 4 9821 9875 9930 9985 900039 900094900149900203900258900312 655 65 900367 900422 900476'900531 0586 0640 0695 0749 0804 0859 55 6 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 55 7 1458 1513 1567 16221 1676 1731 17865 1840 1894 1948 54 8 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 64 9 2547 2601 2655 2710 2764 2818 2873 2927 2981 3036 54 900 903090 903144 903199 903253 903307 903361 903416 903470 903524 903578 54 1 3633 3687 3741 3795 3849 3904 3958j 4012 4066 4120 54 2 4174 4229 4283 4337 4391 4445 44991 4553 4607 4661 64 3 4716 4770 4824 4878 4932 4986 50401 6094 5148 6202 54 4 6256 56310 56364 6418 56472 6526 556680 5634 5688 6742 541 56796 6850 56904 5958 6012 6066 6119 6173 6227 6281 64 6 6335 6389 6443 6497 6551 6604 6658 6712 6766 6820154 7 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 64 8 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 9 7949 8002 8056 8110 8163 8217 8270 8324 8378 8431 64 810 908485 908539 908592 908646'908699[9087531908807 908860 908914i908967 564 1 9021 9074 9128 9181 9235 9289 9342 9396 94491 9503 64 2 95561 9610 9663 9716 9770 9823 9877 9930 99841910037 63 3 i10091 910144 910197 910251 910304 910358 910411 910464 9105181 0571 63 4 0624 0678 0731 0784 0838 0891 0944 0998 1051 1104 63 6 1690 1743 1797 1850 1903 1956 2009 2063 2116 2169 53 7 2222 2275 2328 2381 2435 2488 2541 2594 2647 2700 53 8 2753 2806 28591 2913 2966 3019 3072 3125 3178 3231 63 9 3284 3337 33901 3443 34961 3549 3602 3655 3708 3761153 N._ I ' 2 I 3 4 ' 6 6 | 8 | " I ' "., 7 8 -: 14 LOGARITHMS 0.I 1 2I3 4 6 7 48 '- D. 320 913814 913867 913920 913973 914026 914079 914132 914184 914237 914290 53 1 4343 4396 4449 4502 4555 4608 4660 4713 4766 4819 53 2 4872 4925 4977 6030 6083 5136 5189 6241 6294 5347 53 3 5400 6453 5505 6558 5611 5664 5716 6769 6822 5875 53 4 5927 6980 6033 6085 6138 6191 6243 6296 6349 6401 53 5 6454 6507 6559 6612 6664 6717 6770 6822 6875 6927 53 6 6980 7033 7085 7138 7190 7243 7295 7348 7400 7453 53 7 7506 7558 7611 7663 7716 7768 7820 7873 7925 7978 52 8 8030 8083 8135 8188 8240 8293 8345 8397 8450 8502 52 9 8555 8607. 8659 8712 8764 8816 88691 8921 8973 9026152 830 919078 919130 919183 919235 919287 919340 919392 919444919496 919549 52 1 9601 9653 9706 9758 9810 9862 9914 9967 9200191920071 52 2 920123 920176 920228 920280 920332 920384 920436 920489 0541 0593 52 3 0645 0697 0749 0801 0853 0906 0958 1010 1062 1114 52 4 1166 1218 1270 1322 1374 1426 1478 1530 1582 1634 52 5 1686 1738 1790 1842 1894 1946 1998 2050 2102 2154 52 6 2206 2258 2310 2362 2414 2466 2518 2570 2622 2674 52 7 2725 2777 2829 2881 2933 2985 3037 3089 3140 3192 52 83244 3296 3348 3399 3451 3503 3555 3607 3658 3710 52 9 3762 3814 3865 3917 3969 40S11 4072 4124 4176 4228 52 940 924279 924331 924383 924434 924486 9245381924589 92464 92 4693 924744 52 1 4796 4848 4899 4951 5003 6054 5106 6157 5209 5261 52 2 5312 5364 5415 5467 5518 5570 5621 5673 65725 5776 52 3 5828 5879 5931 5982 6034 6085 6137 6188 6240 6291 51 4 6342 6394 6445 6497 6548 6600 6651 6702 6754 6805 51 5 6857 6908 6959 7011 7062 7114 7165 7216 72G8 7319 51 6 7370 7422 7473 7524 7576 7627 7678 7730 7781 7832 51 7 7883 7935 7986 8037 8088 8140 8191 8242 8293 8345 51 8 8396 8447 8498 8549 8601 8652 8703 8754 8805 8857 51 91 8908 8959 9010 9061 9112 9163 9215 9266 9317 936851 860j929419 929470 929521 929572 929623 9296741929725 929776 929827 929879 51 1 9930 9981 9300321930083 930134 930185i930236 930287 930338 930389 51 2930440930491 0542 0532 0643 0694 0745 0796 0847 0898 51 3 09491 1000 1051 1102 1153 1204 1254 1305 1356 1407 51 4 1458 1509 1560 1610 1661 1712 1763 1814 '1865 1915 51 5 1966 2017 2068 2118 2169 2220 2271 2322 2372 242351 6 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 51 7 981 3031 3082 3133 3183 3234 3285 3335 3386 3437 51 8 3487 3538 3589 3639 3690 3740 3791 3841 3892 3943 51 9.1 3993 4044 4094 4145 4195 42461 4296 4347 4397 4448 51 360 934498 934549 934599 934650 934700 934751 934801 934852 934902 934953 50 1 6003 5054 5104 5154 5205 5255 5306 5356 5406 5457 50 2 5507 5558 5608 5658 5709 5759 5809 5860 5910 5960 50 3 6011 6061 6111 6162 6212 6262 6313 6363 6413 6463 50 4 6514 6564 6614 6665 6715 6765' 6815 6865 6916 6966 50 5 7016 7066 7117 7167 7217 7267 7317 7367 7418 7468 50 6 7518 7568 7618 766, 7718 7769 7819 7869 7919 7969 50 7 8019 8069 8119 8169 8219 8269 8320 8370 8420 8470 50 81 8520 8570 8620 8670 8720 8770 8820 8870 8920 8970 50 9, 9020 9070 9120 9170 9220 9270 9320 9369 9419 9469 50 70 939519 969569 939619193966919397193939769 939819 939869 939918 939968 50 1 940018 940068 940118 940168,940218 940267 940317 940367 940417 940467 50 2 0516 0566 0616 0666| 0716 0765 0815 0865 0915 0964 50 8 1014 1064 1114 1163 1213 1263 1313 1362 1412 1462 50 41 1511 1561 1611 1660 1710 1760 1809 1859 1909 1958 50 5 2008 2058 2107 2157 2207 2256 2306 2355 2405 2455 50 6 2504 2554 2603 2653 2702 2752 2801 2851 2901 2950 50 71 3000 3049 3099 3148 3198 3247 3297 3346 3396 3445 49 8 3445 3544 3593 36431 36921 3742i 3791 38411 4890 3939 49 9 3989 4038 4088 4137 4186 4236 4285433 4384 443349 N.I0 1 I 2 3 4 I 6 7 8 9 D. OF NUMBERS. 15 N.1j 0 1 3 4 6 7I 8 1t I1D. 880 944483 94452 944581 944631 944680 944729 944779 944828 944877 944927 49 1 4976 5025 5074 5124 5173 5222 5272 5321 5370 6419 49 2 6469 5518 6567 5616 5665 5715 5764 6813 58t2 5912 49 3 5961 6010 6059 6108 6157 6207 6256 6305 6354 6403 49 4 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 49 6 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 6 7434 7483 7532 7561 7630 7679 7728 7777 7826 7875 49 7 7924 7973 8022 8070 8119 8168 8217 8266 8315 8364 49 8 8413 8462 8511 8560 8609 8657 8706 8755 8804 8853 49 9 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 890 949390949439 9494881949536 949585 949634 949683 949731 949780 949829 49 1 9878 9926 9975 950C24 950073 950121 950170 19 9150267 950316 49 2950365950414950462 0511 0560 0608 0657 0706 0754 0803 49 3 0851 0900 0949 0997 1046 1095 1143 1192 1240 1289 49 4 1338 1386 1435 1483 1532 1580 1629 1677 1726 1775 49 5 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 48 6 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 7 2792 2841 2889 2938 2986 3034 3083 3131 3180 3228 48 8 3276 3325 3373 3421 3470 3518 35tc6 3615 3663 3711 48 9 3760 3808 3856 3905 3953 4001 4049 4098 4146 4194 48 9001954243 9542 ~!954339 954387 954435 954484 954532 954580 954628 9546771 48 1 4725 4773 4821 4869 4918 4966 5014 5062 5110 5158 48 2 5207 5255 5303 5351 5399 5447 5495 5543 5592 5640 48 3 5688 5736 5784 5832 5880 5928 5976 6024 6072 6120148 4 6168 6216 6265 6313 6361 6409 6457 6505 6553 6601 48 5 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 6 7128 7176 7224 7272 7320 734;8 7416 7464 7512 7559 48 7 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 48 8 8086 8134 8181 8229 8277 8325 8373 8421 8468 8516 48 9 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 48 910 959041 959089 959137 959185 95923211959280 959328 959375 959423 959471 48 1 9518 9566 9614 9661 9709' 9757 9804 9852 9900 9947 48 2 9995 960042 96G090 960138 960185 960233 960281 960328 960376 960423 48 3 960471 0518 0566 0613 066L 0709 0756 0804 0851 0899 48 4 0946 0994 1041 1089 1136 1184 1231 1279 1326 1374 48 5 1421 1469 1516 1563 1611 1658 1706 1753 1801 1848 47 6 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 7 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 81 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 47 9' 3316 3363 3410 3457 3504 3552 3599 3646 3693 3741 47 920 963788 963835 963882 963929r963977 964024 964071 964118 964165 964212 47 1 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 47 2 4731 4778 4825 4872 4919 4966 5013 5061 5108 5155 47 3 5202 5249 5296 534, 5390 5437 5484 5531 5578 5625 47 4 5672 5719 5766 5813 5860 5907 5954 6001 6048 6095 47 5 6142 6189 6236 6283 6329 6376 6423 6470 6517 6564 47 6 6611 6658 6705 6752 6799 6845 6892 6939 6986 7033 47 7 7080 7127 7173 72201 7267 7314 7361 7408 7454 7501 47 8 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 9 8016 8062 8109 8156 8203 8249 8296 8343 8390 8436 47 930968483 968530 968576 96 968623 968670968716 968763 968810 968856 968903 47 1 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 2 9416 9463 9509 9556 9602 9649 9695 9742 9789 9835 47 3 9882 9928 99751970021 970068 970114 970161 970207 9702541970300 47 4 970347 970393 970440 0486 0533 0579 0626 0672 0719 0765 46 6 0812 0858 0904 0951 0997 1044 1090 1137 1183 1229 46 6 1276 1322 1369 1415 14:61 1508 1554 1601 1647 1693 46 7 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 46 8 2203 2249 2295 2342 2388 2434 2481 2527 2573 2619 46 9 2666 2712 2758 2804 2851 2897 2943 2989 3035 3082 46 N.I 0 1 2! 3 4 6_ 6 I 7 8 9 D 16 LOGARITHIS OF NUMBERS. IN. o. 1 2 I3 14 5 I 6 7 " I 9 |D 940 97312897321 74 973220; V 3266 97'3313 473359 97340 97349751 973497973543 46 1 3590 3636 3682 3728 3774! 3820 3866 3913 3959 4005 46 2 4051 4 097 4143 4189 4235 4281 4327 4374 4420 4466 46 3 4512 4558 4404 4650 4696 4742 4788 4834 4880 4926 46 4 4972 5018 5064 6110 5156 5202 6248 6294 5340 5386 46 5 6432 5478 5524 6570 5616 6662 6707 5753 5799 6845 46 6 5891 6937 6983 6029 6075 6121 6167 6212 6258 6304 46 7 6350 6396 6442 6488 6533 6579 6625 6671 6717 6763 46 8 6808 6854 6900 6946 6992 7037 7083 7129 7175 7220 46 9 7266 7312 7358 7403 7449 749 95 7541 7586 7632 76781 46 9 0 977724 977769 977815977861 977906 977952 977998 978043 9780 89 97 b35 46 1 8181 8226 8272 8317 8363 8409 8454 8500 8546 8 591 46 2 8637 8683 8728 8774 8819 8865 8911 8956 9002 9047 46 3 9093 9138 9184 9230 9275 9321 9366 9412 9457 9503 46 4 954 9594 9639 9685 9730 9776 9821 9867 9912 9958 46 6 980003 9800 49 9800 94 980140 980185 980231 980276 980322 980367 980412 45 6 0458 0503 0549 0594 0640 0685 0730 0776 0821 0867 45 7 0912 0957 1003 1048 1093 1139 1184 1229 1275 1320 45 8 1366 1411 1456 1501 1547 1592 1637 1683 1728 17T3 45 9 1819 1864 1909 1954 2000 2045[ 2 090 2135 2181 2226 45 960 982271 982316 823629 82407 982452 98 24 94 982543 9 82588 98 2633 982678 45 1 2723 2769 2814 2859 2904 2949 2994 3040 3 085 3130 45 2 3175 3220 3265 3310 3356 3401 3446 3491 3536 3581 45 3 3626 3671 3716 3762 3807 3852 3897 3942 3987 4032 45 4 4077 4122 41 67 4212 4257 4302 4347 4392- 4437 4482 45 5 4527 4572 4617 4662 4707 4752 4797 4842 4,887 4932 45 6 4977 5 022 56067 56112 5157 6202 5247 292 5337 5382 45 7 546 56471 56516 5561 5606 56651 696 6741 5786 56830 45 8 5875 56920 56965 6010 6055 6100 6144 6189 6234 6279 45 9 6324 6369 6413 6458 6503 65481 6593 6637 6682 6727 45 970 986772 986817 986861 986906 986951 986996 987040 987085 987130 987175 45 1 7219 7264 7309 7353 7398 '1443 7488 7532 7577 7622 45 2 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 3 8113 8157 8202 8247 8291 8336 8381 8425 8470 8514 45 4 8559 8604 8648 8693 8737 8782 8826 8871 8916 8960 45 6 9005 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 6 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 44 7 9895 9939 9983 990028 990072990117 990161 990206 990250 990294 44 8 990339 990383 990428 0472 0516 0561 0605 0650 o 0694 0738 44 9 0783 08271 0871 0916 0960 1004 1049 1093 1137 1182 44 980 991226 991270 991315 991359 991403 991448 991492 991636 991580 991625! 44 1 16(;9 1731 1758 1802 1846 1890 1935 1979 2023 20671 44 22111 2156 2200 2244 2288 2333 2377 2421 2465 2509 44 3 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 44 4 2995 3039 3083 3127 3172 3216 3260 3304 3348 3392 44 6 3436 3480 3524 3568 3613 3657 3701 3745 3789 38331 44 6 3877 3921 3965 4009 4053 4097 4141 4185 4229 4273 44 7 4317 4361 4405 4449 4493 4637 4581 4625 4669 4713 44 8 4757 4801! 4845 4889 4933 4977 56021 6065 6108 5615244 9 6196 62401 6284 56328 56372 56416 56460 6504 5547 56591 44 9901995635 9956791996723 '995767 99681 995854 996898 995942'995986 996030 44 1 6074 6117 6161 6205 6249 6293 6337 6380 6424 6468 44 2 6612 6555 6699 6643 6687 6731 6774 6818 6862 6906 44 3 6949 6993 7037 7080 7124 7168 7212 7256 7299 7343 44 4 7386 7430 7474 7517 7661 7605 7648 76921 7736 7779 44 65 7823 7867 7910 7954 7998 8041 8085 8129 8172 8216 44 6 8259 8303 8347 8390 8434 8477 8521 8564 8608 8652 44 7 8695 8739 8782 8826 8869 8913 8956 9000 9043 9087 44 8 9131 9174 9218 9261 9306 9348 9392 9435 9479 9522 44 91 9565 9609 962 9696 9739 9783 9826 9870 9913 9957 43 " I ~____ 12 1 3 14 U'5 1' 6 8 9 | PROFESSORS AND INSTRUCTORS ARE BE8PECTFULLY INVITED TO EXAMIN] (rt#t fI^f's "nim 'olt twatfi f;.y ENXTIRELY NtEW CO'URSE, Clear, Comprehensive, Practical, Complete, For rSU urde* *f Leuarers. PO2 OOMM00VOT SOBCOOLS. NEW PRIMARY ARITHMETIC, NOW BELEBNTARYT ARITEBRT04 OR, NEW INTELLECTUAL ARITHMETIC, NEW PRACTICAL ARITHMETIC. For High Schools, Academies, & Colleges 15W fl:AGTWAL&ZX'I~Sm- T2 OR, NATIONAL ARITHMETIC. NEEW ELEM:ENTARY ALGEBR'A, Or, Nrew Higher Algebra. GXOYXTR T A.D TRIGOXONT01rT W IACS BOOK COMPLETJE ITBISWP. 1 GREENLIEAF'S Ne Pr*acti(!-C-al Aih et; AN ENTIRELY NEW WORK, FOR I6CHOOLS AND SEMINARIES 836 p.Prkc. 94 eenft. Espeolal attention to invited to Its many DISTINGUISHIG CRA3ACTERISTIOS. 1.- The prominence given to the ENUNCIATION of Gzxzsa" P3miNCULzE, (pp. 13, 18, 26, 35, 44, etc.) 2. The -simple and dlear treatment of NoTATio0. and NuxzlR1'Iom. 8. The early introduction of the DECIMAL POINTr, (pp. 15, 16,) ~nd the ex'pkanation of the values expressed by figures at thi right of the point, (p. 52.) 4. ExCHIANGE Of COMMODITIES, BILLS AND INVOICES8, Accoux's, LEDGER COLUMNS, (pp. 72, 78,) and various useful applications. (pp. 142, 178, 186, 187, 202, 263, 212, etc.) 5. The discusion of FACTORING before Multiplication by F'actors, (p. 83,) or Division by Factors, (p. 84.) 6. The simplification of DIVIION of Comnmon Fractions. (p. 117,) and of DIVISION of Decimals, (p. 139.) 7. The introduction Of COMPOUND DENOMINATE N1uxBzi (p. 162,) afler Practions, -so as to avoid at least eight specal 'isae in Reuduion, A4ddi~ion, and Subtraction of Denominate Numnbers 8. The MmzBc SYSTEM Of WEiGHTS and MEASURES is 115 Ite propr place, in the body of the work, (p. 155,) 'and referemce enode to the same in subsequent parts of the book, (pp. '164, 166, 170, 187, 247, 283, etc.,) as urged by PROF. NEWTrON, of Yake College, PRESI81DENT HILL, of Hartzard University, PROF. 11ZiMi, of Smithsonian Institute, PROF. ROGERS, of University qj Pesnusylvania, CIANCELLOrt CHAUVE.NET, Of Washington Uniwrfp, I~.Louis, andW others.' 4 t I 513 G81