5 A. S. Hallid IE, SAN FRANCISCO. tiPfffiNN ■■I / ROBINSON'S MATHEMATICAL SERIES ELEMENTS GEOMETRY, PLANE AND SPHERICAL TRIGONOMETRY; WITH NUMEROUS PRACTICAL PROBLEMS BY HOEATIO K BOBINSON, LL. D., AUTHOR OP A FUIX COURSE OF MATHEMATICS. OF THE OF NEW YORK: IVISON, PHINNEY & CO., 48 AND 50 WALKER ST. CHICAGO : S. C. GRIGGS & CO., 39 AND 41 LAKE ST. BOSTON : BROWN A TAGGARD. PHILADELPHIA : SOWER, BARNES k CO., AND J. B. LIPPINCOTT & CO. CINCINNATI : MOORE, WILSTACH, KEYS & CO. SAVANNAH : J. M. COOPER & CoT ST. LOUIS : KEITH 4; WOODS. NEW ORLEANS : E. R. STEVENS & CO. DETROIT: RAYMOND & LAPHAM. BALTIMORE : CUSHING & BAILEY. 1860 Robinson's Complete Mathematical Course. •*--*- ROBINSON'S SYSTEM OP MATHEMATICS, Recently revised and enlarged, is now the most extensive, complete, practical and scientific Mathematical Series published in this country. »• i — i » 1. Robinson's Progressive Primary Arithmetic. Illustrated. $0 15 2. Robinson's Progressive Intellectual Arithmetic, for ad- • vanced Classes, with an Original and Comprehensive System of Analysis 25 3. Robinson's Progressive Practical Arithmetic; a complete work for Common Schools and Academies. . . . 56 4. Key to Robinson's Progressive Practical Arithmetic. . 50 5. Robinson's Progressive Higher Arithmetic. . . 75 6. Key to Robinson's Progressive Higher Arithmetic. . 75 7. Robinson's New Elementary Algebra: a clear and simple Treatise for Beginners 75 8. Key to Robinson's New Elementary Algebra. . . 75 9. Robinson's University Algebra : a full and complete Trea- tise for Academies and Colleges 1 25 10. Key to Robinson's University Algebra ; separate. . .100 11. Robinson's Geometry and Trigonometry ; with applications to practical examples . 1 50 12. Robinson's Surveying and Navigation; combining theory with practice 1 50 13. Robinson's Analytical Geometry and Conic Sections ; made clear and comprehensive to common minds 1 50 14. Robinson's Differential and Integral Calculus ; a full and complete Treatise 1 50 15. Robinson's Elementary Astronomy; designed to teach the first principles of this Science 75 16. Robinson's University Astronomy; for advanced classes in Academies and Colleges 1 75 17. Robinson's Concise Mathematical Operations : a book of re- ference for the Teacher, embracing the gems of Mathematical Science 2 25 18. Key to Robinson's University Algebra, Geometry, Survey- ing, and Calculus ; in 1 vol 1 50 Entered, according to Act of Congress, in the year 1860, by H. N. ROBINSON, LL.D., in the Clerk's Office of the District Court of the United States for the Northern District of New York. JOHN PAGAN, 8TERE0TTPEE, PHILADELPHIA. PREFACE. In the preparation of this work, the Author's previous treatise, "Elements of Geometry, Plane and Spherical Trigonometry, and Conic Sections," has formed the ground-work of construction. But in adapting the work to the present advanced state of Mathematical edu- cation in our best Institutions, it was found necessary to so alter the plan, and the arrangement of subjects, as to make this essentially a new work. The demonstrations of propositions have undergone radical changes, many new propositions have been introduced, and the number of Practical Problems greatly increased, so that the work is now be- lieved to be as full and complete as could be desired in an elementary treatise. In view of the fact that the Seventh Book is so much larger than the others, it may be asked why it is not divided into two ? We answer, that classifications and divisions are based upon differences, and that the differences seized upon for this purpose must be determined by the nature of the properties and relations we wish to investigate. There is such a close resemblance between the geometrical properties of the polyedrons and the round bodies, and the demonstrations relating to the former require such slight modifications to become applicable to the latter, that there seems no sufficient reason for separating into two Books that part of Geometry which treats of them. The subject of Spherical Geometry, which has been much extended in the present edition, is placed as before, as an introduction to Spheri- cal Trigonometry. The propriety of this arrangement may be ques- tioned by some ; but it is believed that much of the difficulty which the student meets in mastering the propositions of Spherical Trigonometry, arises from the fact that he is not sufficiently familiar witl^the geome- try of the surface of the sphere ; and that, by having the propositions of Spherical Geometry fresh in his mind when he begins the study of Spherical Trigonometry, he will be as little embarrassed with it as with Plane Trigonometry. (Hi) 1 n^a^i iv PREFACE. Both author and teacher must yield to the demands of the age, and by a judicious combination of the abstract and the concrete, the theo- retical and the practical, make the student feel that what he learns with perhaps painful effort at first, may be made available in import- ant applications. In teaching Geometry and Trigonometry, questions should be asked, extra problems given, and original demonstrations required when the proper occasions arise ; but care should be taken that the pupil's powers are not over-tasked. By helping him through his difficulties in such a way that he shall be scarcely conscious of having received assistance, he will be encouraged to make new and greater efforts, and will finally acquire a fondness for a study that may have been highly repugnant to him in the beginning. A demonstration that is easily followed and comprehended by one, may be obscure and difficult to another ; hence the advantage that will sometimes be gained by giving two or more demonstrations of the same proposition. When the student perceives that the same results may frequently be reached by processes entirely different, he will be stimu- lated to independent exertion, and in no respect can the teacher better exhibit his tact than in directing and encouraging such efforts. Instances will be found throughout the work in which the more im- portant propositions are twice and three times demonstrated ; and as the methods of demonstration are in each case quite different, it is believed that extra space has not been thus occupied unprofitably. Practical rules with applications will be found throughout the work, and in addition to these, there are in both the Geometry and the Trigo- nometry, full collections of carefully selected Practical Problems. These are given to exercise the powers and test the proficiency of the pupil, and when he has mastered the most or all of them, it is not likely that he will rest satisfied with present acquisition, but conscious of augmented strength and certain of reward, he will enter new fields of investigation. The Author has been aided, in the preparation of the present work, by J. F. Quinby, A. M., of the University of Rochester, N. Y., late Professor of Mathematics in the United States Military Academy at West Point, and J. H. French, LL. D., of Syracuse, New York. The thorough Scholarship, and long and successful experience of these gen- tlemen in $ie class-room, rendered them eminently qualified for the task ; and to them the public are indebted for much that is valuable, both in the matter and arrangement of this treatise. October, 1860. CONTENTS. PLANE GEOMETRY. DEFINITIONS. Geometrical Magnitudes Page 9 Plane Angles 10 Plane Figures of Three Sides 12 Plane Figures of Four Sides 13 The Circle 14 Units of Measure 15 Explanation of Terms .' 16 Postulates 16 Axioms 17 Abbreviations , 17 BOOK I. Of Straight Lines, Angles, and Polygons 19 BOOK II. Proportion, and its Application to Geometrical Investigations. ... 59 BOOK III. Of the Circle, and the Investigation of Theorems dependent on its Properties 88 1* (v) vi CONTENTS. BOOK IV. Problems in the Construction of Figures in Plane Geometry Ill BOOK Y. On the Proportionalities and Measurement of Polygons and Circles. 130 Practical Problems 142 BOOK VI. On the Intersections of Planes, the Relative Positions of Planes, ■ and of Planes and Lines 152 BOOK VII, Solid Geometry 172 Practical Problems 229 BOOK VIII. Practical Geometry. — Application of Algebra to Geometry, and also Propositions for Original Investigation 231 Miscellaneous Propositions in Plane Geometry 238 TRIGONOMETRY. PART I. \ PLANE TRIGONOMETRY. SECTION I. Elementary Principles 244 Definitions 245 Propositions 248 Equations for the Sines of the Angles 260 Natural Sines, Cosines, etc 265 Trigonometrical Lines for Arcs exceeding 90° 270 CONTENTS. v ii SECTION II. Plane Trigonometry, Practically Applied 272 Logarithms ; 278 GENERAL APPLICATIONS WITH THE USE OF LOGARITHMS. I. Eight- Angled Trigonometry 288 II. Oblique- Angled Trigonometry 291 Practical Problems 295 SECTION III. Application of Trigonometry to Measuring Heights and Distances. 298 Practical Problems 305 PART II. SPHERICAL GEOMETRY AND TRIGONOMETRY. SECTION I. Spherical Geometry 310 SECTION II. Right- Angled Spherical Trigonometry 330 Napier's Circular Parts 335 • SECTION III. Oblique- Angled Spherical Trigonometry 337 Napier's Analogies 343 viii CONTENTS. SECTION IV. Spherical Trigonometry Applied. — Solution of Right- Angled Spherical Triangles 353 Practical Problems 35G Solution of Quadrantal Triangles 358 Practical Problems 361 Solution of Oblique-Angled Spherical Triangles 362 Practical Problems 367 SECTION V. Spherical Trigonometry applied to Astronomy 370 Application of Oblique- Angled Spherical Triangles 373 Spherical Trigonometry applied to Geography 377 Table of Mean Time at Greenwich 379 SECTION VI. Regular Polyedrons 380 ^^ or the ^ GEOMETEY. DEFINITIONS. 1. Geometry is the science which treats of position, and of the forms, measurements, mutual relations, and pro- perties of limited portions of space. Space extends without limit in all directions, and contains all bodies. 2. A Point is mere position, and has no magnitude. 3. Extension is a term employed to denote that pro- perty of bodies by virtue of which they occupy definite portions of space. The dimensions of extension are length, breadth, and thickness. 4. A Line is that which has extension in length only. The extremities of a line are points. 5. A Eight or Straight Line is one all of whose parts lie in the same direction. 6. A Curved Line is one whose consecutive parts, how- ever small, do not lie in the same direction. 7. A Broken or Crooked Line is composed of several straight lines, joined one to another successively, and extending in different directions. When the word line is used, a straight line is to be understood, unless otherwise^ expressed. 8. A Surface or Superficies is that which has extension in length and breadth only. 9. A Plane Surface, or a Plane, is a surface such that m 10 GEOMETRY. if any two of its points be joined by a straight line, every point of this line will lie in the surface. 10. A Curved Surface is one which is neither a plane, nor composed of plane surfaces. U. A Plane Angle, or simply an Angle, is the difference in the direction of two lines proceeding from the same point. The other angles treated of in geometry will be named and defined in their proper connections. 12. A Volume, Solid, or Body, is that which has exten- sion in length, breadth, and thickness. These terms are used in a sense purely abstract, to denote mere space — whether occupied by matter or not, being a question with which geometry is not concerned. Lines, Surfaces, Angles, and Volumes constitute the different kinds of quantity called geometrical magnitudes. 13. Parallel Lines are lines which have the same direction. Hence parallel lines can never meet, however far they may be produced; for two lines taking the same direction cannot approach or recede from each other. Two parallel lines cannot be drawn from the same point; for if parallel, they must coincide and form one line. PLANE ANGLES. To make an angle apparent, the two lines must meet in a point, as AB and -4(7, which meet in the point A. Angles are measured by degrees. 14. A Degree is one of the three hundred and sixty equal parts of the space about a point in a plane. If, in the above figure, we suppose A C to coincide with AB, there will be but one line, and no angle; but if AB retain its posi- tion, arid A C begin to revolve about the point A, an angle will be formed, an Less than, " " < Thus : B is greater than J., is written . B^>A B is less than 4, " " . B By Th. 5, [_BGH + ]_ / GrHD — two right angles. Al- so, by Th. 1, [_AGH + [_BGH = two right angles. From these equals take away the common angle BGH, and L CHD will be left, equal to \_AGH, (Ax. 3). In like manner, we can prove that the angle CHG is equal to the angle HGB. Hence the theorem ; if a line intersects two parallel lines, the alternate interior angles are equal. Cor. 1. Since |__ A GH = [__ FGB, and [__AGH=l_GHD; Therefore, [_ FGB = \__ GHD (Ax. 1). Also, [__AGF + l_AGH=2H. |___, (Th. 1), and L CHG + [_AGH = 2 K. [_, (Th. 5); Therefore, [_AGF+]_AGH = ]_CHG + [__AGH, (Ax.l); and L.AGF = [_ CHG, (Ax. 3). That is, the exterior angle is equal to the interior opposite angle on the same side of the intersecting line. Cor. 2. Since [__AGH = [__FGB, and \_AGH=[_CHE; Therefore, l__FGB = L CHE. In the same manner it may be shown that [__AGF = [_EHD. Hence, the alternate exterior angles are equal. 24 GEOMETKY. THEOREM VII. If a line intersects two other lines, making the sum of the two interior angles on the same side of the intersecting line equal to two right angles, the two straight lines are parallel. Let the line FF intersect the lines AB and CB, making the two angles BaR + GHD A = to two right angles ; then we are to demonstrate that AB and OB are parallel. C /H ~~b As EF is a right line and B ' BCr meets it, the two angles FGrB and BGrlT are together equal to two right angles, (Th. 1). But by hypothesis, the angles, j?##and aHB, are together equal to two right angles. From these two equals take away the common angle BGrH, and the re- maining angles FGrB and CrffB must be equal, (Ax. 3). 'Now, because GfB and SB make equal angles with the same line EF, they must extend in the same direction ; and lines having the same direction are parallel, (Def. 13). Hence the theorem ; if a line intersects two other lines, making the sum of the two interior angles on the same side of the in- tersecting line equal to two right angles, the two lines must be parallel. Cor. 1, Ifa line intersects two other lines, making the alternate interior angles equal, the two lines intersected must be parallel. Suppose the L -4## = L ff^Z>- Adding \__HGB to each, we have [__AGH + L HGLB = L aEI) + L.B&B- but the first member of this equation, that is, [_AGR-\- |__ HGrB, is equal to two right angles ; hence the second member is also equal to the same ; and by the theorem, the lines AB and CD are parallel. Cor. 2. If a line intersects two other lines, making the BOOK I. 25 opposite exterior and interior angles equal, the two lines intersected must be parallel. Suppose the [__ FGB = |_ &#B. Adding the [__EaB to each, we have L fgb + \_hgb = i ® HI > + EaB - But the first member of this equation is equal to two right angles ; hence the second member is also equal to two right angles ; and by the theorem, the lines AB and, CD are parallel. Cor. 3. If a line intersects two other lines, making the alternate exterior angles equal, the lines must be parallel, Suppose [_BGF=\_CHE, and [_AGF = [_DHE, ByTh.4, \_BGF=[_AGH,^di[_OHE^l_DHG. And since [_BGF = [_OHE, [_AGH=\_DHG. That is, the alternate interior angles are equal; and hence (by Cor. 1) the two lines are parallel. THEOREM VIII. If two angles have their sides parallel, the two angles will be either equal or supplementary. Let A be parallel to BD, and AH parallel to BF or to BG. Then we are to prove that the angle DBF is equal to the angle CAH, and that the angle DBG is supplementary to the angle A. The angle OAH is formed by the differ- ence in the direction of A C and AH; and the angle DBF is formed by the differ- ence in the direction of BD and BF. But AC and AH have the same direc- tions as BD and BF, because they are respectively paral- lel. Therefore, by Def. 11, L CAH= [_DBF. But the line BG has the same direction as BF, and the angle DBG is supplementary to DBF. Hence the theorem; angles whose sides are parallel, form either equal or supple- mentary angles. 3 26 GEOMETRY. THEOREM IX. The opposite angles of any parallelogram are equal. Let AEBGi be a parallel- ogram. Then we are to \ prove that the angle GBE G _\B_ is equal to its opposite angle A. * Produce EB to D, and GB to F; then, since BJ) is par- allel to A G, and BF to AE, the angle DBF is equal to the angle A, (Th. 8). But the angles GBE and DBF, being vertical, are equal, (Th. 4). Therefore, the opposite angles QBE and A, of the parallelogram AEBG, are equal. In like manner, we can prove the angle E equal to the angle G. Hence the theorem ; the opposite angles of any parallelogram are equal, THEOREM X. The sum of the angles of any parallelogram is equal to four right angles. Let ABCB be a parallelo- gram. We are to prove that the sum of the angles A, B, and B, is equal to four right angles, or to 360°. Because AB and BC are parallel lines, and AB inter- sects them, the two interior angles A and B are together equal to two right angles, (Th. 5). And because CD in- tersects the same parallels, the two interior angles C and D are also together equal to two right angles. By addi- tion, we have the sum of the four interior angles of the parallelogram ABCB, equal to four right angles. Hence the theorem ; the sum of the angles of any parallelogram is equal to four right angles. BOOK I. THEOREM XI, The sum of the three angles of any triangle is equal to two right angles. Let A B be a triangle, and through its vertex draw a line parallel to the base AB, and produce the sides AC and BO. Then the angles A and a, being exterior and in- terior opposite angles on the same side of two parallels, are equal, (Th. 6, Cor. 1). For like reasons, \__B = [__£. And the angles and c, being vertical angles, are also equal, (Th. 4). Therefore, the angles A, B, C are equal to the angles a, b, c respect- ively. But the angles around the point (7, on the upper side of the parallel CD, are equal to two right angles, (by Th. 1). Hence the theorem; the sum of the three angles, etc. Second Demonstration. hetAEBG be a parallelogram. Draw the diagonal GE; then the parallelogram is divided into two triangles, and the opposite angles E R €r and E are mutually divided by the diagonal GrE. Because GrB and AE are parallel, the alternate interior angles BGrE and GrEA are equal, (Th. 6). Designate each of these by b. In like manner, because EB and A G- are parallel, the alternate interior angles, BEG- and EGA, are equal. Designate each of these by a. Now we are to prove that the three angles B> b, and a, and also that the three angles A, a, and b, are equal to two right angles. 28 GEOMETRY. Because A and B are opposite angles of a parallelo- gram, they are equal, (Th. 9), and [_A + [_B = 2 \__A. And all the interior angles of the parallelogram are equal to four right angles, (Th. 10). Therefore, 2 A + 2a -f 26 = 4 right angles. Dividing by 2, and A 4- a + b = 2 " That is, all the angles of the triangle AGE are together equal to two right angles. Hence the theorem ; the sum of the three angles, etc. Scholium. — Any triangle, as AGE, may be conceived to be part of a parallelogram. For, let A GE be drawn independently of the paral- lelogram ; then draw EB from the point E parallel to A G, and through the point G draw GB parallel to AE, and a parallelogram will be formed embracing the triangle ; and thus the sum of the three angles of any triangle is proved equal to two right angles. This truth is so fundamental, important, and practical, as to require special attention ; we therefore give a Third Demonstration. Let ABO be a triangle. Then we are to show that the angles A, 0, and ABC, are together equal to two right angles. Let AB be produced to D, and from B draw BE parallel to AC. Then, EBB and OAB being exterior and interior op- posite angles on the same side of the line AB, are equal, (Th. 6, Cor. 1). Also, QBE and ACB, being alternate angles, are equal, (Th. 6). By addition, observing that [__ QBE, added to [_EBB, must make [_ CBD, we have [_CBB = l_A + l_a (1.) To each of these equals add the angle CBA, and we shall have [_CBA + [_CBD= L_^ + l_C+l_CBA. But (by Th. 1), the sum of the first two is equal to two BOOK I. 29 right angles; therefore, the three angles, A, 0, and OB A, \re together equal to two right angles. Hence the theorem ; the sum of the three angles, etc. THEOREM XII. If any side of a triangle is 'produced, the exterior angle is equal to the sum of the two interior opposite angles. Let ABO be a triangle. Pro- duce AB to D; and we are to prove that the angle OBI) is equal to the sum of the two angles A and O. We establish this theorem by a course of reasoning in all respects the same as that by which we obtained Eq. (1.), third demonstration, (Th. 11). Oor. 1. Since the exterior angle of any triangle is equal to the sum of the two interior opposite angles, therefore it is greater than either one of them. Oor. 2. If two angles in one triangle be equal to two angles in another triangle, the third angles will also be equal, each to each, (Ax. 3) ; that is, the two triangles will be mutually equiangular. Oor. 3. If one angle in a triangle be equal to one angle in another, the sum of the remaining angles in the one will also be equal to the sum of the remaining angles in the other, (Ax. 3). Oor. 4. If one angle of a triangle be a right angle, the sum of the other two will be equal to a right angle, and each of them singly will be acute, or less than a right angle. Oor. 5. The two smaller angles of every triangle are acute, or each is less than a right angle. Oor. 6. All the angles of a triangle may be acute, but no triangle can have more than one right or one obtuse angle. 30 GEOMETRY. THEOREM XIII. In any quadrilateral, the sum of the four interior angles is equal to four right angles. Let ABCD be a quadrilateral; then we are to prove that the sum of the four in- terior angles, that is A -f B + + D, is equal to four right angles. Draw the diagonal AC, dividing the quadrilateral into two triangles, ABO, ABO. Now, since the sum of the three angles of each of these triangles is equal to two right angles, (Th. 11), it follows that the sum of all the angles of both triangles which make up the four angles of the quadrilateral, must be equal to four right angles, (Ax. 2). Hence the theorem ; in any quadrilateral, etc. Cor. 1. Hence, if three of the angles of a quadrilateral are right angles, the fourth will also be a right angle. Oor. 2. If the sum of two of the four angles be equal to two right angles, the sum of the remaining two will also be equal to two right angles. And, if the sum of either two of the angles be less than two right angles, the sum of the other two angles will be greater than two right angles. THEOREM XIV. In any polygon, the sum of all the interior angles is equal to twice as many right angles, less four, as the figure has sides. Let ABODE be any polygon ; we are to prove that the sum of all its interior angles, A + B -f -f- D + E, is equal to twice as many right angles, less four, as the figure has sides. From any point, p, within the figure, draw lines pA, pB, pO, etc., to all the angles, BOOK I. 31 thus dividing the polygon into as many triangles as it has sides. Now, the sum of the three angles of each of these triangles is equal to two right angles, (Th. 11) ; and the sum of the angles of all the triangles must be equal to twice as many right angles as the figure has sides. But the sum of these angles contains the sum of four right angles about the point p ; taking these away, and the remainder is the sum of the interior angles of the figure. Therefore, the sum must be equal to twice as many right angles, less four, as the figure has sides. Hence the theorem ; in any polygon, etc. From this Theorem is derived the rule for finding the sum of the interior angles of any right-lined figure : Subtract 2 from the number of sides, and multiply the re- mainder by 2 ; the product will be the number of right angles. Thus, if the number of sides be represented by S, the number of right angles will be represented by (2S — 4). The Theorem is not varied in case of a re-entrant angle, as rep- ^1 resented at d, in the figure ABC- ^^ DEF. ^-------JA Draw lines from the angle d \ /' \ / to the several opposite angles, \ / \ / making as many triangles as the figure has sides, less two, and the sum of the three angles of each triangle equals two right angles. THEOREM XV. From any point without a straight line, but one perpendic- ular can be drawn to that line. From the point A let us suppose A it possible that two perpendiculars, A B and A C, can be drawn. Now, be- cause AB is a supposed perpendicu- lar, the angle ABC is a right angle ; . and because A C is a supposed per- B c 32 GEOMETKY. pendicular, the angle A CB is also a right angle; and if two angles of the triangle ABC are together equal to two right angles, the third angle, BAG, must be infinitely small, or zero ; but this is impossible, for it requires the sum of the three angles of a triangle to make two right angles, (Th. 11). Therefore, the lines AB and A must be identical, or but one perpendicular. Hence the theorem ; from any point without a straight line, etc. Cor. At a given point in a straight line but one per- pendicular can be erected to that line ; for, if there could be two perpendiculars, we should- have unequal right angles, which is impossible. THEOREM XYI. Two triangles which have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each, are equal in all respects. In the two A's, ABC and BEF, on the supposition that AB = BE, AC=BF, and [_A = [_B, we are to prove that BC must = EF, the [__B = L.2J, and the [_C = l_F. Conceive the a ABC cut out of the paper, taken up, and placed on the A BEF in such a manner that the point A shall fall on the point B, and the line AB on the line BE; then the point B will fall on the point E, because the lines are equal. Now, as the [__A = [__B, the line A C must take the same direction as BF, and fall on BF; and as AC = BF, the point C will fall on F. B being on E and C on F, BC must be exactly on EF. (otherwise, two straight lines would enclose a space, Ax. 13), and BC '= EF, and the two magnitudes exactly fill the same space. Therefore, BC = EF, [__B = [_E, L (7= [_F, and the two A's are equal, (Ax. 9). Hence the theorem ; two triangles which have two sides, etc. BOOK I. 33 THEOREM XVII. When two triangles have a side and two adjacent angles in the one, equal to a side and two adjacent angles in the other, each to each, the two triangles are equal in all respects. In two a's, as ABO and DEF, on the supposition that BO = EF,[_B=[_E, and [v — [__ F, we are to prove that AB m BE, AG = DF, andL^- - L-^- Conceive the A ABO taken np and placed on the A BEF, so that the side BO shall exactly coincide with its equal side EF; now, because the angle B is equal to the angle E, the line BA will take the direction of ED, and will fall exactly upon it ; and because the angle is equal to the angle F, the line OA will take the direction of FD, and fall exactly upon it ; and the two lines BA and OA, exactly coinciding with the two lines ED and FD, the point A will fall on D, and the two magnitudes will exactly fill the same space ; therefore, by Ax. 10, they are equal, and AB = DE, AO=DF, and the \_A = [_D. Hence the theorem ; when two triangles have a side and two adjacent angles in the one, equal to, etc. THEOREM XVIII. If two sides of a triangle are equal, the angles opposite these sides are also equal. Let ABO be a triangle; and on the supposition that AO = BO, we are to prove that the [__ -4= the [_B. Conceive the angle divided into two equal angles by the line OD; then we have two A's, ADO and BDO, which have the two sides, AO and OD of the one, equal to the two sides, OB and OD of the other ; and to 34 GEOMETRY. the included angle ACD, of the one, equal to the in- cluded angle BCD of the other: therefore, (Th. 16), AD m BD, and the angle A, opposite to CD of the one tri- angle, is equal to the angle B, opposite to CD of the other triangle ; that is, [_A= [_B. Hence the theorem ; if two sides of a triangle are equal, the angles, etc. Cor. 1. Conversely : if two angles of a triangle are equal, the sides opposite to them are equal, and the triangle is isosceles. For, if A C is not equal to BC, suppose BC to be the greater, and make BE = AE; then will A AEB be isos- celes, and [_EAB = \_EBA ; hence [__EAB = [_ CAB, or a part is equal to the whole, which is absurd ; therefore, CB cannot be greater than AC, that is, neither of the sides AC, BC, can be greater than the other, and conse- quently they are equal. Cor. 2. As the two triangles, ACD and BCD, are in all respects equal, the line which bisects the angle included between the equal sides of an isosceles A also bisects the base, and is perpendicular to the base. Scholium 1. — If in the perpendicular DC, any other point than C be taken, and lines be drawn to the extremities A and B, such lines will be equal, as is evident from Th. 16 ; hence, we may announce this truth : Any point in a perpendicular drawn from the middle of a line, is at equal distances from the two extremities of the line. Scholium 2. — Since two points determine the position of a line, it follows, that the line which connects two points equally distant from the extremities of a given line, is perpendicular to this line at its middle point. THEOREM XIX. The greater side of every triangle has the greater angle opposite to it. Let ABC be a A ; and on the supposition that A C is greater than AB, we are to prove that the angle ABCia BOOK I. 35 greater than the [_ 0. From AO, the greater of the two sides, take AB, equal ^ to the less side AB, and draw BB, thus /\ making two triangles of the original tri- / \ angle. As AB = AID, the [_ABB = / \ the [_ ABB, (Th.18). B \T \ But the L ABB is the exterior angle \\ of the A BBC, and is therefore greater C than O, (Th. 12); that is, the [__ABB is greater than the angle 0. Much more, then, is the angle AB greater than the angle 0. Hence the theorem ; the greater side of every triangle, etc. Cor. Conversely: the greater angle of any triangle has the greater side opposite to it. In the triangle ABO, let the angle B be greater than the angle A ; then is the side A greater than the side BO. Tor, if BO — AO, the angle A must be equal to the angle B, (Th. 18), which is contrary to the hypothesis ; and if BC^>AC, the angle A must be greater than the angle B, by what is above proved, which is also contrary to the hypothesis ; hence BO can be neither equal to, nor greater, than AO; it is therefore less than AO. THEOREM XX. The difference between any two sides of a triangle is less than the third side. A Let JLJ? (7 be a A, in which JL<7 is greater \ than AB; then we are to prove that AO I \ — AB is less than BO. I \ On AO, the greater of the two sides, / ,M) lay off AB equal to AB. ^>sJ \ Now, as a straight line is the shortest ^^^ distance between two points, we have c AB + BO>AO. (1) GEOMETRY. From these unequals subtract the equals AB — AD, and we have BO > AC— AB. (Ax. 5). Hence the theorem ; the difference between any two sides of a triangle, etc. THEOREM XXI. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the two triangles are equil, and the equal angles are opposite the equal sides. In two triangles, as ABC and ABB, on the supposition that the side AB of the one = the side AB of the other, AC— AD, and BC=BD, we are to demonstrate that \_ACB=l_ADB, \_BAC = [_BAD, and \__ABC= \__ABD. Conceive the two triangles to be joined together by their long- est equal sides, and draw the line CD. Then, in the triangle A CD, because A C is equal to AD, the angle ACD is equal to the angle ADC, (Th. 18). In like, manner, in the triangle BCD, because BC is equal to BD, the angle BCD is equal to the angle BDC. Now, the angle ACD being equal to the angle ADC, and the angle BCD to the angle BDC, [_ACD + [__BCD= [_ ADC +[_BDC, (Ax. 2) ; that is, the whole angle A CB is equal to the whole angle ADB. Since the two sides AC and CB are equal to the two sides AD and DB, each to each, and their included angles A CB, ADB, are also equal, the two triangles ABC, ABD, are equal, (Th. 16), and have their other angles equal ; that is, \_BAC= \_BAD, and [_ABC= [_ABD. Hence the theorem ; if two triangles have the three side$ of the one, etc. BOOK I. THEOREM XXII. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles un- equal, the third sides will be unequal, and the greater third side will belong to the triangle which has the greater included angle. In the two A's, ABC and A CD, let AB and AC of the one A he equal to AD and A C of the other A, and the angle BAC greater than the angle DAC; we are to prove that the side BCia greater than the side CD. Conceive the two A's joined together hy their shorter equal sides, and draw the line BD. Now, as AB = AD, ABD is an isosceles A. From the vertex A, draw a line bisecting the angle BAD. This line must be perpendic- ular to the base BD, (Th. 18, Cor. 1). Since the [_BAC is greater than the [_DAC, this line must meet BC, and will not meet CD. From the point E, where the per- pendicular meets BC, draw BD. Now BE = DE, (Th. 18, Scholium 1). Add EC to each ; then BC=DE + EC. But DE + EC is greater than DC. Therefore BC>DC. Hence the theorem ; if two triangles have two sides of one equal to two sides of the other, etc. Cor. Any point out of the perpendicular drawn from the middle point of a line, is unequally distant from the extremities of the line. GEOMETRY. THEOREM XXIII. A perpendicular is the shortest line that can be drawn from any point to a straight line ; and if other lines be drawn from the same point to the same straight line, the longer line will be at a greater distance from the perpendicular; and lines at equal distances from the perpendicular, on opposite sides, are equal. Let A be any point without the line DE ; let AB be the perpen- dicular; and AC, AD, and AE oblique lines : then, if BC is less than BB, and BC= BE, we are to show, 1st. That AB is less than AC. 2d. That AC is less than AB. 3d. That A C= AE. 1st. In the triangle ABC, as AB is perpendicular to BC, the angle ABC is a right angle; \__ C + [_BAC = another right angle, (Th. 11); and the angle BCA is less than a right angle; and, as the greater side is always opposite the greater angle, AB is less than AC; and AC may be any line not identical with AB ; therefore a per- pendicular is the shortest line that can be drawn from A to the line BE. 2d. As the two angles, ACB and ACT), are together equal to two right angles, (Th. 1), and ACB is less than a right angle, ACB must be greater than a right angle ; consequently, the [__ B is less than a right angle ; and, in the A ACB, AB is greater than AC, or AC is less than AD, (Th. 19). 3d. In the A's J.£Cand ABE, AB is common, CB=* BE, and the angles at B are right angles ; therefore, AC = AE, (Th. 16). Hence the theorem ; a perpendicular is the shortest line, etc. Cor. Conversely : if two equal oblique lines be drawn BOOK I. 39 from the same point to a given straight line, they will meet the line at equal distances from the foot of the per- pendicular drawn from that point to the given line. THEOREM XXIV. The opposite sides, and also the opposite angles of any par- allellogram, are equal. Let ABOB be a parallelogram. Then we are to show that AB — BO, AB - BO, [_A = [_0, and \_ABQ = \_ABO. Draw a diagonal, as BB ; now, be- cause AB and BO are parallel, the al- ternate angles ABB and BBO are equal, (Th. 6). For the same reason, as AB and BO are parallel, the angles ABB and BBO are equal. Now, in the two triangles ABB and BOB, the side BB is common, the [_ABB = [__BBO (1) and \__BBO = \_ABB (2) Therefore, the angle A — the angle O, (Th. 11), and the two A's are equal in all respects, (Th. 18) ; that is, the sides opposite the equal angles are equal ; or, AB = BO, and AB = BO. By adding equations ( 1 ) and ( 2 ), we have the angle ABO= the angle ABO, (Ax. 2). Hence the theorem ; the opposite sides, and the opposite angles, etc. Oor. 1. As the sum of all the angles of the quadrilateral is equal to four right angles, and the angle A is always equal to the opposite angle 0; therefore, if A is a right angle, is also a right angle, and the figure is a rect- angle. Oor. 2. As the angle ABO, added to the angle A, gives the same sum as the angles of the A ABB; therefore, the two adjacent angles of a parallelogram are together equal to two right angles. This corresponds to Th. 13, Cor. 2. 40 GEOMETRY. THEOREM XXV. If the opposite sides of a quadrilateral are equal, they are also parallel, and the figure is a parallelogram. Let ABDO be any quadrilateral; on the supposition that AD = BO, and AB = DO, we are to prove that AD is parallel to BO, and AB parallel to DO. Draw the diagonal BD; we now have two triangles, ABD and BOD, which have the side BD common, AD of the one = BO of the other, and AB of the one = OD of the other ; therefore the two A's are equal, (Th. 21), and the angles opposite the equal sides are equal ; that is, the angle ADB = the angle OBD ; but these are alternate angles; and, therefore, AD is parallel to BO, (Th. 7); and because the angle ABD = the angle BDO, AB is parallel to OD, and the figure is a parallelogram. Hence the theorem; if the opposite sides of a quadri- lateral, etc. Oor. This theorem, and also Th. 24, proves that the two A's which make up the parallelogram are equal; and thesame would be true if we drew the diagonal from A to 0; therefore, the diagonal of any parallelogram bisects the parallelogram. THEOREM XXVI. The lines which join the corresponding extremities of two equal and parallel strait lines, are themselves equal and parallel ; and the figure thus formed is a parallelogram. On the supposition that AB is equal and parallel to DO, we are to prove that AD is equal and parallel to BO; and that the figure is a par- allelogram. Draw the diagonal BD ; now, since BOOK I. 41 AB and DC are parallel, and BB joins them, the alter- nate angles ABB and BBC are equal ; and since the side AB = the side BO, and the side BB is common to the two A's ABB and OBB, therefore the two triangles are equal, (Th. 16) ; that is, AB = BO, the angle A = 0, and the |_ ABB = the \_BBO; also AB is parallel to BO; and the figure is a parallelogram. Hence the theorem ; the lines which join the corresponding extremities, etc. THEOREM XXVII. Parallelograms on the same base, and between the same parallels, are equivalent, or equal in respect to area or sur- face. Let ABEO and ABBF be two parallelograms on the same base AB, and between the same paral- lels AB and OB ; we are to prove that these two parallelograms are equal. — !Nbw, OB and FB are equal, be- cause they are each equal to AB, (Th. 24) ; and, if from the whole line OB we take, in succession, OB and FB, there will remain EB = OF, (Ax. 3) ; but BE = AO, and AF= BB, (Th. 24); hence we have two A's, OAF and EBB, which have the three sides of the one equal to the three sides of the other, each to each ; therefore, the two A's are equal, (Th. 21). If, from the whole figure ABBO, we take away the A OAF, the parallelogram ABBF will remain ; and if from the whole figure we take away the other A EBB, the parallelogram ABEO will remain. Therefore, (Ax. 3), the parallelogram ABBF = the parallelogram ABEO. Hence the theorem ; Parallelograms on the same base, etc. 4* 42 GEOMETRY. THEOREM XXVIII. Triangles on the same base and between the same parallels are equivalent. Let the two a's ABE and ABF , have the same base AB, and be be- parallels, AF and DCr ; then we are A to prove that they are equal in area. AB — EF=EGr\ but lines which join equal and parallel lines, are themselves equal and parallel, (Th. 26) ; therefore, if AS and BGr be drawn, the figure ABGffis •a parallelogram = to the parallelogram ABCD, (Th. 27); and if we turn the whole figure over, the two parallelo- grams, GrHEF and GrEAB, will stand on the same base, CrH, and between the same parallels ; therefore, GrHEF = aHAB, and consequently ABCD = EFGH, (Ax. 1). Hence the theorem ; Parallelograms on equal bases, etc. BOOK X. 43 Oor. Triangles on equal bases, and between the same parallels, are equal in area. For, draw BB and EG; the A ABB is one half of the parallelogram AO, and the A EFGr is one half of the equivalent parallelogram FE; therefore, the A ABB = the A EFG, (Ax. 7). THEOREM XXX. If a triangle and a parallelogram are upon the same or equal bases, and between the same parallels, the triangle is equiva- lent to one half the parallelogram. Let ABO be a A, and ABBE a parallelogram, on the same base AB, and between the same parallels ; then we are to prove that the A ABO is equivalent to one half of the parallel- ogram ABBE. Draw the diagonal EB to the parallelogram ; now, because the two A's ABO and ABE are on the same base, and between the same parallels, they are equiva- lent, (Th. 28); but the A ABE is one half the parallel- ogram ABBE, (Th. 25, Cor.) ; therefore the A ABO is equivalent to one half of the same parallelogram, (Ax. 7). Hence the theorem ; if a triangle and a parallelogram, etc* THEOREM XXXI. The complementary parallelograms described about any point in the diagonal of any parallelogram, are equivalent to each other. Let A be a parallelogram, and BB its diagonal ; take any point, as E, in the diagonal, and through this point draw lines parallel to the sides of the parallelogram, thus forming four parallelograms. "We are now to prove that the complementary paral- lelograms, AE and EO, are equivalent. . 44 GEOMETRY. By (Th. 25, Cor.) we learn that the A ABD = A DBC. Also by the same Cor., A a = A b, and A c= A d; there- fore by addition Now, from the whole A ABD take A a + A c, and from the whole A DBC take the equal sum, A b -f- A d, and the remaining parallelograms AE and EC are equiv- alent, (Ax. 3). Hence the theorem ; the complementary parallelograms, etc. THEOREM XXXII. The perimeter of a rectangle is less than that of any rhom- boid standing on the same base, and included between the same parallels. Let ABCD be a rect- angle, and ABEF& rhom- boid having the same base, and their opposite sides in the same line parallel to the base. "We are now to prove that the perimeter ABODA is less than ABEFA. Because AD is a perpendicular from A to the line DE, and AF an oblique line, AD is less than AF, (Th. 23). For the same reason BO is less than BE; hence AD + BC< AF+ BE. Adding the sum, AB + DO, to the first member of this inequality, and its equal AB -f FE to the second member, we have AB + BC + CD + DA, or the perimeter of the rectangle, less than AB -f BE + EF + FA, or the perimeter of the rhomboid. Hence the theorem ; the perimeter of a rectangle, etc. Scholium. — In Theorem 30 it is shown that the triangles ABC, ABE, and DBE, are equal in area, and that each is equal to one half the parallelogram ABBE. This parallelogram also has the same area as the rectangle having an equal base and altitude. BOOK I. 45 Thus far, areas have been considered only relatively and in the abstract. We will now explain how we may pass to the absolute measures, or, more properly, to the numerical expressions for areas. THEOREM XXXIII. The area of any plane triangle is measured by the product of its base by one half its altitude; or one half its base by its altitude, or one half the product of its base by its altitude. Let ABO represent any triangle, AB its base, and AD, at right angles to AB, its altitude ; now we are to show that the area of ABC is equal to the product of AB by one half of AD ; or one half of * * AB by AD ; or one half of the product of AB by AD. On AB construct the rectangle ABED', and the area of this rectangle is measured by AB into AD (Def. 54) ; but the area of the A ABO is equivalent to one half this rectangle, (Th. 30). Therefore, the area of the A is measured by J AB x AD, or one half the product of its base by its altitude. Hence the theorem ; the area of any plane triangle, etc. THEOREM XXXIV. The area of a trapezoid is measured by one half the sum of its parallel sides multiplied by the perpendicular distance between them. JjetABDO represent any trape- zoid; draw the diagonal BO, divid- ing it into two triangles, ABO and BOD: OD is the base of one tri- angle, and AB may be considered as the base of the other ; and EF is the common altitude of the two triangles. Now, by Th. 33, the area of the triangle BOD = \ OD x EF; and the area of the A ABC= \AB x EF; but 46 GEOMETRY. L K I by addition, the area of the two A's, or of the trape- zoid, is equal to J ( AB+ CD) x EF. Hence the theorem ; the area of a trapezoid, etc. THEOREM XXXV. If one of two lines is divided into any number of parts, the rectangle contained by the two lines is equal to the sum of the several rectangles contained by the undivided line and the seve- ral parts of the divided line. Let AB and AD be two lines, and suppose AB divided into any number of parts at the points E, F, Gr, etc. ; then the whole rect- angle contained by the two lines is AH, which is measured by AB A * * u u into AB. But the rectangle AL is measured by f AE into AD ; the rectangle EK is measured by EF into EL, which is equal to EF into AD ; and so of all the other partial rectangles ; and the truth of the proposition is as obvious as that a whole is equal to the sum of all its parts. Hence the theorem ; if one of two lines is divided, etc. THEOREM XXXVI. If a straight line is divided into any two parts, the square described on the whole line is equivalent to the sum of the squares described on the two parts plus twice the rectangle con tained by the parts. Let AB be any Hue divided into any two parts at the point C; now we are to prove that the square on AB is equivalent to the sum of the squares on A C and CB plus twice the rectangle contained by AC and CB. On AB describe the square AD. Through the point C draw CM, par- BOOK I. 47 allel to BB ; take BE m BO, and through E draw EKN, parallel to AB. We now have OE, the square on CB, by direct construction. As AB mm BB, and OB = BE, by subtraction, AB — CB = BB — BE; or AC = EB. But NK= AC, being opposite sides of a parallelogram ; and for the same rea- son, KM = EB. Therefore, (Ax. 1), NK = KM, and the figure NM is a square on NK, equal to a square on A 0. But the whole square on AB is composed of the two squares CE, NM, and the two complements or rectangles u4.iT and KB-, and each of these latter is ACm length, and BO in width ; and each has for its measure AO into OB ; therefore the whole square on AB is equivalent to AC 2 + BC 2 + 2AC x OB. Hence the theorem ; if a straight line is divided into any two parts, etc. This theorem may be proved algebraically, thus : Let w represent any whole right line divided into any two parts a and b ; then we shall have the equation ic = a -f b By squaring, w 2 = a 2 -f b 2 -f 2ab. Oor. If a = b, then w 2 — 4a 2 ; that is, the square de- scribed on any line is four times the square described on one half of it. THEOREM XXXVII. The square described on the difference of two lines is equiv- alent to the sum of the squares described on the two lines di- minished by twice the rectangle contained by the lines. Let AB represent the greater of two lines, OB the less line, and A their difference. We are now to prove that the square described ow AC is equivalent to the sum of the squares on AB and BC diminished by twice the rectangle contained by AB and BO. Conceive the square AF to be described on AB, and 48 GEOMETRY. the square BL on CB ; on A C describe the square ACGM, and produce MG to K. As GC=AC, and CL = CB, by addition, (GC + CL), or GL, is equal to AO+ CB, or 4.B. Therefore, the rectangle GE is J.1? in length, and CB in width, and is measured by AB XBC. Also AE= AB, and AM= AC; hy subtraction, MH = CB; and as MK= AB, the rectangle IZjKT is AB in length, and Ci? in width, and is measured by AB X BC; and the two rectangles GE and HK are together equiva- lent to 2AB x BC. ]$o w, the squares on .Ai? and BC make the whole figure AHFELC; and from this whole figure, or these two squares, take away the two rectangles iTiT and GE, and the square on A C only will remain ; that is, AC 2 =AB 2 + BC — 2AB x BC Hence the theorem ; the square described on the differ- ence of two lines, etc. This theorem may be proved algebraically, thus : Let a represent the greater of two lines, b the less, and d their difference ; then we must have this equation : d = a — b By squaring, d 2 = a 2 -f b 2 — 2ab. a a 2 Cor. If d= b, then d = «", and d 2 = -r ; that is, the square described on one half of any line is equivalent to one fourth of the square described on the whole line. THEOREM XXXVIII. The difference of the squares described on any two lines is equivalent to the rectangle contained by the sum and difference of the lines. Let AB be the greater of two lines, and AC the less, and on these lines describe the squares AD, AM; then, the BOOK I. 49 difference of the squares on AB and AC is the two rect- angles EF and FC. We are now to show that the measure of these rect- angles may be expressed by (A B + AC) x(AB — AC). The length of the rectangle EF is ED, or its equal AB; and the length of the rectangle FC is MC, or its equal AC; therefore, the length of the two together (if we con- ceive them put between the same parallel lines) will be AB + AC; and the common width is CB, which is equal to AB—AC; therefore, AB 2 — -AC 2 = (AB+AC) x (AB --AC). Hence the theorem; the difference of the squares de- scribed on any two lines, etc. This theorem may be proved algebraically: thus, Let a represent one line, and b another ; Then a -f b is their sum, and a — b their difference ; and (a + b) X (a — b) = a 2 — b\ THEOREM XXXIX. The square described on the hypotenuse of any right-angled triangle is equivalent to the sum of the squares described on the other two sides. Let ABC represent any right-angled triangle, the right angle at B; we are to prove that the square on A C is equivalent to the sum of two squares; one on AB, the other on BC. On the three sides of the triangle describe the three squares, AB, AL and BM. Through the point B, draw BNE perpendicular to AC, and produce it to meet the line QI in K; also produce AF to meet Gfl in H, and ML to meet the point in K. Remark. — That the lines, GI and ML, produced, meet at the point K, may be readily shown. As the proof of this fact is not necessary for the demonstration, it is left afl an exercise for the learner. 5 D 50 GEOMETRY. The angle BAG is a right angle, and the angle NAE is also a right angle ; if from these equals we subtract the common angle BAR, the re- maining angle, BAG, must be equal to the re- . maining angle GAH. The angle G is a right angle, equal to the angle ABO; and AB = AG ; therefore, the two A's ABC and AGH are equal, and AH=AC. ButA(7 = AF; therefore, AH= AF. Now, the two parallelograms, AF and ARKB are equivalent, because they are upon equal bases, and between the same paral- lels, FH and FK, (Th. 27). But the square A I, and the parallelogram AHKB, are equivalent, because they, are on the same base, AB, and between the same parallels, AB and GK; therefore, the square Al, and the parallelogram AF, being each equiv- alent to the same parallelogram AHKB, are equivalent to each other, (Ax. 1). ,In the same manner we may prove that the square BD is equivalent to the rectangle ND ; therefore, by addition, the two squares, A I and BM, are equivalent to the two parallelograms, AF and ND, or to the square AD. Hence the theorem ; the square described on the hypote- nuse of a right-angled triangle, etc. Cor. If two right-angled triangles have the hypotenuse, and a side of the one equal to the hypotenuse and a side of the other, each to each, the two triangles are equal. BOOK I. 51 Let ABC and AGHbe the two A's, in which we sup- pose ACfam AH, and BC = GH; then will AG = AB. For, we have AC 2 = AB 2 + BO 2 , or, by transposing, AC 2 — BC* = J.J5 2 , and AH 2 = ~AG 2 + GH 2 , or, by transposing, AH 2 — GH 2 = AG 2 . But by the hypothesis AC 2 —~BC 2 = AH 2 — (T^ 2 ; hence, AB 2 = J. G\ or, Jl J5 = A G. Scholium. — The two sides, AB&nd BC, may vary, while AC remains constant. AB may be equal to BC; then the point iVwill be in the middle of A C. When AB is very near the length of A C, and BC very small, then the point N falls very near to C. Now as AE and AD are right-angled parallelograms, their areas are measured by the product of their bases by their altitudes ; and it is evident that, as they have the same altitude, these areas will vary directly as their bases AN and NC; hence the squares on AB and BC, which are equivalent to those rectangles, vary as the lines AN and NC. The following outline of the demonstration of this pro- position is presented as a useful disciplinary exercise for the student. We employ the same figure, in which no change is made except to draw through the line CP, parallel to BK. The first step is to prove the equality of the triangles AGE and ABO, whence AH = AC. But AC = AF; therefore AH= AF. The parallelograms AFEN and AHKB are equiva- lent. Also, the parallelogram AHKB = the square ABIG, (Th. 27), and the parallelogram KBCP=NEDC= square BCML. . Now, by adding the equals AFEN= ABIG NEBC = BCML we obtain AFD C = ABIG -f B CML. That is, the square on A C is equivalent to the sum of the squares on AB and BC. The great practical importance of this theorem, in the extent and variety of its applications, and the frequency of its use in establishing subsequent propositions, ren- der it necessary that the student should master it com- pletely. To secure this end, we present a 52 GEOMETRY. Second Demonstration, Let ABO be a triangle right-angled at B. On the hypotenuse A 0, describe the square A CUB. From B and E let fall the perpendiculars Bb and Ed, on AB and A B produced. Draw Bn and Oa, making right angles with Ed. "We give an outline only of the demonstration, requiring the pupil to make it complete. First Part. — Prove the four triangles ABO, AbB, BnE, and EaO, equal to each other. The proof is as follows: The A's ABO and BnE are equal, because the angles of the one are equal to the angles of the other, each to each, and the hypotenuse AO of the one, is equal to the hypotenuse BE of the other. In like manner, it may be shown that the a's AbB and EaO are equal. Now, the sum of the three angles about A, is equal to the sum of the three angles of the A ABO) and if, from the first sum, we take [_BAO -f L_ CAB, and from the second we take L^ + L CAB = [_BAO+ [__ OAB, the remaining angles are equal ; that is, [_ Bb A is equal to [_AOB ; hence the A's ABO and BbA have their angles equal, each to each; and since AO — BA, the A's are themselves equal, and the four triangles ABO, AbB, BnE, and EaO, are equal to each other. Second. — Prove that the square bBnd is equal to a square on AB. The square BdaCis obviously on BO. Third. — The area of the whole figure is equal to the square on AO, and the area of two of the four equal right-angled triangles. Also, the area of the whole figure is equal to two other BOOK 1. 53 squares, bDnd and daOB, and two of the four equal tri- angles, DnE and EaO. Omitting or subtracting the areas of two of the four right-angled triangles, in each of the two expressions for the area of the whole figure, there will remain the square on A 0, equal to the sum of the two squares, Dndb and da OB. "2 . TT7v2 -T-F*1 That is, AB' + BO = AG Hence the theorem; the square described on the hypote- nuse of a right-angled triangle, etc. Scholium. — Hence, to find the hypotenuse of a right-angled triangle, extract the square root of the sum of the squares of the two sides about the right angle. THEOREM XL. In any obtuse-angled triangle, the square on the side oppo- site the obtuse angle is greater than the sum of the squares on the other two sides, by twice the rectangle contained by .either side about the obtuse angle, and the part of this side produced to meet the perpendicular drawn to it from the vertex of the opposite angle. Let ABO be any triangle in which the angle at B is obtuse. Produce either side about the obtuse angle, as OB, and from A draw AD perpen- dicular to OB, meeting it produced atD. It is obvious that OD = OB + BD. By squaring, ~OD 2 = OB 2 + 20B x BD + BD 2 , (Th. 36). Adding AD 2 to each member of this equation, we have AD 2 +OD 2 = OB 2 + BD 2 -f AD 2 + 20B x BD. But, (Th. 39), the first member of the last equation is equal to AO 2 , and BD 2 + AD 2 = AB\ 5* 54 GEOMETRY. Therefore, this equation becomes ~AC 2 - CB 2 + AB* + 2CB x BD. That is, the square on A is equivalent to the sum of the squares on CB and AB, increased by twice the rect- angle contained by CB and BD. Hence the theorem; in any obtuse -angled triangle, the square on the side opposite the obtuse angle, etc. Scholium. — Conceive AB to turn about the point A, its intersection with CD gradually approaching D. The last equation above will be true, however near this intersection is to D, and when it falls upon D the triangle becomes right-angled. In this case the line BD reduces to zero, and the equation becomes AC 7 = CB* + AB*, in which CB and AB are now the base and per- pendicular of a right-angled triangle. This agrees with Theorem 39, as it should, since we used the property of the right-angled triangle established in Theorem 39 to demonstrate this proposition ; and in the equation which expresses a property of the obtuse-angled triangle, we have introduced a supposition which changes it into one which is right-angled. THEOREM XLI. In any triangle, the square on a side opposite an acute angle is less than the sum of the squares on the other two sides, by twice the rectangle contained by either of these sides, and the distance from the vertex of the acute angle to the foot of the perpendicular let fall on this side, or side produced, from the vertex of its opposite angle. Let ABC, either figure, represent any triangle ; C an acute angle, CB the base, and AB the perpendicular, which falls either without or on the base. !Now we are to prove that AB>= CB 2 + AC 2 — 2CB x CD. HOOK I 55 From the first figure we get BB=-CB—OB ( 1 ) and from the second BB - OB —OB ( 2 ) Either ono of these equations will give, (Th. 87), BB 2 - OB 2 + OB 2 —20B x OB. Adding AD* to each member and reducing, we obtain, (Th. 89), AB 2 ~Jd 2 +OB 2 — 20Bx OB, which proves the proposition. Hence the theorem. THEOREM XLII. If in any triangle a line be drawn from any angle to the middle of the opposite Me, twice the square of this line, together with twice the square of one half the side bisected, will be equivalent to the sum of the squares of the other two sides. Let AB be a triangle, and M the middle point of its base. Then we are to prove that 2AM 2 + 20M a - A0 2 + AJB*. Draw AB perpendicular to the base, and make AB « p } AO=b, AB=*c, OB**2a } AM = m, and MB =■ x; then OM — a, OB =» a + z, BB e» a — X. Now by, (Th. 39), wo have the two following equations : p' + (a — zy = c* (1) f + {a + xy » y (2) By addition, 2tf + 2x % + 2a % m 6" -f c\ But f -f x 7 — m\ Therefore, 2m 9 + 2a* - b 2 + c\ This equation is tho algebraic enunciation of the theorem. 56 GEOMETRY. THEOREM XLIII. The two diagonals of any parallelogram bisect each other ; and the sum of their squares is equivalent to the sum of the squares of the four sides of the parallelogram. I^etABCD be any parallelogram, and A and BD its diagonals. We are now to prove, 1st. That AE mm EC, and DE = m 2d. That AC 2 + 'BD 2 = AB* + ~BC 2 + CD 2 +~AD\ 1. The two triangles ABU and CDE are equal, be- cause AB = CD, the angle ABE = the alternate angle ODE, and the vertical angles at E are equal ; therefore, AE, the side opposite the angle ABE, is equal to CE, the side opposite the equal angle CDE; also EB, the remaining side of the one A, is equal to EB, the remain- ing side of the other triangle. 2. As AOB is a triangle whose base, AC, is bisected in E, we have, by (Th. 42), . 2AK + 2ED' = AD* + DO 2 ( 1 ) And as A OB is a triangle whose base, AG, is bisected in E, we have 2AE' + 2EB = AB' + BQ* ( 2 ) By adding equations (1) and (2), and observing that EB 2 = ED 2 , we have ±AE 2 + 4ED 2 = ~AD 2 +~DC 2 + AB 2 +~BC 2 But, four times the square of the half of a line is equiv- alent to the square of the whole line, (Th. 36, Corollary) ; therefore 4 AE 2 = AC 2 , and 4ED 2 = DW; and by sub- stituting these values, we have AC + BD' - AB' + £ A B na n and B = na, and — = — = -. A ma m To illustrate, let the line A contain the line A a six times, and let the t t line B contain the same a line a five times : then I j i j i I A=6a and B—5a, which B . B 5a- 5 glYe A=6a = 6- 3. A Proportion is a formal statement of the equality of two ratios. Thus, if we have the four magnitudes A, B, and i>, such that — = — , this relation is expressed by the pro- portion A : B : : : D, or A : B = : 2>, the first of which is read, A is to B as is to B ; and the second, the ratio of A to B is equal to that of to D. 4. The Terms of a proportion are the magnitudes, or more properly the representatives of the magnitudes compared. 5. The Extremes of a proportion are its first and fourth terms. 6. The Means of a proportion are its second and third terms. 7. A Couplet consists of the two terms of a ratio* The BOOK II. 61 first and second terms of a proportion are called the first couplet, and the third and fourth terms are called the second couplet. 8. The Antecedents of a proportion are its first and third terms. 9. The Consequents of a proportion are its second and fourth terms. In expressing the equality of ratios in the form of a proportion, we may make the denominators the ante- cedents, and the numerators the consequents, or the reverse, without affecting the relation between the magni- tudes. It is, however, a matter of some little importance to the beginner to adopt a uniform rule for writing the terms of the ratios in the proportion ; and we shall always, unless otherwise stated, make the denominators of the ratios the antecedents, and the numerators the conse- quents.* 10. Equimultiples of magnitudes are the products arising from multiplying the magnitudes by the same number. Thus, the products, Am and Bm, are equimultiples of A and B. U. A Mean Proportional between two magnitudes is a magnitude which will form with the two a proportion, when it is made a consequent to the first ratio, and an antecedent to the second. Thus, if we have three mag- nitudes A, B, and (7, such that A : B : : B : (7, B is a mean proportional between A and O. 12. Two magnitudes are reciprocally, or inversely pro- portional when, in undergoing changes in value, one is multiplied and the other is divided by the same number. Thus, if A and B be two magnitudes, so related that when B A becomes mA, B becomes — , A and B are said to be m inversely proportional. * For discussion of the two methods of expressing Katio, see Uni- versity Algebra. 6 62 GEOMETRY. 13. A Proportion is taken inversely when the ante- cedents are made the consequents and the consequents the antecedents. Thus 14. A Proportion is taken alternately, or by alternation, when the antecedents are made one couplet and the con- sequents the other. 15. Mutually Equiangular Polygons have the same num- ber of angles, those of the one equal to those of the others, each to each, and the angles like placed. 16. Similar Polygons are such as are mutually equi- angular, and have the sides about the equal angles, taken in the same order, proportional. 17. Homologous Angles in similar polygons are those which are equal and like placed ; and 18. The Homologous Sides are those which are like dis- posed about the homologous angles. THEOREM I. If the first and second of four magnitudes are equal, and also the third and fourth, the four magnitudes may form a proportion. Let A, B, C, and D represent four magnitudes, such that A = B and = D ; we are to prove that A : B : : : D. Now, by hypothesis, A is equal to B, and their ratio is therefore 1 ; and since, by hypothesis, C is equal to D, their ratio is also 1. Hence, the ratio of A to B is equal to that of C to D ; and, (by Def. 3), A : B : : 0:1). Therefore, four magnitudes which are equal, two and two, constitute a proportion. BOOK II. 63 THEOREM II. If four magnitudes constitute a 'proportions the product of the extremes is equal to the product of the means. Let the four magnitudes A, B, C, and D form the pro- portion A : B : : : B ; we are to prove that Ax B = Bx C. The ratio of A to B is expressed by -j = r. The ratio of C to B is expressed by -^ = r. Hence, (Ax. 1), -j = -. Multiplying these equals each by A x C, and we have Bx C=AxB. Hence the theorem ; if four magnitudes are in propor- tion, etc. Cor. 1. Conversely : If we have the product of two mag- nitudes equal to the product of two other magnitudes, they will constitute a proportion of which either of the two may be made the extremes and the other two the means. Let the magnitudes B x 0= A x B. Dividing both members of the equation by A x C, and we have B_B A~C Hence the proportion A : B : : : B. Cor. 2. If we divide both members of the equation Ax B = Bx C by .A, we have B = — -. — . A That is, to find the fourth term of a proportion, mul- tiply the second and third terms together and divide the pro- duct by the first term. This is the Rule of Three of Arithmetic. 64 GEOMETRY. This equation shows that any one of the four terms can be found by a like process, provided the other three are given. THEOREM III. If three magnitudes are continued 'proportionals, the 'product of the extremes is equal to the square of the mean. Let A, B, and represent the three magnitudes : Then A : B : : B : O, (by Def. 11). But, (by Th. 2), the product of the extremes is equal to the product of the means ; that is, A x 0= B\ Hence the theorem ; if three magnitudes, etc. THEOREM IV. Equimultiples of any two magnitudes have the same ratio as the magnitudes themselves ; and the magnitudes and their equimultiples may therefore form a proportion. Let A and B represent two magnitudes, and mA and mB their equimultiples. Then we are to prove that A : B : : mA : mB. The ratio of A to B is —, and of mA to mB is A mB B ., ,. — r = -r, the same ratio. mA A* Hence the theorem; equimultiples of any two magni- tudes, etc. THEOREM V. If four magnitudes are proportional, they will he propor- tional when taken inversely. If A : B : : m A : mB, then B : A : : mB : m A ; For in either case, the product of the extremes and means are manifestly equal ; or the ratio of the couplets is the same. Hence the theorem ; if four quantities are proportional, etc. BOOK II. 65 THEOREM VI. Magnitudes which are proportional to the same propor- tionals, are proportional to each other. If A : B = P : Q \ Then we are to prove that and a : b = P : Q) A : B = a : b. 13 Q From the 1st proportion, — = ^ ; From the 2d " - = ^5 a P Therefore, by (Ax. 1), -j = -, or A : B = a : b. jA. a Hence the theorem ; magnitudes which are proportional to the same 'proportionals, etc. Cor. 1. This principle may be extended through any number of proportionals. Cor. 2. If the ratio of an antecedent and consequent of one proportion is equal to the ratio of an antecedent and conse- quent of another proportion, the remaining terms of the two proportions are proportional. For, if A : B : : C : D and M : N :: P: Q in Which A=M then C = P> hence C : D : : P : Q. THEOREM VII. If any number of magnitudes are proportional, any one of the antecedents will be to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let A, B, C, D, U, etc., represent the several magni- tudes which give the proportions A : B A : B A : B 6* C: D E : F Or : R, etc., etc. 66 GEOMETRY. To which we may annex the identical proportion, A : B : : A : B. Now, (by Th. 2), these proportions give the following equations, A x D = B x Q A x F = B x F A x H= B x a A x B = B x A, etc. etc. From which, by addition, there results the equation, A(B + D + F+H, etc.) = B(A + 0+ F + #, etc.) But the sums B -f J) -f F, etc., and A + C + F, etc., may be separately regarded as single magnitudes ; there- fore, (Th. 2), A : B :: A+C+F+ G, etc. : B -f D + F+ JT, etc. Hence the theorem ; if any number of magnitudes are pro- portional, etc. THEOREM VIII. If four magnitudes constitute a proportion, the first will be to the sum of the first and second as the third is to the sum of the third and fourth. By hypothesis, A : B : : C: D ; then we are to prove that A : A + B :: C : C+ D. By the given proportion, — - = -— . A C Adding unity to both members, and reducing them to the form of a fraction, we have — - — = — ^— . Chang- ing this equation into its equivalent proportional form, we have A : A + B :: O : C+ D. Hence the theorem ; if four magnitudes constitute a pro- portion, etc. Cor. If we subtract each member of the equation -j = BOOK II. 67 -» from unity, and reduce as before, we shall have A : A — B :: C : C—D. Hence also ; if four magnitudes constitute a proportion, the first is to the difference between the first and second, as the third is to the difference between the third and fourth. THEOREM IX. If four magnitudes are proportional, 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, and D be the four magnitudes which give the proportion A : B :: 0:1); we are then to prove that they will also give the propor- tion A + B : A — B :: C+D : C — D. By Th. 8 we have A : A + B = : C+D. Also by Scholium, same Th., A : A — B = : C — D. Now, if we change the order of the means in these pro- portions, which may be done, since the products of ex- tremes and means remain the same, we shall have A i C = A + B : C+D. A : = A — B : C—D. Hence, (Th. 6), we have A + B : C+ D = A — B : O—B. Or, A + B : A — B = C + I) : C—D. Hence the theorem ; if four magnitudes are proportional, etc. THEOREM X. If four magnitudes are proportional, like powers or like roots of the same magnitudes are also proportional. If the four magnitudes, A, B, C, and D, give the pro- portion 68 GEOMETRY. A : B : : C : D, we are to prove that A n : B n :: (7" : D w . 7? 7) The hypothesis gives the equation — = — . Eaising A both members of this equation to the nth. power, we have B n D n -j^ = — , which, expressed in its equivalent proportional form, gives A n : B n :: C n : D n . If n is a wAete number, the terms of the given propor- tion are each raised to a power ; but if n is a fraction having unity for its numerator, and a whole number for its denominator, like roots of each are taken. As the terms of the proportion may be first raised to like powers, and then like roots of the resulting propor- tion be taken, n may be any number whatever. Hence the theorem ; if four magnitudes, etc. THEOREM XI. If four magnitudes are proportional, and also four others, the products which arise from multiplying the first four by the second four, term by term, are also proportional. Admitting that A : B : : 0:1), and X : Y : : M : N, We are to show that AX :BY:: CM: DN. B J) From the first proportion, — . == — ; X M' Multiply these equations, member by member, and BY^DN. AX CM' Or, AX : BY :: CM: DN. The same would be true in any number of proportions. Hence the theorem ; if four magnitudes are, etc. From the second, BOOK II. 69 THEOREM XII. If four magnitudes are proportional, and also four others, the quotients which arise from dividing the first four by the second four, term by term, are proportional. By hypothesis, A : B : : O : D, and X : Y :: M : 1ST. Multiply extremes and means, AD = OB, ( 1 ) and XN=MY. (2) Divide (1) by (2), and ^ x ^= ^ x |. Convert these four factors, which make two equal pro- ducts, into a proportion, and we have A . B _ C 9 D X 1 Y i: M : N' By comparing this with the given proportions, we find it is composed of the quotients of the several terms of the first proportion, divided by the corresponding terms of the second. Hence the theorem ; if four magnitudes are proportional, etc. THEOREM XIII. If four magnitudes are proportional, we may multiply the first couplet, the second couplet, the antecedents or the conse- quents, or divide them by the same quantity, and the results will be proportional in every case. Let the four magnitudes A, B, 0, and I) give the pro- portion A : B : : : D. By multiplying the extremes and means we have A.D = B.O (1) Multiply both members of this equation by any num- ber, as a, and we have aA.D = aB.O By converting this equation into a proportion in four different ways, as follows : 70 GEOMETRY. aA : aB A : B : : a A : B : J. : a.B : : : B aC : aD : aO : B ; C : aB resuming the original equation, (1), and dividing both members by a, we have A.B _ B.O a a This equation may also be converted into a proportion in four different ways, with the following results : J x> : B a a A : B :: . D a " a A : B :: C 1:I> a a A . B .. C:» a a Hence the theorem ; if four magnitudes are in proportion, etc. THEOREM XIV. If three magnitudes are in proportion, the first is to the third as the square of the first is to the square of the second. Let A, B, and C, be three proportionals. Then we are to prove that A : C—A % \B l By(Th. 3) AC=B 2 Multiply this equation by the numeral value of A, and we have A 2 0=AB 2 This equation gives the following proportion : A: C=A 2 :B\ Hence the theorem. Remark. — It is now proposed to make an application of the pre- ceding abstract principles of proportion, in geometrical investigations BOOK II. 71 THEOREM XV. If two parallelograms are equal in area, the base and per- pendicular of either may be made the extremes of a propor- tion, of which the base and perpendicular of the other are the means. Let ABCD, E D F c and NLHM, be two paral- lelograms hav- ing equal areas, by hypothesis ; then we are to prove that AB : LN : : MK : BF, in which MK and BF are the altitudes or perpendiculars of the parallelograms. This proportion is true, if the product of the extremes is equal to the product of the means ; that is, if the equation AB.BF = LN.MKiz true. But AB.BF is the measure of the rectangle ABFE, (B.I., Th. 32, Scholium), and this rectangle is equal in area to the parallelegram ABCD, (B. L, Th. 27). In the same manner, we may prove that ~%N.MK is the measure of the parallelogram NLHM. But these two parallelograms have equal areas by" hypothesis. Therefore, AB.BF = LN.MK is a true equation, and (Th. 2, Cor. 1), gives the proportion AB : LN : : MK : BF. Hence the theorem ; if two parallelograms are equal in area, etc. THEOREM XVI. Parallelograms having equal altitudes are to each other as their bases. Since parallelograms having equal bases and equal altitudes are equal in area, however much their angles 72 GEOMETRY. may differ, we can suppose the two parallelograms under consideration to be mutually equiangular, without in the least impairing the generality of this theorem. There- fore, let ABOB and AEFB be two l — 7 — 7 — ; parallelograms / / / / having equal alti- / / / / / tudes,and let them / / / / / / / / be placed with A B their bases on the same line AE, and let the side, AB, be common. First suppose their bases commensurable, and that AE being divided into nine equal parts, AB contains four of those parts. If, through the points of division, lines be drawn paral- lel to AB, it is obvious that the whole figure, or the parallelogram, AEFB, will be divided into nine equal parts, and that the parallelogram, ABOB, will be com- posed of four of those parts. Therefore, ABOB : AEFB : : AB : AE : : 4 : 9. Whatever be the whole numbers having to each other the ratio of the lines AB and AE, the reasoning would remain the same, and the proportion is established when the bases are commensurable. But if the bases are not to each other in the ratio of any two whole numbers, it remains still to be shown that AEFB : ABOB 11 AE 1 AB (1) If this propor- tion is not true, there must be a line greater or less than AB, to which AE will have the A ~ ~~b~l same ratio that AEFB has to ABOB. Suppose the fourth proportional greater than AB, as AK, then, AEFB : ABOB :: AE : AK (2). C M BOOK II. 73 If we now divide the line AE into equal parts, each less than the line BK, one point of division, at least, will fall between B and K Let L be such point, and draw LM parallel to B 0. This construction makes AE and AL commensura- ble; and by what has been already demonstrated, we have AEFD : ALMD :: AE : AL. (3) Inverting the means in proportions ( 2 ) and ( 3 ), they become AEFD : AE : : ABCD : AK; and AEFD : AE : : ALMD : AL. Hence, (Th. 6), ABCD : AK : : ALMD : AL. By inverting the means in this last proportion, we have ABCD : ALMD : : AK : AL. But AK is, by hypothesis, greater than AL; hence, if this proportion is true, ABCD must be greater than ALMD ; but on the contrary it is less. We therefore conclude that the supposition, that the fourth propor- tional, AK, is greater than AB, from which alone this absurd proportion results, is itself absurd. In a similar manner it can be proved absurd to sup- pose the fourth proportional less than AB. Therefore the fourth term of the proportion ( 1 ) can be neither less nor greater than AB ; it is then AB itself, and parallelograms having equal altitudes are to each other as their bases, whether these bases are commensur- able or not. Hence the theorem ; Parallelograms having equal bases, etc. Cor. 1. Since a triangle is one half of a parallelogram having the same base as the triangle and an equal alti- tude, and as the halves of magnitudes have the same ratio as their wholes ; therefore, 7 74 GEOMETRY. Triangles having the same or equal altitudes are to each other as their bases. Cor. 2. Any triangle has the same area as a right- angled triangle having the same base and an equal alti- tude; and- as either side about the right angle of aright- angled triangle may be taken as the base, it follows that Two triangles having the same or equal bases are to each other as their altitudes. Cor. 3. Since either side of a parallelogram may be taken as its base, it follows from this theorem that Parallelograms having equal bases are to each other as their altitudes. THEOREM XVII. If lines are drawn cutting the sides, or the sides produced, of a triangle proportionally, such secant lines are parallel to the base of the triangle ; and conversely, lines drawn parallel to the base of a triangle cut the sides, or the sides produced, proportionally. Let ABC be any triangle, and draw the line BE dividing the sides AB and AC into parts which give the proportion AD : DB : : AE : EC. "We are to prove that BE is parallel \,oBC. If BE is not a parallel through the point B to the line BC, suppose Bm to be that parallel ; and draw the lines BC and Bm. Now, the two triangles ABm and mBC, have the same altitude, since they have a common vertex, B, and their bases in the same line, A C; hence, they are to each other as their bases, A m and mC, (Th. 16, Cor. 1). BOOK II. 75 That is, A ADm : A mDC : : Am : mC, Also, A Ami) : A DmB : : AD : D#. But, since Dm is supposed parallel to BC, the triangles DBm and D(7m have equal areas, because they are on the same base and between the same parallels, (Th. 28, B.I). Therefore the terms of the first couplets in the two preceding proportions are equal each to each, and conse- quently the terms of the second couplets are also propor- tional, (Th. 6). That is, AT) : DB : : Am : mC But AD : DB : : AE : EC by hypothesis. Hence we again have two proportions having the first couplets, the same in both, and we therefore have AE : EG :: Am : mO By alternation this becomes AE : Am :: EC : mO That is, AE is to Am, a greater magnitude is to a less, as EC is to mO, a less to a greater, which is absurd. Had we supposed the point m to fall between E and 0, our conclusion would have been equally absurd ; hence the suppositions which have led to these absurd results are themselves absurd, and the line drawn through the point D parallel to BO must intersect A in the point E. Therefore the parallel and the line BE are one and the same line. Conversely : If BE be drawn parallel to the base of the triangle, then will AD : DB : : AE : EC For as before, A ADE : a EDO :: AE : EC and A DEB : A AED w DB \ AD Multiplying the corresponding terms of these propor- 76 GEOMETRY. tions, and omitting the common factor, a ADE, in the first couplet, we have A DEB : A EDO :: AE x DB : EO x AD. But the a's DEB and EDO have equal areas, (Th. 28, B. I) ; hence AE x DB = EC x AD, which in the form of a proportion is J.^ : EG : : AD : DB or, AD : DB :: AE : EC and therefore the line parallel to the base of the triangle, divides the sides proportionally. It is evident that the reasoning would remain the same, had we conceived ADE to be the triangle and the sides to be produced to the points B and 0. Hence the theorem; if lines are drawn cutting the sides, etc. Cor. 1. Because DE is parallel to BO, and intersects the sides AB and AC, the angles ADE and ABO are equal. For the same reason the angles AED and A OB are equal, and the A's ADE and ABO are equiangular. Let us now take up the triangle ADE, and place it on ABO; the angle ADE falling on [__ B, the side AD on the side AB, and the side DE on the side BO. Now, since the angle A is common, and the angles AED and A OB are equal, the side AE of the A ADE, in its new position, will be parallel to the side A of the A ABO. But we have the proportion AD : AE :: AB : AO Placing the angle ADE on the angle ABO, and rea- soning as before, we shall have the proportion AD : DE : : AB : BO And in like manner it may be shown that AE : ED :: AC : OB That is, the sides about the equal angles of equiangular triangles, taken in the same order y are proportional, and the triangles are similar, (Def. 16). BOOK II 77 Cor. 2. Two triangles having an angle in one equal to an angle in the other, and the sides about these equal angles pro- portional, are equiangular and similar. For, if the smaller triangle be placed on the larger, the equal angles of the triangles coinciding, then will the sides opposite these angles be parallel, and the trian- gles will therefore be equiangular and similar. THEOREM XVIII. If any triangle have its sides respectively proportional to the like or homologous sides of another triangle, each to each, then the two triangles will be equiangular and similar. Let the triangle abc have its sides pro- portional to the triangle ABO ; that is, ac to A as cb to OB, and ac to A as ah to AB ; then we are to prove that the a's, abc and ABO, are equi- angular and similar. On the other side of the base, AB, and from A, conceive angle BAB to be drawn = to the L *r conceive and from the point B, the angle ABB to be drawn = to the [_ b. Then the third [__ B must be = to the third [_ c, (B. I, Th. 12, Cor. 2) ; and the A ABB will be equiangular to the A abc by construction. Therefore, ac : ab = AB : AB By hypothesis, ac : ab = AO : AB Hence, AB : AB = A : AB, (Th. 6). In this last proportion the consequents are equal; therefore, the antecedents are equal : that is, AB = AO In the same manner we may prove that BB = OB 7* 78 GEOMETRY. But AB is common to the two triangles"; therefore, the three sides of the A ABB are respectively equal to the three sides of the A ABC, and the two a's are equal, (B. I, Th. 21). But the A's ABB, and abc, are equiangular by con- struction ; therefore, the A's, ABO, and abc, are also equiangular and similar. Hence the theorem ; if any triangle have its sides, etc, Second Demonstration, Let abc and ABC be two triangles whose sides are respectively propor- tional, then will the triangles be equi- angular and similar. That is, [__a = l_A, [_b = [_B, and l_e=l_C. If the [__ c be in fact equal to the [_ C, the tri- angle abc can be placed on the triangle ABC, ca taking the direction of CA and cb of CB. The line ab will then divide the sides CA and CB proportionally, and will therefore be parallel to AB, and the triangles will be equiangular and similar, (Th. 17). But if the L c be not equal to the [__ C, then place ac on AC as before, the point c falling on C. Under the present supposition cb will not fall on CB, but will take another direction, CV, on one side or the other of CB. Make CV equal to cb and draw aV. Now, the A abc is represented in magnitude and posi- tion by the A a VC; and if, through the point a, the line ab be drawn parallel to AB, we shall have Ca : CA :: ab : AB; but by (Hy.) Ca : CA : : aV : AB. BOOK II 79 Hence, (Th. 6), ab : AB :: aV : AB; which requires that ab = a V, but (Th. 22, B. I) ab can not be equal to a V; hence the last proportion is absurd, and the supposition that the [_ c is not equal to the [_ (7, which leads to this result, is also absurd. Therefore, the [_ e is equal to the [__ (7, and the triangles are equi- angular and similar. Hence the theorem ; if any triangle ham its sides, etc. THEOREM XIX. If four straight lines are in proportion, the rectangle con- tained by the lines which constitute the 'extremes, is equivalent to that contained by those which constitute the means of the proportion. Let A, B, O, D, represent the four A ' j lines; then we are to show, geo- j . j metrically, that A x D = B x 0. D i i Place A and B at right angles to each other, and draw the hypotenuse. Also place and D at right angles to each other, and draw the hypotenuse. Then bring the two triangles together, so that shall be at right angles to B, as represented in the figure. Now, these two A's have each a E. [_, and the sides about the equal angles are pro- portional ; that is, A : B : : 0:1); there- fore, (Th. 18), the two A's are equiangular, and the acute angles which meet at the extremities of B and C, are together equal to one right angle, and the lines B and are so placed as to make another right angle ; therefore, also, the extremities of A, B, 0, and Z>, are in one right line, (Th. 3, B. I), and that line is the diag- \ B \ BC ! C AD V 80 GEOMETKY. onal of the parallelogram be. By Th. 31, B. I, the complementary parallelograms about this diagonal are equal ; but, one of these parallelograms is B in length, and Q in width, and the other is D in length and A in width; therefore, B x = A x D. Hence the theorem; if four straight lines are in propor- tion, etc. Cor. "When B = Q, then A x D = B\ and B is the mean proportional between A and B. That is, if three straight lines are in proportion, the rectangle contained by the first and third lines is equivalent to the square described on the second line. THEOREM XX. Similar triangles are to one another as the squares of their homologous sides. Let ABC and DBF be two similar triangles, and LQ and MF perpendiculars to the sides AB and DE respectively. Then we are to prove that &ABQ:&BEF = AB*:BE\ By the similarity of the tri- angles, we have, AB : BE = LQ : MF But, AB : DE = AB : BE Hence, AB 2 : TW^~AB x LQ : BE x MF. But, (by Th. 30, B. I), AB x LQ is double the area of the A ABC, and BE x MF is double the area of the A BEF. Therefore, aABC:ABEF::AB x LQ :BExMF And, (Th. 6), A ABQ: A DEF = AW : BE 2 . Hence the theorem ; similar triangles are to one another, etc. BOOK II. 81 The following illustration will enable the learner fully to comprehend this important theorem, and it will also serve to impress it upon his memory. Let abc and ABO represent two equiangular triangles. Suppose the length of the side ac to be two units, and the length of the corresponding side A to be three units. Eow, drawing lines through the points of division of the sides ac and A (7, parallel to the other sides of the triangles, we see that the smaller triangle is com- posed of four equal triangles, while the larger contains nine such triangles. That is, the sides of the triangles are as 2 : 3, and their areas are as 4 : 9 = 2 2 : 3 2 . THEOREM XXI. Similar polygons may be divided into the same number of triangles; and to each triangle in one of the polygons there will be a corresponding triangle in the other polygon, these triangles being similar and similarly situated. Let ABCDUsLnd abcde be two similar polygons. Now it is obvious that we can divide each polygon E into as many triangles as the figure has sides, less two; and as the polygons have the same number of sides, the diagonals drawn from the vertices of the homologous angles will divide them into the same number of tri- angles. 82 GEOMETRY. Since the polygons are similar, the angles EAB and eab, are equal, and EA : AB :: ea : ab. Hence the two triangles, EAB and eab, having an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular and similar, and the angles ABE and abe are equal. But the angles ABO and abc are equal, because the polygons are similar. Hence, [_ABO— [_ABE = [_abc — \_abe; that is, [_EBO<=[_ebc. The triangles, EAB and eab, being similar, their ho- mologous sides give the proportion, AB : BE :: ab : be; (1) and since the polygons are similar, the sides about the equal angles B and b are proportional, and we have AB : BO : : ab : be ; or, BO : AB :: be : ab. (2) Multiplying proportions (1) and (2), term by term, and omitting in the result the factor AB common to the terms of the first couplet, and the factor ab common to the terms of the second, we have BO : BE : : be : be. Hence the A's EBO and ebe are equiangular and similar; and thus we may compare all of the triangles of one polygon with those like placed in the other. Hence the theorem ; similar polygons may be divided, etc. THEOREM XXII. The perimeters of similar polygons are to one another as their homologous sides ; and their areas are to one another as the squares of their homologous sides. Let ABODE and abode be two similar polygons ; then we are to prove that AB is to the sum of all the sides BOOK II. 83 of the polygon ABCB, as B a b ab is to the sum of all / X^^. the sides of the polygon y^^ abed. E ^^\^ -— ^C e^^--- ^"^ ¥e have the identical d d proportion AB : ab : : AB : ab; and since the polygons are similar, we may write the following : AB : ab :: BO : be AB : ab :: OD : ed AB : ab : : BE : de, etc. etc. Hence, (Th. 7), AB : ab : AB+BC+CB+BE, etc.; ab+bc+cd+de 9 etc. Therefore, the perimeters of similar polygons are to one another as their homologous sides. This is the first part of the theorem. Sinoe the polygons are similar, the triangles BAB, eab, are similar, and if the triangle BAB is a part expressed by the fraction -, of the polygon to which it belongs, 71 the triangle eab is a like part of the other polygon. Therefore, EAB i eab :: ABCBEA : abedea. But, (Th. 20)^ EAB : eab : : AB 2 : ab\ Therefore, (Th. 6), ABCBEA : abedea : : AB 2 : ab\ Therefore, the similar polygons are to one another as the squares on their homologous sides. This is the second part of the theorem. Hence the theorem ; the perimeters of similar polygons are to one another, etc. THEOREM XXIII. Two triangles which have an angle in the one equal to an angle in the other, are to each other as the rectangle of the sides about the equal angles. 84 GEOMETRY. Let ABC and def be two triangles having the angles A and d equal. It is to be proved that the areas ABO and def are to each other as AB.AO is to de.df. Conceive the triangle def placed on the tri- angle ABO, so that d shall fall on A, and de on A B ; then df will fall on AC, because the [_'s i and d are equal. On AB, lay off Ae, equal to de ; and on AC, lay off Af, equal to df, and draw ef The tri- angle Aef will then be equal to the triangle def. Join B and/. Now, as triangles having the same altitude are to each other as their bases, (Th. 16, Cor. 1), we have Aef : ABf : : Ae : AB also, ABf : ABC : : Af: AC Multiplying these proportions together, term by term, omitting from the result ABf a factor common to the terms of the first couplet, we have Aef : ABC :: Ae . Af : AB . AC But Aef is equal to def, Ae to de, and Af to df; therefore, def : ABC : : de . df : AB . AC Hence the theorem ; two triangles which have an angle, etc. Scholium. — If we suppose that AB : AC :: de : df the two triangles will be similar ; and if we multiply the terms of the first couplet of this proportion by AC, and the terms of the second couplet by df we shall have AB .AC : AC* : : de ^df : ^ or, AB . AC : de . df : : AC 2 : df BOOK II. 85 Comparing this with the last proportion in this theorem, and we have, (Th.6); _ _ def: ABC :: df : AC 2 Remark. — This scholium is therefore another demonstration of Theorem 20, and hence that theorem need not necessarily have been made a distinct proposition. We require no stronger proof of the cer- tainty of geometrical truth, than the fact that, however different the processes by which we arrive at these truths, we are never led into inconsistencies ; but whenever our conclusions can be compared, they are found to harmonize with each completely, provided our premises are true and our reasoning logical. It is hoped that the student will lose no opportunity to exercise his powers, and test his skill and knowledge, in seeking original demonstrations of theorems, and in deducing consequences and conclusions from those already established. THEOREM XXIV. If the vertical angle of a triangle be bisected, the bisecting line will cut the base into segments proportional to the adja- cent sides of the triangle. Let ABO be any triangle, and the vertical angle, 0, be bi- sected by the straight line OD. Then we are to prove that AD : DB = AC : OB. Produce A O to E, making A OB = OB, and draw EB. The exterior angle A OB, of the A OEB, is equal to the two angles E, and OBB; but the angle E = OBE, because OB = OE, and the tri- angle is isosceles; therefore the angle AOD, the half of the angle A OB, is equal to the angle E, and BO and BE are parallel, (Cor., Th. 7, B. R Now, as ABE is a triangle, and OB is parallel to BO, ~ we have AD : DB = A : OE or OB, (Th. 17). Hence the theorem ; if the vertical angle of a triangle be bisected, etc. 8 86 GEOMETBY. THEOREM XXV. If from the right angle of a right-angled triangle, a 'per- pendicular is drawn to the hypotenuse ; ' ft. The perpendicular divides the triangle into two similar triangles, each of which is similar to the whole triangle. 2. The perpendicular is a mean proportional between the segments of the hypotenuse. 3. The segments of the hypotenuse are in proportion to the squares on the adjacent sides of the triangle. 4. The sum of the squares on the two sides is equivalent to the square on the hypotenuse. Let BAO be a triangle, right an- gled at A ; and draw AD perpendicu- lar to BO. 1. The two A's, ABO and ABB, B have the common angle, B, and the right angle BAO = the right angle BDA; therefore, the third |__ 0= [__ BAB, and the two A's are equiangular, and similar. In the same manner we prove the A AB similar to the A ABO; and the two triangles, ABB, ABO, being similar to the same A ABO, are similar to each other. 2. As similar triangles have the sides about the equal angles proportional, (Th. 17), we have BB : AB :: AB : OB; or, the perpendicular is a mean proportional between the seg- ments of the hypotenuse. 3. Again, BO : BA :: BA : BB hence, BA 2 = BO.BB (1) also, BOj_ OA :: OA : OD hence, OA 2 = BO.OB (2) Dividing Eq. (1) by Eq. (2), member by member, we obtain ~BA 2 BB ~OA 2 " OB BOOK II. 87 which, in the form of a proportion, is ~QJl :~BA 2 :: CD : BD; that is, the segments of the hypotenuse are proportional to the squares on the adjacent sides. 4. By the addition of (1) and (2), we have SI 2 + CA* m BC(BD + CD) = BO 2 ; that is, the sum of the squares on the sides about the right angle is equivalent to the square on the hypotenuse. This is another demonstration of Theorem 39, B. I. Hence the theorem ; if from the right angle of a right- angled triangle, etc. 88 GEOMETKY. BOOK III OF THE CIRCLE, AND THE INVESTIGATION OF THEO- REMS DEPENDENT ON ITS PROPERTIES. V, DEFINITIONS. 1. * A Curved Line is one whose consecutive parts, how- ever small, do not lie in the same direction. 2. A Circle is a plane figure bounded by one uniformly- curved line, all of the points of which are at the same distance from a certain point within, called the center. 3. The Circumference of a cir- cle is the curved line that bounds it. 4. The Diameter of a circle is a line passing through the center, and terminating at both extremities in the circumfer- ence. Thus, in the figure, is the center of the circle, the curved line AGrBD is the cir- cumference, and AB is a diameter. 5. The Radius of a circle is a line extending from the center to any point in the circumference. Thus, CD is a radius of the circle. 6. An Arc of a circle is any portion of the circum- ference. * The first six of the above definitions have been before given among the general definitions of Geometry, but it was deemed advisable to reinsert them here. BOOK III. 89 7. A Chord of a circle is the line connecting the ex- tremities of an arc. 8. A Segment of a circle is the portion of the circle on either side of a chord. Thus, in the last figure, ECrF is an arc, and EF is a chord of the circle, and the spaces bounded by the chord EF, and the two arcs EGrF and EDF, into which it divides the circumference, are segments. 9. A Tangent to a circle is a line which, meeting the circumference at any point, will not cut it on being produced. The point in which the tangent meets the circumference is called the point of tangency. 10. A Secant to a circle is a line which meets the cir- cumference in two points, and lies a part within and a part without the circumference. 11. A Sector of a circle is a portion of the circle included between any two radii and their intercepted arc. Thus, in the last figure, the line HL, which meets the circumference at the point D, but does not cut it, is a tangent, D being the point of tangency; and the line MN, which meets the circumference at the points P and Q, and lies a portion within and a portion without the circle, is a secant. The area bounded by the arc BJD, and the two radii OB, CD, is a sector of the circle. 12. A Circumscribed Polygon is one all of whose sides are tangent to the circumference of the circle ; and conversely, the circle is then said to be inscribed in the polygon. 13. An Inscribed Polygon is one the vertices of whose angles are all formed in the circumference of the circle ; and conversely, the circle is then said to be circumscribed about the polygon. 14. A Regular Polygon is one which is both equiangu- lar and equilateral. 8* 90 GEOMETRY. The last three definitions are illustrated by the last figure. THEOREM I. Any radius perpendicular to a chord, bisects the chord, and also the arc of the chord. Let AB be a chord, the center of the circle, and OE a radius perpen- dicular to AB ; then we are to prove that AB t=4 BB, and A E = EB. Since is the center of the circle, AO= BO, OB is common to the two A's AOB and BOB, and the angles at B are right angles; therefore the two A's ABO and BBO are equal, and AB = BB, which proves the first part of the theorem. !N"ow, as AB = BB, and BB is common to the two spaces, ABB and BBE, and the angles at B are right angles, if we conceive the sector OBE turned over and placed on OAE, OE retaining its position, the point B will fall on the point A, because AB = BB and A — BO; then the arc BE will fall on the arc AE\ otherwise there would be points in one or the other arc unequally distant from the center, which is impossible ; therefore, the arc A E = the arc EB, which proves the second part of the theorem. Hence the theorem. Oor. The center of the circle, the middle point of the chord AB, and of the subtended arc AEB, are three points in the same straight line perpendicular to the chord at its middle point. ISTow as but one perpen- dicular can be drawn to a line from a given point in that line, it follows: 1st. That the radius drawn to the middle point of any arc bisects, and is perpendicular to, the chord of the arc. BOOK III. 91 2d. That the perpendicular to the chord at its middle point passes through the center of the circle and the middle of the subtended arc. THEOREM II. Equal angles at the center of a circle are subtended by equal chords. Let the angle A OE = the angle BEO; then the two isosceles triangles, AOE, and EOB, are equal in all re- spects, and AE — EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords are equally distant from the center. Let AB and EF be equal chords, and the center of the circle. From 0, draw OG and OH, perpendicular to the respective chords. These perpendiculars will bisect the chords, (Th. 1), and we shall have A G = EH. We are now to prove that OG = OH. Since the A's EOH and AOG are right-angled, we have, (Th. 39, B. I), and, EH 2 + HO=~EC i ~AO\ AG' + GO' By subtracting these equations, member from mem- ber, we find that EH 1 — AG 1 -f HO 2 — GO 2 = ~E0 2 — AO 2 (1) But the chords are equal by hypothesis, hence their halves, EH and AG, are equal; also EO = AO, being radii of the circle. "Wherefore, 92 GEOMETRY. EE* — AG* = and, JSO — AO L = 0. These values in Equation ( 1 ) reduce it to EX? -GO_ 2 =0 or, E0 2 =G0 2 and, EO = GO. Hence the theorem. Oor. Under all circumstances we have MS 2 + EQ 2 = AG 2 + GO 2 , because the sum of the squares in either member of the equation is equivalent to the square of the radius of the circle. ISTow, if we suppose HQ greater than GO, then will HO 2 be greater than GO 2 . Let the difference of these squares be represented by d. Subtracting GO 2 from both members of the above equation, we have EE 2 +d = AG 2 whence, A G 2 > JEM 2 , and A G > HE. Therefore, AB, the double of AG, is greater than EF, the double of UE; that is, of two chords in the same or equal circles, the one nearer the center is the greater. The equation, ME 2 + EO 2 = AG 2 + ~G0 2 , being true, whatever be the position of the chords, we may suppose GO to have any value between and A 0, the radius of the circle. When GO becomes zero, the equation reduces to EE 2 +~E0 2 = AG 2 = B*; that is, under this supposition, AG coincides with AO, and AB becomes the diameter of the circle, the greatest chord that can be drawn in it. BOOK III. 93 THEOREM IV. A line tangent to the circumference of a circle is at right angles with the radius drawn to the point of contact. Let A be a line tangent to the circle at the point B, and draw the radius, EB, and the lines, AE and CE. Now, we are to prove that EB is per- pendicular to AC. Because B is the only point in the line AC which meets the circle, (Def. 9, B. II), any other line, as AE or CE, must be greater than EB; therefore, EB is the shortest line that can be drawn from the point E to the line AG; and EB is the perpendicu- lar to AC, (Th. 23, B.I). Hence the theorem. THEOREM V. In the same circle, or in equal circles, equal chords subtend or stand on equal portions of the circumference. Conceive two equal circles, and two equal chords drawn within them. Then, conceive one circle taken up and placed upon the other, center upon center, in such a po- sition that the two equal chords will fall on, and exactly coincide with, each other; the circles must also coin- cide, because they are equal ; and the two arcs of the two circles on either side of the equal chords must also coin- cide, or the circles could not coincide ; and magnitudes which coincide, or exactly fill the same space, are in all respects equal, (Ax. 10). Hence the theorem. 94 GEOMETRY. THEOREM VI, Through three given points, not in the same straight line, one circumference can be made to pass, and but one. Let A, B, and be three given points, not in the same straight line, and draw the lines AB and BO. If a circumference is made to pass through the two points A and B, the line AB will be a chord to such a circle ; and if a chord is bisected by a line at right angles, the bisecting line will pass through the center of the circle, (Cor., Th. 1) ; therefore, if we bisect the line AB, and draw DF, perpendicular to N, at the point of bisection, any circumference that can pass through the points, A and B, must have its center somewhere in the line DF. And if we draw EGr at right angles to BO at its middle point, any circumference that can pass through the points B and must have its center somewhere in the line EG. JS"ow, if the two lines, DF and ECr, meet in a common point, that point will be a center, about which a circumference can be drawn to pass through the three points, A, B, and 0, and DF and EG will meet in every case, unless they are parallel ; but they are not parallel, for if they were, it would follow (Th. 5, B. I) that, since DF is intersected at right angles by the line AB, it must also be intersected at right angles by the line BO, having a direction different from that of AB ; which is impossible, (Th. 7, B. I). Therefore the two lines will meet ; and, with the point H, at which they meet, as a center, and HB — HA = SO as a radius, one circumference, and but one, can be made to pass through the three given points. Hence the theorem. BOOK III. 95 THEOREM VII. If two circles touch each other, either internally or exter- nally, the two centers and the point of contact will he in one right line. Let two circles touch each other internally, as represented at A, and conceive AB to be a tangent at the common point A. Now, if a line, perpendicular to AB, be drawn from the point A, it must pass through the center of each circle, (Th. 4) ; and as but one perpendicular can be drawn to a line at a given point in it, A, C, and B, the point of contact and the two centers must be in one and the same line. Next, let two circles touch each other externally, and from the point of contact conceive the common tangent, AB, to be drawn. Then a line, AG, perpendicular to AB, will pass through the center of one circle, (Th. 4), and a per- pendicular, AB, from the same point, A, will pass through the center of the other circle ; hence, BAO and BAB are together equal to two right angles ; therefore CAB is one continued straight line, (Th. 3, B. I). Cor. "When two circles touch each other internally, the distance between their centers is equal to the difference of their radii ; and when they touch each other extern- ally, the distance between their centers is equal to the sum of their radii. THEOREM VIII. An angle at the circumference of any circle is measured by one half the arc on which it stands. In this work it is taken as an axiom that any angle whose vertex is at the center of a circle, is measured by 96 GEOMETRY. the arc on which it stands ; and we now proceed to prove that when the arcs are equal, the angle at the circumference is equal to one half the angle at the center. Let A OB be an angle at the center, and D an angle at the circumference, and at first suppose D in a line with A 0. We are now to prove that the angle A OB is double the angle D. The A DCB is an isosceles triangle, because OB = OB ; and its exterior angle, A OB, is equal to the two interior angles, B, and OBB, (Th. 12, B. I), and since these two angles are equal to each other, the angle AOB is double the angle at B. But A OB is measured by the arc AB ; therefore the angle B is measured by one half the arc AB. Next, suppose B not in a line with A 0, but at any point in the circum- ference, except on AB ; produce BO toE. Now, by the first part of this theorem, the angle EOB = 2EBB, also, BOA = 2EBA, by subtraction, AOB = 2 ABB. But AOB is measured by the arc AB; therefore ABB or the angle D, is measured by one half of the same arc. Hence the theorem. THEOREM IX. An angle in a semicircle is a right angle ; an angle in a segment greater than a semicircle is less than a right angle ; and an angle in a segment less than a semicircle is greater than a right angle. If the angle AOB is in a semicircle, the opposite seg- ment, ABB, on which it stands, is also a semicircle ; and the angle AOB is measured by one half the arc ABB. BOOK III. 97 (Th. 8) ; that is, one half of 180°, or 90°, which is the measure of a right angle. If the angle ACB is in a segment greater than a semicircle, then the opposite segment is less than a semi- circle, and the measure of the angle is less than one half of 180°, or less than a right angle. If the angle ACB is in a segment less than a semicircle, then the opposite segment, ABB, on which the angle stands, is greater than a semicircle, and its half is greater than 90° ; and, consequently, the angle is greater than a right angle. Hence the theorem. Cor. Angles at the circumference, and standing on the same arc of a circle, are equal to one another ; for all angles, as BAC, BBC, BBC, are equal, because each is measured by one half of the arc BC. Also, if the angle BBC is equal to CEG-, then the arcs BC and CG- are equal, be- cause their halves are the measures of equal angles. THEOREM X. The sum of two opposite angles of any quadrilateral in- scribed in a circle, is equal to two right angles. Let ACBD represent any quadri- lateral inscribed in a circle. The angle ACB has for its measure, one half of the arc ABB, and the angle ABB has for its measure, one half of the arc ACB; therefore, by addition, the sum of the two opposite angles at C and B, are together measured by one half of the whole circumference, or by 180 degrees, = two right angles. Hence the theorem. 9 98 GEOMETRY. THEOREM XI. An angle formed by a tangent and a chord is measured by one half of the intercepted arc. Let AB be a tangent, and AD a chord, and A the point of contact ; then we are to prove that the angle BAD is measured by one half of the arc AED. From A draw the radius A C; and from the center, 0, draw CE per- pendicular to AD. The l_BAD + [_DAO= 90°, (Th. 4). Also, ]^C+l_DAC= 90°, (Cor. 4, Th. 12, B. I). Therefore, by subtraction, BAD — (7=0; by transposition, the angle BAD = 0. But the angle 0, at the center of the circle, is measured by the arc AE, the half of AED ; therefore, the equal angle, BAD, is also measured by the arc AE, the half of AED. Hence the theorem. See Th. 13, for another proof. THEOREM XII. An angle formed by a tangent and a chord, is equal to an angle in the opposite segment of the circle. Let AB be a tangent, and AD a chord, and from the point of contact, A, draw any angles, as AOD, and AED, in the segments. Then we are to prove that [__ BAD =[__ACD, and [_ aAD = L AED. By Th. 11, the angle BAD is meas- ured by one half the arc AED ; and as the angle ACD is measured by one half of the same arc, (Th. 8), we have [_ BAD = [_ACD. BOOK III. 99 Again, as AEBO is a quadrilateral, inscribed in a circle, the sum of the opposite angles, AOB + AEB = 2 right angles. (Th. 10). Also, the sum of the angles BAB + BAG = 2 right angles. (Th. 1, B. I). By subtraction (and observing that BAB has just been proved equal to AOB), we have, AEB — BAG = 0. Or, by transposition, AEB = BAG. Hence the theorem. THEOREM XIII. Arcs of the circumference of a circle intercepted by paral- lel chords, or by a tangent and a parallel chord, are equal. Let AB and OB be parallel chords, and draw the diagonal, AB ; now, be- cause AB and OB are parallel, the angle BAB = the angle ABO (Th. 6, B. I) ; but the angle BAB has for its meas- ure, one half of the arc BB; and the angle ABO has for its measure, one half of the arc A 0, (Th. 8) ; and because the angles are equal, the arcs are equal ; that is, the arc BB = the arc AO. Next, let EF be a tangent, parallel to a chord, OB, and from the point of contact, G, draw GB. Since EF and OB are parallel, the angle OBG = the angle BGF. But the angle OBG has for its measure, one-half of the arc OG, (Th. 8) ; and the angle BGF has for its measure, one half of the arc GB, (Th. 11) ; therefore, these equal measures of equals must be equal ; that is, the arc OG = the arc GB. Hence the theorem. 100 GEOMETRY. THEOREM XIV. When two chords intersect each other within a circle, the angle thus formed is measured by one half the sum of the two intercepted arcs. Let AB and CD intersect each other within the circle, forming the two angles, E and E', with their equal vertical angles. Then, we are to prove that the angle E is measured by one half the sum of the arcs A and BD; and the angle E 1 is measured by one half the sum of the arcs AB and OB. First, draw AF parallel to CD, and FD will be equal to AC, (Th. 13); then, by reason of the parallels, |__ BAF = |_ E. But the angle BAF is measured by one half of the arc BDF; that is, one half of the arc BD plus one half of the arc AC Now, as the sum of the angles B and E' is equal to two right angles, that sum is measured by one half the whole circumference. But the angle E, alone, as we have just proved, is measured by one half the sum of the arcs BD and AC; therefore, the other angle, E 1 , is measured by one half the sum of the other parts of the circumference, AD + OB. Hence the theorem. THEOREM XV, When two secants intersect, or meet each other without a circle, the angle thus formed is measured by one half the dif- ference of the intercepted arcs. BOOK III. 101 Let BE and BE be two secants meeting at E ; and draw A F parallel to OB. Then, by reason of the parallels, the angle E, made by the intersection of the two secants, is equal to the angle BAF. But the angle BAF is measured by one half the arc BF; that is, by one half the difference be- tween the arcs BB and AC. Hence the theorem. THEOREM XVI. The angle formed by a secant and a tangent is measured by one half the difference of the intercepted arc. Let BQ be sl secant, and OB a tan- gent, meeting at 0. We are to prove that the angle formed at 0, is meas- ured by one half the difference of the arcs BB and BA. From A, draw AE parallel to OB ; then the arc AB = the arc BE; BB — BE = BE; and the [_BAE = L 0. But the angle B A E is measured by one half the arc BE, (Th. 8,) that is, by one half the difference between the arcs BB and AB; there- fore, the equal angle, 0, is measured by one half the arc BE. Hence the theorem. THEOREM XVII. When two chords intersect each other in a circle, the rect- angle contained by the segments of the one, will be equivalent to the rectangle contained by the segments of the other. 9* 102 GEOMETRY. Let AB and CD be two chords inter- secting each other in E. Then we are to prove that the rectangle AE x EB = the rectangle OE x ED. Draw the lines AD and CB, forming the two triangles AED and CEB. The angles B and D are equal, because they are each measured by one half the arc, AC. Also the angles A and are equal, because each is measured by one half the arc, DB ; and L AED = [_ CEB, because they are vertical angles ; hence, the triangles, AED and CEB, are equiangular and similar. But equiangular tri- angles have their sides about the equal angles propor- tional, (Cor. 1, Th. 17, B. II); therefore, AE and ED, about the angle E, are proportional to CE and EB, about the same or equal angle. That is, AE : ED : : CE : EB; Or, (Th. 19, B. n), AExEB= CEx ED. Hence the theorem. Cor. When one chord is a diameter, and the other at right angles to it, the rectangle contained by the segments of the diameter is equal to the square of one half the other chord; or one half of the bisected chord is a mean proportional be- tween the segments of the diameter. For, ADxDB*=FD x DE. But, if AB passes through the center, C, at right angles to FE, then FD = DE (Th. 1) ; and in the place of FD, write its equal, DE, in the last equation, and we have ADxDB = DE 2 , or, (Th. 3, B. IT), AD : DE : : DE : DB. Put, DE =x, CD = y, and CE = R, the radius of the circle. BOOK III. 103 Then AD = B—y, and DB = B + y. With this nota- tion, AD x DB = DE 2 becomes, (B — y) (B + y) — x 2 or, B 2 — y 2 = x 2 or, B 2 = x 2 +y 2 That is, the square of the hypotenuse of the right-angled triangle, DOE, is equal to the sum of the squares of the other two sides. THEOREM XVIII. If from a point without a circle, a tangent line be drawn to the circumference, and also any secant line terminating in the concave arc, the square of the tangent will be equivalent to the rectangle contained by the whole secant and its external seg- ment. Let A be a point without the circle DEGr, and let AD be a tangent and AE any secant line. Then we are to prove that AOxAE^AD 2 . In the two triangles, ADE and ADC, the angles ADO and AED are equal, since each is meas- ured by one half of the same arc, DO; the angle A is com-, mon to the two triangles ; their third angles are therefore equal, and the triaugles are equiangular and similar. Their homologous sides give the proportion AE : AD : : AD : AO whence, AE x AO= AD 2 Hence the theorem. Oor. If AE and AF are two secant lines drawn from the same point without the circumference, we shall have 104 GEO ME THY. ACx AE=AD 2 and, ABxAF=AD 2 hence, AC x AE = AB X AF, which, in the form of a proportion, gives AC : AF ::AB : AE. That is, ^6 secants are reciprocally proportional to their ex- ternal segments. Scholium. — By means of this theorem we can determine the diam- eter of a circle, when we know the length of a tangent drawn from a point without, and the external segment of the secant, which, drawn from the same point, passes through the center of the circle. Let Am be a secant passing through the center, and suppose the tangent AD to be 20, and the external seg- ment, An, of the secant to be 2. Then, if D denote the diameter, we shall have whence, Am x An - 2 (2 + D) = 4 + 2D = (20) 2 = 400, 22) =396, and 2) = 198. If An, the height of a mountain on the earth, and AD, the distance of the visible sea horizon, be given, we may determine the diameter of the earth. For example ; the perpendicular height of a mountain on the island of Teneriffe is about 3 miles, and its summit can be seen from ships when they are known to be 154 or 155 miles distant ; what then is the diameter of the earth ? Designate, as before, the diameter by 2>. Then Am = 3 + 2), and Am x An = 9 + 32). AD = 154, 5 ; hence, 9 + 32) = (154, 5) 2 = 23870. 25, from which we find D = 7953.T3, which differs but little from the true diameter of the earth. One source of error, in this mode of computing the diameter of the earth, is atmospheric refraction, the ex- planation of which does not belong here. BOOK III. 105 THEOREM XIX. If a circle he described about a triangle, the rectangle con- tained by two sides of the triangle is equivalent to the rectangle contained by the perpendicular let fall on the third side, and the diameter of the circumscribing circle. Let ABO be a triangle, AO and OB, the sides, OB the perpendicular let fall on the base AB, and OB the diameter of the circumscribing circle. Then we are to prove that AOx OB= OEx OB. The two A's, AOB and OEB, are equiangular, because [_A—[__B, both being measured by the half of the arc OB; also, ABO is a right angle, and is equal to OBB, an angle in a semi- circle, and therefore a right angle ; hence, the third angle, AOB = \_BOE, (Th. 12, Cor. 2, B. I). Therefore, (Cor., Th. 17, B. II), AO : OB :: OB : OB and, AOx BO= OB x OB. Hence the theorem ; if a circle, etc. Oor. The continued product of three sides of a triangle is equal to twice the area of the triangle into the diameter of its circumscribing circle. Multiplying both members of the last equation by AB, and we have, AO x BO x AB = OB x {AB x OB). But OB is the diameter of the circle, and (AB x OB) = twice the area of the triangle ; Therefore, AO X OB x AB = diameter multiplied by twice the area of the triangle. 105 GEOMETRY. THEOREM XX. The square of a line bisecting any angle of a triangle, to* gether with the rectangle of the segments into which it cuts the opposite side, is equivalent to the rectangle of the two sides including the bisected angle. Let ABO he a triangle, and CD a line bisecting the angle C. Then we are to prove that CD' + (AD x DB) = ACx OB. The two A's, AOE and CDB, are equiangular, because the angles E and B are equal, both being in the same segment, and the [_ ACE = BCD, by hypothesis. Therefore, (Th. 17, Cor. 1, B. H), AC : CE n CD : CB. But it is obvious that CE = CD -f DE, and by substi- tuting this value of CE, in the proportion, we have, AC : CD + DE :: CD : CB. By multiplying extremes and means, UD 2 + (DE x CD) = ACx CB. But by (Th. 17), DE x CD = AD x DB, and substituting, we have, CD 2 + (AD x DB) = ACx CB. Hence the theorem. THEOREM XXI. The rectangle contained by the two diagonals of any quad- rilateral inscribed in a circle, is equivalent to the sum of the tivo rectangles contained by the opposite sides of the quadri- lateral. Let ABCD be a quadrilateral inscribed in a circle; then we are to prove that AC x BD = (AB x DC) -f (AD x BC). From C y draw CE, making the angle DCE equal to BOOK III. 107 the angle A OB; and as the angle BAOis equal to the angle ODE, both being in the same seg- ment, therefore, the two triangles, DEC and ABC, are equiangular, and we have (Th. 17, Cor. 1, B. II), AB : AC :: BE : DC (1) The two A's, ABC and BEC, are equiangular; for the \__DAC= [__EBC, both being in the same segment; and the |_ BCA = [_ECB, for BCE = BCA; to each of these add the angle EC A, and BCA = ECB; therefore, (Th. 17, Cor. 1, B. II), AB : AC -.: BE : BO (2). By multiplying the extremes and means in proportions (1) and (2), and adding the resulting equations, we have, (AB x DC) + (AB x BO) = (BE + BE) x AC. But, DE + BE = BB ; therefore, (AB x DO) + (AD x BO) = AC x #Z). , Cor. When two adjacent sides of the quadrilateral aio equal, as AB and BO, then the resulting equation is, (AB x DC) + (AS x AD) = AC x BB; or, AB x (BO + 4i>) = AC x BB; or, AB : AC :: BB : DC+ AB. That is, owe of the two equal sides of the quadrilateral is to the adjoining diagonal, as the transverse diagonal is to the sum of the two unequal sides. THEOREM XXII. If two chords intersect each other at right angles in a cir- cle, the sum of the squares of the four segments thus formed is equivalent to the square of the diameter of the circle. Let AB and CD be two chords, intersecting each other at right angles. Draw BE parallel to EB, and draw DF and A F. Now, we are to prove that ~AE 2 +~EB 2 +~EC 2 + ED* =TAF\ 108 GEOMETRY. As BF is parallel to ED, ABF is a riglit angle, and therefore AFis a diam- eter, (Th. 9). Also, because BF is parallel to CD, CB = DF, (Th. 13). Because QEB is a right angle, OF 2 + FB 2 = OB 2 = DF 2 . Because AFD is a right angle, ~AF 2 +~ED 2 = AD 2 . Adding these two equations, we have, OF 2 +~FB 2 + AF 2 +~ED 2 = DF 2 + ~AD 2 . But, as AF is a diameter, and ADF a right angle, (Th.9), ___ DT+AD 2 = AF 2 ; therefore, OF 2 + i£# 2 + AF 2 + ED 2 - 27P 2 . Hence the theorem. Scholium. — If two chords intersect each other at right angles, in a circle, and their opposite extremities be joined, the two chords thus formed may make two sides of a right-angled triangle, of which the diameter of the circle is the hypotenuse. For, AD is one of these chords, and CB is the other ; and we have shown that CB = DF; and AD and DF are two sides of a right- angled triangle, of which AF is the hypotenuse ; therefore, AD and CB may be considered the two sides of a right-angled triangle, and AF its hypotenuse. THEOREM XXIII. If two secants intersect each other at right angles, the sum of their squares, increased by the sum of the squares of the two segments without the circle, will be equivalent to the square of the diameter of the circle. Let AF and ED be two secants in- tersecting at right angles at the point E. From B, draw BF parallel to CD, and draw AF and AD. !Now we are to prove that FA 2 + ED 2 + EB 2 +~E(f = AF 2 . BOOK III. 109 Because BF is parallel to CB, ABF is a right angle, and consequently A F is a diameter, and BC — BF; and because AF is a diameter, ABF is a right angle. As ABB is a right angle, ~AE 2 +W5 2 =AD 2 Also, FB 2 +JEC 2 =BC 2 =BF 2 By addition, A^ 2 +WD 2 +^ 2 +W 2 =A^ 2 +BF 2 =AF i Hence the theorem. THEOREM XXIV. If perpendiculars be drawn to each of the sides of a plane triangle, they will, when sufficiently produced, meet in a com- mon point. The three angular points of a triangle are not in the same straight line; consequently one circumference, and but one, may be made to pass through them. Conceive a triangle to be thus circumscribed. The sides of the triangle then become chords of the circum- scribing circle, and they are bisected by the perpendicu- lar radii, (Th. 6). Conversely: The perpendiculars bisecting the three sides of a triangle will meet in a common point, and that point will be the center of the circumscribing circle. Hence the theorem. THEOREM XXV. The sums of the opposite sides of a quadrilateral circum- scribing a circle are equal. Let ABCB be a quadrilateral circumscribed about a circle, whose center is 0. Then we are to prove that AB + BC=AB + BC. From the center of the circle draw OF and OF to the points of contact of the sides AB and BC. Then, 10 110 GEOMETRY. the two right-angled triangles, OEB and OFB, are equal, because they have the hypotenuse OB common, and the side OF = OE; therefore, BE = BF, (Cor., Th. 23, B. I). In like manner we can prove that AE=AH, CF= CG, sindDG^DK Now, taking the equation BE = BF, and adding to its first mem- ber CG, and to its second the equal line OF. we have, BE + CG = BF + OF (1) The equation AE=AE, by adding to its first member DG } and to the second the equal line, BE, gives AE+BG=AE+BE (2) By the addition of (1) and (2), we find that BE + AE+ CG + BG = BF + CF+AH+BH. That is, AB + CD +BC+ AD. Hence the theorem. BOOK IV. Ill BOOK IV. PROBLEMS. In this section, we have, in most instances, merely shown the construction of the problem, and referred to the theorem or theorems that the student may use, to prove that the object is attained by the construction. In obscure and difficult problems, however, we have gone through the demonstration as though it were a theorem. PROBLEM I. To bisect a given finite straight line. Let AB be the given line, and from its extremities, A and B, with any radius greater than one half of AB, (Postulate 3), describe arcs, cutting A — each other in n and m. Draw the line nm ; and (7, where it cuts AB, will be the middle of the given line. Proof, (B. I, Th. 18, Sch. 2). PROBLEM II. To bisect a given angle. Let ABO be the given angle. With any radius, and B as a center, describe the arc AC. From A and <7, as centers, with a radius greater than one half of AG, de- scribe arcs, intersecting in n ; join B and n ; the joining line will bisect the given angle. Proof, (Th. 21, B. I). •k x A OF THE ' *V lih!l\/rr>~ 112 GEOMETRY. A n m B Proof, PROBLEM III. From a given point in a given line, to draw a perpendicular to that line. Let AB be the given line, and O the given point. Take n and m, equal distances on opposite sides of 0; and with the points m and n, as centers, and any radius greater than nO or mO, describe arcs cutting each other in S, Draw SO, and it will be the perpendicular required. (B. I, Th. 18, Sch. 2). The following is another method, which is preferable, when the given point, 0, is at or near the end of the line. Take any point, 0, which is mani- festly one side of the perpendicular, as a center, and with 00 as a radius, describe a circum- ference, cutting AB in m and 0. Draw mn through the points m and 0, and meeting the arc again in n ; mn is then a diameter to the circle. Draw On, and it will be the perpendicular required. Proof, (Th. 9, B. III). A m PROBLEM IV. From a given point without a line, to draw a 'perpendicular to that line. Let AB be the given line, and O the given point. From draw any oblique line, as On, Find the mid- dle point of On by Problem 1, and with that point, as a center, describe a semicircle, having On as a diam- eter. From m, where this semi-cir- cumference cuts AB, draw Om, and it will be the perpen- dicular required. Proof, (Th. 9, B. III). m B BOOK IV. 113 PROBLEM V. At a given point in a line, to construct an angle equal to a given angle. Let A be the point given in the line AB, and DOE the given angle. With C as a center, and any radius, OF, draw the arc FD. With A as a center, and the radius A F= OF, describe an indefinite arc ; and with J 7 as a center, and FG- as a radius, equal to FD, describe an arc, cutting the other arc in Gr, and draw A G-; GrAF will be the angle required. Proof, (Th. 5, B. III). PROBLEM VI. From a given point, to draw a line parallel to a given line. Let A be the given point, and BO the given line. Draw A C, making an angle, A OB; and from the given point, A, in the line AC, draw the angle CAD = ACB, by Problem 5. Since AJD and BO make the same angle with AC, they are, therefore, parallel, (B. I, Th. 7, Cor. 1). PROBLEM VII. To divide a given line into any number of equal parts. Let AB represent the given line, and let it be required to di- vide it into any number of equal parts, say live. From one end of the line A, draw AJD, indefinite in both length and position. Take any convenient distance in the di- 10* h 114 GEOMETRY. viders, as Aa, and set it off on the line AD, thus making the parts Aa, ab, be, etc., equal. Through the last point, e, draw EB, and through the points a, b, c, and d, draw parallels to eB, by Problem 6 ; these parallels will divide the line as required. Proof, (Th. 17, Book IT). PROBLEM VIII, To find a third proportional to two given lines. Let AB and A be any two lines. Place them at any angle, and draw CB. On the greater line, AB, take AD — AO, and through D, draw DE parallel to BO', AE is the third proportional required. Proof, (Th. 17, B. n). PROBLEM IX. To find a fourth proportional to three given lines. Let AB, AC, AD, represent the A "~ three given lines. Place the first two at any angle, as BAO, and draw BO. On AB place AD, and from the point D, draw DE parallel to BO, by Problem 6 ; AE will be the fourth proportional required. Proof, (Th. 17, B. II). PROBLEM X. To find the middle, or mean proportional, between two given lines. BOOK IV. 115 Place AB and BC in one right line, and on A C, as a diameter, de- scribe a semicircle, (Postulate 3), and from the point B, draw BD at right angles to AC, (Problem 3); BD is the mean proportional re- quired. Proof, (B. m, Th. 17, Cor.). PROBLEM XI. To find the center of a given circle. Draw any two chords in the given cir- cle, as AB and CD, and from the middle points, m and n, draw perpendiculars to AB and CD ; the point at which these two perpendiculars intersect will be the center of the circle. Proof, (B. m, Th. 1, Cor.). PROBLEM XII. To draw a tangent to a given circle, from a given either in or without the circumference of the circle. When the given point is in the cir- cumference, as A, draw the radius A C, and from the point A, draw AB per- pendicular to AC; AB is the tangent required. Proof, (Th. 4, B. HI). "When the given point is without the circle, as A, draw AC to the center of the circle ; on i(J, as a diameter, describe a semicircle ; and from B, where the semi-circumfer- ence cuts the given circumference, draw AB, and it will be tangent to the circle. Proof, (Th. 9, B. Ill), and, (Th. 4, B. III). point, 116 GEOMETRY. PROBLEM XIII. On a given line, to describe a segment of a circle, that shall contain an angle equal to a given angle. Let AB be the given line, and O the given angle. At the ends of the given line, form angles DAB, DBA, each equal to the given angle, O. Then draw AE and BE perpendiculars to AD and BD ; and with E as a center, and EA, or EB, as a radius, describe a circle ; then AFB will be the segment required, as any angle F, made in it, will be equal to the given angle, O. Proof, (Th. 11, B. HI), and (Th. 8, B. LEI). PROBLEM XIV. ■ From any given circle to cut a segment, that shall contain a given angle. Let be the given angle. Take any point, as A, in the circumfer- ence, and from that point draw the tangent AB ; and from the point A, in the line AB, construct the angle BAD = 0, (Problem 5), and AED is the segment required. Proof, (Th. 11, B. IH), and (Th. 8, B. III). PROBLEM XV. To construct an equilateral triangle on a given straight line. Let AB be the given line; from the extremities A and B, as centers, with a radius equal to AB, describe arcs cutting each other at 0. From 0, the point of intersection, draw OA and CB; ABO will be the triangle required. The construction is a sufficient demonstration. Or, (Ax. 1). BOOK IV. 117 PROBLEM XVI. To construct a triangle, having its three sides equal given lines, any two of which shall be greater than the Let AB, OB, and EF, represent the E three lines. Take any one of them, as c AB, to be one side of the triangle. From A, as a center, with a radius equal to CD, describe an arc ; and from B, as a center, with a radius equal to EF, describe an- other arc, cutting the former in n. Draw An and Bn, and AnB will be the A re- quired. Proof, (Ax. 1). to three third. F D PROBLEM XVII. To describe a square on a given line. Let AB be the given line ; and from the extremities, A and B, draw A and BB per- c pendicular to AB. (Problem 3.) From A, as a center, with AB as radius, strike an arc across the perpendicular at C; t and from O draw OB parallel to AB ; AOBB is the square required. Proof, (Th. 26, B. I). PROBLEM XVIII. To construct a rectangle, or a parallelogram, whose adja- cent sides are equal to two given lines. Let AB and A be the two given A c lines. From the extremities of one A # line, draw perpendiculars to that line, as in the last prob- lem ; and from these perpendiculars, cut off portions equal to the other line ; and, by a parallel, complete the figure. 118 GEOMETRY. When the figure is to be a parallelogram, with oblique angles, describe the angles by Problem 5. Proof, (Th. 26, B. I). PROBLEM XIX. To describe a rectangle that shall be equivalent to a given square, and have a side equal to a given line. Let AB be a side of the given square, c D and CD one side of the required rect- A B angle. E p Find the third proportional, FF, to CD and AB, (Prob- lem 8). Then we shall have CD : AB :: AB : FF. Construct a rectangle with the two given lines, CD and FF, (Problem 18), and it will be equal to the given square, (Th. 3, B. II). PROBLEM XX. To construct a square that shall be equivalent to the differ- ence of two given squares. Let A represent a side of the greater of two given squares, and B a side of the less square. On A, as a diameter, describe a semicircle, and from one extremity, p, as a center, with a radius equal to B, describe an arc, n, and, from the point where it cuts the circumference, — - — draw mn and np ; np is the side of a square, which, when constructed, will be equal to the difference of the two given squares, (Problem 17). Proof, (Th. 9, B. Ill, and Th. S6, B. I.) To construct a square equivalent to the sum of two given squares, we have only to draw through any point two lines at right angles, and lay off on one a distance equal to the side of one of the squares, and on the other BOOK IV. 119 a distance equal to the side of the other. The straight line connecting the extremities of these lines will be the side of the required square, (Th. 36, B. I). PROBLEM XXI. To divide a given line into two parts, which shall be in the ratio of two other given lines. M^ Ni- Let AB be the line A ~ HB to be divided, and M and N the lines hav- ing the ratio of the required parts of AB. From the extremity A draw AZ), making any angle with AB, and take AC = M, and CD = N. Join the points D and B by a straight line, and through C draw Ca parallel to BD. Then will the point Gf divide the line AB into parts having the required ratio. (Proof, Th. 17, B. II). Or, having drawn AD, lay off A C = M, and through B draw B V parallel to AD, making it equal to N, and join C and V by a line cutting AB in the point (7. Then the two triangles ACGr and GrBV are equiangu- lar and similar, and their homologous sides give the proportion, Aa : GB : AC :: BV :: Mi N The line AB is therefore divided, at the point Q-, into parts which are in the ratio of the lines M and N", 120 GEOMETKY. PROBLEM XXII. To divide a given line into any number of parts, having to each other the ratios of other given lines. Let AB be the given M line to be divided, and Nl M, JST, P, etc., the lines p > to which the parts of AB are to be propor- tional. Through the point A draw an indefinite line, making, with AB, any conve- nient angle, and on this line lay off from A the lines M, JV, P, etc., successively. Join the extremity of the last line to the point B by a straight line, parallel to which draw other lines through the points of division of the indefinite line, and they will divide the line AB at the points 0, D, etc., into the required parts. (Proof, Th. 17, B. II). PROBLEM XXIII. To construct a square that shall be to a given square, as a line, M, to a line, N. Place M and N in a line, and on the sum describe a semicir- cle. From the point where the two lines meet, draw a perpen- dicular to meet the circumfer- ence in A. Draw Am and An, and produce them indefinitely. On Am or Am produced, take AB = to the side of the given square ; and from B, draw BO parallel to mn; A is a side of the required square. For, Am :A^ 2 :: AB 2 : ~AC\ (Th. 17, B. II). Also, Am :An ::M : JV, (Th. 25, B.II. Sch.). Therefore, A& : A C 2 : : M : 1ST, (Th. 6, B. II). BOOK IV. 121 PROBLEM XXIV. To cut a line into extreme and mean ratio ; that is, so that the whole line shall be to the greater part, as that greater part is to the less. Remark. — The geometrical solution of this problem is not imme- diately apparent, but it is at once suggested by the form of the equa- tion, which a simple algebraic analysis of its conditions leads to. Bepresent the line to be divided by 2a, the greater part by x, and consequently the other, or less part, by 2a — x. Now, the given line and its two parts are required, to satisfy the following proportion : 2a : x : : x : 2a — x whence, x 2 = 4a 2 — 2ax By transposition, x 2 -f 2ax = 4a 2 = (2a) 2 If we add a 2 to both members of this equation, we shall have, x 2 -f 2ax + a 2 = (2af + a 2 ,or, (x +df = (2af + a 2 This last equation indicates that the lines represented by (x + a), 2a, and a, are the three sides of a right- angled triangle, of which (x + a) is the hypotenuse, the given line, 2a, one of the sides, and its half, a, the other. Therefore, let AB represent the given line, and from the extremity, B, draw BO at right angles to AB, and make it equal to one half of AB. With 0, as a center, and radius CB, describe a circle. Draw A and pro- duce it to F. With A as a center and AB as a radius, describe the arc BE) this arc will divide the line AB, as required. We are now to prove that AB : AB : : AB : EB 11 122 GEOMETRY. By Scholium to Th. 18, B. m, we have, AF x AD = AB? or, AF : AB : : AB : AD Then, (by Cor., Th. 8, Book II), we may have, (AF— AB) : AB :: (AB — AD) : AD Since CB = \AB = JD.F; therefore, AB = DJ 7 . Hence, AF — AB = AF — DF = AD = AF. Therefore, AF : AB :: FB : AF By taking the extremes for the means, we have, AB : AF : : AF : FB. PROBLEM XXV. To describe an isosceles triangle, having its two equal angles each double the third angle, and the equal sides of any given length. Let AB be one of the equal sides of the required triangle; and from the point A, with the radius AB, describe an arc, BD. Divide the line AB into extreme and mean ratio by the last problem, and sup- pose C the point of division, and A the greater segment. From the point B, with AC, the greater segment, as a radius, describe another arc, cutting the arc BD in D. Draw BD, DC, and DA. The triangle ABD is the tri- angle required. As AC — BD, by construction ; and as AB is to A C as A C is to B C, by the division of AB; therefore AB : BD : : BD : BC Now, as the terms of this proportion are the sides of the two triangles about the common angle, B, it follows, (Cor. 2, Th. 17, B. II), that the two triangles, ABD and BOOK IV. 123 BBC, are equiangular; but the triangle ABB is isos- celes; therefore, BBC is isosceles also, and BB = BO; but BB m AC: hence, BC = AC, (Ax. 1), and the tri- angle ACB is isosceles, and the [_ CBA = [_ J.. But the exterior angle, BCB m CBA + A, (Th. 12, B. I). Therefore, [_BCB, or its equal \__B = L CBA ■+■[__ A; or the angle B = 2[__A. Hence, the triangle ABB has each of its angles, at the base, double of the third angle. Scholium. — As the two angles, at the base of the triangle ABD, are equal, and each is double the angle A, it follows that the sum of the three angles is Jive times the angle A. But, as the three angles of every triangle are always equal to two right angles, or 180°, the angle A must be one fifth of two right angles, or 36° ; therefore, BD is a chord of 36°, when AB is a radius to the circle ; and ten such chords would extend exactly round the circle, or would form a decagon. PROBLEM XXVI. Within a given circle to inscribe a triangle, equiangular to a given triangle. Let ABC be the circle, and ale the given triangle. From any point, as A, draw BB tan- gent to the given circle at A, (Problem 12). From the point A, in the line AB, lay off the angle BAC = the angle b, (Problem 5), and the angle BAB = the angle c, and draw BC The triangle ABC is inscribed in the circle; it is equi- angular to the triangle abc, and hence it is the triangle required. Proof, (Th. 12, B. III). 124 GEOMETRY. PROBLEM XXVII. To describe a regular pentagon in a given circle. 1st. Describe an isosceles tri- angle, abc, having each of the equal angles, b and c, double the third angle, a, by Problem 25. 2d. Inscribe the triangle, ABO, in the given circle, equi- angular to the triangle abc, by Problem 26 ; then each of the angles, B and 0, is double the angle A. 3d. Bisect the angles B and 0, by the lines BB and OE, (Problem 2), and draw AE, EB, CB, BA; and the figure AEBCB is the pentagon required. By construction, the angles BAG, ABB, BBC, BOE, EGA, are all equal ; therefore, (B. HI, Th. 9, Scho.), the arcs, BO, AB, BO, AE, and EB, are all equal; and if the arcs are equal, the chords AE, EB, etc., are equal. Scholium. — The arc subtended by one of the sides of a regular pen- 360° tagon, being one fifth of the whole circumference, is equal to — — =72°* PROBLEM XXVIII. To describe a regular hexagon in a circle. Draw any diameter of the circle, as AB, and from one extremity, B, draw BB equal to BO, the radius of the circle. The arc, BB, will be one sixth part of the whole circumference, and the chord BB will be a side of the regu- lar polygon of six sides. In the A OBB, as OB = OB, and BB = OB by con- struction, the A is equilateral, and of course equiangular. Since the sum of the three angles of every A is equal to two right angles, or to 180 degrees, when the E^- — ^p / / \ BOOK IV. 125 three angles are equal to one another, each one of them must be 60 degrees ; but 60 degrees is a sixth part of 360 degrees, the whole number of degrees in a circle ; therefore, the arc whose chord is equal to the radius, is a sixth part of the circumference ; and, if a polygon of six equal sides be inscribed in a circle, each side will be equal to the radius. Scholium. — Hence, as BD is the chord of 60°, and equal to BC qt CD, we say generally, that the chord of 60° is equal to radius. PROBLEM XXIX. To find the side of a regular polygon of fifteen sides, which may be inscribed in any given circle. Let CB be the radius of the given circle; divide it into extreme and mean ratio, (Problem 24), and make BD equal to CB, the greater part; then BD will be a side of a regular polygon of ten sides, (Scholium to Problem 25). Draw BA = to CB, and it will be a side of a polygon of six sides. Draw DA, and that line must be the side of a polygon which cor- responds to the arc of the circle expressed by \ less ^, of the whole circumference ; or J — -^ = g % = T ^ ; that is, one-fifteenth of the whole circumference ; or, DA is a side of a regular polygon of 15 sides. But the 15th part of 360° is 24° ; hence the side of a regular inscribed polygon of fifteen sides is the chord of an arc of 24°. PROBLEM XXX In a given circle to inscribe a regular polygon of any num- ber of sides, and then to circumscribe the circle by a similar polygon. 11* 126 GEOMETRY. Let the circumference of the circle, whose center is 0, be divided into any number of equal arcs, as AmB, Bw(7, OoD, etc. ; then will the polygon abode, etc., bounded by the chords of these arcs, be regu- lar and inscribed ; and the poly- gon ABODE, etc., bounded by the tangents to these arcs at their middle points m, n, o, etc, be a similar circumscribed polygon. First — The polygon abode, etc., is equilateral, because its sides are the chords of equal arcs of the same circle, (Th. 5, B. Ill) ; and it is equi- angular, because its angles are inscribed in equal segments of the same circle, (Th. 8, B. III). Therefore the poly- gon is regular, (Def. 14, B. Ill), and it is inscribed, since the vertices of all its angles are in the circumference of the circle, (Def. 13, B. HI). Second. — If we draw the radius to the point of tangency of the side AB of the circumscribed polygon, this radius is perpendicular to AB, (Th. 4, B. Ill), and also to the chord ah, (B. Ill, Th. 1, Cor.) ; hence AB is parallel to ah, and for the same reason BO is parallel to bo ; therefore the angle ABO is equal to the angle abo, (Th. 8, B. I). In like manner we may prove the other angles of the circumscribed polygon, each equal to the corresponding angle of the inscribed polygon. These polygons are therefore mutually equiangular. Again, if we draw the radii Om and On, and the line OB, the two A's thus formed are right-angled, the one at m and the other at n, the side OB is common and Om is equal to On ; hence the difference of the squares described on OB and Om is equivalent to the difference of the squares described on OB and On. But the first difference is equivalent to the square described on Bm, and the second difference is equivalent to the square described BOOK IV. 127 on Bn ; hence Bm is equal to Bn, and the two right- angled triangles are equal, (Th. 20, B. I), the angle BOm opposite the side Bm being equal to the angle BOn, op- posite the equal side Bn. The line OB therefore passes through the middle point of the arc mbn ; but because m and n are the middle points of the equal arcs amb and bne, the vertex of the angle abe is also at the middle point of the arc mbn. Hence the line OB, drawn from the center of the circle to the vertex of the angle ABO, also passes through the vertex of the angle abc. By pre- cisely the same process of reasoning, we may prove that 00 passes through the point c, OD through the point d, etc. ; hence the lines joining the center with the vertices of the angles of the circumscribed polygon, pass through the vertices of the corresponding angles of the inscribed polygon ; and conversely, the radii drawn to the vertices of the angles of the inscribed polygon, when produced, pass through the vertices of the corresponding angles of the circumscribed polygon. Now, since ab is parallel to AB, the similar A's abO and ABO, give the proportion Ob : OB :: ab : AB, and the A's, bcO and BOO, give the proportion Ob : OB : : be : BO. As these two proportions have an antecedent and con- sequent, the same in both, we have, (Th. 6, B. II), ab : AB : : be : BO. In like manner we may prove that be : BO : : cd : OB, etc., eta The two polygons are therefore not omy equiangular, but the sides about the equal angles, taken in the same order, are proportional ; they are therefore similar, (Def. 16, B. n). 128 GEOMETRY. Cor. 1. To inscribe any regular polygon in a circle, we have only to divide the circumference into as many equal parts as the polygon is to have sides, and to draw the chords of the arcs ; hence, in a given circle, it is possible to inscribe regular polygons of any number of sides whatever. Having constructed any such polygon in a given circle, it is evident, that by changing the radius of the circle without changing the number of sides of the polygon, it may be made to represent any regular poly- gon of the same name, and it will still be inscribed in a circle. As this reasoning is applicable to regular poly- gons of whatever number of sides, it follows, that any regular polygon may be circumscribed by the circumference of a circle. Cor. 2. Since ab, be, cd, etc., are equal chords of the same circle, they are at the same distance from the center, (Th. 3, B. Ill) ; hence, if with as a center, and Ot, the distance of one of these chords from that point, as a radius, a circumference be described, it will touch all of these chords at their middle points. It follows, therefore, that a circle may be inscribed within any regular polygon. Scholium. — The center, 0, of the circle, may be taken as the center of both the inscribed and circumscribed polygons; and the angle A OB, included between lines drawn from the center to the extremities of one of the sides AB, is called the angle at the center. The perpen- dicular drawn from the center to one of the sides is called the Apothem of the polygon. Cor. 3. The angle at the center of any regular polygon is equal to four right angles divided by the number of sides of the polygon. Thus, if n be the number of sides of the polygon, the angle at the center will be expressed . 360° by . n Cor. 4. If the arcs subtended by the sides of any regular inscribed polygon be bisected, and the chords of these semi-arcs be drawn, we shall have a regular BOOK IV. 129 inscribed polygon of double the number of sides. Thus, from the square we may pass successively to regular inscribed polygons of 8, 16, 32, etc., sides. To get the corresponding circumscribed polygons, we have merely to draw tangents at the middle points of the arcs sub- tended by the sides of the inscribed polygons. Cor. 5. It is plain that each inscribed polygon is but a part of one having twice the number of sides, while each circumscribed polygon is but a part of one having one half the number of sides. 130 GEOMETRY. BOOK V ON THE PROPORTIONALITIES AND MEASUREMENT OF POLYGONS AND CIRCLES. PROPOSITION I. — THEOREM. The area of any circle is equal to the product of its radius by one half of its circumference. Let OA be the radius of a circle, and AB a very small portion of its circumference; then A OB will be a sector. "We may conceive the whole circle made up of a great number of such sectors; and when each sector is very small, the arcs AB, BD, etc., each one taken separately, may be considered a right line ; and the sectors CAB, CBD, etc., will be triangles. The triangle, AOB, is measured by the product of the base, AC, multiplied into one half the altitude, AB, (Th. 33, Book I) ; and the triangle BOD is measured by the pro- duct of BO, or its equal, AO, into one half BD; then the area, or measure of the two triangles, or sectors, is the product of AO, multiplied by one half of AB plus one half of BD, and so on for all the sectors that compose the circle ; therefore, the area of the circle is measured by the product of the radius into one half the circumference. BOOK V. 131 PROPOSITION II. — THEOREM. Circumferences of circles are to one another as their radii, and their areas are to one another as the squares of their radii. Let CA be the radius of a circle, and Oa the radius of another circle. Conceive the two circles to be so placed upon each other so as to have a common center. Let AB be such a certain definite portion of the circumference of the larger circle, that m times AB will represent that cir- cumference. But whatever part AB is of the greater circumference, the same part ah' is of the smaller; for the two circles have the same number of degrees, and are of course sus- ceptible of division into the same number of sectors. But by proportional triangles we have, CA : Ca : : AB : ab Multiply the last couplet by m, (Th. 4, B. II), and we have CA : Ca :: mAB : mab. That is, the radius of one circle is to the radius of another, as the circumference of the one is to the circumference of the other. To prove the second part of the theorem, let C repre- sent the area of the larger circle, and c that of the smaller ; now, whatever part the sector CAB is of the circle C, the sector Cab is the corresponding part of the circle c. That is, Cic : : CAB : Cab, but, CAB : Cab : : (CAf : (Ca) 2 , (Th. 20, B. II). Therefore, C : c : : (CAf : (Ca) 2 , (Th. 6, B. H). That is, the area of one circle is to the area of another, a% 132 GEOMETEY. the square of the radius of the one is to the square of the radius of the other. Hence the theorem. Cor. If : e :: (OAf : (Co) 2 , then, : c :: 4 (Oaf : 4 (Ob) 2 . But 4 (OAf is the square of the diameter of the larger circle, and 4 (Oaf is the square of the diameter of the smaller. Denoting these diameters respectively by D and d, we have, O : c : : D 2 : d\ That is, the areas of any two circles are to each other, as the squares of their diameters. Scholium. — As the circumference of every circle, great or small, is assumed to be the measure of 360 degrees, if we conceive the circum- ference to be divided into 360 equal parts, and one such part repre- sented bjAB on one circle, or ab on the other, AB and ab will be very- near straight lines, and the length of such a line as AB will be greater or less, according to the radius of the circle ; but its absolute length cannot be determined until we know the absolute relation between the diameter of a circle and its circumference. PROPOSITION III. — THEOREM. When the radius of a circle is unity, its area and semi- circumference are numerically equal. Let R represent the radius of any circle, and the Greek letter, *, the half circumference of a circle whose radius is unity. Since circumferences are to each other as their radii, when the radius is R, the semi-circumference will be expressed by «R. Let m denote the area of the circle of which R is the radius ; then, by Theorem 1, we shall have, for the area of this circle, *R? = m, which, when R = 1, reduces to ic = m. This equation is to be interpreted as meaning that the semi-circumference contains its unit, the radius, as many BOOK V. 133 times as the area of the circle contains its unit, the square of the radius. Remark. — The celebrated problem of squaring the circle has for its object to find a line, the square on which will be equivalent to the area of a circle of a given diameter ; or, in other words, it proposes to find the ratio between the area of a circle and the square of its radius. An approximate solution only of this problem has been as yet dis- covered, but the approximation is so close that the exact solution is no longer a question of any practical importance. PROPOSITION IV. — PROBLEM. Given, the radius of a circle unity, to find the areas of regular inscribed and circumscribed hexagons. Conceive a circle described with the radius QA, and in this circle inscribe a regular polygon of six sides (Prob. 28, B. IV), and each side will be equal to the radius QA ; hence, the whole perimeter of this poly- gon must be six times the ra- dius of the circle, or three times the diameter. The chord bd is bisected by QA. Produce Ob and Qd, and through the point A y draw BD parallel to bd ; BD will then be a side of a regular polygon of six sides, circumscribed about the circle, and we can compute the length of this line, BD, as follows : The two triangles, Cbd and QBD, are equiangular, by construction ; therefore, Oa : bd :: QA : BD. Now, let us assume QA = QD = the radius of the circle, equal unity ; then bd = 1, and the preceding pro- portion becomes - Qa : 1 :: 1 : BD (1) In the right-angled triangle Qad, we have, Qa 2 + ad 2 = Qd 2 , (Th. 39, B. I). That is, Qa 2 -f J = 1, because Qd = 1, and ad = J. 12 134 GEOMETRY. Whence, Ca = J 2 x 3 = 2^3. V3 3 2 : :2:: 3:4:: 9 : 2^3 = 3.46410161 BOOK V. 135 number of sides. Also denote by p f the area of the inscribed polygon of double the number of sides, and by P f that of the corresponding circumscribed polygon. Now, if the arc KAL be some exact part, as one-fourth, one fifth, etc., of the circumference of the circle, of which Q is the center and QA the radius, then will KL be the side of a regular inscribed polygon, and the triangle KQL will be the same part of the whole polygon that the arc KAL is of the whole circumference, and the triangle QBB will be a like part of the circumscribed polygon. Draw QA to the point of tangency, and bisect the angles ACB and AQB, by the lines QGr and QB, and draw KA. It is plain that the triangle AOK is an exact part of the inscribed polygon of double the number of sides, and that the A EQG- is a like part of the cir- cumscribed polygon of double the number of sides. Repre- sent the area of the A LQK by a, and the area of the A BOB by b, that of the A ACK by x, and that of the A BQGr by y, and suppose the A's, KCL and BBC, to be each the nth part of their respective polygons. Then, na = p, nb = P, 2nx = p f , and, Zny = P f ; But, by (Th. 33, B. I), we have CM.MK=*a (1) QA . AB = b (2) QA . MK=2x (3) Multiplying equations ( 1 ) and ( 2 ) ? member by member, we have (CM . AB) x (QA . MK) =ab (4) 136 GEOMETRY. From the similar A's CMK and CAD, we have CM : MK : : CA : AD whence CM . AD = (Li . ME" But from equation ( 3 ) we see that each member of this last equation is equal to 2x; hence equation (4) becomes 2# . 2# = ab If we multiply both members of this by n* = n . n, we shall have 4n 2 x* = na.nb = p.P or, taking the square root of both members, 2nx = s/pl? That is, the area of the inscribed polygon of double the number of sides is a mean proportional between the areas of the given inscribed and circumscribed polygons p and P. Again, since CE bisects the angle ACD, we have, by, (Th. 24, B. II), AE : ED hence, AE : AE +ED CA : CD CM: CK CM: CA CM: CM+ CA. Multiplying the first couplet of this proportion by CA, and the second by MK, observing that AE -f ED = AD, we shall have AE.CA : AD.CA :: CM.MK : (CM + CA) MK. But AE. CA measures the area of the A CEG-, which we have called y, AD.CA = A CBD = 5, CM.MK = A CKL = a, and (CM + CA)MK = A CMK, and CAK = a 4- 2a:, as is seen from equations ( 1 ) and ( 3 ). Therefore the above proportion becomes y : b : : a : a + 2x. Multiplying the first couplet by 2n, and the second by n, we shall have BOOK V. That is, whence, Zny : 2nb : : na : na + 2nx P' : 2P : : p : p + p' 137 and as the value of p f has been previously found equal to v'Pp, the value of P f is known from this last equation, and the problem is completely solved. PROPOSITION VI. — PROBLEM. To determine the approximate numerical value of the area of a circle, when the radius is unity. We have now found, (Prob. 4), the areas of regular inscribed and circumscribed hexagons, when the radius of the circle is taken as the unit ; and Prob. 5 gives us formulae for computing from these the areas of regular inscribed and circumscribed polygons of twelve sides, and from these we may again pass to polygons of twenty-four sides, and so on, without limit. Now, it is evident that, as the number of sides of the inscribed polygon is increased, the polygon itself will increase, gradually approaching the circle, which it can never sur- pass. And it is equally evident that, as the number of sides of the circumscribed polygon is increased, the poly- gon itself will decrease, gradually approaching the circle, less than which it can never become. The circle being included between any two corres- ponding inscribed and circumscribed polygons, it will differ from either less than they differ from each other ; and the area of either polygon may then be taken as the area of the circle, from which it will differ by an amount less than the difference between the polygons. It is also plain that, as the areas of the polygons ap- proach equality, their perimeters will approach coinci- dence with each other, and with the circumference of the circle. 12* 138 GEOMETKY. Assuming the areas already found for the inscribed and circumscribed hexagons, and applying the formulae of Prob. 5 to them and to the successive results ob- tained, we may construct the following table : NUMBER OF SIDES. INSCRIBED POLYGONS. CIRCUMSCRIBED POLYGONS. 6 ^v^ 2.59807621 2^3=3.46410161 19 12 3 = 3.0000000 7r=—^= 3.2153904 2+v/3 6 24 v^l+^f = 3 - 1058286 3.1596602 48 3.132628T 3.1460863 96 3.1393554 3.1427106 192 3.1410328 3.1418712 384 3.1414519 3.1416616 768 3.1415568 3.1416092 1536 3.1415829 3.1415963 3072 3.1415895 3.1415929 6144 3.1415912 3.1415927 Thus we have found, that when the radius of a circle is 1, the semi-circumference must be more than 3.1415912, and less than 3.1415927 ; and this is as accurate as can be determined with the small number of decimals here used. To be more accurate we must have more decimal places, and go through a very tedious mechanical opera- tion; but this is not necessary, for the result is well known, and is 3.1415926535897, plus other decimal places to the 100th, without termination. This result was dis- covered through the aid of an infinite series in the Dif- ferential and Integral Calculus. The number, 3.1416, is the one generally used in prac- tice, as it is much more convenient than a greater num- ber of decimals, and it is sufficiently accurate for all ordinary purposes. In analytical expressions it has become a general cus- tom with mathematicians to represent this number by BOOK V. 139 the Greek letter *, and, therefore, when any diameter of a circle is represented by D, the circumference of the same circle must be *D. If the radius of a circle is re- presented by M, the circumference must be represented by 2*B. Scholium. — The side of a regular inscribed hexagon subtends an arc of 60°, and the side of a regular polygon of twelve sides subtends an arc of 30° ; and so on, the length of the arc subtended by the sides of the polygons, varying inversely with the number of sides. Angles are measured by the arcs of circles included between their sides ; they may also be measured by the chords of these arcs, or rather by the half chords called sines in Trigonometry. For this purpose, it becomes necessary to know the length of the chord of every possible arc of a circle. PROPOSITION VII. — PROBLEM. Given, the chord of any arc, to find the chord of one half that arc, the radius of the circle being unity. Let FE be the given chord, and draw the radii OA and OE, the first perpen- dicular to FE, and the second to its ex- tremity, E. Denote FE by 2c, and the chord of the half arc AE by x. Then, in the right-angled triangle, DOE, we have ~ DC 2 = OE 2 — BE\ Whence, since OE = 1, _Z><7= ^1 — c\ If from CA = 1 we subtract DO, we shall have AD. That is, AD = 1 — s/\ — c % \ b utAD 2 -f DE 2 = AE\ and AD' = 2 — 2Vl — c 2 — c\ Adding to the first member of this last equation DE 2 , and to the second its value c a , we have AD 2 -f DB 2 = 2^T^ 7. Whence, AE — ^2 — 2^1 — c\ the value sought. By applying this formula successively to any known chord, we can find the chord of one half the arc, that of half of the half, and so on, to the chords of the most minute arcs. 140 GEOMETRY. Application. The greatest chord in a circle is its diameter, which is 2 when the radius is 1 ; therefore, we may commence by making 2c = 2, and c = 1. Then, AE = ^2 — s/T^c* = ^2—2^1^1 = ^2 = 1.41421356, which is the chord of 90°. Now make 2c = 1.41421356, and c =.70710678 = 1^2. We shall then have, chord of 45° = V 2 — 2^t = V2 — 1.41421356 = /l2lZ78l = \/40. Whence, DC = «/51 -f v /40. AC 2 = (v/5l + \/40) 2 + (16) 2 ; AC = 20.89, Ans. The area of the triangle, AB C, can be determined by first find- ing the area of the trapezoid, ABHD, then the area of the trian- gle, BHCj and from their sum subtracting the area of the triangle, ADC 16. Construct a triangle on a base of 400, one of the angles at the base being 80°, and the other 70° ; and 13 k Ans. 146 GEOMETRY. determine the third angle, and the area of the triangle thus constructed. The third angle is 30°, and as nearly as our scale of equal parts can determine for us, the side opposite the angle 80° is 787, and that opposite 70° is 740. The exact solution of problems like the last, except in a few par- ticular cases, requires a knowledge of certain lines depending on the angles of the triangle. The properties and values of these lines are investigated in trigonometry • and as we are not yet supposed to be acquainted with them, we must be content with the approxi- mate solutions obtained by the constructions and measurements made with the plane scale. 17. If we call the mean radius of the earth 1, the mean distance of the moon will be 60 ; and as the mean distance of the sun is 400 times the distance of the moon, its distance will be 400 times 60. The sun and moon appear to have the same diameter; supposing, then, the real diameter of the moon to be 2160 miles, what must be that of the sun ? Let E be the center of the earth, M that of the moon, and S that of the sun, and suppose ENP to be a line from the center of the earth, touching the moon and the sun. Then, EM : MIST : : ES : SP; but 3/iVis the radius of the moon, and SP that of the sun. Mul- tiplying the consequents by 2, the above proportion becomes EM: 2MJST:: ES : 2SP-, or in numbers, 60 : 2160 : : 400 X 60 : 2SP; whence, 2SP = sun's diameter = 864000 miles, Ans. 18. In Problem 15, suppose BO to be drawn on the other side of BIT, what, then, will be the value of A O, and what the area of the triangle ACB1 Ans. l A0 = 16 > 021 ; _ I Area of triangle, 8^51, very nearly. BOOK.V. 147 19. A man standing 40 feet from a building which was 24 feet wide, observed that when he closed one eye, the width of the building just eclipsed or hid from view 90 rods of fence which was parallel to the width of the building; what was the distance from the eye of the observer to the fence ? Ans. 2475 feet. 20. Taking the same data as in the last problem, ex- cept that we will now suppose the direction of the fence to be inclined at an angle of 45° to the side of the building which we see ; what, in this case, must be the distance between the eye of the observer and the remoter point of the fence ? Let HF be the width of the house, E the position of the eye, and AB that of the fence. Draw BD perpendicular to EA produced ; then, since the triangle ABB is right-angled and isosceles, we have AD = DB, and 2AD 2 = AB 2 = (90) 2 ; BD = 63.64 rods, and the similar triangles EFH and EDB give the proportion HF : EF : : BD : ED = 1750.1 feet; and from this we find " EB 2 =ED 2 + BD 2 = (63.64 x 3 5 3 ) 2 + (1750.1)* Whence EB = 2040.94 -f Ans. 21. In a right-angled triangle, ABO, we have AB = 493, AC = 1425, and BC = 1338 ; it is required to divide this triangle into parts by a line parallel to AB, whose areas are to each other as 1 is to 3. How will the sides AC and BO be divided by this line ? (See Th. 20, B. II). Ans, Into equal parts. 22. In a right-angled triangle, ABO, right-angled at B, the base AB is 320, and the angle A is 60° ; required the remaining angle and the other sides. . ( The angle 0- 30°; I AC= 640; BC= 554.24. 148 GEOMETRY. 23. A hunter, wishing to determine his distance from a village in sight, took a point and from it laid off two lines in the direction of two steeples, which he supposed equally distant from him, and which he knew to he 100 rods asunder. At the distance of 50 feet on each line from the common point, he measured the distance be- tween the lines, and found it to be 5 feet 8 inches. How far was he from the steeples ? 5 ft. 8 in. : 100 rods : : 50 ft. : distance. r 14,559 feet, or, 68: lOOxfx 12:: 50: distance. Ans '\ ? r n * arl ? 2 I 3 miles. 24. A person is in front of a building which he knows to be 160 feet long, and he finds that it covers 10 minutes of a degree ; that is, he finds that the two lines drawn from his eye to the extremities of the building include an angle of 10 minutes. What is his distance from the building? Ans ( 50,672 feet, or •\ nearly 10 miles. Remark. — The questions of distance, with which we are at present occupied, depend for their solution on the properties of similar tri- angles. In the preceding example we apparently have but one tri- angle, but we have in fact two ; the second being formed by the dis- tances unity on the lines drawn from the eye of the observer, and the line which connects the extremities of these units of distance. This last line may be regarded as the chord of the arc 10 minutes to the radius unity. We have seen that the length of the arc 180° to the radius 1, is 3.1415926 ; hence the chord of 1° or 60 / is 0.017455, and of 10 / it must be 0.0029088. Therefore, by similar triangles, we have 0.0029088 : 160 : : 1 : Ans. = ^^. 25. In the triangle, ABO, we have given the angles A = 32°, and B = 84°. The side AB is produced, and the exterior angle* CBD thus formed, is bisected by the line BU, and the angle A is also bisected by the line AE, BE and AE meeting in the point E. "What is the angle (7, and what is the relation between the angles C and El Am. (7=64°; E= J O. BOOK V. 149 26. Suppose a line to be drawn in any direction be- tween two parallels. Bisect the two interior angles thus formed on either side of the connecting line, and prove that the bisecting lines meet each other at right angles, and that they are the sides of a right-angled triangle of which the line connecting the parallels is the hypotenuse. 27. If the two diagonals of a trapezoid be drawn, show that two similar triangles will be formed, the parallel sides of the trapezoid being homologous sides of the triangles. What will be the relative areas of these triangles ? f The triangles will be to each other Ans. < as the squares on the parallel sides ( of the trapezoid. 28. If from the extremities of the base of any triangle, lines be drawn to any point within the triangle, forming with the base another triangle ; how will the vertical angle in this last triangle compare with that in the original triangle ? [ It will be as much greater than the angle in the original triangle as the sum of Ans. angles at the base of the new triangle is less than the sum of those at the base of the first. 29. The two parallel sides of a trapezoid are 12 and 20, respectively, and their perpendicular distance is 8. If a line whose length is 14.5 be drawn between the in- clined sides and parallel to the parallel sides, what is the area of the trapezoid, and what the area of each part, respectively, into which the trapezoid is divided ? Area of the whole, 128 square units ; " smaller part, 33J " " larger " 94| " Dividing line at the distance of 2J from shorter parallel side. 30. If we assume the diameter of the earth to be 13* 150 GEOMETRY. 7956 miles, and the eye of an observer be 40 feet above the level of the sea, how far distant will an object be, that is just visible on the earth's surface. (Employ Th. 18, B. Ill, after reducing miles to feet.) Ans. 40992 feet = 7 miles 4032 feet. 31. The diameter of a circle is 4 ; what is the area of the inscribed equilateral triangle? Ans. 3^3. 32. Three brothers, whose residences are at the ver- tices of a triangular area, the sides of which are severally 10, 11, and 12 chains, wish to dig a well which shall be at the same distance from the residence of each. Deter- mine the point for the well, and its distance from their residences. Remark. — Construct a triangle, the sides of which are, respectively, 10, 11, and 12. The sides of this triangle will be the chords of a cir- cle whose radius is the required distance. To find the center of this circle, bisect either two of the sides of the triangle by perpendiculars, and their intersection will be the center of the circle, and the location of the well. Ans, The well is distant 6.25 chains, nearly, from each residence. 33. The base of an isosceles triangle is 12, and the equal sides are 20 each. What is the length of the per- pendicular from the vertex to the base ; and what the area of the triangle ? Ans. Perpendicular, 19.07; area, (19.07) x 6. 34. The hypotenuse of a c right-angled triangle is 45 inches, and the difference be- tween the two sides is 8.45 inches. Construct the triangle. Suppose the triangle drawn and represented by ABC, DC being the difference between the two sides. Now, by inspection, we discover the steps to be taken for the construc- tion of the triangle As AD = AB, BOOK V. 151 the angle ABB, must be equal to the angle DBA, and each equal to 45°. Therefore, draw any line, AC, and from an assumed point in it as D, draw BD, making the angle ABB = 45°. Take from a • scale of equal parts, 8.45 inches, and lay them off from D to C, and with C as a center, and CB = 45 inches as a radius, describe an arc cutting BD in B. Draw CB, and from B, draw BA at right angles to AC', then is ABC the triangle sought. Ans. AB =27.3; A C— 35.76, when carefully constructed. 35. Taking the same triangle as in the last problem, if J%£* **\ we draw a line bisecting the right angle, where will it flo*K meet the hypotenuse ? Ans. 19.5 from B; and 25.5 from C. 36. The diameters of the hind and fore wheels of a carriage, are 5 and 4 feet, respectively ; and their centers are 6 feet asunder. At what distance from the fore wheels will the line, passing through their centers, meet the ground, which is supposed level? Ans. 24 feet. 37. If the hypotenuse of a right-angled triangle is 35, and the side of its inscribed square 12, what are its sides ? Ans. 28 and 21. 38. "What are the sides of a right-angled triangle having the least hypotenuse, in which if a square be in- scribed, its side will be 12 ? c The sides are equal to 24 each, and the Ans. < least hypotenuse is double the diagonal I of the square. 39. The radius of a circle is 25 ; what is the area of a sector of 50° ? Remark. — First find the length of an arc of 50° in a circle whose radius is unity. Then 25 times that will be the length of an arc of the same number of degrees in a circle of which the radius is 25. t ^^ io a- .. 3.14159269 Length of arc 1° radius unity = ^r — . u 50 o u u = 1-047197 63 x 5> Area of sector « iM^i x 125 X f = 54.541, A?is. o A 152 GEOMETRY. BOOK VI ON THE INTERSECTIONS OF PLANES, AND THE REL- ATIVE POSITIONS OF PLANES AND OF PLANES AND LINES. DEFINITIONS. A Plane has been already defined to be a surface, such that the straight line which joins any two of its points will lie entirely in that surface. (Def. 9, page 9.) 1. The Intersection or Common Section of two planes is the line in which they meet. 2. A Perpendicular to a Plane is a line which makes right angles with every line drawn in the plane through the point in which the perpendicular meets it; and, con- versely, the plane is perpendicular to the line. The point in which the perpendicular meets the plane is called the foot of the perpendicular. 3. A Diedral Angle is the separation or divergence of two planes proceeding from a common line, and is meas- ured by the angle included between two lines drawn one in each plane, perpendicular to their common sec- tion at the same point. The common section of the two planes is called the edge of the angle, and the planes are its faces, 4. Two Planes are perpendicular to each other, when their diedral angle is a right angle. 5. A Straight Line is parallel to a plane, when it will not meet the plane, however far produced. BOOK VI. 153 6. Two Planes are parallel, when they will not intersect, however far produced in all directions. 7. A Solid or Polyedral Angle is the separation or diver- gence of three or more plane angles, proceeding from a common point, the two sides of each of the plane angles being the edges of diedral angles formed by these plane angles. The common point from which the plane angles pro- ceed is called the vertex of the solid angle, and the inter- section "of its bounding planes are called its edges. 8. A Triedral Angle is a solid angle formed by three plane angles. THEOREM I. Two straight lines which intersect each other, two parallel straight lines, and three points not in the same straight line, will severally determine the position of a plane. Let AB and AC be two lines intersecting each other at the point A*, then will these lines determine a plane. For, conceive A< ^ a plane to be passed through AB, and turned about AB as an axis until it contains the point in the line AC. The plane, in this position, contains the lines AB and AC, and will contain them in no other. Again, let AB and BE be two parallel straight lines, and take at pleasure two points, A and B, in the one, and two points, D and E, in the other, and draw AE and BD. These last lines, from what precedes, determine the position of a plane which contains the points A, B, JD, and E. And again, if A, B, and be three points not in the same straight line, and we draw the lines AB and AG, it follows, from the first part of this proposition, that these points fix the plane. 154 GEOMETRY. Cor. A straight line and a point out of it determine the position of a plane. THEOREM II. If two planes meet each other, their common points will be found in, and form one straight line. Let B and D be any two of the points common to the two planes, and join these points by the straight line BB ; then will BB contain all the points common to the two planes, and be their intersection. For, suppose the planes have a common point out of the line BB ; then, (Cor. Th. 1), since a straight line and a point out of it determine a plane, there would be two planes determined by this one line and single point out of it, which is absurd. Hence the common section of two planes is a straight line. Remark. — The truth of this proposition is implicitly assumed in the definitions of this Book. THEOREM III. If a straight line stand at right angles to each of two other straight lines at their point of intersection, it will be at right angles to the plane of those lines. Let AB stand at right angles to EF&ndi CB, at their point of intersection A. Then AB will be at right angles to any other line drawn through A in the plane, pass- ing through EF, OB, and, of course, at right angles to the plane itself. (Def. 2.) Through A, draw any line, A G, in the plane EF, CB, and from any point G, draw GH parallel to AB. Take HF = AH, and join F and G and produce FG to B. Because HG is parallel to AB, we have FH : HA :: FG : GB. BOOK VI. 155 But, in this proportion, the first couplet is a ratio of equality; therefore the last couplet is also a ratio of equality, That is, FG = GB, or the line FB is bisected in ff. Draw BB, BG, and BF. Now, in the triangle AFB, as the base FB is bisected in G, we have, JJF 2 +~AB 2 = 2AG 2 + 2GF 2 (l) (Th. 42, B. I). Also, as BF is the base of the A BBF, we have by the same theorem, ~BF 2 +~~BB 2 = 2BG 2 + 2GF 2 (2) By subtracting ( 1 ) from (2 ), and observing that BF — AF = AB ? because BAF is a right angle ; and BB — AB 2 — AB 2 , because BAB is a right angle, we shall have, ~AB* +~AB 2 = 2BG 2 — 2 AG 2 . Dividing by 2, and transposing AG- 2 , and we have, AB 2 + AG 2 = BG\ This last equation shows that BA G is a right angle. But AG is any line drawn through A, in the plane FF, OB ; therefore AB is at right angles to any line in the plane, and, of course, at right angles to the plane itself. Cor. 1. The perpendicular BA is shorter than any of the oblique lines BF, BG, or BB, drawn from the point B to the plane ; hence it is the shortest distance from a point to a plane. Cor. 2. But one perpendicular can be erected to a plane from a given point in the plane; for, if there could be two, the plane of these perpendiculars would intersect the given plane in some line, as AG, and both the per- pendiculars would be at right angles to this intersection at the same point, which is impossible. Cor. 8. But one perpendicular can be let fall from a given point out of a plane on the plane ; for, if there can 156 GEOMETRY. be two, let BGr and BA be such perpendiculars, then would the triangle BAGr be right angled at both A and 6r, which is impossible. THEOREM IV. If from any point of a perpendicular to a plane, oblique lines be drawn to different points in the plane, those oblique lines which meet the plane at equal distances from the foot of the perpendicular are equal; and those which meet the plane at unequal distances from the foot of the perpendicular are unequal, the greater distances corresponding to the longer oblique lines. Take any point B in the perpendicular BA to the plane ST, and draw the oblique lines BO, BD, and BE, the points 0, B, and E, being equally distant from A, the foot of the perpendicular. Produce AE to F, and draw BF; then will BC= BD = BE, and BF> BE. For, the triangles BAG, BAD, and BAE are all right- angled at A, the side BA is common, and AC= AD= AE by construction, hence, (Th. 23, B.I), BC=BB = BE. Moreover, since AF^> AE, the oblique line BF^> BE. Cor. If any number of equal oblique lines be drawn from the point B to the plane, they will all meet the plane in the circumference of a circle having the foot of the perpendicular for its center. It follows from this, that, if three points be taken in a plane equally distant from a point out of it, the center of the circumference passing through these three points will be the foot of the perpendicular drawn from the point to the plane. BOOK VI. 157 THEOREM V. The line which joins any point of a perpendicular to a plane, with the point in which a line in the plane is inter- sected, at right angles, by a line through the foot of the per- pendicular, will be at right angles to the line in the plane. Let AB be perpendic- ular to the plane ST, and AB a line through its foot at right angles to EF, a line in the plane. Connect B with any point, as B, of the perpendicular; and BD will be perpendicular to EF. Make BF= DE, and join B to the points E, D, and F. Since BE = BF, and the angles at B are right angles, the oblique lines, AE and AF, are equal ; and, since AE = AF, we have, (Th. 4), BE = BF; therefore the line BB has its two points, B and B, equally distant from the extremities E and F of the line EF, and hence BB is perpendicular to EF at its middle point B. Cor. Since FB is perpendicular to the two lines AB and BB at their intersection, it is perpendicular to their plane ABB, (Th. 3). Scholium. — The inclination of a line to a plane is measured by the angle included between the given line and the line which joins the point in which it meets the plane and the foot of the perpendicular drawn from any point of the line to the plane ; thus, the angle BFA is the inclination of the line BF to the plane ST. THEOREM VI. If either of two parallels is perpendicular to a plane, the other is also perpendicular to the plane. Let BA and ED be two parallels, of which one, BA, is perpendicular to the plane ST; then will the other also be perpendicular to the same plane. 14 158 GEOMETKY. The two parallels de- termine a plane which intersects the given plane in AB ; through B draw MJV perpendicular to AB; then, (Cor., Th. 5,) will MJSf be perpendicu- lar to the plane BAB, and the angle MBE is therefore a right angle ; but EBA is also a right angle, since BA and ED are parallel, and BAB is a right angle by hypothesis; hence, EB is perpendicular to the two lines MB and AB in the plane ST; it is therefore perpen- dicular to the plane, (Th. 3). Cor. 1. The converse of this proposition is also true , that is, if two straight lines are both perpendicular to the same plane, the lines are parallel. For, suppose BA and EB to be two perpendiculars ; if not parallel, draw through B a parallel to BA, and this last line will be perpendicular to the plane ; but EB is a perpendicular by hypothesis, and we should have two perpendiculars erected to the plane at the same point, which is impossible, (Cor. 2, Th. 3). Cor. 2. If two lines lying in the same plane are each parallel to a third line not in the same plane, the two lines are parallel. For, pass a plane perpendicular to the third line, and it will be perpendicular to each of the others; hence they are parallel, THEOREM VII. A straight line is parallel to a plane, when it is parallel to a line in the plane. Suppose the line MN to be parallel to the line CB, in the plane ST; then will ifefJVbe parallel to the plane ST BOOK VI. 159 For, CB being in the plane ST, and at the same time g parallel to MN, it must be the intersection of the plane of these parallels with the plane ST; hence, if MN meet the plane ST, it must do so in the T line CB, or OB produced ; but MN and CB are parallel, and cannot meet; therefore MN, however far produced, can have no point in the plane ST, and hence, (Def. 5), it is parallel to this plane. THEOREM VIII. If two lines are parallel, they will be equally inclined to any given plane. Let AB and CB be •n tv two parallels, and ST any plane met by them in the points A and V 'C; then will the lines \ AB and CB be equally \ ^ "~c V inclined to the plane \ \ ST. T For, take any distance, AB, on one of these parallels, and make CB = AB, and draw A C and BB. From the points B and B let fall the perpendiculars, BB and BF, on the plane ; join their feet by the line EF, and draw AE and OF. Now, since AB is equal and parallel to CB, ABB Cm a parallelogram, and BB is equal and parallel to A 0, and BB is parallel to the plane ST, (Th. 7) ; and, since BE and BF are both perpendicular to this plane, they are parallel ; but BB and EF are in the plane of these parallels; and as EF is in the plane ST, and BB is parallel to this plane, these two lines must be parallel and equal, and BBFE is also a parallelogram. Now, 160 GEOMETRY. we have shown that BD is equal and parallel to AC, and EF equal and parallel to BD; hence, (Cor. 2, Th. 6), EFw equal and parallel to AC, and ACFE is a parallel- ogram, and AE = CF. The triangles ABE and CDF have, then, the sides of the one equal to the sides of the other, each to each, and their angles are consequently equal; that is, the angle BAE is equal to the angle DCF; but these angles measure the inclination of the lines AB and CD to the plane ST, (Scholium, Th. 5). Scholium. — The converse of this proposition is not generally true ; that is, straight lines equally inclined to the same plane are not neces- sarily parallel. THEOREM IX. The intersections of two parallel planes by a third plane, are parallel. Let the planes QR and ST be intersected by the third plane, AD : then will the intersections, AB and CD, be parallel. Since the lines AB and CD are in the same plane, if they are not parallel, they will meet if sufficiently produced; but they cannot meet out of the planes QR and ST, in which they are respectively found; therefore, any point common to the lines, must be at the same time common to the planes ; and since the planes are parallel, they have no common point, and the lines, therefore, do not intersect ; hence they are parallel. THEOREM X. If two planes are perpendicular to the same straight line, they are parallel to each other. Let QR and ST be two planes, perpendicular to the line AB m , then will these planes be parallel* BOOK VI. 161 For, if not parallel, suppose M to be a point in their line of intersection, and n from this point draw lines to the extremities of the perpendicular m AB, thus forming a tri- "^oii. angle, MAB. .Now, since the line AB is \ \ T perpendicular to both planes, it is perpendicular to each of the lines MA and MB, drawn through its feet in the planes, (Def. 2) ; hence, the triangle has two right angles, which is impos- sible; the planes cannot therefore meet in any point as My and are consequently parallel. Cor. Conversely : The straight line which is perpendicu- lar to one of the parallel planes, is also perpendicular to the other. For, if AB be perpendicular to the plane QB, draw in the other plane, through the point in which the perpendicular meets it, any line, as AC. The plane of tjie lines AB and A will intersect the plane QR in the line BD ; and since the planes are parallel by hypothesis, the lines A O and BD must be parallel, (Th. 9) ; but the angle DBA is a right angle ; hence, BAG must be a right angle, and the line BA is perpendicular to any line what- ever drawn in the plane through the point A ; BA is therefore perpendicular to the plane ST. THEOREM XI. If two straight lines be drawn in any direction through parallel planes, the planes will cut the lines proportionally. Conceive three planes to be parallel, as represented in the figure, and take any points, A and B, in the first and third planes, and draw AB, the line passing through the second plane at E. 14* L 162 GEOMETRY. G\ Also, take any other two points, as O and D, in the first and third planes, and draw OB, the line passing through the second plane at F. Join the two lines by the diagonal AB, which passes through the second plane at #. Draw BB, EQ, 'OJFi and A 0. We are now to prove that, AE : EB :: OF : FB. For the sake of perspicuity, put AGr = X, and GB=Y. As the planes are parallel, BB is parallel EGr ; then, in the two triangles ABB and AEGr, we have, (Th. 17, B.H); AE : EB : : X : Y. Also, as the planes are parallel, GrF is parallel to A C y and we have, OF : FB : : X : Y. By comparing the proportions, and applying Th. 6, B. II, we have AE : EB n OF i FB. THEOREM XII, If a straight line is perpendicular to a plane, all planes passing through that line will be perpendicular to the plane. Let MNhe a plane, and AB a per- pendicular to it. Let BO be any other plane, passing through AB ; this plane will be perpendicular to mjst. Let BB be the common intersec- tion of the two planes, and from the point B, draw BE at right angles to BB. Then, as AB is perpendicular to the plane MJSf, it is perpendicular to every line in that plane, passing through BOOK VI. 163 B; (Def. 2,) ; therefore, ABE is a right angle. But the angle ABU, (Def. 3), measures the inclination of the two planes ; therefore, the plane OB is perpendicular to the plane MN\ and thus we can show that any other plane, passing through AB, will be perpendicular to JOT. Hence the theorem. THEOREM XIII. If two planes are perpendicular to each other, and a line be drawn in one of them perpendicular to their common in- tersection, it will be perpendicular to the other plane. Let the two planes, QB and ST, be perpendicular to each other, and draw in QB the line CD at right angles to their common intersection, B V; then will this line be perpendicular to the plane ST. In the plane iSTdraw ED, perpen- dicular to VB at the point D. Then, since the planes QB and ST are perpendicular to each other, the angle ODE is a right angle, and CD is perpendicular to the two lines, ED and VB, passing through its foot in the plane ST. CD is therefore perpendicular to the plane ST, (Th. 3). Cor. Conversely: if we erect a perpendicular to the plane ST, at any point, D, of its intersection with the plane QB, this perpendicular will lie in the plane QB. For, if it be not in this plane, we can draw in the plane the line CD, at right angles to VB ; and, from what has been shown above, CD is perpendicular to the plane ST, and we should thus have two perpendiculars erected to the plane, ST, at the same point, which is impossible, (Cor. 2, Th. 2>\ 164 GEOMETRY. THEOREM XIV. The common intersection of two planes, loth of which are perpendicular to a third plane, will also be perpendicular to the third plane. Let MN be the common intersection of the two planes, QR and VX, both of which are perpendicular to the plane ST; then will MJSfbe perpendicular to the plane ST. For, if we erect a perpendicular to the plane ST, at the point M, it will lie in both planes at the same time, (Cor. Th. 13); and this perpendicular must therefore be their intersection. Hence the theorem. THEOREM XV. Parallel straight lines included between parallel planes, are equal. Let J. B and D be two parallel lines, included by the two parallel planes, QR and ST; then will AB = BO. For, the plane A 0, of the parallel lines, intersects the planes, QR and ST, in the parallel lines, AB and BO, (Th. 9) ; hence ABBO is a parallelogram, and its oppo- site sides, AB and BO, are equal. Oor. It follows from this proposition, that parallel planes are everywhere equally distant ; for, two perpendiculars drawn at pleasure between the two planes are parallel lines, (Cor. 1, Th. 6), and hence are equal ; but these per- pendiculars measure the distance between the planes. V 1\ s . R V ^^ -\ BOOK VI. 165 THEOREM XVI. Two planes are parallel when two lines not parallel, lying in the one, are respectively parallel to two lines lying in the other. Let QR and ST be two planes, the first containing the two lines AB and CD which intersect each other at U, and the second the two lines LM and NO, respect- ively parallel to AB and OB; then will these planes be par- allel. For, if the two planes are not parallel, they must intersect when sufficiently produced; and their common section lying in both planes at the same time, would be a line of the plane QR. Now, the lines AB and OB intersect each other by hypothesis ; hence one or both of them must meet the common sec- tion of the two planes. Suppose AB to meet this com- mon section ; then, since AB and LM are parallel, they determine a plane, and AB cannot meet the plane ST in a point out of the line LM ; but AB and LM being par- allel, have no common point. Hence, neither AB nor OB can meet the common section of the two planes ; that is, they have no common section, and are therefore par- allel. Oor. Since two lines which intersect each other, deter- mine a plane, it follows from this proposition, that the plane of two intersecting lines is parallel to the plane of two other intersecting lines respectively parallel to the first lines. 166 GEOMETRY. THEOREM XVII. W7ien two intersecting lines are respectively parallel to two other intersecting lines lying in a different plane, the angles formed by the last two lines will be equal to those formed by the first two, each to each, and the planes of the angles will be parallel. Let QR be the plane of the two lines AB and CD, which inter- sect each other at the point E, and ST the plane of the two lines LM and NO, respect- ively parallel to AB and CD ; then will the [_BED - \__MPO, and L BEQ = L MPN, etc., and the planes QR and ST will be parallel. That the plane of one set of angles is parallel to that of the other, follows from the Corollary to Theorem 16 ; we have then only to show that the angles are equal, each to each. Take any points, B and D, on the lines AB and CD, and draw BD. Lay off PM, equal to and in the same direction with EB, and PO, equal to and in the same direction with ED, and draw MO. Kow, since the planes QR and ST are parallel, and ED is equal and parallel to PO, ED OP is a parallelogram, and DO is equal and par- allel to EP. For the same reason, BM is equal and parallel to EP; therefore, BDOM is a parallelogram, and MO is equal and parallel to BD. Hence the A's, EBD and PMO, have the sides of the one equal to the sides of the other, each to each ; they are therefore equal, and BOOK VI. 167 the [_MPO = the \_BED. In the same manner it can be proved that [_BEC = [_MPJST, etc. Cor. 1. The plane of the parallels AB and LM is in- tersected by the plane of the parallels CD and NO, in the line EP. Now, EB and ED are the intersections of these two planes with the plane QB, and PM and PO are the intersections of the same planes with the parallel plane ST. It has just been proved that the \_ BED = [_MPO. Hence, if the diedral angle formed by two planes, be cut by two parallel planes, the intersections of the faces of the diedral angle with one of these planes will include an angle equal to that included by the intersections of the faces with the other plane. Cor. 2. The opposite triangles formed by joining the cor- responding extremities of three equal and parallel straight lines lying in different planes, will be equal and the planes of the triangles will be parallel. Let EP, BM, and DO, be three equal and parallel straight lines lying in different planes. By joining their corresponding extremities, we have the triangles EBD and PMO. Now, since EP and BM are equal and parallel, EBMP is a parallelogram, and EB is equal and parallel to PM; in the same manner, we show that ED is equal and parallel to PO, and BD to MO', hence the triangles are equal, having the three sides of the one, respectively, equal to the three sides of the other. That their planes are parallel, follows from Cor., Theo- rem 16. THEOREM XVIII. Any one of the three plane angles bounding a triedral angle, is less than the sum of the other two. Let A be the vertex of a solid angle, bounded by the three plane angles, BAC, BAD, and DAC; then will any one of these three angles be less than the sum of the 168 GEOMETRY. other two. To establish this proposition, we have only to compare the greatest of the three angles with the sum of the other two. Suppose, then, BAC to be the greatest angle, and draw in its plane B 4 the line AE, making the angle CAB equal to the angle CAD, On 1) AH, take any point, E, and through it draw the line CUB. Take AD, equal to AE, and draw BD and DC. Now, the two triangles, CAD and CAE, having two sides and the included angle of the one equal to the two sides and included angle of the other, each to each, are equal, and CE = CD', but in the triangle, BDC, BCi?. In like manner it may be shown that SO = TF, and BO = FF. That granted, the quadrilateral SAOO is equal to the quadri- lateral TBPF; for, place the angle ASQ upon its equal, BTF, and because SA = 2 r D, and £# = TF, the point J. will fall on B, and the point on jP; and, at the same time, A 0, which is perpendicular to SA, will fall on PB, which is perpendicular to TB, and, in like manner, 00 on PF; wherefore, the point will fall on the point P, and A will be equal to DP. But the triangles, AOB, DPE, are right angled at and P ; the hypotenuse AB == BE, and the side AO = BP; hence, those triangles are equal, (Cor, Th. 39, B. I), and [_A0B=[_PBE. The angle OAB is the inclination of the two planes, ASB, ASO; the angle PBE is that of the two planes, BTE, BTF; conse- quently, those two inclinations are equal to each other. Hence the theorem. Scholium 1. — The angles which form the solid angles at S and T, may be of such relative magnitudes, that the perpendiculars, BO and EP, may not fall within the bases, ASC and BTF; but they will always either fall on the bases, or on the planes of the bases produced, and will have the same relative situation to A y S, and C, as P has to D, T, and jF. In case that and P fall on the planes of the bases produced, the angles BCO and EFP, would be obtuse angles ; but the demonstration of the problem would not be varied in the least. Scholium 2. — If the plane angles bounding one of the triedral angles be equal to those of the other, each to each, and also be simi- larly arranged about the triedral angles, these solid angles will be ab- solutely equal. For it was shown, in the course of the above demon- stration, that the quadrilaterals, SA OC and TDPF, were equal; and on being applied, the point falls on the point P; and since the trian- gles A OB and DPE are equal, the perpendiculars OB and PE are BOOK VI. 171 also equal. Now, because the plane angles are like arranged about the triedral angles, these perpendiculars lie in the same direction ; hence the point B will fall on the point E, and the solid angles will exactly coincide. Scholium 3. — When the planes of the equal angles are not like dis- posed about the triedral angles, it would not be possible to make these triedral angles coincide ; and still it would be true that-the planes of the equal angles are equally inclined to each other. Hence, these triedral angles have the plane and diedral angles of the one, equal to the plane and diedral angles of the other, each to each, without having of themselves that absolute equality which admits of superposition. Magnitudes which are thus equal in all their component parts, but will not coincide, when applied the one to the other, are said to be symmetrically equal. Thus, two triedral angles, bounded by plane angles equal each to each, but not like placed, are symmetrical triedral angles. 172 GEOMETRY. BOOK VII SOLID GEOMETRY. DEFINITIONS. 1. A Polyedron is a solid, or volume, bounded on all sides by planes. The bounding planes are called the faces of the polyedron, and their intersections are its edges. 2. A Prism is a polyedron, having two of its faces, called bases, equal polygons, whose planes and homolo- gous sides are parallel. The other, or lateral faces, are parallelograms, and constitute the convex surface of the prism. The bases of a prism are distinguished by the terms, upper and lower ; and the altitude of the prism is the per- pendicular distance between its bases. Prisms are denominated triangular, quadrangular, pent- angular, etc., according as their bases are triangles, quad- rilaterals, pentagons, etc. 3. A Right Prism is one in which the planes of the lateral faces are perpendicular to the planes of the bases. 4. A Parallelopipedon is a prism whose bases are parallelograms. 5. A Rectangular Parallelopipedon is a right parallelopipedon, with rectangular bases. BOOK VII 178 6. A Cube or Hexaedron is a rectangu- lar parallelopipedon, whose faces are all equal squares. 7. A Diagonal of a Polyedron is a straight line joining the vertices of two solid angles not adjacent. 8. Similar Polyedrons are those which are bounded by the same number of similar polygons like placed, and whose solid angles are equal each to each. Similar parts, whether faces, edges, diagonals, or angles, similarly placed in similar polyedrons, are termed homologous. 9. A Pyramid is a polyedron, having for one of its faces, called the base, any polygon whatever, and for its other faces triangles having a common vertex, the sides opposite which, in the several trian- gles, being the sides of the base of the pyramid. 10. The Vertex of a pyramid is the common vertex of the triangular faces. 11. The Altitude of a pyramid is the perpendicular distance from its vertex to the plane of its base. 12. A Right Pyramid is one whose base is a regular polygon, and whose vertex is in the perpendicular to the base at its center. This perpendicular is called the axis of the pyramid. 13. The Slant Height of a right pyramid is the perpen- dicular distance from the vertex to one of the sides of the base. 14. The Frustum of a Pyramid is a portion of the pyr- amid included between its base and a section made by a plane parallel to the base. Pyramids, like prisms, are named from the forms of their bases. 15* 174 GEOMETRY. 15. A Cylinder is a body, having for its ends, or bases, two equal circles, the planes of which are perpendicular to the line joining their centers ; the remainder of its surface may be con- ceived as formed by the motion of a line, which constantly touches the cir- cumferences of the bases, while it remains parallel to the line which joins their centers. "We may otherwise define the cylinder as a body gen- erated by the revolution of a rectangle about one of its sides as an immovable axis. The sides of the rectangle perpendicular to the axis generate the bases of the cylinder ; and the side opposite the axis generates its convex surface. The line joining the centers of the bases of the cylinder is its axis, and is also its altitude. If, within the base of a cylinder, any polygon be in- scribed, and on it, as a base, a right prism be con- structed, having for its altitude that of the cylinder, such prism is said to be inscribed in the cylinder, and the cylin- der is said to circumscribe the prism. Thus, in the last figure, ABOBEc is an inscribed prism, and it is plain that all its lateral edges are con- tained in the convex surface of the cylinder If, about the base of a cylinder, any polygon be circumscribed, and on it, as a base, a right prism be con- structed, having for its altitude that of the cylinder, such prism is said to be circumscribed about the cylinder, and the cylinder is said to be inscribed in the prism. Thus, ABCBEFc is a circum- scribed prism; and it is plain that BOOK VII 175 the line, w, which joins the points of tangency of the sides, EF and ef, with the circumferences of the bases of the cylinder, is common to the convex sur- faces of the cylinder and prism. 16. A Cone is a body bounded by a circle and the surface generated by the motion of a straight line, which con- stantly passes through a point in the perpendicular to the plane of the circle at its center, and the different points in its circumference. The cone may be otherwise defined as a body gene- rated by the revolution of a right-angled triangle about one of its sides as an immovable axis. The other side of the triangle will generate the base of the cone, while the hypotenuse generates the convex surface. The side about which the generating triangle revolves is the axis of the cone, and is at the same time its altitude. If, within the base of the cone, any polygon be inscribed, and on it, as a base, a pyramid be constructed, having for its vertex that of the cone, such pyramid is said to be inscribed in the cone, and the cone is said to circumscribe the pyramid. Thus, in the accompanying figure, V — ABODE, is an inscribed pyramid, and it is plain that all its lateral edges are contained in the convex surface of the cone. If, about the base of a cone, any poly- gon be circumscribed, and on it, as a base, a pyramid be constructed, having for its vertex that of the cone, such pyramid is said to be circumscribed about the cone, and the cone is said to be inscribed in the pyramid. 176 GEOMETRY. 17. The Frustum of a Cone is the portion of the cone that is included between its base and a section made by a plane parallel to the base. 18. Similar Cylinders, and also Similar Cones, are such as have their axes proportional to the radii of their bases. 19. A Sphere is a body bounded by one uniformly-curved surface, all the points of which are at the same distance from a certain point within, called the center. We may otherwise define the sphere as a body gene- rated by the revolution of a semicircle about its diameter as an immovable axis. 20. A Spherical Sector is that portion of a sphere which is in- cluded between the surfaces of two cones having their verti- ces at the center of the sphere. Or, it is that portion of the sphere which is generated by a sector of the generating semi- circle. 21. The Radius of a Sphere is a straight line drawn from the center to any point in the surface ; and the diameter is a straight line drawn through the center, and limited on both sides by the surface. All the diameters of a sphere are equal, each being twice the radius. 22. A Tangent Plane to a sphere is one which has a single point in the surface of the sphere, all the others being without it. 23. A Secant Plane to a sphere is one which has more than one point in the surface of the sphere, and lies partly within and partly without it. Assuming, what will presently be proved, that the in- tersection of a sphere by a plane is a circle, 24. A Small Circle of a sphere is one whose plane does not pass through its center; and BOOK VII. 177 25. A Great Circle of a sphere is one whose plane passes through the center of the sphere. 26. A Zone of a sphere is the portion of its surface in- cluded between the circumferences of any two of its paral- lel circles, called the bases of the zone. When the plane of one of these circles becomes tangent to the sphere, the zone has a single base. 27. A Spherical Segment is a portion of the volume of a sphere included between any two of its parallel circles, called the bases of the segment. The altitude of a zone, or of a segment, of a sphere, is the perpendicular distance between the planes of its bases. 28. The area of a surface is measured by the product of its length and breadth, and these dimensions are always conceived to be exactly at right angles to each other. 29. In a similar manner, solids are measured by the product of their length, breadth, and height, when all their dimensions are at right angles to each other. The product of the length and breadth of a solid, is the measure of the surface of its base. Let P, in the annexed fig- ure, represent the measuring unit, and A F the rectangular solid' to be measured. A side of P is one unit in length, one in breadth, and one in height ; one inch, one foot, one yard, or any other unit that may be taken. Then, lxlxl==l, the unit cube. Now, if the base of the solid, AC, is, as here repre- sented, 5 units in length and 2 in breadth, it is obvious that (5x2 = 10), 10 units, each equal to P, can be placed on the base of AC, and no more; and as each of these units will occupy a unit of altitude, therefore, 2 units of M 178 GEOMETRY. altitude will contain 20 solid units, 3 units of altitude, 30 solid units, and so on ; or, in general terms, the num- ber of square units in the base multiplied by the linear units in perpendicular altitude, will give the solid units in any rect- angular solid. THEOREM I. If the three plane faces bounding a solid angle of one prism be equal to the three plane faces bounding a solid angle of another, each to each, and similarly disposed, the prisms will be equal. Suppose A and a to be the vertices of two solid angles, bounded by equal and similarly placed faces; then will the prisms, ABODE — iV'and abcde — n, be equal. For, if we place the base, abcde, upon its equal, the base ABODE, they will coincide; and since the solid angles, whose vertices are A and a, are equal, the lines ab, ae, and ap, respectively coincide with AB, AE, and AP ; but the faces, al and ao, of the one prism, are equal, each to each, to the faces, AL and A 0, of the other; therefore pi and po coincide with PL and PO, and the upper bases of the prisms also coincide : hence, not only the bases, but all the lateral faces of the two prisms coincide, and the prisms are equal. Oor. If the two prisms are right, and have equal bases and altitudes, they are equal. For, in this case, the rect- angular faces, al and ao, of the one, are respectively equal to the rectangular faces, AL and AO, of the other ; and hence the three faces bounding a triedral angle in the one, are equal and like placed, to the faces bounding a triedral angle in the other. BOOK VII. 179 THEOREM II. The opposite faces of any parallelopipedon are equal, and their planes are parallel. Let ABOJ) — E be any parallelopipedon ; then will its opposite faces be equal, and their planes will be parallel. The bases ABCB and FEGR are equal, and their planes are parallel, by definitions 2 and 4 of this Book; it remains for us, therefore, only to show that any two of the opposite lateral faces are equal and parallel. Since all the faces of the parallelopipedon are parallel- ograms, AB is equal and parallel to BO, and AH is also equal and parallel to BF; hence the angles HAB and FBO are equal, and their planes are parallel, (Th. 17, B. YI), and the two parallelograms, HABGr and FBCE, having two adjacent sides and the included angle of the one equal to the two adjacent sides and included angle of the other, are equal. Cor. 1. Hence, of the six faces of the parallelopipedon, any two lying opposite may be taken as the bases. Cor. 2. The four diagonals of a parallelopipedon mutu- ally bisect each other. For, if we draw AC and SB, we shall form the parallelogram A CEBl, of which the diago- nals are AE and HC, and these diagonals are at the same time diagonals of the parallelopipedon ; but the diagonals of a parallelogram mutually bisect each other. Now, if the diagonal FB be drawn, it and HC will bisect each other, since they are diagonals of the parallelogram FRBC. In like manner we can show that if BG- be drawn, it will be bisected by AE. Hence, the four diag- onals have a common point within the parallelopipedon. Scholium. — It is seen at once that the six faces of a parallelopipe- don intersect each other in twelve edges, four of which are equal to if A, four to AB, and four to AD. Now, we may conceive the parallel- opipedon to be bounded by the planes determined by the three linea 180 GEOMETRY. AH, AB, and AD, and the three planes passed through the extremi- ties, H, B, and D, of these lines, parallel to the first three planes. THEOREM III. The convex surface of a right prism is measured by the perimeter of its base multiplied by its altitude. Let ABODE — iVbe a right prism, of which AP is the altitude ; then will its convex surface be measured by {AB + BO+CD + DE + JEA) x AP. For, its convex surface is made up of the rectangles AL, BM, ON, etc., and each rectangle is measured by the product of its base by its altitude ; but the altitude of each rectangle is equal to AP, the alti- tude of the prism ; hence the convex sur- face of the prism is measured by the pro- duct of the sum of the bases of the rectangles, or the perimeter of the base of the prism, by the common alti- tude, AP. Cor. Eight prisms will have equivalent convex surfaces, when the products of the perimeters of their bases by their altitudes are respectively equal ; and, generally, their convex surfaces will be to each other as the products of the perimeters of their bases by their altitudes. Hence, when their altitudes are equal, their surfaces will be as the perimeters of their bases ; and when the perimeters of their bases are equal, their convex surfaces will be as their altitudes. THEOREM IV. The two sections of a prism made by parallel planes between its bases are equal polygons. Let the prism ABODE — N be cut between its bases by two parallel planes, making the sections QBS, etc., BOOK VII 181 and TVX, etc. ; then will these sections be equal polygons. For, since the secant planes are paral- lel, their intersections, QR and TV, by the plane of the face UAPO are parallel, (Th. 10, B. VI) ; and being included be- tween the parallel lines, AP and HO, they are also equal. In the same manner we may prove that US is equal and parallel to VX, and so on for the intersections of the secant planes by the other faces of the prism. Hence, these polygonal sections have the sides of the one equal to the sides of the other, each to each. The angles QEjS and TVX are equal, because their sides are parallel and lie in the same direction ; and in like manner we prove |_ RSY = [_ VXZ, and so on for the other corresponding angles of the polygons. Therefore, these polygons are both mutually equilateral and mutually equiangular, and consequently are equal. - Cor. A section of a prism made by a plane parallel to the base of the prism, is a polygon equal to the base. THEOREM V. Two parallelopipedons, the one rectangular and the other oblique, will be equal in volume when, having the same base and altitude, two opposite lateral faces of the one are in the planes of the corresponding lateral faces of the other. Designating the parallelo- pipedons by their opposite diagonal letters, let AGr be the rectangular, and AL the oblique, parallelopipedon, hav- ing the same base, AC, and of the same altitude, namely, the perpendicular distance be- 16 182 GEOMETRY. tween the parallel planes, A and EL. Also let the face, AK, be in the plane of the face, AF, and the face, BL, in the plane of the face, DGr. We are now to prove that the oblique parallelopipedon is equivalent to the rectangular parallelopipedon. As the faces, AF and AK, are in the same plane, and the parallelopipedons have the same altitude, FFK is a straight line, and EF — IK, because each is equal to AB. If from the whole line, EK, we take EF, and then from the same line we take IK= EF, we shall have the re- mainders, Eland FK, equal ; and since AE and BF are parallel, [_AEI = [_BFK; hence the A's, AEI and BFK, are equal. Since HE and MI are both parallel to DA, they are parallel to each other, and EIMH is a par- allelogram; for like reasons, FKLGr is a parallelogram, and these parallelograms are equal, because two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. The parallelograms, BE and OF, being the opposite faces of the parallelopipedon, AGr, are equal. Hence, the three plane faces bounding the triedral angle, E, of the trian- gular prism, EAI — H, are equal, each to each, and like placed, to the three plane faces bounding the triedral, F, of the triangular prism, FBK — Cr, and these prisms are therefore equal, (Th. 1). Now, if from the whole solid, EABK — H, we take the prism, EAI — H, there will remain the parallelopipedon, AL; and, if from the same solid, we take the prism, FBK—Gr, there will remain the rectangular parallelopipedon, AG. Therefore, the oblique and the rectangular parallelopidon are equiva- lent. Cor. The volume of the rectangular parallelopipedon, AGr, is measured by the base, ABCB, multiplied by the altitude, AE, (Def. 29) ; consequently, the oblique paral- lelopipedon is measured by the product of the same base by the same altitude. BOOK VII 183 Scholium. — If neither of the parallelopipedons is rectangular, but they still have the same base and the same altitude, and two opposite lateral faces of the one are in the planes of the corresponding lateral faces of the other, by precisely the same reasoning we could prove the parallelopipedons equivalent. Hence, in general, any two parallelo- pipedons will be equal in volume when, having the same base and altitude, two opposite lateral faces of the one are in the planes of the correspond- ing lateral faces of the other. THEOREM VI. Two parallelopipedons having equal bases and equal alti- tudes, are equivalent Let AG and AL be two paral- lelopipedons, having a common lower base, and their npper bases in the same plane, HF. Then will these parallelopipedons be equivalent. Since their upper bases are in the same plane, the lines IM, KL, UF, and HG, will intersect, when produced, and form the quadrilateral, NOPQ, and this quadrilateral will be a parallelogram, (Cor. 2, Th. 6, B. YI), equal to the common lower base of the two parallelopipedons. Now, if a third parallelo- pipedon be constructed, having BD for its lower base, and OQ for its upper base, it will be equivalent to the par- allelopipedon AG-', and also to the parallelopipedon AL, (Th. 5, Scholium) ; hence, the two given parallelopipe- dons, being each equivalent to the third parallelopipe- don, are equivalent to each other. Hence, two parallelopipedons having equal bases, etc. THEOREM VII. The volume of any parallelopipedon is measured by the product of its base and altitude, or the product of its three dimensions. 184 GEOMETRY. h H r 9 G Let ABCD — Q be any parallelopipedon ; then will its volume be expressed by the product of the area of its base and altitude. If the parallelopipedon is oblique, we may construct on its base a right parallelopipedon, by erecting perpen- diculars at the points A, B, C, and D, and making them each equal to the altitude of the given parallelopipedon ; and the right parallelopipedon, thus constructed, will be equivalent to the given parallelopip- edon, (Th. 6). Now, if the base, ABCD, is a rectangle, the new parallelopipedon will be rectangular, and meas- ured by the product of its base and altitude, (Def. 16). But if the base is not rectangular, let fall the perpen- diculars, Be and Ad, on CD and CD produced, and take the rectangle ABcd for the base of a rectangular paral- lelopipedon, having for its altitude that of the given parallelopipedon. We may now regard the rectangular face, ABFU, as the common base of the two parallelo- pipedons, Ag and AG-', and, as they have a common base, and equal altitude, they are equivalent. Thus we have reduced the oblique parallelopipedon, first to an equivalent right parallelopipedon on the same base, and then the right to an equivalent rectangular parallelopip- edon on an equivalent base, all having the same alti- tude. But the rectangular parallelopipedon, Ag, is measured by product of its base, ABcd, and its altitude ; hence, the given and equivalent oblique parallelopipedon is measured by the product of its equivalent base and equal altitude. Hence, the volume of any parallelopipedon, etc. Cor. Since a parallelopipedon is measured by the pro- duct of its base by its altitude, it follows that parallelo- pipedons of equivalent bases, and equal altitudes, are equiva- lent, or equal in volume. BOOK VII. 185 THEOREM VIII. Parallelopipedons on the same, or equivalent bases, are to each other as their altitudes ; and parallelopipedons having equal altitudes, are to each other as their bases. Let P and p represent two parallelopipedons, whose bases are denoted by B and b, and altitudes by A and a, respectively. Now, P = B x A, and p = b x a, (Th. 7). But magnitudes are proportional to their numerical measures ; that is, P : p : : B x A : b X a. If the bases of the parallelopipedons are equivalent, we have B = b; and if the altitudes are equal, we have A — a. Introducing these suppositions, in succession, in the above proportion, we get P : p : : A : a, and P : p : : B : b. Hence the theorem ; Parallelopipedons on the same, etc. THEOREM IX. Similar parallelopipedons are to each other as the cubes of their like dimensions. Let P and p represent any two similar parallelopipe- dons, the altitude of the first being denoted by h, and the length and breadth of its base by I and n, respect- ively ; and let h', V, and n r , in order, denote the corres- ponding dimensions of the second. Then we are to prove that P : p :: n* : n n :: P : J' 3 :: h* : h'\ We have P == Inh, and p = Vn'h' (Th. 7) ; and by dividing the first of these equations by the second, member by member, we get 16* 186 GEOMETRY. P Ink p Vn'h' ' which, reduced to a proportion, gives P : p :: Inh : Vn'V. But, by reason of the similarity of the parallelopipe- dons, we have the proportions I : V : : n : n' h : y : : n : n'; we have also the identical proportion, n : n' : : n : n'. By the multiplication of these proportions, term by term, we get, (Th. 11, B. II), Inh : Vn'h' : : n* : n' 3 . That is, P : p :: n 3 : n' 3 . By treating in the same manner the three proportions, I : V : : h : h' n : n f : : h : h f h : h! : : h : h', we should obtain the proportion P : p :: h 3 : h' 3 ', and, by a like process, the three proportions, h : h ! : : I : V n : n f : : I : V X V V : : I : % will give us the proportion P : p : : P : V 3 . Hence the theorem; similar parallehpipedons are to each other, etc, THEOREM X. The two triangular prisms into which any parallelopipedon is divided, by a plane passing through its opposite diagonal edges, are equivalent. Let ABCD — F be a parallelopipedon, and through the diagonal edges, BF and DH, pass the plane BH, divi- ding the parallelopipedon into the two triangular prisms, BOOK VII. 187 ABD — E and BOD — G- ; then we are to prove that these prisms are equivalent. Let us divide the diagonal, BD, in which the se- cant plane intersects the base of the parallelopipedon, into three equal parts, a and c being the points of division. In the base, AB CD, con- struct the complementary paral- lelograms, a and a A, and in the parallelogram, badD, construct the complementary parallelograms, cd and cb, and conceive these, to- gether with the parallelograms, Ba, ac, cD, to be the bases of smaller parallelopipedons, having their lateral faces parallel to the lateral faces of, and their altitude equal to the altitude oi\ the given parallelopipedon, AGr. Now it is evident that the triangular prism, BOD — Or, is composed of the parallelopipedons on the bases, aO and cd, and the triangular prisms, on the side of the secant plane with this prism, into which this plane divides the parallelopipedons on the bases, Ba, ac, and cD. The triangular prism, ABD — E, is also composed of the par- allelopipedons on the bases, Aa and be, together with the triangular prisms on the side of the secant plane with this prism, into which this plane divides the parallelopip- edons on the bases, Ba, ac, and cD. But the parallelograms, a and a A, being complement- ary, are equivalent, (Th. 31, B. I) ; and for the same reason the parallelograms, cd and cb, are equivalent ; and since parallelopipedons on equivalent bases and of equal altitudes, are equivalent, (Cor., Th. 7), we have the sum of parallelopipedons on bases a and cd, equivalent to the sum of parallelopipedons on the bases, aA and cb. Hence, the triangular prisms, ABD — E and BOD — #, 188 GEOMETRY. differ in volume only by the difference which may exist between the snms of the triangular prisms on the two sides of the secant plane into which this plane divides the parallelopipedons on the bases, Ba, ac, and cd. Now, if the number of equal parts into which the diag- onal is divided, be indefinitely multiplied, it still holds true that the triangular prisms, ABB — E and BOB — 6r, differ in volume only by the difference between the sums of the triangular prisms on the two sides of the. secant plane into which this plane divides the parallelopipedons constructed on the bases whose diagonals are the equal portions of the diagonal, BB. But in this case the sum of these parallelopipedons themselves becomes an indefi- nitely small part of the whole parallelopipedon, A 6r, and the difference between the parts of an indefinitely small quantity must itself be indefinitely small, or less than any assignable quantity. Therefore, the triangular prisms, ABB — E and BOB — 6r, differ in volume by less than any assignable volume, and are consequently equiv- alent. Hence the theorem ; the two triangular prisms into which, etc. Cor. 1. Any triangular prism, as ABB — E, is one half the parallelopipedon having the same triedral angle, A, and the same edges, AB, AB, and AE. Cor. 2. Since the volume of a parallelopipedon is meas- ured by the product of its base and altitude, and the tri- angular prisms into which it is divided by the diagonal plane, have bases equivalent to one half the base of the parallelopipedon, and the same altitude, it follows that, the volume of a triangular prism is measured by the product of its base and altitude. The above demonstration is less direct, but is thought to be more simple, than that generally found in authors, and which is here given as a BOOK VII. 189 Second Demonstration. Let ABQD — F be a parallelo- pipedon, divided by the diagonal plane, BH, passing through the edges, BF and Dff; then we are to prove that the triangular prisms, ABD—E and BCD— a, thus formed, are equivalent. Through the points B and F, pass planes perpendicular to the edge, BF, and produce the late- ral faces of the parallelopipedon to intersect the plane through B ; then the sections Bcda and Fghe are equal parallelograms. For, since the cutting planes are both perpendicular to BF, they are parallel, (Th. 10, B. VI) ; and because the opposite faces of a parallelo- pipedon are in parallel planes, (Th. 2), and the intersec- tions of two parallel planes by a third plane are parallel, (Th. 9, B. YI), the sections, Bcda and Fghe, are equal parallelograms, and may be taken as the bases of the right parallelopipedon, Bcda — h. But the diagonal plane divides the right parallelopipedon into the two equal tri- angular prisms, aBd — e and Bed — g, (Th. 1). "We will now compare the right prism with the oblique triangular prism on the same side of the diagonal plane. The volume ABB — e is common to the two prisms, ABB — E and aBd — e ; and the volume eFh — E, which, added to this common part, forms the oblique triangular prism, is equal to the volume aBd — A, which, added to the common part, forms the right triangular prism. For, since ABFE and aBFe are parallelograms, AE — ae, and taking away the common part Ae, we have aA=eE; and since BFHD and BFhd are parallelograms, we have BH = dh ; and from these equals taking away the common part Dh, we have dD = hH. Now, if the volume eFh — B 190 GEOMETRY. be applied to the volume aBd — D, the base eFh falling on the equal base aBd, the edges eE and hH will fall upon aA and dD respectively, because they are perpen- dicular to the base aBd, (Cor. 2, Th. 3, B. VI), and the point E will fall upon the point A, and the point H upon the point D ; hence the volume eFh — H exactly coincides with the volume aBd — D, and the oblique triangular prism ABB — E is equivalent to the right triangular prism aBd — e. In the same manner, it may be proved that the oblique triangular prism, BCBG,i8 equivalent to the right tri- angular prism, Bcdg. The oblique triangular prism on either side of the diagonal plane is, therefore, equivalent to the corresponding right triangular prism ; and, as the two right triangular prisms are equal, the oblique trian- gular prisms are equivalent. Hence the theorem ; the two triangular prisms, etc. THEOREM XI. The volume of any prism whatever is measured by the prod- uct of the area of its base and altitude. For, by passing planes through the homologous diag- onals of the upper and lower bases of the prism, it will be divided into a number of triangular prisms, each of which is measured by the product of the area of its base and altitude. Now, as these triangular prisms all have, for their common altitude, the altitude of the given prism, when we add the measures of the triangular prism, to get that of the whole prism, we shall have, for this measure, the common altitude multiplied by the sum of the areas of the bases of the triangular prisms : that is, the product of the area of the polygonal base and the altitude of the prism. Hence the theorem ; the volume of any prism, etc. Cor. If A denote the area of the base, and H the alti- BOOK VII. 191 tude of a prism, its volume will be expressed by A x 11. Calling this volume F, we have V 4 A x H. Denoting by A', W, and V, in order, the area of the base, altitude, and volume of another prism, we have V = A' x H'. Dividing the first of these equations by the second, member by member, we have V_ AxH V ' A 1 x R 1 ' which gives the proportion, V : V : : A x E : A f x H f . If the bases are equivalent, this proportion becomes V : V : : H : H r ; and if the altitudes are equal, it reduces to V : V : : A : A'. Hence, prisms of equivalent bases are to each other as their altitudes ; and prisms of equal altitudes are to each other as their bases. THEOREM XII. A plane passed through a pyramid parallel to its base, divides its edges and altitude proportionally, and makes a section, which is a polygon similar to the base. Let ABODE — V be any pyramid, whose base is in the plane, MN, and vertex in the parallel plane, mn ; and let a plane be passed through the pyramid, parallel to its base, cutting its edges at the points, a, 5, c, d, e, and the altitude, JEF, at the point I. By joining the points, a, b, c, etc., we have the polygon formed by the intersection of the plane and the sides of the pyramid. Now, we are to prove that the edges, VA, VB, etc., and the altitude, FE, are divided proportionally at the points, a, b, etc., and Z; and that the polygon, a, b, c, d, e, is similar to the base of the pyramid. GEOMETRY. Since the cutting plane is parallel to the base of the pyramid, ab is parallel to AB, (Th. 9, B. VI) ; for the same reason, be is parallel to BO, cd to OB, etc. Now, in the triangle VAB, because ab is parallel to the base AB, we have, (Th. 17, B. II), the proportion, VA : Va : : VB : Vb. In like manner, it may be shown that VB : Vb : : VO : Vc, and so on for the other lateral edges of the pyramid. F being the point in which the perpendicular from E pierces the plane mn, and I the point in which the parallel secant plane cuts the perpendicular, if we join the points F and V, and also the points I and e by straight lines, we have in the triangle FFV, the line le parallel to the base FV; hence the proportion VF : Ve : : FE : Fl. Therefore, the plane passed through the pyramid par- allel to its base, divides the altitude into parts \diich have BOOK VII. 193 to each other the same ratio as the parts into which it divides the edges. Again, since ab is parallel to AB, and be to BO, the angle abc is equal to the angle ABO, (Th. 8, B. I); in the same manner we may show that each angle in the polygon, abode, is equal to the corresponding angle in the polygon, ABODE; therefore these polygons are mutually equiangular. But, because the triangles VBA and Vba are similar, their homologous sides give the proportion Vb : VB :: ab : AB; and because the triangles Vbc and VBO are similar, we also have the proportion Vb : VB : : be : BO. Since the first couplet in these two proportions is the same, the second couplets are proportional, and give ab : AB : : be : BO. By a like process, we can prove that be : BO : : cd : OB, and that cd : OD :: de : BE, and so on, for the other homologous sides of the two polygons. Hence, the two polygons are not only mutually equi- angular, but the sides about the equal angles taken in the same order are proportional, and the polygons are there- fore similar, (Def. 16, B. II). Hence the theorem; a plane passed through a pyramid, etc. Oor. 1. Since the areas of similar polygons are to each other as the squares of their homologous sides, (Th. 22, B. II), we have area abode : area ABODE : ab 2 : AB*. But, ab : AB :: Va : VA :: Fl : FE; hence, ab 2 : AB 2 i:~Ff : FE 2 : therefore, area abode : area ABODE : Fl 2 : FE . 17 N 194 GEOMETRY. That is, the area of the section made by a plane passing through a pyramid parallel to its base, is to the area of the base, as the perpendicular distance from the vertex of the pyramid to the section, is to the altitude of the pyramid. Cor. 2. Let V— ABODE and X—RST be two pyra- mids, having their bases in the plane MN, and their ver- tices in the parallel plane mn ; and suppose a plane to be passed through the two pyramids parallel to the common plane of their bases, making in the one the section abcde, and in the other the section rst. Now, arenABCDE: area abcde ::AB : ab , (Th.22,B.II), and " RST: " rst ::RS 2 :rs\ But, AB : ab : : VB : Vb, and RS : rs : : XR : Xr. Because the plane which makes the sections is parallel to the planes MN and mn, we have, (Th. 11, B. VI), VB : Vb :: XR : Xr; therefore, (Cor. 2, Th. 6, B.II), AB i ab : : RS : rs. :2 ~T2 ~~ 777*2 By squaring, AB : ab : RS : rs ; hence, area ABCDE : area abcde : : area RST : area rst. That is, if two pyramids having equal altitudes, and their bases in the same plane, be cut by a plane parallel to the com- mon plane of their bases, the areas of the sections will be proportional to the areas of the bases ; and if the bases are equivalent, the sections will also be equivalent. THEOREM XIII. If two triangular pyramids have equivalent bases and equal altitudes, they are equal in volume. Let V — ABC and v — abc be two triangular pyramids, having the equivalent bases, ABC and abc, and let the altitude of each be equal to CX; then will these two pyramids be equivalent. BOOK VII 195 - fI // V ' /It \l 1 JT^ 1 ''' ' 1 1 r \ 1 1 // V / h/\ i 1 / 7 \ / x l 1/ VH A // S\ let Place the bases of the pyramids on the same plane, with their vertices in the same direction, and divide the altitude into any number of equal parts. Through the points of division pass planes parallel to the plane of the bases ; the corresponding sections made in the pyramids by these planes are equivalent, (Th. 12, Cor. 2) ; that is, the triangle DEI? is equivalent to the triangle def, the triangle GrHI to the triangle ghi, etc. Now, let triangular prisms be constructed on the tri- angles ABO, DEF, etc., of the pyramid V— ABO, these prisms having their lateral edges parallel to the edge, VO, of the pyramid, and the equal parts of the altitude, OX, for their altitudes. Portions of these prisms will be exterior to the pyramid V — ABO, and the sum of their volumes will exceed the volume of the pyramid. On the bases def, ghi, etc., in the other pyramid, con- struct interior prisms, , as represented in the figure, their lateral edges being parallel to vc, and their alti- tudes also the equal parts of the altitude, OX. Portions of the pyramid, v — abe, will be exterior to these prisms, 196 GEOMETRY. and the volume of the pyramid will exceed the sum of the volumes of the prisms. Since the sum of the exterior prisms, constructed in connection with the pyramid V — ABO, is greater than the pyramid, and the sum of the interior prisms, con- structed in connection with the pyramid v — abc, is less than this pyramid, it follows that the difference of these sums is greater than the difference of the pyramids them- selves. But the second exterior prism, or that on the base DEF, is equivalent to the first interior prism, or that on the base def, and the third exterior prism is equivalent to the second interior prism, (Th. 10, Cor. 2), and so on. That is, beginning with the second prism from the base of the pyramid, V — ABO, and taking these prisms in order towards the vertex of the pyramid, and comparing them with the prisms in the pyramid, v — abc, beginning with the lowest, and taking them in order toward the vertex of this pyramid, we find that to each exterior prism of the pyramid, V — ABO, exclusive of the first or lowest, there is a corresponding equivalent interior prism in the pyramid, v — abc. Hence the prism, ABODEF, is the difference between the sum of the prisms constructed in connection with the pyramid, V— ABO, and the sum of the interior prisms constructed in the pyramid, v — abc. But the first sum being a volume greater than the pyramid, V — ABO, and the second sum a volume less than the pyramid, v — abc, it follows that the volumes of the pyramids differ by less than the prism, ABODEF. Now, however great the number of equal parts into which the altitude, OX, be divided, and the correspond- ing number of prisms constructed in connection with each pyramid, it would still be true that the difference between the volumes of the pyramids would be less than the volume of the lowest prism of the pyramid V— ABO', but when we make the number of equal parts into which BOOK VII. . 197 the altitude is divided indefinitely great, the volume of this prism becomes indefinitely small : that is, the differ- ence between the volumes of the pyramids is less than an indefinitely small volume ; or, in other words, there is no assignable difference between the two pyramids, and they are, therefore, equivalent. Hence the theorem ; if two triangular pyramids, etc. THEOREM XIV. Any triangular pyramid is one third of the triangular prism having the same base and equal altitude. Let F — ABC be a triangular pyramid, and through F pass a plane parallel to the plane of the base, ABC. In this plane, through F, construct the triangle, FDE, having its sides, FD, E DF, and FF, parallel and equal to B C, 7vnT ~y 7 \ CA, and AB, respectively. The tri- / \/*\ / angle, FDF, may be taken as the / /\ \ \ upper base of a triangular prism of \/ / \\/ which the lower base is ABC. ^\ ~\ / IsTow, this triangular prism is com- 13 posed of the given triangular pyramid, F — ABC, and of the quadrangular pyramid, F — A CDF. This last pyramid may be divided by a plane through the three points, C, F, and F, into the two triangular pyra- mids, F—DFC and F—ACF. But the pyramid, jP— BFC, may be regarded as having the triangle, FFB, equal to the triangle, ABC, for its base, and the point, C, for its vertex. The two pyramids, F—AB C and C—JDFF, have equal bases and equal altitudes ; they are therefore equivalent, (Th. 13). Again, the two pyramids, F—DFC and F — ACE, have a common vertex, and equivalent bases in the same plane, and they are also equivalent. There- fore, the triangular prism, ABCDEF, is composed of 17* 198 GEOMETRY. three equivalent triangular pyramids, one of which is the given triangular pyramid, F — ABC. Hence the theorem; any triangular pyramid is one third of the triangular prism, etc. Cor. The volume of the triangular prism being meas- ured by the product of its base and altitude, the volume of a triangular pyramid is measured by one third of the product of its base and altitude. THEOREM XV, The volume of any pyramid whatever is measured by one third of the product of its base and altitude. Let V — ABCDU be any pyramid ; then will its volume be measured by one third of the product of its base and altitude. In the base of the pyramid, draw the diagonals, AB and AC, and through its vertex and these diagonals, pass planes, thus dividing the pyramid into a number of triangular pyramids having the common vertex V, and the altitude of the given pyramid for their common altitude. Now, each of these triangular pyra- mids is measured by one third of the product of its base and altitude, (Cor., Th. 14), and their sum, which constitutes the polygonal pyramid, is therefore measured by one third "of the product of the sum of the trian- gular bases and the common altitude ; but the sum of the triangular bases constitutes the polygonal base, ABCDE. Hence the theorem ; the volume of any pyramid what- ever, etc. Cor. 1. Denote, by B, H, and V, respectively, the base, altitude, and volume of one pyramid, and by B', W, and BOOK VII. 199 F 7 , the base, altitude, and volume of another ; then we shall have V = $B x IT, and V = \B' x W. Dividing the first of these equations by the second, member by member, we have V - B x H V f B' x W' which, in the form of a proportion, gives V : V : : B X H : B' X W. From this proportion we deduce the following conse- quences : 1st. Pyramids are to each other as the products of their bases and altitudes. 2d. Pyramids having equivalent bases are to each other as their altitudes. 3d. Pyramids having equal altitudes are to each other as their bases. Cor. 2. Since a prism is measured by the product of it's base and altitude, and a pyramid by one third of the product of its base and altitude, we conclude that any pyramid is one third of a prism having an equivalent base and equal altitude. THEOREM XVI. The volume of the frustum of a pyramid is equivalent to the sum of the volumes of three pyramids, each of which has an altitude equal to that of the frustum, and whose bases are, respectively, the lower base of the frustum, the upper base of the frustum, and a mean proportional between these bases. Let V— ABODE and X—RST be two pyramids, the one polygonal and the other triangular, having equiva- lent bases and equal altitudes ; and let their bases be placed on the plane MN, their vertices falling on the parallel plane mn. Pass through the pyramids a plane GEOMETRY. parallel to the common plane of their bases, cutting out the sections abode and rst ; these sections are equivalent, (Th. 12, Cor. 2), and the pyramids, V — abode and X — rst, are equivalent, (Th. 13). Now, since the pyramids, V— ABODE and X—EST, are equivalent, if from the first we take the pyramid, V — abode, and from the second, the pyramid, X — rst, the remainders, or the frusta, ABODE — a and BST—r, will be equivalent. If, then, we prove the theorem in the case of the frus- tum of a triangular pyramid, it will be proved for the frustum of any pyramid whatever. Let ABO—D be the frustum of a triangular pyramid. Through the points D, B, and 0, pass a plane, and through the points D, C, and E, pass another, thus dividing the frustum into three triangular pyra- mids, viz., D—ABO, O—DEF, and D—BEO. Now, the first of these has, for its BOOK VII. 201 base, the lower base of the frustum, and for its altitude the altitude of the frustum, since its vertex is in the upper base ; the second has, for its base, the upper base of the frustum, and for its altitude the altitude of the frustum, since its vertex is in the lower base. Hence, these are two of the three pyramids required by the enunciation of the theorem ; and we have now only to prove that the third is equivalent to one having, for its base, a mean proportional between the bases of the frus- tum, and an altitude equal to that of the frustum. In the face ABED, draw HB parallel to BE, and draw HE and HO. The two pyramids, B — BEO and H—BEO, are equivalent, since they have a common base and equal altitudes, their vertices being in the line BH, which is parallel to the plane of their common base, (Th. 7, B. VI). We may, therefore, substitute the pyramid, H—BEO, for the pyramid, D—BEC. But the triangle, BOH, may be taken as the base, and E as the vertex of this new pyramid ; hence, it has the required altitude, and we must now prove that it has the required base. The triangles, ABO and HBO, have a common vertex, and their bases in the same line ; hence, (Th. 16, B. II), A ABO : A HBO : : AB : HB :: AB : BE. (1) In the triangles, BEE and HBO, [__ E = L -#, and BE=HB', hence, if BEE be applied to HBO, [__ E fil- ing on [_ B, and the side BE on HB, the point B will fall on H, and the triangles, in this position, will have a common vertex, H, and their bases in the same line ; hence A HBO : A BEE : : BO : EF. (2) But, because the triangles, ABO and BEE, are similar, we have AB : BE :: BO : EF. (3) From proportions (1), (2) ? and (3), we have, (Th. 6, B. II), 202 GEOMETRY. A ABO : A HBO : : A HBO : A DBF; that is, the base, HBO, is a mean proportional between the lower and upper bases of the frustum. Hence the theorem ; the volume of the frustum of a pyra- mid, etc. THEOREM XVII. The convex surface of any right pyramid is measured by the perimeter of its base, multiplied by one half its slant height. Let S— ABODE F be a right pyramid, of which SH is the slant height ; then will its convex surface have, for its measure, ±SH{AB + BO+ OD + DB+EF+ FA). Since the base is a regular polygon, and the perpendicular, drawn to its plane from S, passes through its center, the edges, SA, SB, SO, etc., are equal, (Cor. Th. 4, ah~b B. VI), and the triangles SAB, SBO, etc., are equal, and isosceles, each having an altitude equal to SH. Now, AB x %SH measures the area of the triangle, SAB ; and BO x %SH measures the area of the triangle, SBO; and so on, for the other triangular faces of the pyramid. By the addition of these different measures, we get iSH(AB + BO+OD + DH+UF+ FA), as the measure of the total convex surface of the pyramid. Hence the theorem; the convex surface of any right pyramid, etc. THEOREM XVIII. The convex surface of the frustum of any right pyramid is measured by the sum of the perimeters of the two bases, mul- tiplied by one half the slant height of the frustum. Let ABODEF — d be the frustum of a right pyramid ; then will its convex surface be measured by iHh{AB+BC+CD+DI!+EF+FA+ab+b<>±cd+de+tf+fa). BOOK VII. 203 e / Xffl i d hbV W 1 |] 1 l V For, the upper base, abcdef, of the frustum is a section of a pyramid by a plane parallel to the lower base, (Def. 14), and is, therefore, similar to the lower base, (Th. 12). But the lower base is a regular polygon, (Def. 12); hence, the up- per base is also a regular polygon, of the same name; and as ab and AB are intersections of a face of the pyramid by two parallel planes, A ST B they are parallel. For the same reason, be is parallel to BO, cd to OB, etc., and the lateral faces of the frustum are all equal trapezoids, each having an altitude equal to ITJi, the slant height of the frustum. The trapezoid ABba has, for its measure, %Hh(AB+ab), (Th. 34, Book I) ; the trapezoid BCcb has, for its meas- ure, %Hh(BC + be), and so on, for the other lateral faces of the frustum. Adding all these measures, we find, for their sum, which is the whole convex surface of the frustum, \Rh {AB+BC+ CD+DE+EF+ FA+db+bc+cd+de+ef+fa). Hence the theorem ; the convex surface of the frustum, etc. THEOREM XIX. The volumes of similar triangular prisms are to each other as the cubes constructed on their homologous edges. Let ABC— I 7 and abp— /be two similar triangular prisms ; then will their vol- umes be to each other as the cubes, whose edges are the homologous edges 204 GEOMETRY. AB and ab, or as the cubes, whose edges are the homol- ogous edges BE and be, etc. Since the prisms are similar, the solid angles, whose vertices are B and b, are equal; and the smaller prism, when so applied to the larger that these solid angles coincide, will take, within the larger, the position represented by the dotted lines, In this position of the prisms, draw EH perpendicular to the plane of the base ABO, and join the foot of the perpen- dicular to the point B, and in the triangle BEH draw, through e, the line eh, parallel to EH; then will EH represent the altitude of the larger prism, and eh that of the smaller. J^ow, as the bases ABC and aBc, are homologous faces, they are similar, and we have, (Th. 20, Book II), A ABC : A aBc :: AB* :~a~B 2 (1) But the A's BEH and Beh are equiangular, and there- fore similar, and their homologous sides give the propor- tion BE : Be :: EH : eh (2) and from the homologous sides of the similar faces, ABED and aBed, we also have BE : Be :: AB : aB (3) Proportions (2) and (3 ), having an antecedent and con- sequent the same in both, we have, (Th. 6, B. II), EH : eh :: AB : aB (4) By the multiplication of proportions (1) and (4) ? term by term, we get A ABC X EH: A aBc X eh:: AB 3 : aB 3 But A ABC x EH measures the volume of the larger prism, and A aBc x eh measures the volume of the smaller. Hence the theorem; the volumes of similar triangular prisms, etc. BOOK VII. 205 Cor. 1. The volumes of two similar prisms having any bases whatever, are to each other as the cubes constructed on their homologous edges. For, if planes be passed through any one of the lateral edges, and the several diagonal edges, of ,one of these prisms, this prism will be divided into a number of smaller triangular prisms. Taking the homologous edge of the other prism, and passing planes through it and the seve- ral diagonal edges, this prism will also be divided into the same number of smaller triangular prisms, similar to those of the first, each to each, and similarly placed. Kow, the similar smaller prisms, being triangular, are to each other as the cubes of their homologous edges ; and being like parts of the larger prisms, it follows that the larger prisms are to each other as the cubes of the homologous edges of any two similar smaller prisms. But the homologous edges of the similar smaller prisms are to each other as the homologous edges of tlie given prisms ; hence we conclude that the given prisms are to each other as the cubes of their homologous edges. Cor. 2. The volumes of two similar pyramids having any bases whatever, are to each other as the cubes constructed on their homologous edges. For, since the pyramids are similar, their bases are similar polygons ; and upon them, as bases, two similar prisms may be constructed, having for their altitudes, the altitudes of their respective pyramids, and their lateral edges parallel to any two homologous lateral edges of the pyramids. Now, these similar prisms are to each other as the cubes of their homologous edges, which may be taken as the homologous sides of their bases, or as their lateral edges, which were taken equal and parallel to any two arbi- trarily assumed homologous lateral edges of the two pyramids ; hence the pyramids are to each other as the cubes constructed on any two homologous edges. 18 206 GEOMETRY. Cor. 3. The volumes of any two similar polyedrons are to each other as the cubes constructed on their homologous edges. For, by passing planes through the vertices of the homologous solid angles of such polyedrons, they may both be divided into the same number of triangular pyramids, those of the one similar to those of the other, each to each, and similarly placed. Now, any two of these similar triangular pyramids are to each other as the cubes of their homologous edges ; and being like parts of their respective polyedrons, it follows that the polyedrons are to each other as the cubes of the homologous edges of any two of the similar tri- angular pyramids into which they may be divided. But the homologous edges of the similar triangular pyramids are to each other as the homologous edges of the poly- edrons ; hence the polyedrons are to each other as the cubes of their homologous edges. THEOREM XX. !! J~£X The convex surface of the frustum of a cone is measured by the product of the slant height and one half the sum of the circumferences of the bases of the frustum. Let ABOB — abed be the frustum of a cone ; then will its convex surface be t , A (circ. 00 4- circ. oc) measured by Aa x ~ '-, in which the expression, circ. 00, de- notes the circumference of the circle of which 00 is the radius. Inscribe in the lower base of the frustum, a regu- lar polygon haviug any number of sides, and in the upper base a similar polygon, having its sides parallel to those of the polygon in the lower base. These polygons BOOK VII. 207 may be taken as the bases of the frustum of a right pyramid inscribed in the frustum of the cone. ]STow, however great the number of sides of the in- scribed polygons, the convex surface of the frustum of the pyramid is measured by its slant height multiplied by one half the sum of the perimeters of its two bases, (Th. 18) ; but when we reach the limit, by making the number of sides of the polygon indefinitely great, the slant height, perimeters of the bases, and convex surface of the frustum of the pyramid become, severally, the slant height, circumferences of the bases, and convex sur- face of the frustum of the cone. Hence the theorem ; the convex surface of the frustum, etc. Cor. 1. If we make oc — 00, and, consequently, circ. oc = circ. 0(7, the frustum of the cone becomes a cylin- der, and the half sum of the circumferences of the bases becomes the circumference of either base of the cylinder, and the slant height of the frustum, the altitude of the cylinder. Hence, the convex surface of a cylinder is meas- ured by the circumference of the base multiplied by the alti- tude of the cylinder. Cor. 2. Kwe make oc = 0, the frustum of the cone becomes a cone. Hence, the convex surface of a cone is measured by the circumference of the base multiplied by one half the slant height of the cone. Cor. 3. If through E, the middle point of Co, the line Ff be drawn parallel to Oo, and Em perpendicular to Go, the line oc being produced, to meet Ff at/, we have, because the A's EFC and Efc are equal, w OC + oo Em = _^-_ . If we multiply both members of this equation by 2*, we have 2«.Em = 2*-0Q+2™. 208 GEOMETEY. that is, circ. Em is equal to one half the sum of the cir- cumferences of the two bases of the frustum. Hence, the convex surface of the frustum of a cone is measured by the circumference of the section made by a plane half way between the two bases, and parallel to them, multiplied by the slant height of the frustum. Cor. 4. If the trapezoid, OCco, be revolved about Oo as an axis, the inclined side, Cc, will generate the con- vex surface of the frustum of a cone, of which the slant height is Cc, and the circumferences of the bases are circ. OC and circ. oc. Hence, if a trapezoid, one of whose sides is perpendicular to the two parallel sides, be revolved about the perpendicular side as an axis, it will generate the frustum of a cone, the inclined side opposite the axis generating the convex surface, and the parallel sides the bases of the frustum. THEOREM XXI. The volume of a cone is measured by the area of its base multiplied by one third of its altitude. Let V — ABC, etc., be a cone; then will its volume be measured by area ABC, etc., multiplied by \VO. Inscribe, in the base of the cone, any regular polygon, as ABCDEF, which may be taken as the base of a right pyra- mid, of which V is the vertex. The volume of this inscribed pyramid will AJ have, for its measure, (Th. 15), polygon ABCDEF x \VO. Now, however great the number of sides of the poly- gon inscribed in the base of the cone, it will still hold true that the pyramid of which it is the base, and whose vertex is V, will be measured by the area of the poly- gon, multiplied by one third of VO; but when we reach the limit, by making the number of sides indefi- BOOK VII. 209 nitely great, the polygon becomes the circle in which it is inscribed, and the pyramid becomes the cone. Hence the theorem ; the volume of a cone, etc. Cor. 1. If R denote the radius of the base of a cone, and H its altitude, or axis, its volume will be expressed by hence, if Fand V designate the volumes of two cones, of which R and R ' are the radii of the bases, and H and H r the altitudes, we have V: V :: iKx«R 2 : ±H f x «R n :: Hx«R 2 : H f X :: R : i2'; J2 7 R r , E 2f R f * s = r ; hence > M*'^' By substituting for the factors, in the second member of eq. ( 1 ), their values successively, and resolving into a proportion, we get V : V :: R* : R*'; and V : V :: jEP': E'\ Hence, similar cones are to each other as the cubes of the radii of their bases, and also as the cubes of their altitudes. Cor. 3. A cone is equivalent to a pyramid having an equiv- alent base and an equal altitude. 18* o 210 GEOMETRY. THEOREM XXII. The volume of the frustum of a cone is equivalent to the sum of the volumes of three cones, having for their common altitude the altitude of the frustum, and for their several bases, the bases of the frustum and a mean proportional be- tween them. Let ABQD — abed be the frustum of a cone ; then will its volume be equiva- lent to the sum of the volumes, having Oo for their common altitude, and for their bases, the circles of which, OG, oc, and a mean proportional between 00 and oc, are the respective radii. Inscribe in the lower base of the frus- tum any regular polygon, and in the upper base a similar polygon, having its sides parallel to those of the first. These polygons may be taken as the bases of the frustum of a right pyra- mid inscribed in the frustum of the cone. The volume of the frustum of the pyramid is equiva- lent to the sum of the volumes of three pyramids, having for their common altitude the altitude of the frustum, and for their several bases the bases of the frustum, and a mean proportional between them, (Th. 16). !Now, however great the number of sides of the poly- gons inscribed in the bases of the frustum of the cone, this measure for the volume of the frustum of the pyra- mid, of which they are the bases, still holds true ; but when we reach the limit, by making the number of the sides of the polygon indefinitely great, the polygons be- come the circles, the frustum of the pyramid becomes the frustum of the cone, and the three partial pyramids, whose sum is equivalent to the frustum of the pyramid, become three partial cones, whose sum is equivalent to the frustum of the cone. BOOK VII. 211 Hence the theorem ; the volume of the frustum of a cone, etc. Cor. 1. Let R denote the radius of the lower base, R ■' that of the upper base, and ^Tthe altitude of the frustum of a cone ; then will its volume be measured, (Th. 21), by ±H x *R 2 + ±R x «R n + i&x *Rx R r , since *R x R r expresses the area of a circle which is a mean proportional between the two circles, whose radii are R and R f . E"ow, if the bases of the frustum become equal, or R = R', the frustum becomes a cylinder, and each of the last two terms in the above expression for the volume of the frustum of a cone will be equal to the first ; hence, the volume of a cylinder, of which H is the altitude, and R the radius of the base, is measured by H x «R 2 . Therefore, the volume of a cylinder is measured by the area of its base multiplied by its altitude. Cor. 2. By a process in all respects similar to that pur- sued in the case of cones, it may be shown that similar cylinders are to each other as the cubes of the radii of their bases, and also as the cubes of their altitudes. Cor. 3. A cylinder is equivalent to a prism having an equivalent base and an equal altitude. THEOREM XXIII. If a plane be passed through a sphere, the section will be a circle. Let be the center of a sphere through which a plane is passed, making the section AmBn ; then will this section be a circle. From let fall the perpendic- ular Oo upon the secant plane, and draw the radii OA, OB, and Om, to the different points in the intersection of the plane with the surface of the sphere. Now. 212 GEOMETRY. the oblique lines OA, OB, Om, are all equal, being radii of the sphere; they therefore meet the plane at equal dis- tances from the foot of the perpendicular Oo, (Cor., Th. 4, B.VI); hence oA, oB, om, etc., are equal: that is, all the points in the intersection of the plane with the surface of the sphere are equally distant from the point 0. This intersection is therefore the circumference of a circle of which o is the center. Hence the theorem; if a plane be passed through a sphere, etc. Cor. 1. Since AB, the diameter of the section, is a chord of the sphere, it is less than the diameter of the sphere ; except when the plane of the section passes through the center of the sphere, and then its diameter becomes the diameter of the sphere. Hence, 1. All great circles of a sphere are equal. 2. Of two small circles of a sphere, that is the greater whose plane is the less distant from the center of the sphere. 3. All the small circles of a sphere whose planes are at the same distance from the center, are equal. Cor. 2. Since the planes of all great circles of a sphere pass through its center, the intersection of two great circles will be both a diameter of. the sphere and a com- mon diameter of the two circles. Hence, two great circles of a sphere bisect each other. Cor. 3. A great circle divides the volume of a sphere, and also its surface, equally. For, the two parts into which a sphere is divided by any of its great circles, on being applied the one to the other, will exactly coincide ; otherwise all the points in their convex surfaces would not be equally distant from the center. Cor. 4. The radius of the sphere which is perpendicular to the plane of a small circle, passes through the center of the circle. BOOK VII. 213 Cor. 5. A plane passing through the extremity of a radius of a sphere, and perpendicular to it, is tangent to the sphere. For, if the plane intersect the sphere, the section is a circle, and all the lines drawn from the center of the sphere to points in the circumference are radii of the sphere, and are therefore equal to the radius which is per- pendicular to the plane, which is impossible, (Cor. 1, Th. 3, B. VI). Hence the plane does not intersect the sphere, and has no point in its surface except the extremity of the perpendicular radius. The plane is therefore tangent to the sphere by Def 22. THEOREM XXIY. If the line drawn through the center and vertices of two opposite angles of a regular polygon of an even number of sides, be taken as an axis of revolution, the perimeter of either semi-polygon thus formed will generate a surface whose measure is the axis multiplied by the circumference of the inscribed circle. Let ABCDEF be a semi-polygon cut off from a regular polygon of an even number of sides by drawing the line AF through the center 0, and the vertices A and F, of two opposite angles of the poly- gon ; then will the surface generated by the perimeter of this semi-polygon re- volving about AF as an axis, be meas- ured by AF X circumference of the in- scribed circle. From m, the middle point, and the extremities B and of the side 2? (7, draw mn, BK, and OL, perpendicular to AF; join also m and 0, and draw BH perpendicular to CL. The surface of the frustum of the cone generated by the trapezoid BKLO, has for its measure circ. mn X BO, (Cor. 3, Th. 20). Since mO is perpendicular to BO, and mn to BH, the two A's, BOH and mnO, are similar, and their homologous sides give the proportion 214 GEOMETRY. mn : mO :: BH (= jK£) : BO and as circumferences are to each other as their radii, we have circ. mn : circ. mO :: KL : BO Hence, circ. mn X BO = circ. mO X KL. But mO is the radius of the circle inscribed in tne polygon. Hence, the surface generated by BO during the revolution of the semi-polygon, is measured by the cir- cumference of the inscribed circle multiplied by KL, the part of the axis included between the two perpendicu- lars let fall upon it from the extremities B and 0. The surface generated by any other side of the semi-polygon will be measured, in like manner, -by the circumference of the inscribed circle multiplied by the corresponding part of the axis. By adding the measures of the surfaces generated by the several sides of the semi-polygon, we get Circ. mO x {AK + KL + LN + NM+ MF) for the measure of the whole surface. Hence the theorem ; if the line drawn through the cen- ter, etc. Oor. It is evident that the surface generated by any portion, as OB and BF, of the perimeter, is measured by circ. mO x LM. THEOREM XXV. The surface of a sphere is measured by the circumference of one of its great circles multiplied by its diameter. Let a sphere be generated by the revolution of the semi-circle, ARF, about its diameter, AF; then will the surface of the sphere be measured by Circ. AO x AF. Inscribe in the semi-circle any regular semi-polygon, and let it be revolved, with the semi-circle, about the axis BOOK VII. 215 AF; the surface generated by its perim- eter will be measured by Circ. mOx AF, (Th. 24), and this measure will hold true, how- ever great the number of sides of the in- H| scribed semi-polygon. But as the num- ber of these sides is increased, the radius mO, of the inscribed semi-circle, increases and approaches equality with the radius, AO; and when we reach the limit, by making the number of sides indefinitely great, the radii and semi-circles become equal, and the surface generated by the perimeter of the inscribed semi-polygon becomes the surface of the sphere. Therefore, the surface of the sphere has, for its measure, Girc. AO x AF. Hence the theorem ; the surface of a sphere is meas- ured, etc. Cor. 1. A zone of a sphere is measured by the circumfer- ence of a great circle of the sphere multiplied by the altitude of the zone. For, the surface generated by any portion, as CD and DF, of the perimeter of the inscribed semi-polygon has, for its measure, circ. mO X LM, (Cor. Th. 24) ; and as the number of the sides of the semi-polygon increases, LM remains the same, the radius mO alone changing, and becoming, when we reach the limit, equal to AO) hence, the surface of the zone is expressed by Circ. AO x LM, whether the zone have two bases, or but one. Cor. 2. Let H and H' denote the altitudes of two zones of spheres, whose radii are R and R ' ; then these zones will be expressed by 2*R x H and 2*R ' x H f ; and if the surfaces of the zones be denoted by Z and Z r , we have 216 GEOMETRY. Z : Z' : : 2«R x E : 2«R f x E' : : R x E : R f x E'. Hence, 1. Zones in different spheres are to each other as their altitudes multiplied by the radii of the spheres. 2. Zones of equal altitudes are to each other as the radii of the spheres. 3. Zones in the same, or equal spheres, are to each other as their altitudes. Cor. 3. Let R denote the radius of a sphere; then will its diameter be expressed by 2R, and the circumference of a great circle by 2*R ; hence its surface will be ex- pressed by 2«R x2E = ±«R\ That is, the surface of a sphere is equivalent to the area of four of its great circles. Cor. 4. The surfaces of spheres are to each other as the squares of their radii. THEOREM XXVI. If a triangle be revolved about either of its sides as an axis, the volume generated will be measured by one third of the prod- uct of the axis and the area of a circle, having for its radius the perpendicular let fall from the vertex of the opposite angle on the axis, or on the axis produced. First. Let the triangle ABC, in which the perpendicular from C falls on the opposite side, AB, be revolved about AB as an axis; then will *Yol. A ABC have, for its measure, %AB x *CD . The two A's into which A ABC is divided by the perpendicular DC, are right-angled, and during the rev- olution 'they will generate two cones, having for their * Vol. A ABC, cone A ABC, are abbreviations for volume gener- ated by A ABC, cone generated by A ABC', and surfaces of revolu- tion generated by lines will hereafter be denoted by like abbreviations. BOOK VII. 217 common base the circle, of which DO is the radius, and for their axes the parts DA and DB, into which AB is divided. Now, *Cone A ADO is measured by \AD x mj>(jy (Th. 21), and cone A BDO, by ±BD x iDQ* ; but these two cones compose Yol. A ABO; and by adding their measures, we have, for that of Yol. A ABO, iAD x *D0 2 + iBD x r~D0 2 = \AB x _ DE x \ODx %. b^a* — b* Ans. R = 5 — . a + b Prob. XXXII. — From a point without a circle whose diameter is d, a line equal to d is drawn, terminating in the concave arc, and this line is bisected at the first point in which it meets the circumference. What is the distance of the point without from the center of the circle? It is not deemed necessary to multiply problems in the application of algebra to geometry. The preceding will be a sufficient exercise to give the student a clear con- ception of the nature of such problems, and will serve as a guide for the solution of others that may be proposed to him, or that may be invented by his own ingenuity. MISCELLANEOUS PROPOSITIONS. We shall conclude this book, and the subject of Geom- etry, by offering the following propositions, — some the- orems, others problems, and some a combination of both, — not only for the purpose of impressing, by application, the geometrical principles which have now been estab- lished, but for the not less important purpose of culti- vating the power of independent investigation. After one or two propositions in which the beginner will be assisted in the analysis and construction, we shall leave him to his own resources, with the caution that a BOOK VIII. 239 — D patient consideration of all the conditions in each case, and not mere trial operation, is the only process by which he can hope to reach the desired result. 1. From two given points, to draw two equal straight lines, which shall meet in the same point in a given straight line. Let A and B be the given points, and CD the given straight line. Pro- duce the perpendicular to the straight line AB at its middle point, until it meets CD in G. It is then easily proved that G is the point in CD in which the equal lines from A and B must meet. That is, that AG = BG. If the points A and B were on opposite sides of CD, the directions for the construction would be the same, and we should have this fig- ure; but the reasoning by which we prove AG = BG would be un- changed. 2. From two given points on the same side of a given straight line, to draw two straight lines which shall meet in the given line, and make equal angles with it. Let CD be the given line, and A and B the given points. From B draw BE perpendicular to CD, and produce the perpen- dicular to F, making EF equal to BE) then draw AF, and from the point G, in which it intersects CD, draw GB. Now, [__B GE = l_EGF=[_AGC Hence, the angles B GD and A G C are equal, and the lines AG and BG meet in a common point in the line CD, and made equal angles with that line. 240 GEOMETRY. 3. If, from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the center, the parts of these lines which are intercepted within the circle, are equal. 4. Ka circle be described on the radius of another circle, any straight line drawn from the point where they meet, to the outer circumference, is bisected by the interior one. 5. From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and be terminated in that line. 6. If, from any point without a circle, lines be drawn touching the circle, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the diameter drawn through one of them. 7. If, from any two points in the circumference of a circle, there be drawn two straight lines to a point in a tangent to that circle, they will make the greatest angle when drawn to the point of contact. 8. From a given point within a given circle, to draw a straight line which shall make, with the circumference, an angle, less than any angle made by any other line drawn from that point. 9. If two circles cut each other, the greatest line that can be drawn through either point of intersection, is that which is parallel to the line joining their centers. 10. If, from any point within an equilateral triangle, perpendiculars be drawn to the sides, their sum is equal to a perpendicular drawn from any of the angles to the opposite side. 11. If the points of bisection of the sides of a given tri- angle be joined, the triangle so formed will be one fourth of the given triangle. 12. The difference of the angles at the base of any tri- angle, is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisect- ing the angle at the vertex. BOOK VIII. 241 13. If, from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point. 14. The three straight lines which bisect the three angles of a triangle, meet in the same point. 15. The two triangles, formed by drawing straight lines from any point within a parallelogram to the ex- tremities of two opposite sides, are, together, one half the parallelogram. 16. The figure formed by joining the points of bisection of the sides of a trapezium, is a parallelogram. 17. If squares be described on three sides of a right- angled triangle, and the extremities of the adjacent sides be joined, the triangles so formed are equal to the given triangle, and to each other. 18. If squares be described on the hypotenuse and sides of a right-angled triangle, and the extremities of the sides of the former, and the adjacent sides of the others, be joined, the sum of the squares of the lines joining them will be equal to five times the square of the hypotenuse. 19. The vertical angle of an oblique-angled triangle inscribed in a circle, is greater or less than a right angle, by the angle contained between the base and the diam- eter drawn from the extremity of the base. 20. If the base of any triangle be bisected by the diam- eter of its circumscribing circle, and, from the extremity of that diameter, a perpendicular be let fall upon the longer side, it will divide that side into segments, one of which will be equal to one half the sum, and the other to one half the difference, of the sides. 21. A straight line drawn from the vertex of an equi- lateral triangle inscribed in a circle, to any point in the opposite circumference, is equal to the sum of the two lines which are drawn from the extremities of the base to the same point. 22. The straight line bisecting any angle of a triangle 21 Q 242 GEOMETRY. inscribed in a given circle, cuts the circumference in a point which is equi-distant from the extremities of the side opposite to the bisected angle, and from the center of a circle inscribed in the triangle. 23. If, from the center of a circle, a line be drawn to any point in the chord of an arc, the square of that line, together with the rectangle contained by the segments of the chord, will be equal to the square described on the radius. 24. If two points be taken in the diameter of a circle, equidistant from the center, the sum of the squares of the two lines drawn from these points to any point in the cir- cumference, will be always the same. 25. If, on the diameter of a semicircle, two equal circles be described, and in the space included by the three cir- cumferences, a circle be inscribed, its diameter will be § the diameter of either of the equal circles. 26. If a perpendicular be drawn from the vertical angle of any triangle to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base. 27. The square described on the side of an equilateral triangle, is equal to three times the square of the radius of the circumscribing circle. 28. The sum of the sides of an isosceles triangle is less than the sum of any other triangle on the same base and between the same parallels. 29. In any triangle, given one angle, a side adjacent to the given angle, and the difference of the other two sides, to construct the triangle. 30. In any triangle, given the base, the sum of the other two sides ; and the angle opposite the base, to con- struct the triangle. 31. In any triangle, given the base, the angle opposite to the base, and the difference of the other two sides, to construct the triangle. PLANE TRIGONOMETRY AND SPHERICAL GEOMETRY AND TRIGONOMETRY. (243) TKIGONOMETKY. PAET I. PLANE TRIGONOMETRY. SECTION I. ELEMENTARY PRINCIPLES. Trigonometry, in its literal and restricted sense, has for its object the measurement of triangles. When it treats of plane triangles it is called Plane Trigonometry. In a more enlarged sense, trigonometry is the science which investigates the relations of all possible arcs of the circumference of a circle to certain sti%ight lines, termed trigonometrical lines or circular functions, connected with and dependent on such arcs, and the relations of these trigonometrical lines to each other. The measure of an angle is the arc of a circle inter- cepted between the two lines which form the angle — the center of the arc always being at the point where the . two lines meet. The arc is measured by degrees, minutes, and seconds; there being 360 degrees to the whole circle, 60 minutes in one degree, and 60 seconds in one minute. Degrees, minutes, and seconds, are designated by °, ', " ; thus, 27° 14' 21", is read 27 degrees 14 minutes 21 seconds. The circumferences of all circles contain the same number of degrees, but the greater the radius the greater (244) SECTION I. -. 245 is the absolute length of a degree. The circumference of a carriage wheel, the circumference of the earth, or the still greater and indefinite circumference of the heavens, has the same number of degrees ; yet the same number of degrees in each and every circumference is the meas- ure of precisely the same angle. DEFINITIONS. 1. The Complement of an arc is 90° minus the arc. 2. The Supplement of an arc is 180° minus the arc. 3. The Sine of an angle, or of an arc, is a line drawn from one end of an arc, perpendicular to a diameter drawn through the other end. Thus, BF is the sine of the arc AB, and also of the arc BBE. BK is the sine of the arc BB. 4. The Cosine of an arc is the per- pendicular distance from the center of the circle to the sine of the arc ; or, it is the same in magnitude as the sine of the complement of the arc. Thus, OF is the cosine of the arc AB; but CF= KB, is the sine ojLBB. 5. The Tangent of an arc is a line touching the circle in one extremity of the arc, and continued from thence, to meet a line drawn through the center and the other ex- tremity. Thus, AH is the tangent to the arc AB, and BL is the tangent of the arc BB. 6. The Cotangent of an arc is the tangent of the com- plement of the, arc. Thus, BL, which is the tangent of the arc BB, is the cotangent of the arc AB. Remark. — The co is but a contraction of the word complement. 7. The Secant of an arc is a line drawn from the center of the circle to the extremity of the tangent. Thus, CH is the secant of the arc AB, or of its supplement BBE. 8. The Cosecant of an arc is the secant of the comple- ment. Thus, CL y the secant of BB, is the cosecant of AB. 21* PLANE TRIGONOMETRY. 9. The Versed Sine of an arc is the distance from the extremity of the arc to the foot of the sine. Thus, AF is the versed sine of the arc AB, and DK is the versed sine of the arc DB. For the sake of brevity, these technical terms are con- tracted thus : for sine AB, we write sin. AB ; for cosine AB, we write cos. AB; for tangent AB, we write tan. AB, etc. From the preceding definitions we deduce the follow- ing obvious consequences : 1st. That when the arc AB becomes insensibly small, or zero, its sine, tangent, and versed sine are also nothing, and its secant and cosine are each equal to radius. 2d. The sine and versed sine of a quadrant are each equal to the radius ; its cosine is zero, and its secant and tangent are infinite. 3d. The chord of an arc is twice the sine of one half the arc. Thus, the chord, BCr, is double the sine, BF. 4th. The versed sine is equal to the difference between the radius and the cosine. 5th. The sine and cosine of any arc form the two sides of a right-angled triangle, which has a radius for its hypotenuse. Thus, OF and FB are the two sides of the right-angled triangle, CFB. Also, the radius and tangent always form the two sides of a right-angled triangle, which has the secant of the arc for its hypotenuse. This we observe from the right-angled triangle, CAR. To express these relations analytically, we write sin. 2 + cos. 2 = B 2 (1) E l + tan. 2 = sec. 2 (2) From the two equiangular triangles CFB, CAR, we have OF : FB = CA : AH. SECTION I. That is, cos. : sin. = Also, = R : OF tan. ; : OB whence, = OA : tan. OR. .R.sin. cos. That is, • cos. : R = = R : : sec. ; whence, COS. sec. = I 247 (3) The two equiangular triangles, OAR and ODD, give OA : AH = DL : DO. That is, R : tan. = cot. : R; whence, tan. cot. = R\ (5) Also, OF : FB = DL : DO. That is, cos. : sin. = cot. : R; whence, cos. R = sin. cot. (6) From equations (4) and (5), we have cos. sec. m tan. cot. (7) Or, cos. : tan. = cot. : sec. ver. sin. = 1 — cos. (8) The ratios between the various trigonometrical lines are always the same for arcs of the same number of degrees, whatever be the length of the radius ; and we may, therefore, assume radius of any length to suit our convenience. The preceding equations will be more con- cise, and more readily applied, by making the radius equal unity. This supposition being made, we have, for equations 1 to 6, inclusive, sin. 2 + cos. 2 m 1. (1) 1 + tan. 2 = sec. 2 (2) sin. , o , tan. = (3) cos. 1 cos. = sec. (4) tan. = 1 (5) cos. == sin. cot. (6) cot Let the circumference, AFDH, be divided into four equal parts by the diameters, AD and FIT, the one hori- 248 PLANE TRIGONOMETRY. w R '/? n m/ m, C \ n' B" ^^ [3/// zontal arid the other vert- ical. These equal parts are called quadrants, and they may be distinguished as the first, second, third, and fourth quadrants. The center of the circle is taken as the origin of distances, or the zero point, and the different directions in which distances are esti- mated from this point are indicated by the signs + and — . If those from to the right be marked +, those from to the left must be marked — ; and if distances from O upwards be considered plus, those from down- wards must be considered minus. If one extremity of a varying arc be constantly at A, and the other extremity fall successively in each of the several quadrants, we may readily determine, by the above rule, the algebraic signs of the sines and cosines of all arcs from 0° to 360°. Now, since all other trigo- nometrical lines can be expressed in terms of the sine and cosine, it follows that the algebraic signs of all the circular functions result from those of the sine and cosine. We shall thus find for arcs terminating in the ein. 1st quadrant, + 2d " + 3d " — 4th " — PROPOSITION I. The chord of 60° and the tangent of 45° are each equal to radius ; the sine of 30°, the versed sine of 60°, and the co- sine of 60° are each equal to one half the radius. COS. tan. cot. sec. cosec. vers. + + + + + + — — — + + — + + — + + — + 4- SECTION I. 249 With C as a center, and CA as a radius, describe the arc ABF, and F from A lay off the arcs AD = 45°, AB = 60°, and AE = 90° ; then is EB = 30°. 1st. The side of a regular in- scribed hexagon is the radius of the circle, (Prob. 28, B. IV), and as the arc subtended by each side of the hexagon contains 60°, we have the chord of 60° equal to the radius. 2d. The triangle OAH is right-angled at A, and the angle O is equal to 45°, being measured by the arc AD ; hence the angle at H is also equal to 45°, and the trian- gle is isosceles. Therefore AH = CA = radius of the circle. 3d. The triangle ABC is isosceles, and Bn is a per- pendicular from the vertex upon the base ; hence An = n C — Bm, But Bm is the sine of the arc BE, Cn is the cosine of the arc AB, and An is the versed sine of the same arc, and each is equal to one half the radius. - Hence the proposition ; the chord of 60°, etc. PROPOSITION II. Given, the sine and the cosine of two arcs, to find the sine and ' the cosine of the sum and of the difference of the sanv arcs expressed by the sines and cosines of the separate arcs. Let Gr be the center of the circle, CD the greater arc, and DF the less, and denote these arcs by a and b re- spectively. Draw the radius GrD ; make the arc DE equal to the arc DF, and draw the chord EF, From F and E, the extremi- ties, and J, the middle point G M NO 250 PLANE TRIGONOMETRY. of the chord, let fall the perpendiculars FM, FP, and IN, on the radius GO. Also draw DO, the sine of the arc CD, and let fall the perpendiculars Iff on FM, and FK on IK Now, by the definition of sines and cosines, DO = sin.a; GrO = coa.a; FI = sin.6; GI = cos.5. "We are to find M" = sin. (a + b); GM = cos. {a -f 5); ^P = sin. (a — 6) ; GP = cos. (a — b). Because IN is parallel to DO, the two a's, GDO, GIN, are equiangular and similar. Also, the A FHI is similar to the A GIN; for the angles, FIG and HIN, are right angles ; from these two equals, taking away the common angle HIL, we have the angle FIB. = the angle GIN The angles at H and N are right angles ; there- fore, the A's FHI, GIN, and GDO, are equiangular and similar; and the side HI is homologous to IN and DO. Again, as FI = IF, and IK is parallel to FM 7 FH= IK, and HI = KF. By similar triangles we have GD : DO = #7 : 7ZV". That is, R : sin.a = cos.5 : IN; or, 7^= sin -yggj. ( l ) Also, GD : GO = FI : Fff That is, i£ : cos.a = sin.6 : HF; or, FH- Also, GD : GO = GI : GN That is, R : cos.a = cos.5 : GN; or, GN= Also, GD : DO = FI : IH. That is, R : sin.a = sin.5 : 7#; or, 75*= 8m ' a sin ' h . (4) By adding the first and second of these equations, we have IN+ FH= FM= sin. (a + b). That is, R : cos.a = sin.6 : HF; or, FH= 2£^HL_ 5 . (2) That is, R : coa.a = cos.6 : #iV; or, GN= C08, ^ C — 5 . ( 3 ) SECTION I. 251 mi ,. • / , \ sin. a cos.5 + cos.a sin. b That is, sin. (a + ) = ^ By subtracting the second from the first, since IN— FH= IN— 1K= UP, we have . r TN sin.fl cos.5 — cos.tf sin.5 bib. (a— I) = ^ By subtracting the fourth from the third, we have G-N — IH = GrM = cos. (a + b) for the first member. rr / , tn cos.tf cos. b — sin. a sin. b , KX Hence, cos. (a + o) = s . (5) By adding the third and fourth, we have GN+ I&= GN+NP=GP = cos.(a— b). t-t / tn cos.a cos.5 + sin.a sin.6 ; a4 Hence, cos. (a — b) = s . ( 6 ) Collecting these four expressions, and considering the radius unity, we have {sin. (a + b)=- sin.a eos.b + cos.a sin.5 ( 7 ) sin.(a — b)— sin.a cos.6 — cos.a sin.5 (8) cos.(a + b) — cos.a cos.6 — sin.a sin. b ( 9 ) cos.(a — b) = cos.a cos.5 -f sin.a sin. b ( 10 ) Formulae (A) accomplish the objects of the proposi- tion, and from these equations many useful and import- ant deductions can be made. The following are the most essential : By adding ( 7 ) to ( 8 ), we have ( 11 ) ; subtracting ( 8 ) from ( 7 ) gives ( 12 ). Also, ( 9 ) added to ( 10 ) gives ( 13 ) ; ( 9 ) taken from ( 10 ) gives ( 14 ). r sin.(a + b) + sin. (a — b) — 2sin.a cos.5 ( 11 ) (*) sin.(a + 5) — sin. (a — b) = 2cos.a sin. b (12) cos.(a + b) + cos.(a — b) — 2cos.a cos.5 ( 13 ) cos. (a — b) — cos.(a + 5) = 2sin.a sin.5 (14) If we put a + b = A, and a — b = B, then ( H ) become* (15), (12) becomes (16), (13) becomes (17), and (14) be- comes ( 18 ). 3 sin.^ + sm.5=2sin.(^i-?)cos.(^— — ) (15) (0) 252 PLANE TRIGONOMETRY. A + B\ /A- ) COS. ( 2 } > 2 J. - sin.£ = 2cos. (^t^) sin. (^— ?) ( 16 ) .A + cos.^ = 2cos. (^—^ cos. (^— ?) ( 17 ) cos.^— cos. J. = 2sin. (^-^) sin. (^—?) ( 18 ) If we divide ( 15 ) by (I 6 ), (observing tbat 55l = tan., cos. cos and - 7 — 1 = cot. = as we learn by equations (6) and tan. - u sm cos sin. ( 5 ) 7 we sball bave sin.^1 + sin.i? sm, rA+B fA—B^ A+B> fJL+lf\ (A—B\ . /A+B\ (__)cos.(.- r -)tan.(- T -) sin.^1 — sin.i? /A+B\ . /A—B\~ ,A—B (19) COS. /A+B\~. (A—B\ . /A—Bx Whence, sin.A-f sin.i? : sin.J. — sin.i? = tan. L ) : tan. (— — -) That is : The sum of the sines of any two arcs is to the dif- ference of the same sines, as the tangent of one half the sum of the same arcs is to the tangent of one half their difference. By operating in the same way with the different equa- tions in formulae (<7), we find, fsin.J. + sin.i? / A -f B\ -^tan^-^— ) A-B^ COS.B—C08.A ™ cot ' V - 2 (■*>) COS. J. + COS. sin.A + sin.i? /A — B\ sin.vl — sin.i? /A — B\ cos. A + cosT5 "■ tan * V 2~ /. sin.JL — sin.i? ,A + B\ co^B^co^A " cot \ 2/ cos. J. -f cos.ff __ cot * \ 2/ cos.i? — cos. A "~ /A-^B tan. *.—$ ) (20) (21) (22) (23) (24) SECTION I. 253 These equations are all true, whatever be the value of the arcs designated by A and B ; we may, therefore, assign any possible value to either of them, and if in equations (20) ? (21), and (24), we make B = 0, we shall have, 8in ^ tan.^ = — L, (25) 1 + cos.^. 2 cot. \A sin. A , A 1 , OA * cot. -^ = — r ( 26 ) @ 1 — cos. A 2 tan. J J. 1 + cos. A __ cot. \A (27) 1 — cos. A tan. J J. tan 2 . \ A If we now turn back to formulae (A), and divide equa- tion ( 7 ) by ( 9 ), and ( 8 ) by ( 10 ), observing at the same sin time that — - = tan., we shall have, cos. ' tan.(a + 5) = Sin ^ cos ' 5 + cos ^ sin * 5 tan.(# — 5) = cos. a cos. b — sin.a sin. b sin.# cos.5 — cos.a sin.5 cos.a cos.5 -fsin.a sin. b By dividing the numerators and denominators of the second members of these equations by (cos.a cos.5), we find, sin.a cos.5 cos.a sin.5 , . , x cos.a cos.5 cos.a cos.5 tan.a-ftan.5 tan.(a+5)= , — i *ri""i — r + i ( 28 ) v J cos.a cos.5 sin. a sin. 5 1 — tan.atan.5 cos.a cos.5 cos.a cos.5 sin.a cos.5 cos.a sin.5 , , N cos.a cos.5 cos.a cos.5 tan. a— tan. 5 tan.(a— 5)= 1 , . — ,=r-— -—. , (29 ) x ' cos.a cos.5 sm.fl sm.5 1-ftan.a tan.5 cos.a cos.5 cos.a cos.5 If in equation ( 11 ), formulae ( B ), we make a = 5, we shall have, sin. 2a = 2sin.a cos.a (30) Making the same hypothesis in equation ( 13 ), gives, cos.2a + 1= 2c<3s 2 .a (31) 22 254 PLANE TRIGONOMETRY. The same hypothesis reduces equation (14) to 1 — cos.2a = 2sin 2 .a (32) The same hypothesis reduces equation ( 28 ) to , 2tan.# ,oo\ tan.2a = - — - — -— ( 33 ) 1 — tan\a If we substitute a for 2a in ( 31 ) and ( 32 ) y we shall have 1 + cos.a = 2cos. 2 Ja. (34) and 1 — cos.a = 2sin. 2 Ja. ( 35 ) PROPOSITION III. In any right-angled plane triangle, we may have the fol- lowing proportions : 1st. The hypotenuse is to either side, as the radius is to the sine of the angle opposite to that side. 2d. One side is to the other side, as the radius is to the tan- gent of the angle adjacent to the first side. 3d. One side is to the hypotenuse, as the radius is to the secant of the angle adjacent to that side. Let CAB represent any right- angled triangle, right-angled at A. (Here, and in all cases hereafter, we shall represent the angles of a triangle by the large letters A, B, C, and the sides opposite to them, by the small letters a, b, c.) From either acute angle, as 0, take any distance, as CD, greater or less than CB, and describe the arc BF. This arc measures the angle C. From B, draw BF par- allel to BA ; and from JE, draw EG, also parallel to BA oyBF. By the definitions of' sines, tangents, secants, etc, BF is the sine of the angle C; EG is the tangent, CG the secant, and OF the cosine. SECTION I. 255 Now, by proportional triangles we have, OB:BA= OD:DF or, a : c = R : sin.C OA:AB=OF:FG or, b : c = i2 : tan. OA: OB = OB: OG or, b : a = R : sec.<7 Hence the proposition. Scholium. — If the hypotenuse of a triangle is made radius, one side is the sine of the angle opposite to it, and the other side is the cosine of the same angle. This is obvious from the triangle CDF, PROPOSITION IV. In any triangle, the sines of the angles are to one another as the sides opposite to them. Let ABO be any tri- angle. From the points A and B, as centers, with any radius, de- scribe the arcs meas- uri ng these angles, and j* draw pa, OD, and mn, perpendicular to AB. Then, pa = sin.JL, and mn = sin.i?. By the similar A's, Apa and A OB, we have, R : sin. J. = b : OD ; or, R{OD) = b sin.J. (1) By the similar a's, Bmn and BOD, we have, R : sin.B = a: OD; or, R(OD) = a sin.B (2) By equating the second members of equations ( 1 ) and (2) b sin. A = a sin.i?. Hence, sin.A : sin.i? — a :b Or, a : b = biii.A : sin.i?. Scholium 1. — When either angle is 90°, its sine is radius. Scholium 2. — When CB is less than AC, and the angle B, acute, the triangle is represented by A CB. When the angle B becomes B', it is obtuse, and the triangle is ACB / ; but the proportion is equally 256 PLANE TRIGONOMETRY. true with either triangle ; for the angle CB'D = CBA, and the sine of CB'D is the same as the sine of AB / C. In practice we can deter- mine which of these triangles is proposed, by the side AB being greater or less than AC; or, by the angle at the vertex C being large, as A CB, or small, as A CB'. In the solitary case in which AC, CB, and the angle A, are given, and CB less than AC, we can determine both of the A's ACB and A CB / ; and then we surely have the right one. PROPOSITION V. If from any angle of a triangle, a perpendicular he let fall on the opposite side, or base, the tangents of the segments of the angle are to each other as the segments of the base. Let ABC be the triangle. Let fall the perpendicular CD, on the side AB. Take any radius, as Cn, and de- scribe the arc which measures the A QX angle C. From n, draw qnp parallel to AB. Then it is obvious that np is the tangent of the angle DCB, and nq is the tangent of the angle ACB. Now, by reason of the parallels AB and qp, we have, qn : np = AB : BB That is, tan.^OZ) : tsm.BCB = AB : BB. PROPOSITION VI. If a perpendicular be let fall from any angle of a triangle to its opposite side or base, this base is to the sum of the other two sides, as the difference of the sides is to the difference of the segments of the base. (See figure to Proposition 5.) Let AB be the base, and from C, as a center, with the shorter side as radius, describe the circle, cutting AB in (x, and AC in F; produce AC to E. SECTION I. 257 It is obvious that AE is the sum of the sides AG and OB, and AF is their difference. Also, AD is one segment of the base made by the per- pendicular, and BB = BGr is the other; therefore, the difference of the segments is AG. As A is a point without a circle, by Cor. Th. 18, B. Ill, we have AE x AF = AB x AG Hence, ^LB : AE - AF : J.#. PROPOSITION VII. 2%£ mm 0/ awy free szefes 0/ a triangle is to their difference, as the tangent of one half the sum of the angles opposite to these sides, is to the tangent of one half their difference. Let ABO be any plane triangle. ^ Then, by Proposition 4, we have, BO: A = sin.J. : sin.^. Hence, A~ ~~B BC + A C: BC— A 0= sin.A+sm.B : sin.^— -sin.^ (Th. 9,B. II). But, tan. ( — ^ — j : tan. ( — - — ) == sin. A + sin.i? : sin. J. — sin.B, (eq. (19), Trig.) Comparing the two latter proportions, (Th. 6, B. H), we have, BO+AO:BO— AO=t&n. (f-j^) : tan. (^-^) Hence the proposition. PROPOSITION VIII. Given, the three sides of any plane triangle, to find some relation which they must bear to the sines and cosines of the respective angles. 22* r 258 PLANE TRIGONOMETRY. Let ABO be the triangle, and let the perpen- dicular fall either upon, or without the base, as shown c in the figures. C « b x By recurring to Th. 40, B. I, we shall find a 2 -f b 2 — c* OB = 2a (1) £fow, by Proposition 3, we have R : cos. C = b : CD. Therefore, OB = b cos. ~~R (2) Equating these two values of OB, and reducing, we have 2ab (m) In this expression we observe, that the part c, whose square is found in the numerator with the minus sign, is the side opposite to the angle ; and that the denominator is twice the rectangle of the sides adjacent to the angle. From these observations we at once draw the following expressions for the cosine A, and cosine B : A R(b 2 + c 2 — a?) cos. J. = - v cos. B 2bc R(a 2 + c* b 2 ) 2ac (n) (P) As these expressions are not convenient for logarith- mic computation, we modify them as follows : If we put 2a = JL, in equation ( 31 ), we have cos. A + 1 = 2cos. 2 % A. In the preceding expression, (»), if we consider radius unity, and add 1 to both members, we shall have SECTION I. 259 COS. ^+1 = 1+ _JT_ Therefore, 2cos.' \A = S^fcHl^ 26<* L+ ° + "~H 2 2fo Or, cos. 2 1 j. - : ? Li I r By putting — - m s, and extracting square root, A the final result for radius unity is cos. \A = \1WE& v bo For any other radius we must write cos. |,i= y/ifo»-«), By inference, cos.JJ? = \j ?L^Z^1. Also, cos. J(7 i v/ ^^^ l). In every triangle, the sum of the three angles is equal to 180° ; and if one of the angles is small, the other two must be comparatively large ; if two of them are small, the third one must be large. The greater angle is always opposite the greater side ; hence, by merely inspecting the given sides, any person can decide at once which is the greater angle ; and of the three pre- 260 PLANE TRIGONOMETRY. ceding equations, that one should be taken which applies to the greater angle, whether that be the particular angle required or not ; because the equations bring out the cosines to the angles ; and the cosines to very small arcs vary so slowly, that it may be impossible to decide, with sufficient numerical accuracy, to what particular arc the cosine belongs. For instance, the cosine 9.999999, carried to the table, applies to several arcs ; and, of course, we should not know which one to take ; but this difficulty does not exist when the angle is large ; there- fore, compute the largest angle first, and then compute the other angles by Proposition 4. But we can deduce an expression for the sine of any of the angles, as well as the cosine. It is done as fol- lows: EQUATIONS FOR THE SINES OF THE ANGLES. Resuming equation ( m ), and considering radius unity, we have cos. = __ Subtracting each member of this equation from unity, gives Make 2a = O, in equation (32) ; then a = \Q y and 1 — cos. = 2sin. 2 |<7. (2) Equating the second members of (1) and (2), 2ab — a? — b 2 + e* 2sin. 2 i<7 = 2ab „+ e +l> — a c+b+a iC+a — b c+a+b , But.__= _ o.and—g— = — ^ h. Put = s, as before; then, sin.i<7= J BS v ab By taking equation (p ), and proceeding in the same manner, we have sin.ji?= \J¥E^VEh. v ao From t4 sin. JJ. = \JEEM3, v c6 The preceding results are for radius unity ; for any other radius, we must multiply by the number of units in such radius. For the radius of the tables we write H; and if we put it under the radical sign, we must write B 2 ; hence, for the sines corresponding with our logarithmic table, we must write the equations thus, v be v ac sm. 10 = J '#<»=ME3. v ab A large angle should not be determined by these equations, for the same reason that a small angle should not be determined from an equation expressing the cosine. In practice, the equations for cosine are more gener- ally used, because more easily applied. 262 PLANE TRIGONOMETRY. The formulae which we have thus analytically devel- oped, express nearly all the important relations between the sines, cosines, and tangents of arcs or angles ; and we have also demonstrated all the theorems required for the determination of the unknown parts of any plane triangle, three of the parts of which are given, one at least being a side. Such relations might be indefinitely multiplied, but those already established are sufficient for most practical purposes, and when others are required, no difficulty will be found in deducing them from these. The following geometrical demonstrations of many of the preceding relations, are offered, in the belief that they will prove useful disciplinary exercises to the stu- dent. 1st. Let the arc AD=A; then 2) #= sin. J.; CG-=cos.A; J9J=sin.}J.;^2)=2sin.JJL; OZ=cos.|J.; CI=DO; and Z>£=2DO=2cos.}A The angle, DBA, is measured by one half the arc AD ; that is, by \A. Also, AD a = DBA = J A. Now, in the triangle, BD Gr, we have sm.DBG : 2>#=sin.90° : BD. That is, sin. J A : sin. J.=l : 2cos. \A. Or, sin.J=2sin.JJ. cos.JJ.; which corresponds to equation ( 30 ). In the same triangle, sin.90° : BD=sin.BDG : BG; and sin.BDG=co8.DBG. That is, 1 : 2cos.|J=eos.|J. : 1+cos.JL Or, 2cos. 2 %A=l+co8.A y same as equation (34). In the triangle, DGrA, we have, sin.90° : AD = sin. GDA : a A. That is, 1 : 2sin. JJ. = sin.JJ. : 1— cos. J.. Or, 2sin. 2 \A = 1 — cos.^4, same as equation ( 35 ). SECTION I. 263 2 : 2sin. \A = 2sin.JJ. : versed sin. A. versed sin. A = 2sin. 2 \A. By similar triangles, we have, BA : AD = AD: AG. That is, Or, 2d. From O as the center, with OA as the radius, describe a circle. Take any arc, AB, and call it A ; and AD a less arc, and call it B ; then BD is the difference of the two arcs, and must be designated by (A — B) ; arc AG = arc AB ; therefore, arcD# = A + B; FG = sm.A; En ss= sin. B ; Gn = sin. J. -f sin. B ; Bn = sin. J. — sin.i?. Fm = mD = OJI= cos.B ; mn = cos. J. ; therefore, Fm + mn= cos. J. -f cos.i? = Fn ; mD — mn = eos.B — cos. A = nD ; A + B> and Because, therefore, -or, I># = 2sin. pm JVF=AD; AB + NF=A + B; 180° — (J. + i?) = arc ZE; 90°-(£±*)~larcJ®. But the chord, FB, is twice the sine of J arc _Fi? ; that is, FB = 2sin. (90° — A ± ffi ) = 2cos. (^-^). The L_7i6rD = LjSZZ), because both are measured by one half of the arc BD; that is, by ( ~ \ and the two triangles, 6rftD and .Fwi?, are similar. The angle, GFn, is measured by f — - — \ In the triangle, FBG, Fn m drawn from an angle per- OF THE ^A 264 PLANE TRIGONOMETRY. pendicular to the opposite side ; therefore, by Proposition 5, we have, Gn : nB = tan. GFn : t&n.BFn. That is, sin.-4.-f sin.i? : sin. A — sin.JB=tan.( — i — ) : tan. I — - — ). This is equation (19 ). In the triangle, GnB, we have, sin.90° : BG = am.nBG : Gn; &m.nDG=cos.nGD. That is, 1 : 2sin. (^~) = cos. ( A ~ B ) : sin.^l-f sin.J?. Or, sin. J. -f sin.2? = 2sin. ( — - — ) cos. ( — - — ), the same as equation (15). 3d. In the triangle, FnB, we have, sin.90 : FB = sin.^Frc : Bn. That is, 1 : 2cos.(^±^) = sin.(4ip?) - s i n . A— sin.#. Or, sin.JL-sin.£ = 2cos. (^~~) sin. ( A ~ B ), the same as equation (16). 4th. In the triangle, FBn, we have, sin.90 : FB — cos.BFn : Fn. That is, 1 : 2cos. (^jp ) = cos.(^2?) : cos.^L+cos.5. Or, cos. J. + cos.5 = 2cos. ( — - — \ cos. ( ~ V the same as equation ( 17 ). 5th. In the triangle, GnB, we have, sin.90° : GD = mi.nGB : nB. That is, 1 : 2sin. (_ — ) = sin. (— ~^— \ : cos.B—cos.A, the same as equation (18). 6th. In the triangle, FGn, we have, sin. GFn : Gn = cos. GFn : Fn. SECTION I. 265 That is, sin. — ^— : sin. A+sin.B — cos. — — : cos.A+ 2 2i COS.B. Or, (sin. J. -f ain.B) cos. (—J — ) = (cos. A -f cos.i?) sift. . A + B Or, sin 'f ± si ^| = _* - tan. (*±*\ the cos.^ + cos..B J. + .S V 2 /' cos. — o — same as equation (20). 7th. In the triangle, FnB, we have, Fn : nB :: 1 : tan.BFn. That is, cos.J5+cos.JL : sin. A — sin.i? :: 1 : tan.|(A — B), r. sin. A — sin.i? , /A — B\ .* Or, = - = tan. ( — - — ), the same cos. A + cos._# V 2 /' as equation (22). 8th. In the triangle, GrnD, we have, Gn : nB : : 1 : tan.w6rD. That is, sin. J. -f sin. B : cos. B — cos. J. :: 1 : tanY — - — V cos. B — cos. A , /A — B\ sin. A + sin . B = ^ (~J~ ).' NATURAL SINES, COSINES, ETC. When the radius of the circle is taken as the unit of measure, the numerical values of the trigonometrical lines belonging to the different arcs of the quadrant, be- come natural sines, cosines, etc. They are then, in fact, but numbers expressing the number of times that these lines contain the radius of the circle in which they are taken. The tables usually contain only the sines and cosines, because these are generally sufficient for practi- 23 266 PLANE TRIGONOMETRY. cal purposes, and the others, when required, are readily expressed in terms of them. We proceed to explain a method for computing a table of natural sines and cosines. It was shown, in Book V, that the linear value of the arc 180°, in a circle whose radius is unity, is 3.141592653. This divided by 180 x 60, the number of minutes in 180°, will give the length of one minute of arc, which is .00029088820867. But there can be no sensible difference between the length of the arc V and its sine ; and, within narrow limits, that sine will increase directly with the arc. Hence, sin. V = .0002908882. sin. 2' = .0005817764. sin. 3' = .0008726646. sin. 4' = .0011635528. sin. 5' = .0014544410. sin. 6' = .0017453292. sin. V = .0020362175. sin. 8' = .0023271057. sin. 9' = .0026179938. sin. 10' = .0029088811. Beyond this, the error which would arise from taking the arc for its sine, upon which the above proceeds, would affect the final decimal figures; and we must, therefore, continue the computation of the series by other processes. To find the values of the cosines of arcs, from V to 10', we have cos. = *S\ — sin. 2 = 1 — J sin. 2 , nearly. That is, when the sines are very small fractions, as is the case for all arcs below 10', we can find the cosine by subtracting one half of the square of the sine from unity. SECTION I. 267 Whence, cos. 1' = .9999999577. cos. 2' = .9999998308. cos. 3' - .9999993204. cos. 4' = .99999932304. cos. 5' = .99999894290. cos. 6' = .99999847753. cos. V = .99999792735. cos. 8' - .9999973035. cos. 9' = .9999965730. cos. 10' = .9999957703. The natural sines of arcs, differing by 1', from 10' up to 1°, may be computed from those of arcs less than 10', by means of equation ( 11 ), group B, which is sin. (a -f b) = 2sin. a cos. b — sin. (a -f b) ; And when a = 5, this equation becomes sin. 2a = 2sin. a cos. b. Eq. ( 30 ). To find the sine of 11', we make a = 6', and 5 = 5'; then sin. 11' = 2sin. 6' cos. 5'— sin. 1'= .00319976913. 0=6 = 6', sin. 12' = 2sin. 6' cos. 6'. a = 7', b = 6', sin. 13' = 2sin. T cos. 6' — sin. 1'. a = b = 7, sin. 14' = 2sin. 7' cos. 7'. a = 8, b = 7, sin. 15' = 2sin. 8 ; cos. 7' — sin. V And so on to the sin. 30' = 2sin.l5'cos.l5'. sin.l° = sin. 60' = 2sin.30'cos.30'. sin. 2° = 2sin. 1° cos. 1°. sin. 3° = 2sin. 2° cos. 1° — sin. 1°, etc., etc., etc. This process may be continued until we have found the sines and cosines of all arcs differing by V, from to 90°, the values of the cosines being deduced success- ively from those of the sines by means of the formula, cos. = */l — sin. 2 . In this calculation, we began by assuming that, for small arcs, the sines and the arcs were sensibly equal. 268 PLANE TRIGONOMETRY. It must be remembered that this is but an approxima- tion ; and although the error in the early stages of the process is not sufficient to affect any of the decimal fig- ures which enter the tables, it will finally become so, since it is constantly increased in the operations by which the sines and cosines of the larger arcs are de- duced from those of the smaller. "When the error has been thus increased until it reaches the order of the last decimal unit of the table, which assigns our limit of error, we must have the means of detecting and correct- ing it. • This consists in calculating the sines and cosines of certain arcs by independent processes, and comparing them with those found by the above method. "We have seen, for example, (Prop. 7, B. V), that the chord of 30° = .517638090; whence, sin. 15° = .258819045. 15° = .2610523842; " " 7° 15' =.130526192. 7° 15' = .1308062583; " " 3° 7' 30"= .0654031291. And so on to sin. 14' 3" 45'" = .004090604. etc. etc. etc. The following elegant method of deducing, from the sine of an arc, the sine and cosine of one half the arc, is given, assuming that the student is familiar with the simple algebraic principles upon which it depends. Let us take the natural sine of 18°, which is .3090170, 18° and make x = sine, and y the cosine of 9° = — . A Then, x 2 + y 2 = 1; (1) and 2xy = .3090170 (2); Eq. (30). Adding, we have z 2 -f 2xy + f = 1.3090170; SECTION I. 269 Taking tlie square root, we have x + y = 1.144123. (3) Subtracting ( 2 ) from ( 1 ), x 2 — 2xy + y 2 = .690983; taking the square root, x — y - —.831254* (4) Adding (3) and (4), 2z = .312869, hence, a; = sin.9° = .1564345 Subtracting (4) from (3), 2y = 1.97537T hence, y = cos.9° = .9876885 Now, by making # = the sine of 4° 30', and y = cosine of 4° 30', and as before x 2 + y 2 = 1 and 2xy = .1564345, we obtain the sine and cosine of 4° 30' ; and another ope- ration will give the sine and cosine 2° 15', etc., etc. We may in this manner compute the sines and cosines of all arcs resulting from the division of 18° by 2, and we may make their values accurate to any assigned deci- mal figure. This has been carried far enough to show how a table of natural sines, etc., could be computed ; but in conse- quence of the tedious numerical operations which the process requires, other methods are resorted to in the actual construction of the table. The Calculus furnishes formulae giving the values of the sines and cosines of arcs developed into rapidly con- verging series, and from these the sines and cosines of all arcs from 0° to 90°, can be determined with great * When an arc is less than 45°, the cosine exceeds the sine ; and ■when the arc is between 45° and 90°, the sine exceeds the cosine. Hence, when the arc is 9°, y, its cosine, exceeds x, its sine ; and we therefore placed the minus sign before the second member of Eq. (4). 23* 270 PLANE TRIGONOMETRY. accuracy and with comparatively little labor. In the last two columns on each page of Table II, will be found the values thus computed of the sines and cosines of every degree and minute of a quadrant. TRIGONOMETRICAL LINES FOR ARCS EXCEEDING 90°. X// From the annexed figure, the construction of which needs no explanation, are deduced by simple inspec- tion the results given in the following TABLE 90° -f a° 270° — a° sin. = cos. a, cos. = — sin. a sin. = — cos. a, cos. = — sm. a tan. = — cot. a, cot. = — tan. a tan. = cot. a, cot. = tan. a sec. = — cosec. a, cosec. = sec. a sec. — — cosec. a, cosec. = — sec. a 180° — a° 270° + a° sm. = sin. a, cos. = — cos. a sin. = — cos. a, cos. = sin. a tan. = — tan. a, cot. = — cot. a tan. = — cot. a, cot. = — tan. a sec. = — sec. a, cosec. = cosec. a sec. = cosec. a, cosec. = — sec. a 180° + a 360° — a° sin. = — sin. a, cos. = — cos. a sin. = — sin. a, cos. = cos. a tan.= tan. a, cot. = cot. a tan. = — tan. a, cot. = — cot. a sec. = — sec. a, cosec.=— cosec. a sec. = sec. a, cosec. = — cosec. a By means of this table, the values of the trigonomet- rical lines of any arc between 90° and 360°, can be ex- pressed by those of arcs less than 90°. If, for example, the arc is 118°, we have SECTION I. 271 sin. 118° = sin. (90° + 28°) = cos.28° ; tan.H8° = tan.(90° + 28°) = — cot.28° ; etc., etc., etc. For the arc 230°, we have sin. 230° = sin. (270° — 40°) = — cos. 40° ; sec.230° = sec.(270° — 40°) = — cosec.40°; etc., etc., etc. In many investigations, it becomes necessary to con- sider the functions of arcs greater than 360° ; but since the addition of 360° any number of times to the arc a, will give an arc terminating in the extremity of a, it is obvious that the arc resulting from such addition will have the same functions as the arc a. And hence it fol- lows that the functions of arcs, however great, may be expressed in terms of the functions of arcs less than 90°. 272 PLANE TKIGONOMETEY. SECTION II. PLANE TRIGONOMETRY, PRACTICALLY APPLIED. In the preceding section, the theory of Trigonometry has been quite fully developed, and the student should now be prepared for its various applications, were he acquainted with logarithms. But logarithms are no part of Trigonometry, and serve only to facilitate the numeri- cal operations. Trigonometrical computations can be made without logarithms, and were so made long before the theory of logarithms was understood. For this reason, we proceed at once to the solution of the following triangles. 1. The hypotenuse of a right-angled triangle is 21, and the base is 17 ; required the perpendicular and the acute angles. Let CAB be the triangle, in which CB - 21, and CA = 17. With C as a center, and CD = 1 as a radius, describe the arc DE, of which the sine is DFj the tangent is EG, and the cosine is OF, By similar triangles we have CB : CA that is, 21 : 17 CD 1 CF; cos. C. Hence, 17 cos. C = -- = .80952+. SECTION II. 273 We must now turn to Table II, and find in the last two columns the cosine nearest to .80952, and the corresponding degrees and minutes will be the value of the angle C. On page 56, of Tables, near the bottom of the page, and in the column with cosine at the top, we find .80953, which corresponds to 35° 56' for the angle C. The angle B is, therefore, 54° 4'. This Table is so arranged, that the sum of the degrees at the top and bottom of the page, added to the sum of the minutes which are found on the same horizontal line in the two side columns of the page, make 90°. Thus, in finding the angle (7, the number .80953 was found in the column with cosine at its foot. We therefore took the degrees from the bottom of the page, and the minutes were taken from the right hand column, counting upwards. For the side AB } we have the proportion CF : FD :: CA : AB; or, cos. C : sin. C : : 17 : AB; that is, .80953 : .58708 : : 17 : AB. From which we find AB = .58708 X 17 -J- .80953; whence, AB = 12.328. If we had formed a table of natural tangents, as well as of natu- ral sines, AB could have been found by the following proportion • CE : EG : : CA : AB or, 1 : tan. C : : 17 : AB) whence, AB = 17 tan. C. The perpendicular AB may also be found by the proportion CD : DF n CB : AB) or, 1 : sin. C : : 21 : AB; whence, AB = 21 sin. C = 21 x .58708 = 12.32868. 2. The two sides of a right-angled triangle are 150 pnd 125 ; required the hypotenuse and the acute angles. Let CAB be the triangle, which is the same as in the pre- . ceding problem. Then, from the similar trian- gles, CFD and CAB } we get CF : FD :: CA i AB) 274 PLANE TRIGONOMETRY. that is, cos. G : sin. G : : 150 : 125 : : 6 : 5, which gives 6 sin. (7=5 cos. C; hence, 36 sin. 2 (7 = 25 cos.*<7. Adding member to member, 86 cos. 2 6 r = 36 cos. 2 C. we have 36 (sin. 2 C + cos. 2 C) == 61 cos.* G. But sin. 2 C-f cos. 2 = 14.2763+. And hence, .£D = CB — CD = 24 — 14.2763 = 9.7237. In the triangle ADB, we have BA : £Z> :: 1 : cos. 5 or, 18 : 9.7237 : : 1 : cos. B = .54020 rr u tt x> kq f cos. 57° 18' = .54024) Table H, Page 53, { cos< 5? o 1Q , = .54000 } diff. ^ 24 : 60" : : 4 : 10" hence, \__B = 57° 18' 10". It will be observed that Examples 5 and 6 refer to the same tri- angle, and that in Example 5 the angle B was 57° 18' 15". This slight discrepancy in the results should be expected, on account of the small number of decimal places used in the computations. Second. By Prop. 8. Sum of the sides, , =62.815, half sum denoted by S, = 31.4075 a = 24 S—a = 7.4075 Formula, cos. £ A = \ / — — - -, radius being unity. S(S—a) = 31.4075 x 7.4075 = 232.65105625 be = 20.815 X 18 = 374.67 £(ff — a) ^ 620Q5 very near] ^ V 152095 = .78800. Hence, cos. \A = .78800, and \A (Table II, page 59) = 38° very nearly ; the angle A is therefore equal to 76°, which agrees with Example 5. 7. Given, the three sides, 1425, 1338, and 493, of a tri- angle ; required, the angle opposite the greater side, using the formula for the sine of one half an angle. 24 278 PLANE TRIGONOMETRY. Make a = 1425, b = 1338, and c = 493 ; then the [__ A is opposite the side a, and the formula is oc in which s denotes the half sum of the three sides. Then we have s == 1628, s — b = 290, s — c = 1135, (s — 6) 0_. c ) = 329150, 6c = 659634, ( g — ft ) (* — c ) = .498988. Hence, sin. JJ. = v/,498988 = .70632. In the table we find sin. 44° 56' 12" = .70632. Therefore, \A = 44° 56' 12", and A = 89° 52' 24";— but little less than a right angle. In these seven examples we have shown that it is possi- ble to solve any plane triangle, in which three parts, one at least being a side, are given, without the aid of loga- rithms. But, when great accuracy is required, and the number of decimal places employed is large, the necessary multiplications and divisions, the raising to powers, and the extraction of roots, become very tedious. All of these operations may be performed without impairing the cor- rectness of results, and with a great saving of labor, by means of logarithms ; but, before using them, the student should be made acquainted with their nature and pro- perties. LOGARITHMS. Logarithms are the exponents of the powers to which a fixed number, called the base, must be raised, to pro- duce other numbers. The exponent of a number is also a number express- ing how many times the first number is taken as a factor. Thus, let a denote any number ; then a 3 indicates that a has been used three times as a factor, a 4 that it has been used four times as a factor, and a n that it has been thus used n times. SECTION II. 279 Now, instead of calling these numbers 3, 4, w, exponents, we call them the logarithms of the powers a% a\ a n . To multiply a 2 by a 5 , we have simply to write a, giving it an exponent equal to 2 + 5 ; thus, a 2 X a 5 = a\ Hence, the sum of the logarithms of any number of factors is equal to the logarithm of the product. To divide a 12 by a 9 , we have only to write a, giving it an exponent equal to 12 — 9 ; thus, a 12 -*- a 9 = a 3 ; and, generally, the quotient arising from the division of a m by a% is equal to a m ~ n . Hence, the logarithm of a quotient is the logarithm of the dividend diminished by the logarithm of the divisor. If it is required to raise a number denoted by a 3 , to the fifth power, we write a, giving it an exponent equal to 3x5; thus, (a 3 ) 5 =a 15 , and, generally, (a n ) m =a nm . Hence, the logarithm of the power of a number is equal to the logarithm of the number multiplied by the exponent of the power. v To extract the 5th root of the number a s , we write «, giving it an exponent equal to f ; thus, ^/a~ s =a^ and, generally, to extract any root of a number, we divide the exponent of the number by the index of the root, and the quotient will be the exponent of the required root. Hence, the logarithm of a root of a number is equal to the quotient obtained by dividing the logarithm of the number by the index of the root Now, understanding that by means of a table of loga- rithms we may find. the numbers answering to given logarithms, with as much facility as we can find the loga- rithms of given numbers, we see from what precedes that multiplications, divisions, raising to powers, and the ex- traction of roots, may be performed by logarithms ; and the utility of logarithms, in trigonometrical computations, mainly consists in the simplicity and abridgment of these operations as executed by them. 280 PLANE TRIGONOMETRY. The common logarithms are those of which 10 is the base ; that is, they are the exponents of 10. Thus, lO 1 ^ 10 Hence the logarithm L 10 = 1. 10 2 = 100 U u u 100 = 2. 10 3 = 1000 u <■ a 1000 =3. 10* = 10000 a it a 10000 = 4. etc. etc. etc. etc. etc. Since Jq = 1 = 10 1 -' = 10°, and a m generally — = a = 1, it follows that in this, as in all other systems, the loga- rithm or 1 = 0. From what precedes, it is evident that the logarithm of any number between 10 and 100 must be found between 1 and 2 ; that is, its logarithm is 1 plus a number less than 1; and any number between 100 and 1000, will have for its logarithm 2 plus some number less than 1, and so on. The fractional part of the logarithms of numbers are expressed decimally. The entire number belonging to a logarithm is called its index. The index is never put in the tables, (except from 1 to 100), and need not be put there, because we always know what it is. It is always one less than the number of digits in the whole number. Thus, the num- ber 3754 has 3 for the index to its logarithm, because the number consists of 4 digits ; that is, the logarithm is 3 and some decimal. The number 347.921 has 2 for the index of its loga- rithm, because the number is between 347 and 348, and 2 is the index for the logarithms of all numbers over 100, and less than 1000. All numbers consisting of the same figures, whether integral, fractional, or mixed, have logarithms consisting of the same decimal part. The logarithms would differ only in their indices. 24* SECTION II. 281 Thus, the number 7956. has 3.900695 for its log. the number 795.6 has 2.900695 the number 79.56 has 1.900695 the number 7.956 has 0.900695 the number .7956 has —1.900695 the number .07956 has —2.900695 From this we perceive that we must take the logarithm out of the table for a mixed number or a decimal, the same as if the figures expressed an entire number; and then, to prefix the index, we must consider the value of the number. The decimal part of a logarithm is always positive; but the index becomes negative when the number is a decimal; and the smaller the decimal, the greater the negative index. Hence, To prefix the index to a decimal, count the decimal point as 1, and every cipher as 1, up to the first significant figure, and this is the negative index. For example, find the logarithm of the decimal .0000831. Num. .0000831; log. —5.919601. The point is counted one, and each of the ciphers is counted one ; therefore the index is minus five. •The smaller the decimal, the greater the negative index ; and when the number becomes 0, the logarithm is negatively infinite. Hence, the logarithmic sine of 0° is negatively infinite, however great the radius. A number being given, to find its corresponding logarithm. The logarithm of any number consisting of four figures, or less, is taken out of the table directly, and without the least difficulty. Thus, to find the logarithm of the number 3725, we 24* 282 PLANE TRIGONOMETRY. find 372 at the side of the table, and in the column marked 5 at the top, and opposite 372, we find .571126, for the decimal part of the logarithm. Hence, the logarithm of 3725 is 3.571126. the logarithm of 37250 is 4.571126. the logarithm of 37.25 is 1.571126, etc. Find the logarithm of the number 834785. This number is so large that we cannot find it in the table, but we can find the numbers 8347 and 8348. The logarithms of these numbers are the same as the loga- rithms of the numbers 834700 and 834800, except the indices. 834700 log. 5.921530 834800 log. 5.921582 Difference, 100 52 Now, our proposed number, 834785, is between the two assumed numbers ; and, of course, its logarithm lies between the logarithms of the two assumed numbers; and, without further comment, we may proportion it thus, 100 : 85 = 52 : 44.2 Or, 1. : .85 = 52 : 44.2 Hence, for finding from the table the logarithm of a number consisting of more than four places of figures, we have the following RULE. Take from the table the log. of the number expressed by the the four superior figures ; this, with the proper index, is the approximate logarithm. Multiply the number expressed by the remaining figures of the number, regarded as a decimal, by the tabular difference, and the product will be the correction to be added to the approximate log. to obtain the true log. SECTION II. 283 EXAMPLES. 1. What is the log. of 35T.32514? The log. of 357.3 is 2.553033 No. not included, .2514 Tabular diff., 122 Prod., 30.6708; correction, 31 log. sought, 2.553064 The log. of 35732.514 is 4.553064 " .035732514" —2.553064. 2. What is the log. of 7912532 ? Approximate log., 6.898286 .532 x 55 = correction, 29 True log. = 6.898315. A logarithm being given, to find its corresponding number. For example, what number corresponds to the log. 6.898315 ? The index 6 shows that the entire part of the number must con- tain seven places of figures. With the decimal part, .898315, of the log., we turn to the table, and find the next less decimal part to be .898286, which corresponds to the superior places, 7912. The difference between the given log. and the one next less is 29. This we divide by the tabular difference, 55, because we are working the converse of the preceding problem. Thus, 29 -f- 55 = 52727+. Place the quotient to the right of the four figures before found, and we shall have 7912527.27 for the number sought. This example was taken from the preceding case, and the number found should have been 7912532 ; and so it would have been, had we used the true difference, 29.26, in place of 29. When the numbers are large, as in this example, the 284 PLANE TRIGONOMETRY. result is liable to a small error, to avoid which the loga- rithms should contain a great number of decimal places ; but the logarithms in our table contain a sufficient num- ber of decimal places for most practical purposes. Hence, for finding the number corresponding to any given logarithm, we have the following RULE. Look in the table for the decimal part of the given loga- rithm, and if not found, take the decimal next less, and take out the four corresponding figures. Take the difference between the given log. and the next less in the table ; divide that difference by the tabular difference, and write the quotient on the right of the four superior fig- ures, and the result is the number sought. Point off the whole number required by the given index. EXAMPLES. 1. Given, the logarithm 3.743210, to find its corres- ponding number true to three places of decimals. Ans. 5536.182. 2. Given, the logarithm 2.633356, to find its corres- ponding number true to two places of decimals. Ans. 429.89. 3. Given, the logarithm — 3.291746, to find its corres- ponding number. Ans. .0019577. 4. What number corresponds to the log. 3.233568 ? Ans. 1712.25. 5. What is the number of which 1.532708 is the log. ? Ans. 34.0963. 6. Find the number whose log. is 1.067889. Ans. 11.692. EXPLANATION OP TABLE II. Table I is merely a table of numbers and their corres- ponding logarithms, and requires no explanation other SECTION II. 285 than that which has been given in connection with the subject of logarithms. Table II, with the exception of the last two columns, which contain natural sines and cosines, is a table in which are arranged the logarithms of the numerical values of the several trigonometrical lines corresponding to the different angles in a quadrant. The values of these lines are computed to the radius 10,000,000,000, and their logarithms are nothing more than the loga- rithms, each increased by 10, of the natural sines, co- sines, and tangents, of the same angles; because the values of these lines, for arcs of the same number of de- grees taken in different circles, are directly proportional to the radii of the circles. The natural sines are made to the radius of unity; and, of course, any particular sine is a decimal fraction, expressed by natural numbers. The logarithm of any natural sine, with its index increased by 10, will give the logarithmic sine. Thus, the natural sine of 8° is .052336. The logarithm of this decimal is — 2.718800 To which add 10. The logarithmic sine of 3° is, therefore, 8.718800 In this manner we may find the logarithmic sine of any other arc, when we have the natural sine of the same arc. If the natural sines and logarithmic sines were on the same radius, the logarithm of the natural sine would be the logarithmic sine, at once, without any increase of the index. The radius for the logarithmic sines is arbitrarily taken so large that the index of its logarithm is 10. It might have been more or less ; but, by common consent, it is settled at this value ; so that the sines of the smallest arcs ever used shall not have a negative index. 286 PLANE TRIGONOMETRY. In our preceding equations, sin. a, cos. a, etc., refei to natural sines; and by such equations we determine their values in natural numbers ; and these numbers are put in Table II, under the heads of nat. sine and nat. co- sine, as before observed. When we have the sines and cosines of an arc, the tangent and cotangent are found by Eq. ( 3 ) ; that is, , R sin. /A ; R cos. tan. = (6) cot. = — : : cos. sm. and the secant is found by equation (4); that is, R 2 sec. = cos. For. example, the logarithmic sine of 6° is 9.019235, and its cosine 9.997614. From these it is required to find the logarithmic tangent, cotangent, and secant. R sin. 19.019235 Cos. subtract 9.997614 Tan. is 9.021621 R cos. 19.997614 Sin. subtract 9.019235 Cotan. is 10.978379 R* is 20.000000 Cos. subtract 9.997674 Secant is 10.002326 The secants and cosecants of arcs are not given in our table, because they are very little used in practice ; and if any particular secant is required, it can be deter- mined by subtracting the cosine from 20 ; and the cose- cant can be found by subtracting the sine from 20. The sine of every degree and minute of the quadrant is given, directly, in the table, commencing at 0°, and extending to 45°, at the head of the table ; and from 45° to 90°, at the bottom of the table, increasing backward. SECTION II. 287 The same column that is marked sine, at the top, is marked cosine at the bottom ; and the reason for this is apparent to any one who has examined the definitions of sines. The difference of two consecutive logarithms is given, corresponding to ten seconds. Removing the decimal point one figure, will give the difference for one second ; and if we multiply this difference by any proposed num- ber of seconds, we shall have a difference corresponding to that number of seconds, above the logarithm corres- ponding to the preceding degree and minute. For example, find the sine of 19° 17' 22". The sine of 19° 17', taken directly from the table, is 9.518829 The difference for 10" is 60.2 ; for 1", is 6.02 ; and • 6.02 X 22 = 133 Hence, 19° 17' 22" sine is 9.518962 From this it will be perceived that there is no difficulty in obtaining the sine or tangent, cosine or cotangent, of any angle greater than 30'. Conversely : Given, the logarithmic sine 9.982412, to find its corresponding arc. The sine next less in the table is 9.982404, which gives the arc 73° 48'. The differ- ence between this and the given sine is 8, and the dif- ference for 1" is .61 ; therefore, the number of seconds corresponding to 8, must be discovered by dividing 8 by the decimal .61, which gives 13. Hence, the arc sought is 73° 48' 13". These operations are too obvious to require a rule. When the arc is very small, — and such arcs as are sometimes required in Astronomy, — it is necessary to be very accu- rate ; for this reason we omitted the difference for seconds for all arcs under 30'. Assuming that the sines and tan- gents of arcs under 30' vary in the same proportion as the arcs themselves, we can find the sine or tangent of any very small arc, with great exactness, as follows : 288 PLANE TRIGONOMETRY. The sine of V, as expressed in the table, is 6.463726 Divide this by 60; that is, subtract logarithm 1.778151 The logarithmic sine of 1", therefore, is 4.685575 Now, for the sine of 17", add the logarithm of 17 1.230449 Logarithmic sine of 17", is 5.916024 In the same manner we may find the sine of any other small arc. For example, find the sine of 14' 21J"; that is, 861"5. The logarithmic sine of 1" is 4.685576 Add logarithm of 861.5, 2.935254 Logarithmic sine of 14' 21£", 7.620830 Two lines drawn, the one from the surface and the other from the center of the earth, to the center of the sun, make with each other an angle of 8.61". What is the logarithmic sine of this angle ? The log. of the sine 1" is 4.685575 Log. of 861, 0.935003 Log. sine of sun's horizontal parallax = 5.620578 GENERAL APPLICATIONS WITH THE USE OF LOGARITHMS. I. RIGHT-ANGLED TRIGONOMETRY. One figure will be sufficient to represent the triangle in all of the following examples ; the right angle being at£. PRACTICAL PROBLEMS. 1. In a right-angled triangle, ABC, given the base AB, 1214, and the angle A, 51° 40' 30", to find the other parts. SECTION II. 289 To find BC. Kadius, 10.000000 : tan. .4, 51° 40' 30", 10.102119 :: AB,12U, 3.084219 : BC, 1535.8, 3.186338 Remark. — When the first term of a logarithmic proportion is radius, the required logarithm is found by adding the second and third loga- rithms, rejecting 10 in the index, which is dividing by the first term. In all cases we add the second and third logarithms together ; which, in logarithms, is multiplying these terms together ; and from that sum we subtract the first logarithm, whatever it may be, which is dividing by the first term. To find AC. Sin. C, or cos. A, 51° 40' 30", 9.792477 : AB, 1214, 3.084219 :: Radius, 10.000000 : AC, 1957.7, 3.291742 To find this resulting logarithm, we subtracted the first logarithm from the second, conceiving its index to be 13. Let ABO represent any plane triangle, right-angled at B. 2. Given, AO 73.26, and the angle A, 49° 12' 20"; required the other parts. Ans. The angle 0, 40° 47' 40" ; BO, 55.46 ; and AB, 47.87. 8. Given, AB 469.34, and the angle A, 51° 26' 17", to find the other parts. Ans. Theangletf, 38° 33' 43"; B (7,588.7; and^LC, 752.9. 4. Given, AB 493, and the angle C, 20° 14' ; required, the remaining parts. Ans. The angle A, 69° 46'; BO, 1338 ; and AC, 1425.5. 5. Let AB = 331, and the angle A = 49° 14' ; what are the other parts ? Ans. AC, 506.9; BC, 383.9; and the angle O, 40° 46'. 6. If AC=4:5, and the angle (7=37° 22', what are the remaining parts ? Ans. AB, 27.31 ; BC, 35.76 ; and the angle A, 52° 38'. 25 t 290 PLANE TRIGONOMETRY. 7. Given, AC= 4264.3, and the angle A = 56° 29' 13", to find the remaining parts. Ans.AB, 2354.4; BO, 3555.4; and the angle 0, 33° 30'47". 8. If AB = 44.2, and the angle A = 31° 12' 49", what are the other parts ? Ans. AC, 49.35 ; BO, 25.57 ; and the angle 0, 58° 47' 11". 9. If ^LB - 8372.1, and BO = 694.73, what are the other parts ? An8 (AC, 8400.9; the angle 0, 85° 15'; and the ' \ angle A, 4° 45'. 10. If AB be 63.4, and AC be 85.72, what are the other parts? A f BO, 57.7 ; the angle C, 47° 42'; and the angled, ' I 42° 18'. 11. Given, AC = 7269, and AB = 3162, to find the other parts. A (BO, 7546; the angle C, 25° 47' 7"; and the m ' \ angle A, 64° 12' 53". 12. Given, AC = 4824, and BO = 2412, to find the other parts. A ( The angle A = 30° 00', the angle C = 60° 00', m ' I and AB - 4178. 13. The distance between the earth and sun is 94,770,000 miles, and at that distance the semi-diameter of the sun subtends an angle of 16' 6". "What is the diameter of the sun in miles ? Ans. 887,700 miles. In this example, let E be the center of the earth, S that of the sun, and EB a tangent to the sun's surface. Then the A EBS is right-angled at B, and BJS is the semi-diameter of the sun. The value of 2BS is required. SECTION II. 291 14. The semi-diameter of the earth is 3956 miles, and the distance of the sun 94.770000 miles. What angle will the semi-diameter of the earth subtend, as seen from the sun? Ans. 8.60". This angle is called, in astronomy, the sun's horizontal parallax. The preceding figure applies to this example, by supposing E to be the center of the sun, S that of the earth,' and BS equal to 3956 miles. 15. The mean distance of the moon from the earth is 60.3 times 3960 miles, and at. this distance the semi- diameter of the moon subtends an angle of 15' 32". What is the diameter of the moon in miles ? Ans. 2159 miles. H. OBLIQUE-ANGLED TRIGONOMETRY. PROBLEM I. In a plane triangle, given a side and the two adjacent angles, to find the other parts. ' In the triangle ABC, let AB = c 376, the angle A = 48° 3', and the angle B - 40° 14', to find the other parts. As the sum of the three angles of every B triangle is always 180°, the third angle, C, must be 180° — 88° 17' = 91° 43'. To find A a Sin. 91° 43', : AB, 376, : : sin. B 40° 14', 9.999805 2.575188 9.810167 12.385355 : .AC, 243, 2.385550 Observe, that the sine of 91° 43' is the same as the cosine of 1° 43'. 292 PLANE TRIGONOMETRY. To find BC. Sin. 91° 43', 9.999805 : A5, 376, 2.575188 : : sin. A, 48° 3', 9.871414 12.446602 : sin. 5 0,279.8, 2.446797 PROBLEM II. In a plane triangle, given two sides and an angle opposite one of them, to determine the other parts. Let AD = 1751 feet, one of the given sides ; the angle D = 31° 17' 19" ; and the side opposite, 1257.5. From these data, we are required to find the other side and the other two angles. In this case we do not know whether A G or AE represents 1257.5, because AG = AE. If we take AG for the other given side, then D G is the other required side, and DA G is the vertical angle. If we take AE for the other given side, then DE is the required side, and DAE is the vertical angle. In such cases we determine both triangles. To find the angle U = G. (Prop. 4.) AG = AE= 1257.5, log. 3.099508 : D, 31° 17' 19", sin. 9.715460 : : AD, 1751, log. 3.243286 12.958746 < E = G, 46° 18', sin. 9.859238 From 180° take 46° 18', and the remainder is the angle DGA = 133° 42'. The angle DAG = AGE — D, (Th. 11, B. I) y that is, DAG = 46° 18' — 31° 17' 19" = 15° 0'41". The angles D and E, taken from 180°, give DAE = 102° 24' 41". SECTION II. 293 To find DO. Sin. D, 31° 17' 19", log. 9.715460 : ^1(7, 1257.5, log. 3.099508 : : sin. DAG 15° 0' 41", log. 0.413317 12.512825 : DC, 626.86, 2.797165 To find DE. Sin. D, 31° 17' 17", 9.715460 : AE, 1257.5, 3.099508 : : sin. DAE, 102° 24' 41", 9.989730 13.089238 : DE, 2364.7, 3.373778 Remark. — To make the triangle possible, A C must not be less than AB, the sine of the angle D, when DA is made radius. PROBLEM III. In any plane triangle, given two sides and the included angle, to find the other parts. Let AD = 1751, (see last figure), DE = 2364.5, and the included angle D = 31° 17' 19". "We are required to find AE, the angle DAE, and the angle E. Observe that the angle E must be less than the angle DAE, be- cause it is opposite a less side. From 180° Take D, 31° 17' 19", Sum of the other two angles, = 148° 42' 41", (Th. 11, B. I), * sum = 74° 21' 20". By Proposition 7, DE+DA : DE— DA = tan. 74° 21' 20" : tan. \{DAE— E). That is, 4115.5 : 613.5 = tan. 74° 21' 20" : *a*k(DAE— JP> 25* 294 PLANE TRIGONOMETRY. Tan. 74° 21' 20", 613.5, 4115.5 log. (subtracted), 10.552778 2.787815 13.340593 3.614423 tMi.i(DAEE-) tan.28° 1' 36", 9.726170 But the half sum plus the half difference of any two quantities is equal to the greater of the two; and the half sum minus the half difference is equal the less. Therefore, to 74° 21' 20", Add 28° V 36", DAE = E = 102° 22' 56", 46° 19' 45", To find AE. Sin. E, 46° 19' 45", : DA, 1751, : : sin. D, 31° 17' 19", : XE, 1257.2, 9.859323 3.243286 9.715460 12.958746 3.099423 PROBLEM IV. Given, the three sides of a plane triangle, to find the angles. Let AQ m 1751, OB = 1257.5, AB = 2364.5, to find the angles A, B, and O. If we take the formula for cosines, we will compute the greatest angle, which is C. To correspond with the formula, cos. ab ' A B we must take a = 1257.5, b = 1751, and c = 2364.5. The half sum of these is, s == 2686.5; and s — c = 322. SECTION J [I. R* 20.000000 s = 2686.5 3.429187 s — c = 322 2.507856 Numerator, log. 25.937043 a 1257.5 3.099508 b 1751. 3.243286 295 Denominator, log. 6.342794 6.342794 2 )19.594249 ' \C=* 51° 11' 10" cos. 9.797124 C=102 22 20 The remaining angles may now be found by Problem 4. PRACTICAL PROBLEMS. Let ABC represent any oblique-angled triangle. 1. Given, AB 697, the angle A 81° 30' 10", and the angle B 40° 30' 44", to find the other parts. Am. AC, 534; BC, 813; and [__C, 57° 59' 6". 2. If A C = 720.8, [_A=7Q° 5' 22", [__B m 59° 35' 36", required the other parts. Am. AB, 643.2; BC, 785.8; and [_C, 50° 19' 2". 3. Given, BC 980.1, the angle A 7° 6' 26", and the angle B 106° 2' 23", to find the other parts. Am. AB, 7284; AC, 7613.3; and [_C, 66° 51' 11". 4. Given, AB 896.2, B C 328.4, and the angle (7113° 45' 20", to find the other parts. A ( AC, 712; LA 19° 35' 48"; ' I and [_B, 46° 38' 52". 5. Given, AC r = 4627, BC= 5169, and the angle A = 70° 25' 12", to find the other parts. A (AB, 4328; L^, 57° 29' 56"; ' \ and [_C, 52° 4' 52". 296 PLANE TRIGONOMETRY. 6. Given, AB 793.8, BO 481.6, and AO 500.0, to find the angles. An jLA, 35° 15' 32"; IB, 36° 49' 18"; and [__0, •\ 107° 55' 10". 7. Given, .45 100.3, 5(7 100.3, and AC 100.3, to find the angles. A i The angle A, 60°; the angle 5, 60°; and the m ' \ angle (7, 60°. 8. Given, AB 92.6, 5(7 46.3, and AC 71.2, to find the angles. , f LA 29° IT' 22"; L*> 48° 47' 31"; and [_0, ^'1 101° 55' 8". 9. Given, AB 4693, BO 5124, and AO 5621, to find the angles. A j [_A, 57° 30' 28"; \_B, 67° 42' 36"; and [__0, ^ nS '\ 54° 46' 56". 10. Given, AB 728.1, BO 614.7, and JL(7 583.8, to find the angles. A ] \jkm 54° 32' 52", \_B = 50° 40' 58", and \__0 #H =74° 46' 10". 11. Given, AB 96.74, BO 83.29, and AO 111.42, to find the angles. A j L^ = 46° 30' 45", [_B = 76° 3' 45", and [_0 ^'\ =57° 25' 30". 12. Given, AB 363.4, BO 148.4, and the angle B 102° 18' 27", to find the other parts. A ( [_A = 20° 9' 17", the side A = 420.8, and [__0 13. Given, .45 632, 5(7 494, and the angle A 20° 16', to find the other parts, the angle being acute. ([_ (7= 26° 18' 19", [__B = 133° 25' 41", and ns '\ ^4(7=1035.86. 14. Given, AB 53.9, .4(7 46.21, and the angle B 58° 16', to find the other parts. Arts. \__A = 38° 58', [__0= 82° 46', and 5(7= 34.16. SECTION II. 29T 15. Given, AB 2163, BC 1672, and the angle 112° 18' 22", to find the other parts. Ana. AC, 877.2; [_B, 22° 2' 16"; and [_A, 45° 39' 22". 16. Given, AB 496, BC 496, and the angle B S8° 16', to find the other parts. Ans. AC, 325.1; [_A, 70° 52'; and [__fc for refraction = 25.7 miles. 12. From the top of a tower, by the seaside, 143 feet high, it was observed that the angle of depression of a SECTION III. 307 ship's bottom, then at anchor, measured 35° ; what, then, was the ship's distance from the foot of the tower ? Arts. 204.22 feet. 13. Wanting to know the "breadth of a river, I meas- ured a base of 500 yards in a straight line on one bank; and at each end of this line I found the angles subtended by the other end and a tree on the opposite bank of the river, to be 53° and 79° 12'. What, then, was the per- pendicular breadth of the river? Ans. 529.48 yards. 14. What is the perpendicular height of a hill, its angle of elevation, taken at the bottom of it, being 46°, and 200 yards further off, on a level with the bottom, 31° ? Am. 286.28 yards. 15. Wanting to know the height of an inaccessible tower, at the least accessible distance from it, on the same horizontal plane, I found its angle of elevation to be 58° ; then going 300 feet directly from it, I found the angle there to be only 32° ; required the height of the tower, and my distance from it at the first station. A /Height, 307.53 feet. \ Distance, 192.15 " 16. Two ships of war, intending to cannonade a fort, are, by the shallowness of the water, kept so far from it, that they suspect their guns cannot reach it with effect. In order, therefore, to measure the distance, they separate from each other a quarter of a mile, or 440 yards, and then each ship observes and measures the angle which the other ship and fort subtends ; these angles are 83° 45', and 85° 15'. What, then, is the distance between each ship and the fort ? , / 2292.26 yards. Jin8 ' \ 2298.05 " 17. A point of land was observed by a ship, at sea, to bear east-by-south ;* and after sailing north-east 12 miles, * That is, one point south of east. A point of the compass is 11° 15'. 308 PLANE TRIGONOMETRY. it was found to bear south-east-by-east. It is required to determine the place of that headland, and the ship's dis- tance from it at the last observation. Ans. Distance, 26.0728 miles. 18. "Wishing to know my distance from an inaccessible object, 0, on the opposite side of a river, and having a chain or chord for measuring distances, but no instru- ment for taking angles ; from each of two stations, A and B, which were taken at 500 yards asunder, I meas- ured in a direct line from the object, 0, 100 yards, viz., A O and BD, each equal to 100 yards; and I found that the diagonal AD measured 550 yards, and the diagonal BO 560. What, then, was the distance of the object from each station A and B1 A ( AO, 536.25 yards. Ji.ns. 1^500.09 " 19. A navigator found, by observation, that the summit of a certain mountain, which he supposed to be 45 min- utes of a degree distant, had an altitude above the sea horizon of 31' 20". Now, on the supposition that the earth's radius is 3956 miles, and the observer's dip was 4' 15", what was the height of the mountain ? Ans. 3960 feet. Remark. — This should be diminished by about one eleventh part of itself, for the influence of horizontal refraction. 20. From two ships, A and B, which are anchored in a bay, two objects, O and i>, on the shore, can be seen. These objects are known to be 500 yards apart. At the ship A, the angle subtended by the objects was measured, and found to be 41° 25' ; and that by the object D and the other ship was found to be 52° 12'. At the other ship, the angle subtended by the objects on shore was found to be 48° 10'; and that by the object O, and the ship A, to be 47° 40'. Eequired the distance between Ans. SECTION III. 309 the ships, and the distance from each ship to the objects on shore. Distance between ships, 395.6 yards. From ship A to object D, 743.5 " From ship A to object 0, 467.7 " .From ship B to object D, 590.5 " To solve this problem, suppose the distance between the ships to be 100 yards, and determine the several distances, including the distance between the objects, C and D, under this supposition; then multiply the values thus found for the required distances by the quotient obtained by dividing the given value of CD by the com- puted value. PART II. SPHERICAL GEOMETRY AND TRIGONOMETRY. SECTION I. SPHERICAL GEOMETRY. DEFINITIONS. 1. Spherical Geometry has for its object the investiga- tion of the properties, and of the relations to each other, of the portions of the surface of a sphere which are bounded by the arcs of its great circles. 2. A Spherical Polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles, called the sides of the polygon. 3. The Angles of a spherical polygon are the angles formed by the bounding arcs, and are the same as the angles formed by the planes of these arcs. 4. A Spherical Triangle is'a spherical polygon having but three sides, each of which is less than a semi-circum- ference. 5. A Lime is a portion of the surface of a sphere in- cluded between two great semi-circumferences having a common diameter. 6. A Spherical Wedge, or TJngula, is a portion of the surface of a sphere included between two great semi-cir- cles having a common diameter. (310) SECTION I. 3H 7. A Spherical Pyramid is a portion of a sphere bounded by the faces of a solid angle having its vertex at the center, and the spherical polygon which these faces inter- cept on the surface. This spherical polygon is called the base of the pyramid. 8. The Axis of a great circle of a sphere is that diameter of the sphere which is perpendicular to the plane of the circle. This diameter is also the axis of all small circles parallel to the great circle. 9. A Pole of a circle of a sphere is a point on the sur- face of the sphere equally distant from every point in the circumference of the circle. 10. Supplemental, or Polar Triangles, are two triangles on a sphere, so related that the vertices of the angles of either triangle are the poles of the sides of the other. PROPOSITION I. Any two sides of a spherical triangle are together greater than the third side. Let AB, AC, and BO, be the three sides of the triangle, and D the center of the sphere. The arcs AB, AC, and BO, are meas- ured by the angles of the planes that form the solid angle at D. But any two of these angles are together greater than the third angle, (Th. 18, B. VI). Therefore, any two sides of the triangle are, together, greater than the third side. Hence the proposition. PROPOSITION II. The sum of the three sides of any spherical triangle is less than the circumference of a great circle. Let ABO be a spherical triangle ;. the two sides, AB and A 0, produced, will meet at the point which is diame- trically opposite to A, and the arcs, ABB and AOB are 312 SPHERICAL GEOMETRY. together equal to a great circle. But, by the last proposition, BO is less than the two arcs, BD and D 0. There- fore, AB + BC + AC, is less than ABB + A CD ; that is, less than a great circle. Hence the proposition. PROPOSITION III. The poles of a great circle of a sphere are the extremities of its axis, and these points are also the poles of all small circles parallel to the great circle. Let be the center of the sphere, and BD the axis of the great circle, Cm Am" ; then will B and D, the extremities of the axis, be the poles of the circle, and also the poles of any parallel small cir- cle, as FnJEJ. For, since BD is per- pendicular to the plane of the circle, Cm Am", it is perpendicular to the lines OA, Om', Om", etc., passing through its foot in the plane, (Th. 3, B. VI); hence, all the arcs, Bm, Bm', etc., are quadrants, as are also the arcs Dm, Dm 1 , etc. The points B and D are, therefore, each equally distant from all the points in the circumfer- ence, Cm Am" ; hence, (Def. 9), they are its poles. Again, since the radius, OB, is perpendicular to the plane of the circle, Cm Am", it is also perpendicular to the plane of the parallel small circle, FnE, and passes through its center, f . Now, the chords of the arcs, BF, Bn, BF, etc., being oblique lines, meeting the plane of the small circle at equal distances from the foot of the ■p y^- — 0'~ --N.F -Jr »""■""" i in m" \ o ^^A mi> SECTION I. 313 perpendicular, BO f , are all equal, (Tli. 4, B. VI); hence, the arcs themselves are equal, and B is one pole of the circle, FnF. In like manner we prove the arcs, BF, Bn, BF, etc., equal, and therefore B is the other pole of the same circle. Hence the proposition, etc. Cor. 1. A point on the surface of a sphere at the distance of a quadrant from tivo points in the arc of a great circle, not at the extremities of a diameter, is a pole of that arc. For, if the arcs, Bm, Bm ! , are each quadrants, the angles, BOm and BOm f , are each right angles; and hence, BO is perpendicular to the plane of the lines, Om and 0m\ which is the plane of the arc, m m> f ; B is therefore the pole of this arc. Cor. 2. The angle included between the arc of a great circle and the arc of another great circle, connecting any of its points with the pole, is a right angle. For, since the radius, BO, is perpendicular to the plane of the circle, Cm Am", every plane passed, through this radius is perpendicular to the plane of the circle ; hence, the plane of the arc Bm is perpendicular to that of the arc Cm', and the angle of the arcs is that of their planes. PROPOSITION IV. The angle formed by two arcs of great circles ivhich inter- sect each other, is equal to the angle included between the tan- gents to these arcs at their point of intersection, and is meas- ured by that arc of a great circle whose pole is the vertex of the angle, which is limited by the sides of the angle or the sides produced. Let AM and AJSf be two arcs intersecting at the point A, and let AE and AF be the tangents to these arcs at this point. Take A C and AB, each quadrants, and draw the arc CD, of which A is the pole, and 00 and OB are the radii. 27 314 SPHERICAL GEOMETRY. !Now, since the planes of the arcs intersect in the radius OA, and AE is a tangent to one arc, and AF a tangent to the other, at the common point A, these tangents form with each other an angle which is the measure of the angle of the planes of the arcs ; but the angle of the planes of the arcs is taken as the angle included by the arcs, (Def. 4). Again, because the arcs, AC and AD, are each quadrants, the angles, A 00, A OD, are right angles ; hence the radii, 00 and OD, lie, the one in one face, and the other in the other face, of the diedral angle formed by the planes of the arcs, and are perpendicular to the common intersection of these faces at the same point. The angle, OOD, is therefore the angle of the planes, and consequently the angle of the arcs ; but the angle OOD is measured by the arc OD, Hence the proposition. Oor. 1. Since the angles included between the arcs of great circles on a sphere, are measured by other arcs of great circles of the same sphere, we may compare such angles with each other, and construct angles equal to other angles, by processes which do not differ in principle from those by which plane angles are compared and con- structed. Oor. 2. Two arcs of great circles will form, by their in- tersection, four angles, the opposite or vertical ones of which will be equal, as in the case of the angles formed by the intersection of straight lines, (Th. 4, B. I). PROPOSITION V. The surface of a hemisphere may be divided into three right- angled and four quadrantal triangles, and one of these right- angled triangles will be so related to the other two, that two of its sides and one of its angles will be complemental to the SECTION I. 315 Bides of one of them, and two of its sides supplemental to two of the sides of the other. Let ABO be a right-angled spherical triangle, right angled at B. Produce the sides, AB and A 0, and they will meet at A', the opposite point on the sphere. Produce BO, both ways, 90° from the point B, to P and P', which are, therefore, poles to the arc AB, (Prop. 3). Through A, P, and the center of the sphere, pass a plane, cutting the sphere into two equal parts, forming a great circle on the sphere, which great circle will be represented by the circle PAP' A 1 in the figure. At right angles to this plane, pass another plane, cutting the sphere into two equal parts ; this great circle is represented in the figure by the straight line, POP'. A and A' are the poles to the great circle, POP' \ and P and P' are the poles to the great circle, ABA'. Now, OPB is a spherical triangle, right-angled at D, and its sides OP and OB are complemental respectively to the sides £6 7 and AO of the A ABO, and its side PB is .complemental to the arc BO, which measures the [_B A of the same triangle. Again, the A A'B is right- angled at B, and its sides A'O, A'B, are supplemental respectively to the sides AO, AB, of the a ABO. There- fore, the three right-angled A's, ABO, OPB, and A'BO, have the required relations. In the A AOP, the side AP is a quadrant, and for this reason the A is called a quad- ran tal triangle. So also, are the A's A' OP, AOP', and P'OA', quadrantal triangles. Hence the proposition. Scholium. — In every triangle there are six elements, three sides and three angles, called the parts of the triangle. Now, if all the parts of the triangle ABC are known, the parts of each of the A' s > PCD and A'BC, are as completely known. And when the parts of the ^ PCD are known, the parts of the A's A CP 316 SPHERICALTRIGONOMETRY. and A'CP are also known ; for, the side PD measures each of the | 'a PAC and PA'C, and the angle CPD, added to the right angle A'PD, gives the [_A / PC, and the | CPA is supplemental to this. Hence, the solution of the A ABC is a solution of the two right-angled and four quadrantal A's, which together with it make up tne surface of the hemisphere. PROPOSITION VI. If there be three ares of great circles whose poles are the angular points of a spherical triangle, such arcs, if produced, will form another triangle, whose sides will be supplemental to the angles of the first triangle, and the sides of the first triangle will be supplemental to the angles of the second. Let the arcs of the three great cir- cles be GH, PQ, KL, whose poles are respectively A, B, and Q. Produce the three arcs until they meet in D, E, and F. We are now to prove that E is the pole of the arc AO; D the pole of the arc BO; F the pole to the arc A B. Also, that the side EF, is supplemental to the angle A; ED to the angle 0; and DF to the angle B; and also, that the side A is supplemental to the angle E, etc. A pole is 90° from any point on the circumference of its great circle ; and, therefore, as A is the pole of the arc GH, the point A is 90° from the point E. As is the pole of the arc LK, is 90° from any point in that arc; therefore, is 90° from the point E; and E being 90° from both A and O, it is the pole of the arc AQ. In the same manner, we may prove that D is the pole of BO, and F the pole of AB. Because A is the pole of the arc G H, the arc GH measures the angle A, (Prop. 4) ; for a similar reason, PQ measures the angle B, and LK measures the angle O. Because E is the pole of the arc AO, EH = 90° Or, EG+GH= 90° For a like reason, FH -f GH = 90° SECTION I. 317 Adding these two equations, and observing that GH = A, and afterward transposing one A, we have, EG + GH+FH=1SQ° — A. Or, EF= 180°— A j) In like manner, JF!Z> = 180° — B \ («) And, BE = 180° — J But the arc (180°— A), is a supplemental arc to A, by the definition of arcs ; therefore, the three sides of the triangle BEF, are supplements of the angles A, B, 0, of the triangle ABO. Again, as E is the pole of the arc A (7, the whole angle E is measured by the whole arc LE. But, AC + CH = 90° Also, AC + AL = 90° By addition, AC+AC+CE + AL = 180° By transposition, 1(T+ CE+AL = 180°— A O That is, LIT, or .#= 180°— A O ^ In the same manner, .F = 180°— ^LB M 6 ) And, E=1S0° — BO J That is, the sides of the first triangle are supplemental to the angles of the second triangle. PKOPOSITION VII. The sum of the three angles of any spherical triangle, is greater than two right angles, and less than six right angles. Add equations (a), of the last proposition. The first member of the equation so formed will be the sum of the three sides of a spherical triangle, which sum we may designate by S. The second member will be 6 right angles (there being 2 right angles in each 180°) less the three angles A, B, and O. That is, S = 6 right angles — (A+B+C) By Prop. 2, the sum S is less than 4 right angles; 318 SPHERICAL GEOMETRY. therefore, to it add s, a sufficient quantity to make 4 right angles. Then, 4 right angles = 6 right angles — (A-hB+0)-\-s Drop or cancel 4 right angles from both members, and transpose (A + B + 0). Then, A + B + = 2 right angles + s. That is, the three angles of a spherical triangle make a greater sum than two right angles by the indefinite quantity s, which quantity is called the spherical excess, and is greater or less according to the size of the triangle. Again, the sum of the angles is less than 6 right angles. There are but three angles in any triangle, and each one of them must be less than 180°, or 2 right angles. For, an angle is the inclination of two lines or two planes ; and when two planes incline by 180°, the planes are parallel, or are in one and the same plane ; therefore, as neither angle can be equal to 2 right angles, the three can never be equal to 6 right angles. . PROPOSITION VIII. On the same sphere, or on equal spheres, triangles which are mutually equilateral are also mutually equiangular ; and, conversely, triangles which are mutually equiangular are also mutually equilateral, equal sides lying opposite equal angles. First— "Let ABO and DBF, in which AB = BE, AO= DF, and BO = EF, be two triangles on the sphere whose center is 0; then will the [_ A, opposite the side BO, in the first triangle, be equal the [_JD, opposite the equal side EF, in the second; also l_B=[__E, and \__0=[_F. SECTION I. 319 For, drawing the radii to the vertices of the angles of these triangles, we may conceive to be the common vertex of two triedral angles, one of which is hounded by the plane angles A OB, BOO, and A 00, and the other by the plane angles DOE, EOF, and DOE. But the plane angles bounding the one of these triedral angles, are equal to the plane angles bounding the other, each to each, since they are measured by the equal sides of the two triangles. The planes of the equal arcs in the two triangles are therefore equally inclined to each other, (Th. 20, B. VI) ; but the angles included between the planes of the arcs are equal to the angles formed by the arcs, (Def. 3). Hence the [_A, opposite the side J?0, in the A ABO, is equal to the [__ B, opposite the equal side EF, in the other triangle ; and for a similar reason, the [__B= \__E, and the [_0=[_F. Second. — If, in the triangles ABO and BEE, being on the same sphere whose center is 0, the |__ A = [__ B, the [_B = [_E, and the [_0= [_E; then will the side AB, opposite the [__ 0> in * ne first? be equal to the side BE, opposite the equal L_ E, in the second ; and also the side AO equal to the side BE, and the side BO equal to the side EF. For, conceive two triangles, denoted by A'B'O' and B'E'F', supplemental to ABO and BEE, to be formed; then w T ill these supplemental triangles be mutually equi- lateral, for their sides are measured by 180° less the opposite and equal angles of the triangles ABO and BEF, (Prop. 6) ; and being mutually equilateral, they are, as proved above, mutually equiangular. But the triangles ABO and BEE are supplemental to the tri- angles A'B'O' and B'E'F' '; and their sides are therefore measured severally by^l80° less the opposite and equal angles of the triangles A'B'O' and B'E'E', (Prop. 6). 320 SPHERICAL GEOMETRY. Hence the triangles ABO and BEF, which are mutually equiangular, are also mutually equilateral. Scholium. — With the three arcs of great circles, AB, AC, and BC, either of the two triangles, ABC, BEF, may be formed ; hut it is evi- dent that these two triangles cannot be made to coincide, though they are both mutually equilateral and mutually equiangular. Spherical triangles on the same sphere, or on equal spheres, in which the sides and angles of the one are equal to the sides and angles of the other, each to each, but are not themselves capable of superposition, are called symmetrical triangles. PROPOSITION IX. On the same sphere, or on equal spheres, triangles having two sides of the one* equal to two sides of the other, each to each, and the included angles equal, have their remaining sides and angles equal. Let ABO and DEF be two triangles, in which AB — BE, AO = BF, and the angle A — the angle B ; then will the side BO be equal to the side FE, the [_B = theL^andLtf = L^. For, if BE lies on the same side of BF that AB does of AO, the two triangles, ABO and BEF, may be applied the one to the other, and they may be proved to coincide, as in the case of plane tri- angles. But, if BE does not lie on the same side of BF that AB does of AO, we may construct the triangle which is symmetrical with BEF; and this symmetrical triangle, when applied to the triangle ABO, will exactly coincide with it. But the triangle BEF, and the triangle sym- metrical with it, are not only mutually equilateral, but also are mutually equiangular, the equal angles lying opposite the equal sides, (Prop. 8) ; and as the one or the other will coincide with the triangle ABO, it follows that SECTION I. 321 the triangles, ABC and BJEF, are either absolutely or symmetrically equal. Cor. On the same sphere, or on equal spheres, triangles having two angles of the one equal to two angles of the other, each to each, and the included sides equal, have their remain- ing sides and angles equal. For, if [__A = L Pi L.B = [_U, and side AB = side BE, the triangle BEF, or the triangle symmetrical with it, will exactly coincide with A ABO, when applied to it as in the case of plane triangles ; hence, the sides and angles of the one will be equal to the sides and angles of the other, each to each. PROPOSITION X. In an isosceles spherical triangle, the angles opposite the equal sides are equal. A Let ABO be an isosceles spherical tri- angle, in which AB and A are the equal sides ; then will [__ B = Q 0. For, connect the vertex A with B, the i middle point of the base, by the arc of a / great circle, thus forming the two mutu- ^4—. ally equilateral triangles, ABB and ABO. They are mutually equilateral, because AB is common, BB = DC by construction, and AB=AObj supposition; hence they are mutually equiangular, the equal angles being opposite the equal sides, (Prop. 8). The angles B and 0, being opposite the common side AB, are there- fore equal. Cor. The arc of a great circle which joins the vertex of an isosceles spherical triangle with the middle point of the base, is perpendicular to the base, and bisects the ver- tical angle of the triangle ; and, conversely, the arc of a v 322 SPHERICAL GEOMETRY. great circle which bisects the vertical angle of an isosceles spherical triangle, is perpendicular to, and bisects the base. PROPOSITION XI. If two angles of a spherical triangle are equal, the opposite sides are also equal, and the triangle is isosceles. In the spherical triangle, ABC, let the \__B = [__C; then will the sides, AB and AC, opposite these equal angles, be equal. For, let P be the pole of the base, BO, and draw the arcs of great circles, PB, PC; these arcs will be quadrants, and at right angles to BC, (Cor. 1, Prop. 3). Also, produce CA and BA to meet PB and PC, in the points E and F. Now, the angles, PBF and PCE, are equal, because the first is equal to 90° less the [_ABC, and the second is equal to 90° less the equal [_ACB; hence, the A's, PBF and PCE, are equal in all their parts, since they have the [_P common, the \_PBF = [_PCE, and the side PB equal to the side PC, (Cor., Prop. 9). PE is therefore equal to PF, and [_PEC= [__PFB. Taking the equals PF and PE, from the equals PC and PB, we have the remainders, FC and EB, equal ; and, from 180°, taking the [_'s PFB and PEC, we have the remaining L_'s, AFC and AEB, equal. Hence, the A's, AFC and AEB, have two angles of the one equal to two angles of the other, each to each, and the included sides equal; the remaining sides and angles are therefore equal, (Cor., Prop. 9). Therefore, A C is equal to BA> and the A ABC is isosceles. Cor. An equiangular spherical triangle is also equilat- eral, and the converse. SECTION I. 323 Remark. — In this demonstration, the pole of the base, BC, is sup- posed to fall without the triangle, ABC. The same figure may be used for the case in which the pole falls within the triangle ; the modifi- cation the demonstration then requires is so slight and obvious, that it would be superfluous to suggest it. PROPOSITION XII. The greater of two sides of a spherical triangle is opposite the greater angle ; and, conversely, the greater of two angles of a spherical triangle is opposite the greater side. Let ABO be a spherical triangle, in which the angle A is greater than the angle B; then is the side BO greater than the side A 0. Through A draw the arc of a great circle, AD, making, with AB, the angle BAB equal to the angle ABB. The triangle, BAB, is isos- celes, and DA = BB, (Prop. 11). In the A AOD, AO< OD + AD, (Prop. 1) ; or, substituting for AD its equal DB, we have, AO < OB + DB. Inverting the members of the inequality, and writing OB for OB + DB, it becomes OB > OA. Conversely ; if the side OB be greater than the side OA, then is the [_A > the [_B. For, if the [_A is not greater than the [__B, it is either equal to it, or less than it. The \_A is not equal to the [_B ; for if it were, the triangle would be isosceles, and OB would be equal to OA, which is contrary to the hypothesis. The [_A is not less than the [___B; for if it were, the side OB would be less than the side OA, by the first part of the proposition, which is also contrary to the hypothesis ; hence, the [_A must be greater than the L^. 324 SPHERICAL GEOMETRY. PROPOSITION XIII. Two symmetrical spherical triangles are equal in area. Let ABO and DEF be two A's on the same sphere, having the sides and angles of the one equal to the sides and angles of the other, each to each, the triangles themselves not admitting of superposition. It is to be proved that these A's have equal areas. Let P be the pole of a small circle passing through the three points, ABO, and connect P with each of the points, A, B, and O, by arcs of great circles. Next, through E draw the arc of a great circle, UP', making the angle DEP 1 equal to the angle ABP. Take EP' = BP, and draw the arcs of great circles, P'D, P'F. The A's, ABP and DEP', are equal in all their parts, because AB=DE, BP=EP f , and the [_ABP=[_DEP f , (Prop. 9). Taking from the [_ ABO the [_ABP, and from the [_DEF the [_DEP f , we have the remaining angles, PBO and P'EF, equal; and therefore the A's, BOP and EFP' , are also equal in all their parts. Now, since the a's, ABP and DEP', are isosceles, they will coincide when applied, as will also the A's, BOP and EFP' , for the same reason. The polygonal areas, ABOP and DEEP', are therefore equivalent. If from the first we take the isosceles triangle, PAO, and from the second the equal isosceles triangle, P'DF, the remainders, or the triangles ABO and DEF, will be equivalent. Remark. — It is assumed in this demonstration that the pole P falls without the triangle. Were it to fall within, instead of without, no other change in the above process would be required than to add the isosceles triangles, PAO, P'DF, to the polygonal areas, to get the areas of the triangles, ABC, DEF. SECTION I. 325 Cor. Two spherical triangles on the same sphere, or on equal spheres, will be equivalent — 1st, when they are mutually equilateral; — 2d, when they are mutually equi- angular ; — 3d, when two sides of the one are equal to two sides of the other, each to each, and the included angles are equal ; — 4th, when two angles of the one are equal to two angles of the other, each to each, and the included sides are equal. PROPOSITION XIV. If two arcs of great circles intersect each other on the sur- face of a hemisphere, the sum of either two of the opposite tri- angles thus formed will be equivalent to a lune whose angle is the corresponding angle formed by the arcs. Let the great circle, AEBC, be the base of a hemi- sphere, on the surface of which the semi-great circumfer- ences, BBA and CBE, inter- sect each other at B ; then will the sum of the opposite tri- angles, BBC and BAB, be equivalent to the lune whose angle is BBC; and the sum of the opposite triangles, CB A and BBE, will be equiv- alent to the lune whose angle is CBA. Produce the arcs, BBA and CBE, until they intersect on the opposite hemisphere at H; then, since CBB and BEH are both semi-circumferences of a great circle, they are equal. Taking from each the common part BE, we have CB =HE. In the same way we prove BB = HA, and AE = BC. The two triangles, BBC and HAE, are therefore mutually equilateral, and hence they are equivalent, (Prop. 13). But the two tri- angles, HAE and ABE, together, make up the lune 28 326 SPHERICAL GEOMETRY. DEHAD-, hence the sum of the a's, B DO and ADE, is equivalent to the same lune. By the same course of reasoning, we prove that the sum of the opposite A's, DAO and DBJE, is equivalent to the lune BOH AD, whose angle is ABC. PROPOSITION XV. The surface of a lune is to the whole surface of the sphere, as the angle of the lune is to four right arigles ; or, as the arc which measures that angle is to the circumference of a great circle. Let ABFOA be a lune on the surface of a sphere, and BOB an arc of a great circle, whose poles are A and F, the vertices of the angles of the lune. The arc, BO, will then measure the angles of the lune. Take any arc, as BD, that will be con- tained an exact number of times in BO, and in the whole circum- ference, BOJEB, and, beginning at B, divide the arc and the circumference into parts equal to BB, and join the points of division and the poles, by arcs of great circles. We shall thus divide the whole surface of the sphere into a number of equal lunes. Now, if the arc BO con- tains the arc BB m times, and the whole circumference contains this arc n times, the surface of the lune will contain m of these partial lunes, and the surface of the sphere will contain n of the same ; and we shall have, Surf, lune : surf, sphere : : m : n. But, m : n : : BO : circumference great circle ; • hence, surf, lune : surf, sphere : : BO : cir. great circle; or, surf, lune : surf, sphere :: [__BOO : 4 right angles. SECTION I. 327 This demonstration assumes that BD is a common measure of the arc, BC, and the whole circumference. It may happen that no finite common measure can be found ; but our reasoning would remain the same, even though this common measure were to become indefinitely small. Hence the proposition. Cor. 1. Any two lunes on the same sphere, or on equal spheres, are to each as their respective angles. Scholium. — Spherical triangles, formed by joining the pole of an arc of a great circle with the extremities of this arc by the arcs of great circles, are isosceles, and contain two right angles. For this reason they are called bi-redangular. If the base is also a quadrant, the vertex of either angle becomes the pole of the opposite side, and each angle is measured by its opposite side. The three angles are then right angles, and the triangle is for this reason called tri-rectangular. It is evident that the surface of a sphere contains eight of its tri- rectangular triangles. Car. 2. Taking the right angle as the unit of angles, and denoting the angle of a lune by A, and the surface of a tri-rectangular triangle by T, we have, surf, of lune : ST :: A : 4; whence, surf, of lune = 2 A x T. Cor. 3. A spherical ungula bears the same relation to the entire sphere, that the lune, which is the base of the ungula, bears to the surface of the sphere ; and hence, any two spherical ungulas in the same sphere, or in equal spheres, are to each other as the angles of their re- spective lunes. PROPOSITION XVI. The area of a spherical triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle. Let AB C be a spherical triangle, and DEFLK the cir- cumference of the base of the hemisphere on which this triangle is situated. 328 SPHERICAL GEOMETRY. Produce the sides of tlie tri- angle until they meet this cir- cumference in the points, D, U, F, L, K, and P, thus forming the sets of opposite triangles, FAF, AKL ; BFF, BFK; OFF, OFF. Now, the triangles of each of these sets are together equal to a lune, whose angle is the cor- responding angle of the triangle, (Prop. 14) ; hence we have, A FAF + A AKL = 2 A x T, (Prop. 15, Cor. 2). ABFF + ABFK=2B x T A OFF + A CDF = 2(7 x T If the first members of these equations be added, it is evident that their sum will exceed the surface of the hemisphere by twice the triangle ABO; hence, adding these equations member to member, and substituting for the first member of the result its value, 4T -f 2 A ABO, we have 4T + 2aAB0 = 2A.T -f 2B.T + 20.T or, 2T + AABO= A.T + B.T + O.T whence, A ABO = A.T + B.T + O.T—2T. That is, AABO = (A -f B + 0— 2) T. But A -f B + (7 — 2 is the excess of the sum of the angles of the triangle over two right angles, and T de- notes the area of a tri-rectangular triangle. Hence the proposition ; the area, etc. SECTION I. 329 PROPOSITION XVII. The area of any spherical polygon is measured by the excess of the sum of all its angles over two right angles, taken as many times, less two, as the polygon has sides, multiplied by the tri-rectangular triangle. Let AB CJDE be a spherical poly- £ gon; then will its area be meas- j\^^*^ I ured by the excess of the sum of /\ / the angles, A, B, 0, D, and E, over / two right angles taken a number / N. x of times which is two less than c / """" ""^>E the number of sides, multiplied by \ z' T, the tri - rectangular triangle. \. / Through the vertex of any of the ^ angles, as E, and the vertices of the opposite angles, pass arcs of great circles, thus divi- ding the polygon into as many triangles, less two, as the polygon has sides. The sum of the angles of the several triangles will be equal to the sum of the angles of the polygon. Now, the area of each triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle. Hence the sum of the areas of all the triangles, or the area of the polygon, is measured by the excess of the sum of all the angles of the triangles over two right angles, taken as many times as there are triangles, multiplied by the tri- rectangular triangle. But there are as many triangles as the polygon has sides, less two. Hence the proposition ; the area of any spherical poly- gon, etc. Cor. If S denote the sum of the angles of any spherical polygon, n the number of sides, and T the tri-rectan- gular triangle, the right angle being the unit of angles ; the area of the polygon will be expressed by IS— 2 (w-2)]x T= (S— 2n + 4) T. 28* 330 SPHERICAL TRIGONOMETRY. SECTION II. SPHERICAL TRIGONOMETRY. A Spherical Triangle contains six parts — three sides and three angles — any three of which being given, the other three may be determined. Spherical Trigonometry has for its object to explain the different methods of computing three of the six parts of a spherical triangle, when the other three are given. It may be divided into Right-angled Spherical Trigonome- try, and Oblique-angled Spherical Trigonometry ; the first treating of the solution of right-angled, and the second of oblique-angled spherical triangles. RIGHT-ANGLED SPHERICAL TRIGONOMETRY. PROPOSITION I. With the sines of the sides, and the tangent of ONE SIDE of any right-angled spherical triangle, two plane triangles can be formed that will be similar, and similarly situated. Let ABO be a spherical triangle, right-angled at B ; and let D be the center of the sphere. Because the angle OB A is a right angle, the plane OBD is perpendicular to the plane DBA. From let fall OR, perpen- dicular to the plane DBA ; and as the SECTION II. 331 plane CBD is perpendicular to the plane DBA, CE will lie in the plane OBI), and be perpendicular to the line DB, and perpendicular to all lines that can be drawn in the plane DBA, from the point E (Def. 2, B. VI). Draw EG perpendicular to DA, and draw GrC; GC will lie wholly in the plane CDA, and CEG is a right- angled triangle, right-angled at E. We will now demonstrate that the angle DGfC is a right angle. The right-angled ACEG, gives CE 2 +EG 2 = CG 2 fel) The right-angled AD GE, gives DG 2 +EG 2 =DE 2 (2) By subtraction, CH 2 — DG 2 = CG 2 — DE 2 ( 3 ) By transposition, OH 2 + DH 2 = CG 2 + DG 2 (4) But the first member of equation (4) ? is equal to CD 2 , because ODE is a right-angled triangle; Therefore, CD 2 = CG 2 + DG 2 Hence, CD is the hypotenuse of the right-angled tri- angle DGC, (Th. 39, B. I). , From the point B, draw BE at right angles to DA, and BF at right angles to DB, in the plane CDB ex- tended ; the point F will be in the line DC. Draw EF, and as F is in the plane CDA, and i? is in the same plane, the line EF is in the plane CDA. Now we are to prove that the triangle CEG is similar to the triangle BEF, and similarly situated. As EG and BE are both at right angles to DA, they are parallel ; and as EC and BF are both at right angles to DB, they are parallel ; and by reason of the parallels, the angles GEC and EBF are equal ; but GEC is a right angle ; therefore, EBF is also a right angle. Now, as GE and BE are parallel, and CE and BF are also parallel, we have, DE i DB = EG: BE And, DE : DB =EC : BF 332 SPHERICAL TRIGONOMETRY. Therefore, HO : BE = HO : BF (Th. 6, B. II), Or, Ha : HO = BE : BF. Here, then, are two triangles, having an angle in the one equal to an angle in the other, and the sides about the equal angles proportional; the two triangles are therefore equiangular, (Cor. 2, Th. 17, B. II); and they are similarly situated, for their sides make equal angles at H and B with the same line, DB. Hence the proposition. Scholium. — By the definition of sines, cosines, and tangents, we perceive that CH is the sine of the arc BC, DH is its cosine, and BF its tangent; CG is the sine of the arc AC, and DG its cosine. Also, BE is the sine of the arc AB, and BE is the cosine of the same arc. With this figure we are prepared to demonstrate the following propo- sitions. PROPOSITION II. In any right-angled spherical triangle, the sine of one side is to the tangent of the other side, as radius is to the tangent of the angle adjacent to the first-mentioned side. Or, the sine of one side is to the tangent of the other side, as the cotangent of the angle adjacent to the first-mentioned side is to the radius. For the sake of brevity, we will represent the angles of the triangle by A, B, 0, and the sides or arcs opposite to these angles, by a, b, c, that is, a opposite A, etc. In the right-angled plane triangle EBF, we have, EB : BF = B : tzn.BEF That is, sin.c : tan.a = R : tan.J., which agrees with the first part of the enunciation. By reference to equation (5), Section I, Plane Trigonometry, we shall find that, tan.JL cot. A = B 2 ; B 2 therefore, tan. J. = -. cot. A SECTION II. 333 Substituting this value for tangent A, in the preceding proportion, and dividing the last couplet by R, we shall have, sin.c : tan.a = 1 : -. cot.JL Or, sin.c : tan.a = cot. J. : R. Or, R sin.c = tan.a cot. JL, (1) which answers to the second part of the enunciation. Cor. By changing the construction, drawing the tan- gent to AB, in place of the tangent to BC, and proceed- ing in a similar manner, we have, R sin. a = tan.c cot. C. ( 2 ) PROPOSITION III. In any right-angled spherical triangle, the sine of the right angle is to the sine of the hypotenuse, as the sine of either of the other angles is to the sine of the side opposite to that angle. , The sine of 90°, or radius, is designated by R. In the plane triangle, CMGr, we have, Bm.CMG- : Ca = sin.CG-H : CM That is, R : sin.6 = sin. J. : sin.a Or, R sin. a = sin. b sin. A ( 3 ) Cor. By a change in the construction of the figure, drawing a tangent to AB, etc., we shall have, R : sin.6 = sin.<7 : sin.c Or, R sin.c = sin.5 sin. C. ( 4 ) Scholium. — Collecting the four equations taken from this and the preceding proposition, we have, ( 1 ) iJTsin.c = tan.a cot. J. ( 2 ) R sin. a = tan.c cot. C ( 3 ) B sin.a = sin.6 sin. A ( 4 ) E sin.c = sin. b sin. C 334 SPHERICAL TRIGONOMETRY. These equations refer to the right-angled triangle, ABC; but the principles are true for any right-angled spherical triangle. Let us apply them to the right-angled triangle, PDC, the com piemen tal triangle to ABC. Making this application, equation ( 1 ) becomes R sin. CD = tan.PD cot. C ( 2 ) becomes R a'm.PD = tan. CD cot.P ( 3 ) becomes R sin. PD = sin. PC sin. C ( 4 ) becomes R sin. CD = ain.PC sin.P By observing that sin. CD = cos. J. C = cos.6. And that tan.PD = cot.DO = cot. J., etc.; and by running equa- tions ( n ), (m ), ( o ), and (p ), back into the triangle, ABC, we shall have, ( 5 ) R cos. 6 = cot. J. cot. C ( 6 ) R cos. A = cot.6 tan.c (7) R cos. A = cos.a sin. C ( 8 ) R cos.b —-- cos.a cos.c By observing equation ( 6 ), we find that the second member refers to sides adjacent to the angle A. The same relation holds in respect to the angle C, and gives, (9) Rco9.C= cot.6 tan.a. Making the same observations on ( 7 ), we infer, (10) R cos. C = cos.c sin.X Observation 1. Several of these equations can be de- duced geometrically without the least difficulty. For example, take the figure to Proposition 1. The parallels in the plane, DBA, give, DB : VH= DU : Da. That is, R : cos.a = cos.c : cos.5. A result identical with equation ( 8 ) ? and in words it is expressed thus : Radius is to cosine of one side, as the cosine of the other side is to the cosine of tiie hypotenuse. Observation 2. The equations numbered from (1) to (10) cover every possible case that can occur in right- angled spherical trigonometry ; but the combinations are SECTION II. 335 too various to be remembered, and readily applied to prac- tical use. "We can remedy this inconvenience, by taking the com- plement of the hypotenuse, and the complements of the two oblique angles, in place of the arcs themselves. Thus, b is the hypotenuse, and let V be its complement. Then, 5 + ^=90°; or, b= 90° — 6'; and, sin.6 = cos.6', cos.6 = sin. V ; tan.5 = cot.5 r . In the same manner, if A 1 is the complement to A, Then, sin. A = cos. J/; cos. A = sin. J/; and, tan. J. = cot. J/; and similarly, sin. C— cos. C; cos. Q— sin.C; and tan. (7= cot. C. Substituting these values for b, A, and 0, in the fore- going ten equations (a and c remaining the same), we have, (11) (12) (13) (14) (15) (16) (17) (18) (19) (20) NAPIE Rsin.c = Rsm.a = Rsin.a = R sin. \S) cos.c = cos. a cos.5 + sin.a sin.5 cos.C J From these, by simple transposition and division, we deduce the following formulas for the cosines of the angles of any spherical triangle, viz : cos.a — cos.5 cos.c^ cos. A = cos.i? = cos. = sin.&' sin.c cos. b — cos.a cos.c sin. a sin.c cos.c — cos.a cos. b sin. a sin. b (S>) By means of these equations we can find the cosine of any of the three angles of a spherical triangle in terms of the functions of the sides ; but in their present form they are not suited for the employment of logarithms, and we should be compelled to use a table of natural sines and cosines, and to perform tedious numerical ope- rations, to obtain the value of the angle. They are, however, by the following process, trans- formed into others well adapted to the use of logarithms. In Eq. 34, Sec. I, Plane Trig., we have 1 + cos. A = 2cos. 2 J^4. m , p a ,-, a -t , cos.a — COS.6 cos.c Therefore, 2cos. 2 \A = 1 + : — ; — : •. sm.6 sin.c (sin. b sin.c — cos. b cos.c) + cos.a , » sm.o sin.c But, cos.(6 + c) = cos.5 cos.c = sin.c sin. b, (Equation 9, Section I, Plane Trig.). By comparing this equation SECTION III. 343 with the second member of equation ( »* ), we perceive that equation ( m ) is readily reduced to sin. b sin.c Considering (b+c) as one are, and then making appli- cation of equation ( 18 ), Plane Trigonometry, we have, But, 2cos. 2 |J.= b -\- c — a b -f c + a 2 2 W (—,— ) ».n. (—- ') sin. 6 sin. c — a; and if we put S to A -J- /» _j_ /y represent — , we shall have, A __ sin.# sin.(# — a) COS.' -— = Or, COS. 2 sin.6 sin.c A _ * /sm.S sm.(S — a) 2 Y sin. 6 sin.c ' The second member of this equation gives the value of the cosine when the radius is unify. To a greater radius, the cosine would be greater; and in just the same proportion as the radius increases, all the trigonometrical lines increase ; therefore, to adapt the above equation to our tables where the radius is M, we must write R in the second member, as a factor; and if we put it under the radical sign, we must write TC\ For the other angles we shall have precisely similar equations : That is, cos. - = \ / -fl 2 sin.ff sin. ( -#— a) 2 v sin. b sin. sm.6 sin.c when radius is unity. SECTION III When radius is B, we have v sin. b sin.c Similarly, sin.J* = ^H^EfeLtl) v sin.fl sin.c . ; „ v /S 2 sin.(iS' — «)sin.(^— 6 sin.J(7 = y — 345 And, sin. a sin.6 IV) To apply to our tables, B 2 must be put under the radi- cal sign. We shall show the application of these form- ulae, and those in group (T), hereafter. PROPOSITION VIII. The cosine of any of the angles of a spherical triangle, is equal to the product of the sines of the other two angles mul- tiplied by the cosine of the included side, minus the product of the cosines of these other two angles. • Let ABO be a spherical triangle, and A r B r Q r its supplemental or polar tri- angle, the angles of the first being de- noted by A, B, and 0, and the sides opposite these angles by a, 5, c, respect- ively ; A', B f , Q r , a', V , c f , denoting the angles and corresponding sides of the second. By Prop. 5, Sec. I, we have the following relations be- tween the sides and angles of these two triangles. A' m 180° — a,B f = 180° — b, C = 180° — c; a' - 180° — A,V= 180° — B, c 1 = 180° — C. The first of formulae (#), Prop. 7, when applied to the polar triangle, gives cos. a' = cos.5' cos.c' + sin.5 ; sin.c' cos. A r (1) 346 SPHERICAL TRIGONOMETRY. which, by substituting the values of a', b f , c', and A\ becomes cos.(180° — A) = cos.(180° — B) cos.(180° — C) + sin.(180° — B) sin.(180° — 0) cos.(180° — a), ( 2 ) But, cos.(180°— A) = — cos.^4, etc., sin.(180°— B) = sin.B, etc. ; and placing these values for their equals in eq. ( 2 ), and changing the sines of both members of the resulting equation, we get cos. J. = sin.i? sm.O cos.a — cos.i? cos. O, which agrees with the enunciation. By treating the other two of .formulae (£), Prop. 7, in the same manner, we would obtain similar values for the cosines of the other two angles of the triangle ABC; or we may get them more easily by a simple permuta- tion of the letters A, B, C, a, etc. Hence, we have the three equations cos. A = sin.i? sin. (7 cos.a — cos. B cos. C^ C08.B mm sin. A sm.O cos. b — cos. A cos. V CO cos. Q = sin.-d sin.i? cos.c? — cos. A cos.B ) By transposition and division, these equations become cos. A -f cos.B cos. ,o\ cos. a = {o) sm.B sin. C , cos.B 4- cos. A cos. sin. A sm.C 7 cos. + cos. A cos.B cos.c = r — — , — i_ S111..A sinJ From these we can find formulae to express the sine or the cosine of one half of the side of a spherical triangle, in terms of the functions of its angles ; thus : Add 1 to each member of eq. (3), and we have cos. J. + cos.i? cos. O -f sin. B sin. 1 + cos.a = sin.i? sin.6 Y SECTION III. 347 cos. A + cos.(J5 — 0) sin.2? sin.G 7 But, 1 + cos.a = 2cos. 2 \a ; hence, , , cos.^L + cos.fi? — C) 2cos. 2 i« - 4 tf }>„ i sin.J? sin. (7 and since cos. J. 4- cos.(2? — C) = 2cos.J(J. + i? — (7)cos.J (4+a—JB) (Eq.17, Sec. I, Plane Trig.), we have 2cos. 2 la = 2cos -*(^ + B-C)coB.i(A + (7-i?) 2 sin.J5 sin.6 7 Make A + B + 0=2S; then A + B—C=2S—2C, A+C—B = 2S—2B, i(A + B—C) = jS—C,an& ±{A + C—B) = S—B\ whence sm.jo sm.(7 4 /cos.^— <7)cos.(/SCTg5 or, cos.fa = \/ ^— = — £? . > ^ c 1 v sm.2?sin.<7 , Similarly, cos.|5 = V ^/^ - ; / and, cos.Jc - V^S^E! v sin. J. sinJ .To find the sin.Ja in terms of the functions of the angles, we must subtract each member of eq. ( 3 ) from 1, by which we get H _, cos. J. + cos.i? cos. (7 1 — cos.a = l r . . sui.jd sm.G But, 1 — cos.a = 2sin. 2 Ja ; hence we have, o- 2i _(sin.J5 sin. (7 — cos. 5 cos. (7) — cos.JL smJ sin.C T Operating upon this in a manner analogous to that by which cos.Ja was found, we get, 348 SPHERICAL TRIGONOMETRY. t smJ sm. J . l4 f— cos.# cos.(S-B) l $ sin. lb = \ . — —A-jy — J - V 2 ) (W) \ sin. J. sin. (7 J . i f — cos.#cos.(#— O)} i Sin. Jc = ■{ : — r- L_- ' I 2 t sin. J. sin.j? j If the first equation in ( W) be divided by the first in ( V ), we shall have, tan i a = / -cos.ff co s.(S-A) Y J * 2 \cos.(^~ B)cos.(S— C)) And corresponding expressions may be obtained for tan.JS and tan.Jc. NAPIER'S ANALOGIES. If the value of cos.c, expressed in the third equation of group (#), Prop. 7, be substituted for cos.c, in the second member of the first equation of the same group, we have, cos.a = cos.a cos. 2 b -f sin. a sin. b cos.6 cos.(7-f- sin.6 sin.c cos. J.; which, by writing for cos. ? 5 its equal, 1 — sin. 2 £, becomes, cos.a=cos.a — cos.asin. 2 6-J-sin.cz sin. 6 cos. 6 cos. C+sin.b sin.c cos.-4. Or, = — cos.a sin. 2 5-|-sin.a sin.6 cos. b cos. C+ sin. b sin.c cos. J.. Dividing through by sin. b, and transposing, we find, cos.JL sin.c = cos.a sin.6 — sin.a cos.5 cos. C; , A cos.a sin.6 — sin.a cos.5 cos. (7 , ... hence, cos. A = : — . . ( 1 ) sin.c By substituting the value of cos.c, in the second of the equations of group (S), Prop. 7; or, more simply, by writing B for J., and b for a, in the above value, for cos.^., we obtain, -r> eos.b sin.a — sin.5 cos.a cos. (7 , ON COS.i* = ; . (2) sin.c SECTION III. 349 Adding equations (1) and (2), member to member, we have, . . ^ sin.fa+6) — sin. (a 4- b) cos. C0S..A + C0S.2? = a 1 . 1 L- ; , sin,(? by remembering that sin.a cos.5 + cos.a sin.5 = sin.(a+5). (See Eq. (7 ), Sec. I, Plane Trig.). Whence, cos.J. + cos.£ = (1 — cos. C) E^L±3 m ( 3 ) In any spherical triangle we have, (Prop. I), sin.A : sin.i? : : sin.a : sin.5 ; And therefore, sin.J. + sin.i? : sin.jS :: sin.a + sin.6 : sin.5. -,-r - a , - t> (sin.a + sin. b) sin.i? Hence, sm. A + smJ = i : — ~-l . sin. 6 t, , sin.i? sin. i . -. i n sin.i? . ,, , But, — = — — , which value of — ; , in the above sm.6 sm.c sm.6 equation, gives A . . r> (sin.a-f sin.5) sin. (7 , A , sm. A + sinJ = i ■ '- . ( 4 ) sm.c Dividing equation (4) by equation (3) ? member by member, we obtain, sin. J. + sin.i? sin. sin.a + sin. b , - v == x . ( 5 ) cos. A + cos.i? 1 — cos. sin.(a + 5) Comparing this equation with Equations (20) and (26), Sec. I, Plane Trigonometry, we see that it can be re- duced to , i / a i t>\ j. i n sin.a-f sin.5 , nx tan.}(J. + j£) = cot.i(7x - . /" - (6) sin.(a -f b) Again, from the proportion, sin. A : sin.i? : : sin.a : sin.5, we likewise have, sin. A — sin.J5 : sin.I? :: sin.a — sin.5 : sin.5; 30 . 350 SPHERICAL TRIGONOMETRY. hence, sin. J. — sin.i? = (sin.a — sin. b) -.-— = (sin.a — sin. b . , N sin. sin. b) — . sm.)=l(b-a\ \(a> : + V) = 180°- ^±^ 9 j(a' - b') = i(B - A\ \C = 90° - \c. 352 SPHERICAL TRIGONOMETRY. By the substitution of these values in Eq. (™), that equation becomes or, tan. J(a _ b) = gj^f j tan. \c, (p) since tan. J(5 — a) = — tan. J(a — b\ and sin. J(i? — A = — sin.|(J.— £). By applying Eq. (8) to the polar triangle, and treating the resulting equation in a manner similar to the above, we find Equations {p) and (q) maybe resolved into the fol- lowing proportions. sin. l(A + B) : sin. %{A — B) : : tan. \c : tan. J(a — 6); cos. }(JL + J5) : cos. J(J. — i?) : : tan. \c : tan. J(a -f b). These proportions are called Napier's 3d and 4th Analogies, and when expressed in words become the fol- lowing rules : 1. The cosine of the half sum of any two angles of a spherical triangle is to the cosine of the half difference of the same angles, as the tangent of half the included side is to the tangent of the half sum of the other two sides, 2. The sine of the half sum of any two angles of a spheri- cal triangle is to the sine of the half difference of the same angles, as the tangent of half the included side is to the tan- gent of the half difference of the other two sides. The half sum, and the half difference of two sides of a spherical triangle, may be found by these rules, when two angles and the included side are given ; and by add- ing the half sum to the half difference, we get the greater of these sides, and by subtracting the half difference from the half sum, we get the smaller. * SECTION IV. 353 SECTION IV. SPHERICAL TRIGONOMETRY APPLIED. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. A good general conception of the sphere is essential to a practical knowledge of spherical trigonometry, and this conception is best obtained by the examination of an artificial globe. By tracing out upon its surface the various forms of right-angled and oblique-angled tri- angles, and viewing them from different points, we may soon acquire the power of making a natural representa- tion of them on paper, which will be found of much as- sistance in the solution and interpretation of problems. For instance, suppose one side of a right-angled spherical triangle to be 56°, and the angle between this side and the hypotenuse to be 24°. What is the hypote- nuse, and what the other side and angle ? A person might solve this problem by the application of the proper equations or proportions, without really comprehending it ; that is, without being able to form a distinct notion of the shape of the triangle, and of its relation to the surface of the sphere on which it is situated. If we refer this triangle to the common geographical globe, the side 56° may be laid off on the equator, or on a meridian. In the first case, the hypotenuse will be the arc of a great circle drawn through one extremity of the side 56°, above or below the equator, and making with 30* x 354 SPHERICAL TRIGONOMETRY. it an angle of 24° ; the other side will be an arc of a meridian. In the second case, the side 56° falling on a meridian, the hypotenuse will be the arc of a great circle drawn through one extremity of this side, on the right or left of the meridian, and making with it an angle of 24° ; the other side will be the arc of a great circle, at right angles to the meridian in which the given side lies. Generally speaking, the apparent form of a spherical triangle, and consequently the manner of representing it on paper, will differ with the position assumed for the eye in viewing it. From whatever point we look at a sphere, its outline is a perfect circle in the axis of which the eye is situated; and when the eye is, as will be here- after supposed, at an infinite distance, this circle will be a great circle of the sphere. All great circles of the sphere whose planes pass through the eye, will seem to be diameters of the circle which represents the outline of the sphere. "We will now suppose the eye to be in the plane of the equator, and proceed to construct our triangle on paper. Let the great circle, PAS A', represent the out- line of the sphere, the di- ameter AA' the equator, and the diameter PS the central meridian, or the meridian in whose plane the eye is situated. Let AB = 56°, represent the given side, and A (7,making with AB the angle B A (7= 24°, the hypotenuse, then will BO, the arc of a meridian, be the other side at right angles to AB, and the triangle, ABO, corresponds in all respects to the given triangle. Again, measure off 5Q° from P to Q, draw the radius DQ, make the arc A'G- equal to 24°, and draw the quad- rant PRGr. The triangle PQR will also represent the given triangle in every particular. SECTION IV. 355 We know from the construction, that D V, = 24°, is greater than BO, and that A is greater than AB, that is, greater than 56°. In like manner, we know that A', = 24°, is greater than QB, and that PB is greater than PQ, because PB is more nearly equal toPG, =90°, thanP# is to PA, =90°. For illustration and explanation, we also give the fol- lowing example : In a right-angled spherical triangle, there are given, the hypotenuse equal to 150° 33' 20", the angle at the base, 23° 27' 29", to find the base and the perpendicular. Let A 1 BO in the last figure, represent the triangle in which A'0 = 150° 33' 20", the L BA f O= 23° 27' 29", and the sides A'B and BO are required. This problem presents a right-angled spherical tri- angle, whose base and hypotenuse are each greater than 90° ; and in cases of this kind, let the pupil observe, that the b^ase is greater than the hypotenuse, and the oblique angle opposite the base, is greater than a right angle. In all cases, a spherical triangle andits supplemental triangle make a lune. It is 180° from one pole to its opposite, whatever great circle be traversed. It is 180° along the equator ABA', and also 180° along the ecliptic AOA'. The lune always gives two triangles; and when the sides of one of them are greater than 90°, we take the triangle having supplemental sides ; hence in this case we operate on the triangle ABO. A O is greater than AB, therefore A'B is greater than the hypotenuse A'C. The [_AOB is less than 90°; therefore, the adjacent angle A' OB is greater than 90°, the two together being equal to two right angles. These facts are technically expressed, by saying, that the sides and opposite angles are of the same affection.* * Same affection : that is, both greater or both less than 90°. Dif- ferent affection : the one greater, the other less than 90°. 356 SPHERICAL TRIGONOMETRY. !Now, if the two sides of a right-angled spherical triangle are of the same affection, the hypotenuse will be less than 90° ; and if of different affection, the hypotenuse will be greater than 90°. If, in every instance, we make a riatural construction of the figure, and use common judgment, it will be im- possible to doubt whether an arc must be taken greater or less than 90°. We will now solve the triangle A OB. AO = 180° — 150° 33' 20" = 29° 26' 40". To find BO, we use Eq. (3) or (13), Prop. 3, Sec. II., thus: b, sin. 29° 26' 40" . 9.691594 A, sin. 23°_2T_29^ .- 9.599984 a,sin. 11° 17' 7" . 9.291578 To find AB, we use equation ( 1 ) or ( 11 ), thus : a, tan. 11° 17' 7" . 9.300016 A, cot. 2 3° 27' 29" . 10.362674 c,sin. 27° 22' 32" . 9.662690 180 • A'B=lte° 37' 28" PRACTICAL PROBLEMS IN RIGHT-ANGLED SPHERICAL TRIGONOMETRY. 1. In the right-angled spherical triangle ABO, given AB = 118° 21' a- 4", and the angle A = 23° 40' 12", to find the other parts. "~B A (AG, 116° 17' 45"; the angle O, 100° 59' 26"; I and BO, 21° 5' 42". 2. In the right-angled spherical triangle ABO, given AB 53° 14' 20", and the angle A 91° 25' 53", to find the other parts. A (AO, 91° 4' 9"; the angle O, 53° 15' 8"; ^ n *'\ and BO, 91° 47' 11". SECTION IV. 357 3. In the right-angled spherical triangle ABC, given AB 102° 50' 25", and the angle A 113° 14' 87", to find the other parts. A (AC, 84° 51' 36"; the angle C, 101° 46" 57"; \ and BC, 113° 46' 27". 4. In the right-angled spherical triangle ABC, given AB 48° 24' 16", and BC 59° 38' 27", to find the other parts. A (AC, 70° 23' 42"; the angled, 66° 20' 40"; s I and the angle (7, 52° 32' 55". 5. In the right-angled spherical triangle ABC, given AB 151° 23' 9", and ^(7 16° 35' 14", to find the other parts. An§ (AC, 147° 16' 51"; the angle C, 117° 37' 21"; * I and the angle A, 31° 52' 50". 6. In the right-angled spherical triangle ABC, given AB 73° 4' 31", and AC 86° 12' 15," to find the other parts. Am (BC, 76° 51' 20"; the angled, 77° 24' 23"; m ' \ and the angle C, 73° 29' 40". ' 7. 'In the right-angled spherical triangle ABC, given AC 118° 32' 12", and AB 47° 26' 35", to find the other parts. A (BC, 134° 56 f 20"; the angle A, 126° 19' 2"; m ' \ and the angle C, 56° 58' 44". 8. In the right-angled spherical triangle ABC, given AB 40° 18' 23", and AC 100'° 3' 7", to find the other parts. A ( The angle A, 98° 38' 53" ; the angle ^' t C, 40° 4' 6" ; and BC, 103° 13' 52". 9. In the right-angled spherical triangle ABC, given AC 61° 3' 22", and the angle A 49° 28' 12", to find the other parts. Ang ( AB, 49° 36' 6" ; the angle C, 60° 29' 19" ; U ' I and BC, 41° 41' 32". 10. In the right-angled spherical triangle ABC, given 358 SPHERICAL TRIGONOMETRY. AB 29° 12' 50", and the angle 37° 26' 21", to find the other parts. r Ambiguous ; the angle A, 65° 27' 58", or its Ans. < supplement; J. (7, 53° 24/ 13", or its sup- ( plement; BO, 46° 55' 2", or its supplement. 11. In the right-angled spherical triangle ABC, given AB 100° 10' 3", and the angle 90° 14' 20", to find the other parts. cAO, 100° 9' 5b ff , or its supplement; BO, Ans.-l 1° 19' 53", or its supplement; and the L angle A, 1° 21' 8", or its supplement. 12. In the right-angled spherical triangle ABO, given AB 54° 21' 35", and the angle (7 61° 2' 15", to find the other parts. cBO, 129° 28' 28", or its supplement; AO, Ans.l 111° 44' 34", or its supplement; and the I angle A, 123° 47' 44", or its supplement. 13. In the right-angled spherical triangle ABO, given AB 121° 26' 25", and the angle O 111 14' 37", to find the other parts. rThe angle A, 136° 0' 3", or its supplement; Ans. i AO, 66° 15' 38", or its supplement; and L BO, 140° 30' 56", or its supplement. QUADRANTAL TRIANGLES The solution of right-angled spheri- cal triangles includes, also, the solu- tion of quadr anted triangles, as may be seen by inspecting the adjoining fig- ure. When we have one quadr antal triangle, we have four, which with one right-angled triangle, fill up the whole hemisphere. To effect the solution of either of the four quadrantal triangles, APO, AP'O, A ! PO, oxA'P'O, it is sufficient to solve the small right-angled spherical triangle ABO. SECTION IV. 359 To the half lune AP'B, we add the triangle ABO, and we have the quadrantal triangle AP'Q', and hy sub- tracting the same from the equal half lune APB, we have the quadrantal triangle PAC. When we have the side, AC, of the same triangle, we have its supplement, A'C, which is a side of the triangles A'PC, and A'P'C. "When we have the side, CB, of the small triangle, by adding it to 9.0°, we have P'C, a side of the triangle A'P'C; and subtracting it from 90°, we have PC, & side of the triangles APC, and AP'C. PROBLEM I. In a quadrantal triangle, there are given the quadrantal side, 90°, a side adjacent, 42° 21', and the angle opposite this last side, equal to 36° 31'. Required the other parts. By this enumeration we cannot decide whether the triangle APC or AP'C, is the one required, for AC — 42° 21' belongs equally to both triangles. The angle APC = AP'C = 36° 31' = AB. We operate wholly on the triangle ABC. To find the angle A, call it the middle part. Then, R cos. CAB = R sin.iM C = cot. J. C tan.^5. cot. AC = 42° 21' tzn.AB = 36° 31' 10.040231 9.869473 cos. CAS = 35° 40' 51" 90° 9.909704 PAC = 54° 19' 9" P r AC = 125° 40' 51" To find the angle C, call it the middle part. R cos. ACB = sin. CAB cos. AB. sin. CAB = 35° 40' 51" 9.765869 zoz.AB = 36° 31' . 9.905085 cos. ACB = 62° 2' 45" 9.670954 180° ACP « A'CP' «= 117° 57' 15" 360 SPHERICAL TRIGONOMETRY. To find the side BO, call it the middle part. Rs'm.BC = tan.XB cot.ACB. tsm.AB = 36° 31' 0" cotACB = 62° 2' 45" 9.869473 9.724835 sin.BG = 23° 8' 11" 90° 9.594308 PC = 66° 51' 49" PC = 113° 8' 11" "We now have all the sides, and all the angles of the four triangles in question. PROBLEM II. In a quadr anted spherical triangle, having given the quad- rantal side, 90°, an adjacent side, 115° 09', and the included angle, 115° 55', to find the other parts. This enunciation clearly points out the particular triangle A'P'O. A'P' = 90° ; and conceive A'C= 115° 09'. Then the angle P'A f O = 115° bb f = P'D. From the angle PM/(7take 90°, or P f A'B, and the remainder is the angle OA f D = BAO = 25° 55'. We here again operate on the triangle ABO. A'O, taken from 180°, gives 64° 51' = AO. To find BO, we call it the middle part. R sin. 5(7= sin. A C sin. B A C. sin. AC = 64° 51' . 9.956744 sin.BAC = 25° 55' . 9.640544 sm.BC = 23° 18' 19" . 9.597288 90° PC = 113° 18' 19" SECTION IV. 361 To find AB, we call it the middle part. Rsm.AB = t&n.BC cot.BAC. t^ixi.BC = 23° 18' 19" . 9.634251 cot.BAC = 25° 55' . 9.313423 sin.AB = 62° 26' 8" . 8.947674 180° A'B = 117° 33' 52" = the angle AIP'C. To find the angle 0, we call it the middle part. E cos. G = cot. J. C tan.^a cot. AQ = 64° 51' . 9.671634 tan.^BC = 23° 18' 19" . 9.634251 cos. C = 78° . 9.305885 180° 19' 53" . FCA! = 101° 40' 7" Thus we have found the side FC = 113° 18' 19" -\ The angle A'F O = 117° 33' 52" I An*. " FCA' = 101° 40' 7") PRACTICAL PROBLEMS. 1. In a quadrantal triangle, given the quadrantal side, 90°, a side adjacent, 67° 3', and the included angle, 49° 18', to find the other parts. c The remaining side is 53° & 46" ; the angle Ans. < opposite the quadrantal side, 108° 32' 27" ; I and the remaining angle, 60° 48' 54". 2. In a quadrantal triangle, given the quadrantal side, 90°, one angle adjacent, 118° 40' 36", and the side op- posite this last-mentioned angle, 113° 2 ; 28", to find the other parts. c The remaining side is 54° 38' 57" ; the angle Ans. < opposite, 51° 2' 35"; and the angle opposite I the quadrantal side 72° 26' 21". 3. In a quadrantal triangle, given the quadrantal side, 31 362 SPHERICAL TRIGONOMETRY. 90°, and the two adjacent angles, one 69° 13' 46", the other 72° 12' 4", to find the other parts. r One of the remaining sides is 70° 8' 39", the Arts. < other is 73° 17' 29", and the angle opposite I the quadrantal side is 96° 13' 23". 4. In a quadrantal triangle, given the quadrantal side, 90°, one adjacent side, 86° 14' 40", and the angle oppo- site to that side, 37° 12' 20", to find the other parts. ( The remaining side is 4° 43' 2" ; the angle op- Ans. < posite, 2° 51' 23" ; and the angle opposite I the quadrantal side, 142° 42' 2". 5. In a quadrantal triangle, given the quadrantal side, 90°, and the other two sides, one 118° 32' 16", the other 67° 48' 40", to find the other parts — the three angles. rThe angles are 64° 32' 21", 121° 3' 40", and Arts. < 77° 11' 6" ; the greater angle opposite the I greater side, of course. 6. In a quadrantal triangle, given the quadrantal side, 90°, the angle opposite, 104° 41' 17", and one adjacent side, 73° 21' 6", to find the other parts. m ( Eemaining side, 49° 42' 18" ; remaining U8 ' 1 angles, 47° 32' 39", and 67° 5V 13". SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. All cases of oblique-angled spherical trigonometry may be solved by right-angled Trigonometry, except two ; because every oblique-angled spherical triangle is composed of the sum, or the difference, of two right- angled spherical triangles. When a side and two of the angles, or an angle and two of the sides are given, to find the other parts, conform to the following directions : Let a perpendicular be drawn from an extremity of a given side, and opposite a given angle or its supplement; this will form two right-angled spherical triangles ; and SECTION IT. 363 one of them will have its hypotenuse and one of its ad- jacent angles given, from which all its other parts can be computed ; and some of these parts will become as known parts to the other triangle, from which all its parts can be computed. To facilitate these computations, we here give a sum- mary of the practical truths demonstrated in the fore- going propositions. 1. The sines of the sides of spherical triangles are propor- tional to the sines of their opposite angles. 2. The sines of the segments of the base, made by a per- pendicular from the opposite angle, are proportional to the cotangents of their adjacent angles. 3. The cosines of the segments of the base are proportional to the cosines of the adjacent sides of the triangle. 4. The tangents of the segments of the base are reciprocally proportional to the cotangents of the segments of the vertical angle. 5. The cosines of the angles at the base are proportional to the sines of the corresponding segments of the vertical angle. 6. The cosines of the segments of the vertical angle are proportional to the cotangents of the adjoining sides of the triangle. The two cases in which right-angled spherical triangles are not used, are, 1st. When the three sides are given to find the angles ; and, 2d. "When the three angles are given to find the sides. The first of these cases is the most important of all, and for that reason great attention has been given to it, and two series of equations, (2* and Z7, Prop. 7, Sec. Ill), have been deduced to facilitate its solution. As heretofore, let AB represent any triangle whose angles are denoted by A, B, and (7, and sides by a, b, 364 SPHERICAL TRIGONOMETRY. and c ; the side a being opposite |__ A, the side b oppo- site [__ B> etc * EXAMPLES. 1. In the triangle ^LS (7, a = 70° 4' 18"; b = 63° 21' 27"; and e, 59° 16' 23" ; required the angle A. The formula for this is the first equation in group (T, Prop. 7, Sec. Ill), which is cos. A _ ,R 2 sm.Ssm.(S—a\% ' 2 \ sin.6 sin. subtracted from 10; and of - — is that of sin. B = 41 ° n/ 6 "> ° = 134 ° 54 ' SECTION IV. S69 9. Given, a*= 89° 21' 87", 6== 97° 18' 39", _ (cos.L sin._D + sin.Z cos.D) — sin. A coa.L sin.2) _ sin.(2y 4- D) — sin. A cos.L sin.i) Considering (L + D) as a single arc, and (applying Equation 16, Sec. I, Plane Trig.), we have, after dividing bj 2, ( L + D + A\ . ( L + 2> — A\ cos, sin.JP = cos.xe sm.x> and if we assume S = g , A , ,, , • o ir » cos.aS' sin.(/S r — A) we shall have, sin.* IP = =*A — H — J - cos.L sm.D Or, sin.JP . /cos.S sin.(ff — A) V cos.X sin.i) This is the final result, when the radius is unity ; and when the radius is greater by B, then the sin. JP will be greater by R ; and, therefore, the value of this sine, cor- responding to our tables, is, sin.JP = \J(J*\ (-JL>j cos.tfsin.^— A). v Vcos.iy/ Vsin.i)/ 376 SPHERICAL TRIGONOMETRY. PRACTICAL PROBLEMS. 1. In lat. 39° 6' 20" North, when the sun's declination was 12° 3' 10" North, the true altitude* of the sun's cen- ter was observed to be 30° 10' 40", rising. What was the apparent time ? ait 30° 10' 30" Lat. 39° 6' 20" RD. 77° 56' 50" 2 ) 147° 13' 40" S = 73° 36' 50" cos.com. .110146 sin.com. .009680 cos. 9.450416 tr- - A) = 43° 26' 20" 30° 22' 5" 2 sin. 9.837299 2 ) 19.407541 . sin. 9.703770 P = 60° 44' 10" This angle, converted into time at the rate of 15° to one hour, or 4 minutes to 1°, gives 4* 2 m 56' from appa- rent noon; and as the sun was rising, it was before noon or If to this the equation of time were applied, we should have the mean time ; and if such time were compared with that of a clock or watch, we could determine its error. A good observer, with a good instrument, can, in this manner, determine the local time within 4 or 5 seconds. 2. In lat. 40° 21' North, the true altitude of the sun, in the forenoon, was found to be 36° 12', when the declina- * The instrument used, the manner of taking the altitude, its cor- rection for refraction, semi-diameter, and other practical or circum- stantial details, do not belong to a work of this kind, but to a work on Practical Astronomy or Navigation. SECTION V. 377 tion of the sun was 3° 20' South. "What was the appa- rent time ? Ans. 9 A 43 m 44* a. m. 3. In latitude 21° 2' South, when the sun's declination was 18° 32' North, the true altitude, in the afternoon, was found to be 40° 8'. What was the apparent time of day? Ans. 2 A 2 m p. m. SPHERICAL TRIGONOMETRY APPLIED TO GEOGRAPHY. If we wish to find the shortest distance between two places over the surface of the earth, when the dis- tance is considerable, we must employ Spherical Trigo- nometry. Suppose the least distance between Rome and New Orleans is required ; we would first find the distance in degrees and parts of a degree, and then multiply that distance by the number of miles in one degree. In the solution of this problem, it is supposed that we have the latitude and longitude of both places. Then the distances, in degrees, from the north pole of the earth to Rome and to New Orleans are the two sides of a spherical triangle, the difference of longitude of the two places is the angle at the pole included between these sides, and the problem is, to determine the third side of a spherical triangle, when w r e have two sides and the included angle given. Let P be the north pole, B the position of Rome, and N that of New Orleans. Lat. Long. New Orleans, 29° 57' 30" N. 90° "W. Rome, 41° 53' 54" N. 12° 28' 40" E. Whence, PR m 48° 6' 6", pjjf = 60° 2' 30". Angle NPR = 102° 28' 40". 32* 378 SPHERICAL TRIGONOMETRY. We now employ Na- pier's 1st and 2d Analo- gies, and find the dis- tance, in degrees, to be 101° 31' 30". This re- duced to miles, at the rate of 69.16 miles to I \ the degree, will make \ / the distance 7021.469 \ j miles. \ / The angle at JV is 47° 49', and at B, 59° \^ ^"' 35' 40". ^ '"' The third side of a spherical triangle can be found by a single formula, as we shall see by inspecting formulae (£') Prop. 7, Sec. III. Let be the included angle, and c the unknown side opposite ; then, „ cos.c — cos.a cos.5 cos. C = ^ r— i sin.a sin. 6 Adding 1 to each member, and reducing, observing at the same time that 1 -j- cos. (7= 2cos. 2 J (7, we have, «... ~ sin.a sin.5 — cos.a cos.5 + cos.e 2COS. 2 \ C = : : siu.a sm.6 Whence, 2cos. 2 J(7 sin.a sin.6 = cos.c — cos.(a-f b); or, co8.c = cos. (a -f b) + 2cos. 2 \Q sin.a sin.6. The second member of this equation is the algebraic sum of two decimal fractions, and expresses the value of the natural cosine of the side sought. This case of Spherical Trigonometry, namely, that in which two sides and the included angle are given, to find the third side, is very extensively used in practical astronomy, in finding the angular distance of the moon from the sun, stars, and planets. For this purpose, the right ascension and declination of each body must be SECTION V. 3T9 found for the same moment of absolute time. Their difference in right ascen- sion gives the included angle, P, at the celestial pole. The declination subtracted from 90°, if it be north, and added to 90°, if it be south, will give the sides, PZ and PS. In the following exam- ples, we give the right ascension and declination of the bodies, and from these the student is required to compute the distance between them. The right ascensions are given in time. Their differ- ence must be changed to degrees for the included angle. June 24, 1860. MEAN TIME GREENWICH. moon's R. A. Dec. h. m. e. •:* " At noon, 10 51 36.5 3 33 24 N. « 3 h., 10 58 1 2 47 43 " 6 h., 11 4 24.6 1 59 56.2 " 9 h., 11 10 47.6 1 12 6 JUPITER'S R. A. m. s. 4 27.6 4 34.2 4 40.8 4 47.2 Dec. o t it 20 51 36.8 N. 20 51 17.8 20 50 58.7 20 50 39.6 Distance. o t tf 44 8 12 45 53 47 47 39 18 49 24 43 October 6, 1860. ) R.A. h. m. s. At noon, 5 41 21.8 " 3 h., 5 48 30.1 " 6 h., 5 55 40 6 2 50.5 6 10 1.2 " 9 h., " 12 h., Dec. O R.A. Oil! h. m. s. 26 8 ON. 12 49 27.4 26 3 20 12 49 54.8 25 57 19 12 50 22.2 25 49 58 12 50 49.6 25 41 15.8 12 51 11.9 Dec. Q I II 5 18 31 S. 5 20 13.7 5 21 56.4 5 23 38.1 5 25 20.8 Distance. o I If 107 37 2 106 8 19 104 39 19 103 10 101 40 23 380 SPHERICAL TRIGONOMETRY SECTION VI REGULAR POLYEDRONS. A Regular Polyedron is a polyedron having all its faces equal and regular polygons, and all its polyedral angles equal. The sum of all the plane angles bounding any polyedral angle is less than four right angles ; and as the angle of the equilateral tri- angle is | of a right angle, we have f x 3<4, | x 4<[4, and | x 5<4 ; but | x 6=4, | x 7]>4, and so on. Hence, it follows that three, and only three, polyedral angles may be formed, having the equi- lateral triangle for faces; namely, a triedral angle and polyedral angles of four and of five faces. There are, therefore, three distinct regular polyedrons bounded by the equilateral triangle. 1. The Tetraedron, having four faces and four solid angles. 2. The Octaedron, having eight faces and six solid angles. 3. The Icosaedron, having twenty faces and twenty solid angles. With right plane angles we can form only a triedral angle ; hence, with equal squares we may bound a solid having six faces and eight equal triedral angles. This solid is called the Hexaedron. The angle of the regular pentagon being f of a right angle, we have |x3<[4; but |x4>4; hence, with plane angles equal to those of the regular pentagon, we can form only a triedral angle. The solid bounded by twelve regular pentagons, and having twenty solid angles, is called the Dodecaedron. There are, then, but five regular polyedrons, viz. : The tetraedron, the octaedron, and the icosaedron, each of which has the equilateral triangle for faces ; the hexaedron, whose faces are equal squares, and the dodecaedron, whose faces are equal regular pentagons. It is obvious that a sphere may be circumscribed about, or in- scribed within, any of these regular solids, and conversely : and SECTION VI. 381 that these spheres will have a common center, which may also be taken as the center of the polyedron. Any regular polyedron may be regarded as made up of a number of regular pyramids, whose bases are severally the faces of the polyedron, and whose common vertex is its center. Each of these pyramids will have, for its altitude, the radius of the inscribed sphere; and since the volume of the pyramid is measured by one third of the product of its base and altitude, it follows that the volume of any regular polyedron is measured by its surface multi- plied by one third of the radius of the inscribed sphere. PROBLEM. Given, the name of a regular 'polyedron, and the side of the hound- ing polygon, to find the inclination of its faces; the radii of the in- scribed and circumscribed spheres ; the area of its surface ; and its volume. Let AB be the intersection of two adjacent faces of the polye- dron, and G and D the centers of these faces, being the center of the polyedron. Draw the radii, OG and OD, of the inscribed, and ]bhe radii OA and OB,of the circum- scribed sphere ; also from G and D let fall the perpendiculars CE and DE, on the edge AB, and draw OE; then will the angle DEG measure the inclination of the faces of the polyedron, and the angle DEO is one half of this inclination. Let i" denote the inclination of the faces, m the number of faces which meet to form a polyedral angle, n the number of sides in each face, and suppose the edge of the polyedron to be unity. The surface of the sphere of which is the center, and radius unity, will form, by its intersections with the planes, AOE, AOD, DOE, the right-angled spherical triangle dae, right-angled at e. In the right-angled triangle DEO, the angle DOE is equal to ^oil ua^j OF THF 382 SPHERICAL TRIGONOMETRY. 90°— -DEO = 90° — \I, and is measured by the are de. The angle due, of the spherical * • i • i * 360° v..V , , 360° triangle, is equal to , and the angle ade = — — . 2m 2n Now, by Napier's Rules we have cos.c?ae = sin. ade cos.de. -, cos.dae , -. x or, cos.de = _, ; ( 1 ) sin. ade and, cos.ae? = cot.dae cot.ade (2) Substituting in eq. ( 1 ), for the angles dae and ade, their values, we find cos.360 2m /o\ Sin -2 J = ;sof- [ ] 2n . Equation (3 ) gives the value of the sine of one half of the incli- nation of the planes ; and by means of this equation we may readily find the radii of the inscribed and circumscribed spheres. In the triangle BED, we have DE = BE cot.BDE = Jcot. „, 2n since AB = 1, and BE= \AB. In the triangle DOE, we have OD = DE tan. \1 = Jcot. _ tan. J 7 (4) 2n From the triangle A OD, we find cos.DOA : 1 :: OD : OA whence OA = cos.DOA But the angle DO A is measured by the arc ad) hence, substi- tuting in this last equation the values of cos.DOA and OD, taken from eqs. (2) and (4), we have O4=itan.l/cot.???! X l X —i— 2 2n cot360° cot. 360° "ST 2w == 4 tan. J J tan. I!*!?!, (5) 2m by writing tan. for — , and reducing, cot. SECTION IV. 383 Equation ( 4 ) gives the value of OD, the radius of the inscribed sphere, and equation (5) gives that of OA, the radius of the cir- cumscribed sphere. The area of one of the faces of the polyedron is equal to one half of the apothegm multiplied by the perimeter. The apothegm, as found above, is equal to £ cot. ; hence, we 2n 360° have JjixI cot , for the area of one of the faces; and multi- 2n plying this by the number of faces of the polyedron, we will have the expression for its entire area. The expression for the surface multiplied by one third of the radius of the inscribed sphere, gives the measure of the volume of the polyedron. In what precedes, we have supposed the edge of the polyedron to be unity. Having found the radii of the inscribed and circum- scribed spheres, the surfaces, and the volumes of such polyedrons, to determine the radii, surfaces, and volumes of regular polyedrons having any edge whatever, we have merely to remember that the homologous dimensions of similar bodies are proportional; their surfaces are as the squares of these dimensions ; and their volumes as the cubes of the same. Formula (3) gives, for the inclination of the adjacent faces of The Tetraedron, 70° 31' 42" " Hexaedron, 90° 00' 00" " Octaedron, 109° 28 / 18" " Dodecaedron, 116° 33' 54" " Icosaedron, 138° 11' 23" The subjoined table gives the surfaces and volumes of the regular polyedrons, when the edge is unity. Surfaces. Volumes. Tetraedron, 1.7320508 0.1178513 Hexaedron, 6.0000000 1.0000000 Octaedron, 3.4641016 0.4714045 Dodecaedron, 20.6457288 7.6631189 Icosaedron, 8.6602540 2.1816950 LOGARITHMIC TABLES; ALSO A TABLE OF NATURAL AND LOGARITHMIC SINES, COSINES, AND TANGENTS, TO EVERY MINUTE OF TEE QUADRANT. LOGARITHMS OF NUMBERS FROM 1 to lOOOO, N. Log. N. Log. N. Log. N. Log. 1 000000 26 1 414973 51 1 707570 76 1 880814 2 301030 27 1 431364 52 1 716003 77 1 888491 3 477121 28 1 447158 53 1 724276 78 1 892095 4 602030 29 1 462398 54 1 732394 79 1 897627 5 698970 30 1 477121 55 1 740363 80 1 903090 6 778151 31 1 491362 56 . 1 748188 81 1 908485 7 845098 32 1 505150 57 1 755875 82 1 913814 8 903090 33 1 518514 58 1 763428 83 1 919078 9 954243 34 1 531479 59 1 770852 84 1 924279 10 1 000000 35 1 544068 60 1 778151 85 1 929419 11 1 041393 36 1 556303 61 1 785330 86 1 934498 12 1 079181 37 1 568202 62 1 792392 87 1 939519 13 1 113943 38 1 579784 63 1 799341 88 1 944483 14 1 146128 39 1 591035 64 1 803180 89 1 949390 15 1 176091 40 1 602030 65 1 812913 90 1 954243 16 1 204120 41 1 612784 66 1 819544 91 1 959041 17 1 230449 42 1 623249 67 1 826075 92 1 963788 18 1 255273 43 1 633468 68 1 832509 93 1 968483 19 1 278754 44 1 643453 69 1 838849 94 1 9/3128 20 1 301030 45 1 653213 70 1 845098 95 1 977724 21 1 322219 46 1 662578 71 1 851258 96 1 982271 22 1 342423 47 1 672098 72 1 857333 97 1 986772 23 1 361728 48 1 681241 73 1 863323 98 1 991226 24 1 380211 49 1 690196 74 1 869232 99 1 995635 25 1 397940 50 1 698970 75 1 875081 100 2 000000 Si )TE. In the following table, in the last ni he columns of each p age, where the irst or leading figures change from 9's » to 0's, points or do ts are now intri jduced instead of the 0's through the r sst of the line, to cat< ih the eye, and to indicate that from thence the corr ssponding natural r mmber in the irst column stands in the next lower line, and its annexe i first two figu res of the Logarithms in the second co lumn. N. ; LOGARITHMS OF NUMBERS. 3 1 0434 0S68 3 4 5 6 7 8 9 100 000000 1301 1734 2166 2598 3029 3461 3891 101 4321 4750 5181 5609 6038 6468 6894 7321 7748 8174 102 8600 9026 9451 9876 .300 .724 1147 1570 1993 2415 j 103 012837 3259 3680 4100 4521 4940 5360 5779 6197 6616 104 7033 7451 7888 8284 8700 9116 9532 9947 .361 .775 105 021189 1603 2016 2428 2841 3252 3664 4075 4486 4896 103 530S 5715 6125 6533 6942 7350 7757 8164 8571 8978 107 9384 9789 .195 .600 1004 1408 1812 2216 £619 3021 108 033424 3826 4227 4628 5029 5430 5830 6230 6629 7028 109 7426 7825 8223 8620 9017 9414 9811 .207 .602 .998 110 041393 1787 2182 2576 2969 3362 3755 4148 4540 4932 111 5323 5714 6105 6495 6885 7275 7664 8053 8442 8830 112 9218 9608 9993 .380 .766 1153 1538 1924 2309 2694 113 053078 3483 3846 4230 4813 4996 5378 5760 6142 6524 114 6905 7286 7666 8046 8426 8805 9185 9563 9942 .320 115 030398 1075 1452 1829 2206 2582 2958 3333 3709 4083 116 4458 4832 5203 5580 5953 6326 6699 7071 7443 7815 117 8186 8557 8928 9298 9668 ..38 .407 .776 1145 1514 118 071882 2250 2617 2985 3352 3718 4085 4451 4816 5182 119 5547 5912 6276 6640 7004 7368 7731 8094 8457 8819 120 9181 9543 9904 .266 .626 .987 1347 1707 2067 2426 121 082785 3144 3503 3861 4219 4576 4934 5291 5647 6004 122 6360 6716 7071 7426 7781 8136 8490 8845 9198 9552 123 9905 .258 .611 .963 1315 1667 2018 2370 2721 3071 124 093422 3772 4122 4471 4820 5169 5518 5866 6215 6562 125 6910 7257 7604 7951 8298 8644 8990 9335 9681 1026 126 100371 0715 1059 1403 1747 2091 2434 2777 3119 3462 127 3804 4146 4487 4828 5169 5510 5851 6191 6531 6871 128 7210 7549 7S88 8227 8565 8903 9241 9579 9916 .253 129 110590 0926 1263 1599 1934 2270 2605 2940 3275 3609 130 3943 4277 4611 4944 5278 5611 5943 6276 6608 6940 131 7271 7603 7934 8265 8595 8926 9256 9586 9915 0245 132 120574 0903 1231 1560 1888 2216 2544 2871 3198 3525 133 3852 4178 4504 4830 5158 5481 5806 6131 6456 6781 134 7105 7429 7753 8076 8399 8722 9045 9368 9690 ..12 135 130334 0655 0977 1298 1619 1939 2260 2580 2900 3219 136 3539 3858 4177 4496 4814 5133 5451 5769 6086 6403 137 6721 7037 7354 7671 7987 8303 8618 8934 9249 9564 138 9879 .194 .508 .822 1136 1450 1763 2076 2389 2702 1 139 143015 3327 3630 3951 4263 4574 4885 5196 5507 5818 140 6128 6438 6748 7058 7367 7676 7985 8294 8603 8911 141 9219 9527 9835 .142 .449 .756 1063 1370 1676 1982 142 152288 2594 2900 £205 3510 3815 4120 4424 4728 5032 143 5336 5640 5943 6246 6549 6852 7154 7457 7759 8061 144 8362 8664 8965 9266 9567 9868 .168 .469 .769 1068 145 161368 1667 1967 2266 2564 2863 3161 3460 3758 4055 146 4353 4650 4947 5244 5541 5838 6134 6430 6726 7022 147 7317 7613 7908 8203 8497 8792 9086 9380 9674 9968 148 170262 0555 0848 1141 1434 1726 2019 2311 2603 2895 149 3186 3478 3769 4060 4351 4641 4932 5222 5512 5802 1* 4 LOGARITHMS N. 1 2 3 4 5 6 7 8 9 150 176091 6381 6670 6959 7248 7536 7825 8113 8401 8689 151 8977 9264 9552 9839 .126 .413 .699 .985 1272 1558 152 181844 2129 2415 2700 2985 3270 3555 3839 4123 4407 153 4691 4975 5259 5542 5825 6108 6391 6674 6956 7239 154 7521 7803 8084 8366 8647 281 1451 8928 9209 9490 9771 ..51 155 190332 0812 0892 1171 1730 2010 2289 2567 2846 156 3125 3403 3681 3959 4237 4514 4792 5069 5346 5623 157 5899 6176 6453 6729 7005 7281 7556 7832 8107 8382 158 8657 8932 9206 9481 9755 ..29 .303 .577 .850 1124 159 201397 1670 1943 2216 2488 273 2761 3033 3305 3577 3848 160 4120 4391 4663 4934 5204 5475 5746 6016 6286 6556 161 6826 7096 7365 7634 7904 8173 8441 8710 8979 9247 162 9515 9783 ..51 .319 .586 .853 1121 1388 1654 1921 163 212188 2454 2720 2986 3252 3518 3783 4049 4314 4579 164 4844 5109 5373 5638 5902 264 6166 6430 6694 6957 7221 165 7484 7747 8010 8273 8536 8798 9060 9323 9585 9846 166 220108 0370 0831 0892 1153 1414 1675 1936 2196 2456 167 2716 2976 3236 3496 3755 4015 4274 4533 4792 5051 168 5309 5568 5fe26 6084 6342 6600 6858 7115 7372 7630 169 7887 8144 8400 8657 8913 257 9170 9426 9682 9938 .193 170 230449 0704 0960 1215 1470 1724 1979 2234 2488 2742 171 2996 3250 3504 3757 4011 4264 4517 4770 5023 5276 172 5528 5781 6033 6285 6537 6789 7041 7292 7544 7795 173 8046 8297 8548 8799 9049 9299 9550 9800 ..50 .300 174 240549 0799 1048 1297 1546 249 1795 2044 2293 2541 2790 175 3038 3285 3534 3782 4030 4277 4525 4772 5019 5266 176 5513 575!) 6006 6252 6499 6745 6991 7237 7482 7728 177 7973 8219 8464 8709 8954 9198 9443 9687 9932 .176 178 250420 0664 0908 1151 1395 1638 1881 2125 2368 2610 179 2853 3096 3338 3580 3822 242 6237 4064 4306 4548 4790 5031 180 5273 5514 5755 5996 6477 6718 6958 7198 7439 181 7679 7918 8158 8398 8637 8877 9116 9355 9594 9833 182 260071 0310 0548 0787 1025 1263 1501 1739 1976 2214 183 2451 2688 2925 3162 3399 3636 3873 4109 4346 4582 184 4818 5054 5290 5525 5761 235 8110 5996 6232 6467 6702 6937 185 7172 7406 7641 7875 8344 8578 8812 9046 9279 186 9513 9746 9980 .213 .446 .679 .912 1144 1377 1609 18/ 271842 2074 2306 2538 2770 3001 3233 3464 3696 3927 188 4158 4389 4620 4850 5081 5311 5542 5772 6002 6232 189 6462 6692 6921 7151 7380 229 7609 7838 8067 8296 8525 190 8754 8982 9211 9439 9667 9895 .123 .351 .578 .806 191 281033 1261 1488 1715 1942 2169 2396 2622 2849 3075 192 3301 3527 3753 3979 4205 4431 4656 4882 5107 5332 193 5557 5782 6007 6232 6456 6681 6905 7130 7354 7578 194 7802 8026 8249 8473 8696 224 8920 9143 9366 9589 9812 195 290035 0257 0480 0702 0925 1147 1369 1591 1813 2034 196 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 197 4466 4687 4907 5127 5347 5567 5787 6007 6226 6446 198 6665 6884 7104 7323 7542 7761 7979 8198 8416 8635 199 8853 9071 9289 9507 9725 9943 .161 .378 .595 .813 I OF NUMBERS. 5 N. 1 2 3 4 5 6 7 8 9 200 301030 1247 1464 1681 1898 2114 2331 2547 2764 2980 201 3196 3412 3628 3844 4059 4275 4491 4706 4921 5136 202 5351 5566 5781 5996 6211 6425 6639 6854 7038 7282 203 7496 7710 7924 8137 8351 8564 8778 8991 9204 9417 204 9630 9843 ..56 .268 .481 212 2600 .693 .906 1118 1330 1542 205 311754 1966 2177 2389 2812 3023 3234 3445 3656 206 3867 4078 4289 4499 4710 4920 5130 5340 5551 5760 207 5970 6180 6390 6599 6809 7018 7227 7436 7646 7854 208 8063 8272 8481 8689 8898 9106 9314 9522 9730 9938 209 320146 0354 0562 0769 0977 207 1184 1391 1598 1805 2012 210 2219 2426 2633 2839 3046 3252 3458 3665 3871 4077 211 4282 4488 4694 4899 5105 5310 5516 5721 5926 6131 212 6336 6541 6745 6950 7155 7359 7563 7767 7972 8176 213 8380 8583 8787 8991 9194 9398 9301 9805 ...8 .211 214 330414 0617 0819 1022 1225 202 1427 1630 1832 2034 2236 215 2438 2640 2842 3044 3246 3447 3649 3850 4051 4253 216 4454 4655 4856 5057 5257 5458 5658 5859 6059 6260 217 6460 6660 6860 7060 7260 7459 7659 7858 8058 8257 218 8450 8656 8855 9054 9253 9451 9650 9849 . .47 .246 219 340444 0642 0841 1039 1237 198 1435 1632 1830 2028 2225 220 2423 2620 2817 3014 3212 3409 3306 3802 3999 4196 221 4392 4589 4785 4981 5178 5374 5570 5766 5932 6157 222 6353 6549 6744 6939 7135 7330 7525 7720 7915 8110 223 8305 8500 8694 8889 9083 9278 9472 9666 9860 . .54 224 350248 0442 0636 0829 1023 193 1216 1410 1603 1796 1989 225 2183 2375 2568 2761 2954 3147 3339 3532 3724 3916 226 4108 4301 4493 4685 4876 5038 5260 5452 5643 5834 227 6026 6217 6408 6599 6790 6981 7172 7363 7554 7744 228 7935 8125 8316 8506 8696 8886 9076 9266 9456 9646 229 9835 ..25 .215 .404 .593 190 .783 .972 1161 1350 1539 230 361728^ 1917 2105 2294 2482 2671 2859 3048 3236 T424 231 3612 3800 3988 4176 4363 4551 4739 4926 5113 5301 "232 5488 5675 5862 6049 6236 6423 6610 6796 (983 i 7169 233 7356 7542 7729 7915 8101 8287 8473 8659 8845 9030 234 9216 9401 9587 9772 9958 185 .143 .328 .513 .698 .883 235 371068 1253 1437 1622 1806 1991 2175 2330 2544 2128 236 2912 3096 3280 3464 3647 3831 4015 4198 4382 ! 4566 237 4748 4932 5115 5298 5481 5664 5846 6029 6212 ! 6r94 238 6577 6759 6942 7124 7306 7488 7670 7852 8034 ! 8216 239 8398 8580 8761 8943 9124 182 9306 9487 9668 9849 J . .30 240 380211 0392 0573 0754 0934 1115 1296 1476 1656 1837 241 2017 2197 2377 2557 2737 2917 3097 3277 3456 3636 242 3815 3995 4174 4353 4533 4712 4891 5070 5249 5428 243 5606 5785 5964 6142 6321 6499 6677 6856 7034 7212 244 7390 7568 7746 7923 8101 178 8279 8456 8634 8811 ; 8989 245 9166 9343 9520 9698 9875 ..51 .228 .405 .582 ! .759 246 390935 1112 1288 1464 1641 1817 1993 2169 2345 2521 247 2697 2873 3048 3224 3400 3575 3751 3926 4101 4277 248 4452 4627 4802 4977 5152 5326 5501 5676 5850 6025 249 6199 6374 6548 6722 6896 7071 7245 7419 7592 , 7766 6 LOGARITHMS N. 1 2 8287 3 4 5 6 7 8 9 250 397940 8114 8461 8634 8808 8981 9154 9328 9501 251 9874 9847 ..20 .192 .365 .538 .711 .883 1056 1228 252 401401 1573 1745 1917 2089 2261 2433 2605 2777 2949 253 3121 3292 3464 3635 3807 3978 4149 4320 4492 4683 254 4834 5005 5176 5346 5517 171 6688 5858 6029 6199 6370 255 6540 6710 6881 7051 7221 7391 7561 7731 7901 8070 256 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 257 9933 .102 .271 .440 .609 .777 .946 1114 1283 1451 258 411620 1788 1958 2124 •2293 2461 2629 2796 2964 3132 259 3300 3467 3635 3803 3970 4137 4305 4472 4639 4806 260 4973 5140 5307 5474 5641 6808 5974 6141 6308 6474 261 6641 6807 6973 7139 1303 7472 7638 7804 7970 8135 262 8301 8467 8633 8798 8964 9129 9295 9460 9625 9791 263 9956 .121 .286 .451 .616 .781 .945 1110 1275 1439 264 421604 1788 1933 ■2097 2261 2426 2590 2754 2918 3082 265 3246 3410 3574 3737 3901 4085 4228 4392 4555 4718 266 4882 5045 6208 1371 £634 5697 5860 6023 6186 6349 267 6511 6874 6836 6999 7161 7324 7486 7648 7811 7973 268 8135 8297 8459 8621 8783 8944 9108 9268 9429 9591 269 9752 9914 ..75 .236 .398 .559 .720 .881 1042 1203 270 431364 1525 1685 1846 2007 2167 2328 2488 2649 2809 271 2969 3130 3290 3450 3610 3770 3930 4090 4249 4409 272 4589 4729 4888 5048 5207 6357 5526 5885 5844 6004 273 6163 6322 6481 6640 6800 6957 7116 7275 7433 7592 274 7751 7909 8087 8226 8384 158 8542 8701 8859 9017 9175 275 9333 9491 9648 9805 9984 .122 .279 .437 .594 .752 276 440909 1086 1224 1381 1538 1695 1852 2009 2166 2323 277 2480 2637 2793 2950 3103 3263 3419 3576 3732 3889 278 4045 4201 -357 4513 4669 4825 4981 5137 5293 5449 279 5604 5760 5915 6071 6226 6382 6537 6692 6848 7003 280 7158 7313 7468 7623 7778 7933 8088 8242 8397 8552 281 8703 8861 9015 9170 9324 9478 9633 9787 9941 . .95 282 450249 0403 0557 0711 0865 1018 1172 1326 1479 1633 283 1786 1940 2093 2247 2400 2553 2706 2859 3012 3165 284 3318 3471 3624 3777 3930 4082 4235 4387 4540 4892 2 : 5 4845 4997 5150 5302 5454 5603 5758 5910 6082 6214 286 6366 6518 6670 6821 6973 7125 ,276 7428 7579 7731 9242 287 7882 8033 8184 8338 8487 8638 8789 8940 9091 288 9392 9543 9094 9845 9995 .146 .296 .447 .597 .748 2248 289 460898 1048 1198 1348 1499 1649 1799 1948 2098 290 2398 2548 2697 2847 2997 3146 3298 3445 3594 3744 291 3893 4042 4191 4340 449C 4839 4788 4936 5085 5234 292 5383 5532 5880 5829 5977 6126 6274 6423 6571 6719 293 6868 7018 7164 7312 7460 7608 7756 7904 8052 8200 294 8347 8495 8843 8790 8938 147 9085 9233 9380 9527 9676 295 9822 9989 .116 .263 .410 .557 .704 .851 .998 1145 298 471292 1438 1585 1732 1878 2025 2171 2318 2464 2810 297 2758 L>903 3049 3195 3341 3487 3633 3779 £925 40?1 298 4216 4362 4508 4653 4799 4944 5090 5235 5381 5526 299 5871 6816 5962 610/ 6252 6397 6542 6687 6832 6976 OF NUMBERS. 7 N. 1 2 3 4 5 6 7 8 9 300 477121 7266 7411 7555 7700 7844 7989 8133 8278 8422 301 8566 8711 8855 8999 9143 9287 9481 9575 9719 9863 302 480007 0151 0294 0438 0582 0725 0869 1012 1156 1299 303 1443 1586 1729 1872 2016 2159 2302 2445 2588 2731 304 2874 3016 3159 3302 3445 142 3587 3730 3872 4015 4157 305 4300 4442 4585 4727 4869 5011 5153 5295 5437 5579 308 5721 5863 6005 6l47 6289 6430 6572 6714 6855 6997 307 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 308 8551 8692 8833 8974 9114 9255 9396 9537 9667 9818 309 9959 ..99 .239 .380 .620 .661 .801 .941 1081 1222 310 491362 1502 1642 1782 1922 2062 2201 2341 2481 2621 311 2760 2900 3040 3179 3319 3458 3597 3737 3876 4015 312 4155 4294 4433 4572 4711 4850 4989 6128 5267 5406 313 5544 5683 5822 5960 6099 6238 6376 6515 6653 6791 314 6930 7068 7206 7344 7483 7621 7759 7897 8035 8173 315 8311 8448 8586 8724 8862 8999 9137 9275 9412 8550 316 9687 9824 9962 ..99 .236 .374 .611 .648 .786 .922 317 501059 1196 1333 1470 1607 1744 1880 2017 2154 2291 • 3i8 2427 2564 2700 2837 2973 3109 3246 3382 3518 3655 319 3791 3927 4083 4199 4335 4471 4607 4743 1878 5014 320 5150 5283 5421 5557 5693 5828 5964 6099 6234 6370 321 6505 6640 6776 6911 7046 7181 7316 7451 7588 7721 322 7856 7991 8123 8260 8395 8530 8664 8799 8934 9008 323 9203 9337 9471 9606 9740 9874 ...9 .143 .2/7 .411 324 510545 0679 0813 0947 1081 134 1215 1349 1482 1616 1750 325 1883 2017 2151 2284 2418 2551 2684 2818 2951 3084 326 3218 3351 3484 3617 3750 3883 4018 4149 4282 4414 327 4548 4681 4813 4946 5079 5211 5344 5476 5609 5741 328 5874 6008 6139 6271 6403 t535 6668 6800 6932 7084 329 7196 7328 7460 7592 7724 7855 7987 8119 8251 8382 330 8514 8646 8777 8909 9040 9171 9303 9434 9566 9697 331 9828 9959 ..90 .221 .353 .484 .615 .745 .876 1007 .332 521138 1289 1400 1531) 1661 1792 1922 2053 2183 2314 333 2444 2575 2705 2836 2966 3096 3226 3356 3486 3616 334 3746 3876 4006 4136 4266 4396 4526 4656 4785 4915 335 5045 5174 5304 5434 5563 5693 5822 5951 6081 6210 336 6339 6469 6598 6727 6856 6985 7114 7243 7372 7501 337 7630 7759 7888 8016 8145 8274 8402 8531 8660 8788 338 8917 9045 9174 9302 9430 9559 9687 9815 9943 ..72 339 530200 0328 0456 0584 0712 0840 0968 1096 1223 lc51 340 1479 1607 1734 1862 1980 2117 2245 2372 2500 2627 341 2754 2882 3009 3138 3264 3391 3518 3645 3772 3899 342 4026 4153 4280 4407 4534 4661 4787 4914 5041 5167 343 5294 6421 5547 5874 5800 5927 6053 6180 6306 6432 344 6558 6685 6811 6937 7080 129 7189 7315 7441 7567 7693 345 7819 7945 8071 8197 8322 8448 8574 8699 8825 8951 346 9076 9202 9327 9452 9578 9703 9829 9954 ..79 .204 347 540329 0455 0580 0705 0830 0955 1080 1205 1330 1454 348 1579 1704 1829 1953 2078 2203 2327 2452 2576 2701 I 349 2825 2950 3074 3199 3323 3447 3571 3696 3820 3944 ! 8 LOGARITHMS N. 350 1 2 3 4 5 6 7 8 9 544068 4192 4316 4440 4564 4688 4812 4936 5060 5183 351 6307 5431 5555 6578 5805 5925 6049 6172 6296 6419 352 6543 6666 6789 6913 7036 7159 7282 7405 7529 7652 353 7776 7898 8021 8144 8267 8389 8512 8635 8758 8881 354 9003 9126 9249 9371 9494 122 0717 9616 9739 9861 9984 .196 355 550228 0351 0473 0595 0840 0962 1084 1206 1328 356 1450 1572 1694 1816 1938 2060 2181 2303 2425 2547 357 2668 2790 2911 3033 3155 3276 3393 3519 3640 3762 358 3883 4004 4126 4247 4368 4489 4610 4731 4852 4973 359 6094 5215 5346 5457 5578 5699 5820 5940 6061 6182 360 6303 6423 6544 6664 6785 6905 7026 7146 7267 7387 361 7507 7627 7748 7868 7988 8108 8228 8349 8469 8589 362 8709 8829 8948 9068 9188 9308 9428 9548 9667 9787 363 9907 . .26 .146 .265 .385 .504 .624 .743 .863 .982 364 561101 1^21 1340 1459 1578 1698 1817 1936 2056 2173 365 2293 2412 2531 2650 2769 2887 300S 3125 3244 3362 366 3481 3600 3718 3837 3955 4074 4192 4311 4429 4548 367 4666 4784 4903 5021 5139 5257 5376 5494 6612 5730 368 6848 5966 6084 6202 6320 6437 6555 6673 6791 6909 369 7026 7144 7262 7379 7497 7614 7732 7849 7967 8084 370 8202 8319 8436 8554 8671 8788 8905 9023 9140 9257 371 9374 9491 9608 9725 9882 9959 ..76 .193 .309 .426 372 570543 0660 0776 0893 1010 1126 1243 1359 1476 1592 373 1709 1825 1942 2058 2174 2291 2407 2522 2639 2755 374 2872 2988 3104 3220 3336 116 3452 3568 36o4 3800 £915 375 4031 4147 4263 4379 4494 4610 4726 4341 4957 5072 376 5188 5303 5419 5534 5650 5765 5880 6996 6111 6226 377 6341 6457 6572 6687 6802 6917 7032 7147 7262 7377 378 7492 7607 7722 7836 7951 8066 8181 8295 8410 8525 379 8639 8754 8868 8983 9097 9212 9326 9441 9555 9669 380 9784 9898 ..12 .126 .241 .355 .469 .683 .697 .811 381 580925 1039 1153 1267 1381 1495 1608 1722 1836 1950 382 2063 2177 2291 2404 2518 2631 2745 J858 2972 3085 383 3199 3312 3426 3539 3652 3765 3879 J992 4105 4218 384 4331 4444 4557 4670 4783 4896 5009 <)122 5235 5348 385 5461 5574 5686 5799 5912 6024 6137 0250 6362 6475 386 6587 6700 6812 6925 7037 7149 7262 7374 7486 7599 387 7711 7823 7935 8047 8160 8272 8384 8496 8608 8720 388 8832 8944 9056 9167 9279 9391 9603 9615 9726 9834 389 9950 ..61 .173 .284 .396 .507 .619 .730 .842 .953 390 591065 1176 1287 1399 1510 1621 1732 1843 1955 2066 391 2177 2288 2399 2510 2621 2732 2843 2954 3064 317k 392 3286 3397 3508 3618 3729 3840 3950 4061 4171 4282 393 4393 4503 4614 4724 4834 4945 5055 5165 5276 6386 394 ■ 5496 5608 5717 5827 5937 110 6047 6157 6267 6377 6487 395 6597 6707 6817 6927 7037 7146 7256 7366 7476 7586 396 7695 7805 7914 8024 8134 8243 8353 8462 8572 8681 397 8791 8900 9009 9119 9228 £337 9446 . 536 9666 f 774 398 9883 9992 .101 .210 .319 .428 .537 .646 755 .864 399 600973 1082 1191 1299 1408 1517 1625 1734 1843 1951 F NUMBERS. 9 N. 1 2 3 4 2494 5 6 7 8 9 400 6020S0 2169 2277 2386 2603 2711 2819 2928 3036 401 3144 3253 3361 3469 3573 3686 3794 3902 4010 4118 402 4226 4334 4442 4550 4658 4766 4874 4982 5089 5197 403 5305 5413 5521 5628 5736 5844 5951 6059 6166 6274 404 6381 6489 6596 6704 6811 108 6919 7026 7133 7241 7348 405 7455 7562 7669 7777 7884 7991 8098 8205 8312 8419 406 8526 8633 8740 8847 8954 9061 9167 9274 9381 9488 40? 9594 9701 9808 9914 ..21 .128 .234 .341 .447 .554 408 610860 0767 0873 0979 1086 1192 1298 1405 1511 1617 409 1723 1829 1936 2042 2148 2254 2360 2466 2672 2678 410 2784 2890 2996 3102 3207 3313 3419 3525 3630 3736 411 3842 3947 4053 4159 4264 4370 4475 4581 4686 4792 412 4897 5003 5108 5213 5319 5424 5529 5634 5740 5845 413 5950 6055 6160 6265 6370 6476 6581 6686 6790 6895 414 7000 7105 7210 7315 7420 7525 7629 7734 7839 7943 415 8048 8153 8257 8362 8466 8571 8676 8780 8884 8989 416 9293 9198 9302 9406 9511 9615 9719 9824 9928 ..32 417 620136 0140 0344 0448 0552 0656 0760 0864 0968 1072 418 1176 1280 1384 1488 1592 1695 1799 1903 2007 2110 419 2214 2318 2421 2525 2628 2732 2835 2939 3042 3146 420 3249 3353 3456 3559 3663 3766 3869 3973 4076 4179 421 4282 4385 4488 4591 4695 4798 4901 5004 5107 5210 422 5312 5415 5518 5621 5724 5827 5929 6032 6135 6238 423 6340 6443 6546 6648 6751 6853 6956 7058 7161 7263 424 7366 7468 7571 7673 7775 103 7878 7980 8082 8185 8287 ; 425 8389 8491 8593 8695 8797 8900 9002 9104 9206 9308 426 9410 9512 9613 9715 9817 9919 ..21 .123 .224 .326 427 630428 0530 0631 0733 0835 0936 1038 1139 1241 1342 428 1444 1545 1647 1748 1849 1951 2052 2153 2255 2356 429 2457 2559 2660 2761 2862 2963 3064 3165 3266 3367 430 3468 3569 3670 3771 3872 3973 4074 4175 4276 4376 431 4477 4578 4679 4779 4880 4981 5081 5182 5283 5383 432 6484 5584 5685 5785 5886 5986 6087 6187 6287 6388 433 6488 6588 6688 6789 6889 6989 7089 7189 7290 7390 434 7490 7590 7690 7790 7890 7990 8090 8190 8290 8389 435 8489 8589 8689 8789 8888 8988 9088 9188 9287 9387 436 9486 9586 9686 9785 9885 9984 ..84 .183 .283 .382 437 640481 0581 0680 0779 0879 0978 1077 1177 1276 1375 438 1474 1573 1672 1771 1871 1970 2069 2168 2267 2366 439 2465 2563 2662 2761 2860 2959 3058 3156 3255 3354 440 3453 3551 3650 3749 3847 3946 4044 4143 4242 4340 441 4439 4537 4636 4734 4832 4931 5029 5127 5226 5324 442 5422 5521 5619 5717 5815 5913 6011 6110 6208 6306 443 6404 6502 6600 6698 6796 6894 6992 70S9 7187 •7285 444 7383 7481 7579 7676 7774 93 8750 7872 7969 8067 8165 8262 445 8360 8458 8555 8653 8848 8945 9043 9140 9237 446 9335 9432 9530 9627 9724 9821 9919 ..16 .113 .210 447 650308 0405 0502 0599 0696 0793 0890 0987 1084 1181 448 1278 1375 1472 1569 1666 1762 1859 1956 2053 2150 449 2246 2343 2440 2530 2633 2730 2826 2923 3019 3116 10 LOGARITHMS N. 653213 1 2 3 4 5 6 7 8 9 450 3309 3405 3502 3598 3695 3791 3888 3984 4080 451 4177 4273 4369 4465 4562 4658 4754 4850 4946 5042 452 5138 5235 5331 5427 5526 5619 5715 5810 5906 6002 453 6098 6194 6290 6386 6482 6577 6673 6769 6864 6960 454 7056 7152 7247 7343 7438 96 8393 7534 7629 7725 7820 7916 455 8011 8107 8202 8298 8488 8584 8679 8774 8870 456 8965 9U60 9155 9250 9346 9441 9536 9631 9726 9821 457 9916 . .11 .106 .201 .296 .391 .486 .581 .676 .771 458 660865 0960 1055 1150 1245 1339 1434 1529 1623 1718 459 1813 190/ 2002 2096 2191 2286 2380 2475 2569 2663 460 2758 2852 2947 3041 3135 3230 3324 3418 3512 3607 461 3701 3795 3889 3983 4078 4172 4266 4360 4454 4548 462 4642 4736 48-J0 4924 5018 5112 5206 5299 5393 5487 463 5581 56/5 5769 5862 5956 6050 6143 6237 6331 6424 464 6518 6612 6705 6799 6892 6986 7079 7173 7266 7360 465 7453 7546 7640 7733 7826 7920 8013 8106 8199 8293 466 8386 8479 8572 8665 8759 8852 8945 9038 9131 9324 467 9-317 9410 9503 9596 9689 9782 9875 9967 .60 .153 468 670241 0339 0431 0524 0617 0710 0802 0895 0988 1080 469 1173 1265 1358 1451 1543 1636 1728 1821 1913 2005 470 2098 2190 2283 2375 2467 2560 2652 2744 2836 2929 471 3021 3113 3205 3297 3390 3482 3574 3666 3758 3850 472 3942 4034 4126 4218 4310 4402 4494 4586 4677 4769 473 4861 4953 5045 5137 5228 6320 5412 6503 5595 5687 474 5778 5870 5962 6053 6145 91 7059 6236 6328 6419 6511 6602 475 6694 6785 6876 6968 7151 7242 7333 7424 7516 476 7607 7698 7789 7881 7972 8063 8154 8245 8336 8427 477 8518 8609 8700 8791 8882 8972 9064 9155 9246 9337 478 9428 9519 9610 9700 9791 9882 9973 ..63 .154 .245 479 680336 0426 0517 0607 0698 0789 0879 0970 1060 1151 480 1241 1332 1422 1513 1603 1693 1784 1874 1964 2055 481 2145 2235 2326 2416 2508 2596 2686 2777 2867 2957 482 3047 3137 3227 3317 3407 3497 3587 3677 3767 3857 483 3947 4037 4127 4217 4307 4396 4486 4576 4666 4756 484 4854 4935 5025 5114 5204 5294 5383 5473 5563 5652 485 5742 5831 5921 6010 6100 6189 6279 6368 6458 6547 486 6636 6726 6815 6904 6994 7083 7172 7261 7351 7440 487 7529 7618 7707 7796 7886 7975 8064 8153 8242 8331 488 8420 8509 8593 8687 8776 8865 8953 9042 9131 9220 489 9309 9398 9486 9575 9664 9753 9841 9930 ..19 .107 490 690196 0285 0373 0362 0550 0639 0728 0816 0905 0993 491 1081 1170 1258 1347 1435 1524 1612 1700 1789 1877 492 1:65 2053 2142 2230 2318 2406 2494 2583 2671 2759 493 2817 2935 3023 3111 3199 3287 3375 3463 3551 3639 494 3727 3815 3903 3991 4078 88 4956 4166 4254 4342 4430 4517 495 4605 4693 4781 4868 5044 5131 5210 530? 5394 496 5482 5569 5657 5744 5832 5919 6007 6094 6182 6269 497 6356 5444 6531 6618 6708 6793 6880 6968 7055 7142 498 7229 7317 7404 7491 7578 7665 7752 7839 7926 8014 499 8101 8188 8275 8362 8449 8535 8622 8709 8796 8883 OF NUMBERS. 11 N. 1 2 3 4 5 6 7 8 9 500 698970 9057 9144 9231 9317 9404 9491 9578 9664 9751 501 9338 9924 . .11 ..98 .184 .271 .358 .444 .531 .617 502 700704 0790 0877 0963 1050 1136 1222 1309 1395 1482 503 1568 1654 1741 lfc27 1913 1999 2086 2172 2258 '2344 504 2431 2517 2603 2(89 2775 86 3635 2861 2947 3033 3119 3205 505 3291 3377 3463 3549 3721 3807 3895 3979 4085 503 4151 4236 4322 4408 4494 4579 4665 4751 4837 4922 507 5003 5094 5179 5265 5350 5436 5522 5607 5693 5778 503 5864 5949 61)35 6120 6206 6291 6376 6462 6547 6632 509 6718 6803 6888 6974 7059 7144 7229 7315 7400 7485 510 7570 7655 7740 7826 7910 7996 8081 8166 8251 8336 511 8421 8506 8591 8676 8761 8846 8931 9015 9100 9185 512 9270 9355 9440 9524 9609 9694 9/79 9863 9948 . .33 513 710117 0202 0287 0371 0456 0540 0825 0710 0794 0879 614 0963 1048 1132 1217 1301 1385 1470 1554 1639 1723 515 1807 1892 1976 2030 2144 2229 2313 2397 2481 2566 516 2650 2734 2818 2902 2986 3070 3154 3238 3326 3407 517 3491 3575 3659 3742 3826 3910 3994 4078 4162 4246 518 4330 4414 4497 4581 4665 4749 4833 4916 5000 5084 519 6167 5251 5335 5418 5502 5586 5869 5753 5838 5920 520 6003 6087 6170 6254 6337 6421 6504 6588 6671 6754 621 6838 6921 7004 7088 7171 7254 7338 7421 7504 7587 522 7671 7754 7837 7929 8003 8086 8169 8253 8336 8419 623 8502 8585 8668 8751 8834 8917 9000 9083 9165 9248 624 9331 9414 9497 9580 9663 82 0490 9745 9828 9911 9994 ..77 , 625 720159 0242 0325 0407 0573 0655 0738 0821 0903 526 0986 1068 1151 1233 1316 1398 1481 1563 1646 1728 527 1811 1893 .975 2058 2140 2222 2305 2387 2469 2552 528 2634 2716 2798 2881 2963 3045 3127 3209 3291 3374 529 3456 3538 3620 3702 3784 3866 3948 4030 4112 4194 530 4276 4358 4440 4522 4604 4685 4767 4849 4931 5013 531 5095 5176 5258 5340 5422 5503 5585 5667 5748 5830 532 5912 5993 6075 6156 6238 6320 6401 6483 6564 6646 533 6727 6809 6890 6972 7053 7134 7216 7297 7379 7460 534 7541 7623 7704 7785 7866 7948 8029 8110 8191 8273 535 8354 8435 8516 8597 8678 8759 8841 8922 9003 9084 536 9165 9246 9327 9403 9489 9570 9651 9732 9813 9893 537 9974 ..55 .136 .217 .298 .378 .459 .440 .621 .702 538 730782 0863 0944 1024 1105 1186 1266 1347 1428 1508 539 1589 1669 1750 1830 1911 1991 2072 2152 2233 2313 540 2394 2474 2555 2635 2715 2796 2876 2956 3037 3117 541 3197 3278 3358 3438 3518 3598 3679 3759 3839 3919 542 3999 4079 4160 4240 4320 4400 4480 4560 4640 4720 643 4800 4S80 4960 5040 5120 5200 5279 5359 5439 5519 541 5599 5679 5759 5838 5918 . 80 6715 5998 6078 6157 6237 6317 545 6397 6476 6556 6636 6795 6874 6954 7034 7113 546 7193 7272 7352 7431 7511 7590 7670 7749 7829 7908 517 7987 8037 8146 8225 8305 8384 8463 8543 8622 8701 548 8781 8860 8939 9018 9097 9177 9256 9335 9414 9493 549 9572 9651 9731 9810 9889 9968 ..47 .126 .205 .284 12 LOGARITHMS N. 1 2 3 4 5 6 7 8 9 550 740363 0442 0521 0560 0678 0757 0836 0915 0994 1073 551 1152 1230 1309 1388 1467 1546 1624 1703 1782 1860 552 1939 2018 2096 2175 2254 2332 2411 2489 2568 2646 553 2725 2804 2882 2961 3039 3118 3196 3275 3353 3431 554 3510 3558 3667 3745 3823 19 4608 3902 3980 4058 4136 4215 555 4293 4371 4449 4528 4684 4762 4840 4919 4997 556 5075 5153 5231 5309 5387 5465 5543 5621 5699 5777 557 5855 5933 6011 6089 6167 6245 6323 6401 6479 6556 558 6634 6712 6790 6868 6945 7023 7101 7179 7256 7334 559 7412 7489 7567 7645 7722 7800 7878 7955 8033 8110 560 8188 8266 8343 8421 8498 8576 8653 8731 8808 8885 561 8963 9040 9118 9195 9272 9350 9427 9504 9582 9659 562 9736 9814 9891 9968 ..45 .123 .200 .277 .354 .431 563 750508 0586 0663 0740 0817 0894 0971 1048 1125 1202 564 1279 1356 1433 1510 1587 1664 1741 1818 1895 1972 565 2048 2125 2202 2279 2356 2433 2509 2586 2663 2740 566 2816 2893 2970 3047 3123 3200 3277 3353 3430 3506 567 3582 3660 3736 3813 3889 3966 4042 4119 4195 4272 568 4348 4425 4501 4578 4654 4730 4807 4883 4960 5036 569 5112 5189 5265 5341 5417 5494 5570 5646 5722 5799 570 5875 5951 6027 6103 61F0 6256 6332 6408 6484 6560 571 6636 6712 6788 6864 6940 7016 7092 7168 7244 7320 572 7396 7472 7548 7624 7700 7775 7851 7927 8003 8079 573 8155 8230 8308 8382 8458 8533 8609 8685 8761 8836 574 • 8912 8988 9068 9139 9214 74 9970 9290 9366 9441 9517 9592 575 9638 9743 9819 9894 ..45 .121 .196 .272 .347 576 760422 0498 0573 0649 0724 0799 0875 0950 1025 1101 577 1176 1251 1326 1402 1477 1552 1627 1702 1778 1853 578 1923 2003 2078 2153 2228 2303 2378 2453 2529 2604 579 ^679 2754 2829 2904 2978 3053 3128 2203 3278 3353 580 3428 3503 3578 3653 3727 3802 3877 3952 4027 4101 581 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 582 4923 4998 5072 5147 5221 5296 5370 5445 5520 5594 583 5669 5743 5818 5892 5966 6041 6115 6190 6264 6338 584 6413 6487 6562 6636 6710 6785 6859 6933 7007 7082 585 7156 7230 7304 7379 7453 7527 7601 7675 7749 7823 586 7898 7972 8046 8120 8194 8268 8342 8416 8490 8564 587 8638 8712 8786 8860 8934 9008 9082 9156 9230 9303 588 9377 9451 9525 9599 9673 9746 9820 9894 9968 ..42 589 770115 0189 0263 0336 0410 0484 0557 0631 0705 0778 590 0852 0926 0999 1073 1146 1220 1293 1367 1440 1514 591 1587 1661 1734 1808 1881 1955 2028 2102 2175 2248 592 2322 2395 2468 3542 2615 2688 2762 2835 2908 2981 593 3055 3128 3201 3274 3348 3421 3494 3567 3640 3713 594 3786 3860 3933 4006 4079 73 4809 4152 4225 4298 4371 4444 595 4517 4590 1663 4736 4882 4955 5028 5100 6173 596 5246 5319 5392 5465 5538 5610 5683 5756 5829 5902 597 5974 6047 61120 6193 6265 6338 6411 6483 6556 6629 598 6701 6774 6846 6919 6992 7064 7137 7209 7282 7354 599 7427 7499 7572 7644 7717 7789 7862 7934 8006 8079 OF NUMBERS. 13 N. 1 2 3 4 5 6 7 8 9 600 778151 8224 8296 8368 8441 8513 8585 8658 8730 8802 601 8874 8947 9019 9091 9163 9236 9308 9380 9452 9524 602 9596 6669 9741 9813 9885 9957 ..29 .101 .173 .245 603 780317 0389 0461 0533 0605 0677 0749 0821 0893 0965 604 1037 1109 1181 1253 1324 n 2042 1396 1468 1540 1612 1684 605 1755 1827 1899 1971 2114 2186 2258 2329 2401 606 2473 2544 2616 2688 2759 2831 2902 2974 3046 3117 607 3189 3260 3332 3403 3475 3546 3618 3689 3761 3832 608 3904 3975 4046 4118 4189 4261 4332 4403 4475 4546 609 4617 4689 4760 4831 4902 4974 5045 5116 5187 5259 610 5330 5401 5472 5543 5615 5686 5757 5828 5899 5970 611 6041 6112 6183 6254 6325 6396 6467 6538 6609 6680 612 6751 6822 6893 6964 7035 7106 7177 7248 7319 7390 613 7460 7531 7602 7673 7744 7815 7885 79^6 8027 8098 614 8168 8239 8310 8381 8451 8522 8593 8663 8734 8804 615 8875 8946 9016 9087 9157 9228 9299 9369 9440 9510 616 9581 9651 9722 9792 9863 9933 ...4 ..74 .144 .215 617 790285 0356 0426 0496 0567 0637 0707 0778 0848 0918 618 0988 1059 1129 1199 1269 1340 1410 1480 1550 1620 619 1691 1761 1831 1901 1971 2041 2111 2181 2252 2322 620 2392 2462 2532 2602 2672 2742 2812 2882 2952 3022 621 3092 3162 3231 3301 3371 3441 3511 3581 3651 3721 622 3790 3860 3930 4000 4070 4139 4209 4279 4349 4418 623 4488 4558 4627 4697 4767 4836 4906 4976 5045 5115 624 5185 5254 5324 5393 5463 69 6158 5532 5602 5672 5741 5811 ■ 625 5880 5949 6019 6088 6227 6297 6366 6436 6505 626 6574 6644 6713 6782 6852 6921 6990 7060 7129 7198 627 7268 7337 7406 7475 7545 7614 7683 7752 7821 7890 628 7960 8029 8098 8167 8236 8305 8374 8443 8513 8582 629 8651 8720 8789 8858 8927 8996 9065 6134 9203 9272 630 9341 9409 9478 9547 9610 9685 9754 9823 9892 9961 631 800026 0098 0167 0236 0305 0373 0442 0511 0580 0648 ; 632 0717 0786 0854 0923 0992 1061 1129 1198 1266 1335 633 1404 1472 1541 1609 1678 1747 1815 1884 1952 2021 634 2089 2158 2226 2295 2363 2432 2500 2568 2637 2705 635 2774 2842 2910 2979 3047 3116 3184 3252 3321 3389. 636 3457 3525 3594 3662 3730 3798 3867 3935 4003 4071 637 4139 4208 4276 4354 4412 4480 4548 4616 4685 4753 638 4821 4889 4957 5025 5093 5161 5229 5297 5365 5433 639 5501 5669 5637 5705 5773 5841 5908 5976 6044 6112 640 6180 6248 6316 6384 6451 6519 6587 6655 6723 6790 641 6858 6926 6994 7061 7129 7157 7264 7332 7400 7467 642 7535 7603 7670 7738 7806 7873 7941 8008 8076 8143 643 8211 8279 8346 8414 8481 8549 8616 8684 8751 8818 644 8886 8953 9021 9088 9156 9223 9290 9358 9425 9492 645 9560 9627 9694 9762 9829 9896 9964 ..31 ..98 .165 646 810233 0300 0367 0434 0501 0596 0636 0703 0770 0837 647 0904 0971 1039 1106 1173 1240 1307 1374 1441 1508 648 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 1 649 2245 2312 2379 2445 2512 2579 2646 2713 2780 2847 14 LOGARITHMS N. 650 812913 1 o 3 4 5 6 7 8 9 2980 3047 3114 3181 3247 3314 3381 3448 3514 651 3581 3648 3714 3781 3848 3914 3981 4048 4114 4181 652 4248 4314 4381 4447 4514 4581 4647 4714 4780 4847 653 4913 4980 5046 5113 5179 5246 5312 5378 5445 5511 654 5578 5644 5711 5777 5843 67 6508 5910 5976 6042 6109 6175 655 6241 6308 6374 6440 6573 6639 6705 6771 6838 656 6904 6970 7036 7102 7169 7233 7301 7367 7433 7499 657 7565 7631 7698 7764 7830 7896 7962 8028 8094 8160 658 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 659 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 660 9544 9610 9676 9741 9807 9873 9939 ...4 ..70 .136 661 820201 0267 0333 0399 0464 0530 0595 0661 0727 0792 662 0858 0924 0939 1055 1120 1186 1251 1317 1382 1448 663 1514 1579 1645 1710 1775 1841 1906 1972 2037 2103 664 2168 2233 2299 2364 2430 2495 2560 2626 2691 2756 665 2822 2887 2952 3018 3083 3148 3213 3279 3344 3409 666 3474 3539 3605 3670 3735 3800 3865 3930 3996 4081 667 4126 4191 4256 4321 4386 4451 4516 4581 4646 4711 668 4776 4841 4906 4971 5036 5101 5166 5231 5296 5361 669 5426 5491 5556 5621 5686 5751 5815 5880 5945 6010 670 6075 6140 6204 6269 6334 6399 6464 6528 6593 6658 671 6723 6787 6852 6917 6981 7046 7111 7175 7240 7305 672 7369 7434 7499 7563 7628 7692 7757 7821 7886 7951 673 8015 8080 8144 8209 8273 8338 8402 8467 8531 8595 674 8660 8724 8789 8853 8918 65 9561 8982 9046 9111 9175 9239 675 9304 9368 9432 9497 9625 9690 9754 9818 9882 676 9947 ..11 ..75 .139 .204 .268 .332 .396 .460 .525 677 830589 0853 0717 0781 0845 0909 0973 1037 1102 1166 678 1230 1294 1358 1422 1486 1550 1614 1678 1742 1808 679 1870 1934 1998 2062 2126 2189 2253 2317 2381 2445 680 2509 2573 2637 2700 2764 2828 2892 2956 3020 3083 681 3147 3211 3275 3338 3402 3466 3530 3593 3657 3721 682 3784 3848 3912 3975 4039 4103 4166 4230 4294 4357 683 4421 4484 4548 4611 4675 4739 4802 4866 4929 4993 684 5056 5120 5183 5247 5310 5373 5437 5500 5564 5627 685 5691 5754 5817 5881 5944 6007 6071 6134 6197 6261 686 6324 6387 6451 6514 6577 6641 6704 6767 6830 6894 687 6957 7020 7083 7146 7210 7273 7336 7399 7462 7525 688 7588 7652 7715 7778 7841 7904 7967 8030 8093 8156 689 8219 8282 8345 8408 8471 8534 8597 8660 8723 8786 690 8849 8912 8975 9038 9109 9164 9227 9289 9352 9415 691 9478 9541 9604 9667 9729 9792 9855 9918 9981 ..43 692 840106 0169 0232 0294 0357 0420 0482 0545 0608 0671 693 0733 0796 0859 0921 0984 1046 1109 1172 1234 1297 694 1359 1422 1485 1547 1610 62 2235 1672 1735 1797 1860 1922 695 1985 2047 2110 2172 2297 2360 2422 2484 2547 696 2609 2672 2734 2796 2859 2921 2983 3046 3108 3170 697 3233 3295 3357 3420. 3482 3544 3606 3669 3731 3/93 698 3855 3918 3980 4042 4104 4166 4229 4291 4353 4415 699 4477 4539 4601 4664 4726 4788 4850 4912 4974 5036 OF NU MBERS. 15 N. 1 2 3 4 5 6 7 8 9 700 845098 5160 5222 5284 5346 5408 5470 5532 5594 5656 701 5718 5780 5842 5904 5966 6028 6090 6151 6213 6275 702 6337 6399 6461 6523 6585 6646 6708 6770 6832 6894 703 6955 7017 70/9 7141 7202 7264 7326 7388 7449 7511 704 7573 7634 7676 7758 7819 62 8435 7831 7943 8004 8066 8128 705 8189 8251 8312 8374 8497 8559 8620 8682 8743 70S 8805 8866 8928 8989 9051 9112 9174 9235 9297 9358 707 9419 9481 9542 9604 9665 9726 9788 9849 9911 9972 708 850033 0095 0156 0217 0279 0340 0401 0462 0524 0585 709 0646 0707 0769 0330 0891 0952 1014 1075 1136 1197 710 1258 1320 1381 1442 1503 1564 1625 1686 1747 1809 711 1870 1931 1992 2053 2114 2175 2236 2297 2358 2419 712 2480 2541 2602 2663 2724 2785 2846 2907 2968 3029 713 3090 3150 3211 3272 3333 3394 3455 3516 3577 3637 714 3698 3759 3820 3881 3941 4002 4063 4124 4185 4245 715 4308 4367 4428 4488 4549 4610 4670 4731 4792 4852 716 4913 4974 5034 5095 5156 5216 5277 5337 5398 5459 717 5519 5580 5640 5701 5761 5822 5882 5943 6003 6064 718 6124 6185 6245 6306 6366 6427 6487 6548 6608 6668 719 6729 6789 6850 6910 6970 7031 7091 7152 7212 7272 720 7332 7393 7453 7513 7574 7634 7694 7755 7815 7875 721 7935 7995 8056 8116 8176 8236 8297 8357 8417 8477 722 8537 8597 8657 8718 8778 8838 8898 8958 9018 9078 723 9138 9198 9258 9318 9379 9439 9499 9559 9619 9679 724 9739 9799 9859 9918 9978 60 0578 ..38 ..98 .158 .218 .278 725 860338 0398 0458 0518 0637 0697 0757 0817 0877 720 0937 0996 1056 1116 1176 1236 1295 1355 1415 1475 727 1534 1594 1654 1714 1773 1833 1893 1952 2012 2072 728 2131 2191 2251 2310 2370 2430 2489 2549 2608 2668 729 2728 2787 2847 2906 2966 3025 3085 3144 3204 3263 730 3323 3382 3442 3501 3561 3620 3680 3739 3799 3858 . 731 3917 3977 4036 4096 4155 4214 4274 4333 4392 4452 732 4511 4570 4630 4689 4148 4808 4867 4926 4985 5045 733 5104 5163 5222 5282 5341 5400 5459 5519 5578 5637 734 5696 5755 5814 5874 5933 5992 6051 6110 6169 6228 735 6287 6346 6405 6465 6524 6583 6642 6701 6760 6819 736 6878 6937 6996 7055 7114 7173 7232 7291 7350 7409 737 7467 7526 7585 7644 7703 7762 7821 7880 7939 7998 738 8056 8115 8174 8233 8292 8350 8409 8468 8527 8586 739 8644 8703 8762 8821 8879 8938 8997 9056 9114 9173 740 9232 9290 9349 9408 9466 9525 9584 9642 9701 9760 741 9818 9877 9935 9994 ..53 .111 .170 .228 .287 .345 742 870404 0462 0521 0579 0638 0696 0755 0813 0872 0930 743 0989 1047 1106 1164 1223 1281 1339 1398 1456 1515 744 1573 1631 1690 1748 1806 59 2389 1865 1923 1981 2040 2098 745 2156 2215 2273 2331 2448 2506 2564 2622 2681 746 2739 2797 2855 2913 2972 3030 3088 3146 3204 3262 747 3321 3379 3437 3495 3553 3611 3669 3727 3785 3844 748 3902 3960 4018 4076 4134 4192 4250 4308 4360 4424 749 4482 4540 4598 4656 4714 4772 4830 4888 4945 5003 16 LOGARITHMS N. 1 2 3 4 5 6 7 8 9 750 875031 5119 5177 5235 5293 5351 5409 5466 5524 5582 751 5640 5698 5756 5813 5871 5929 5987 6045 6102 6160 752 6218 6276 6333 6391 6449 6507 6564 6622 6680 6737 753 6795 6853 6910 6968 7026 7083 7141 7199 7256 7314 754 7371 7429 7487 7544 7602 57 8177 7659 7717 7774 7832 7889 755 7947 8004 8062 8119 8234 8292 8349 8407 8464 756 8522 8579 8637 8694 8752 8809 8866 8924 8981 9039 757 9096 9153 9211 9268 9325 9383 9440 9497 9555 9612 758 9669 9726 9784 9841 9898 9956 ..13 ..70 .127 .185 759 880242 0299 0356 0413 0471 0528 0580 0642 0699 0756 760 0814 0871 0928 0985 1042 1099 1156 1213 1271 1328 761 1385 1442 1499 1556 1613 1670 1727 1784 1841 1898 762 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 763 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 764 3093 3150 3207 3264 3321 3377 3434 3491 3548 3605 765 3661 3718 3775 3832 3888 3945 4002 4059 4115 4172 766 4229 4285 4342 4399 4455 4512 4569 4625 4682 4739 767 4795 4852 4909 4965 5022 5078 5135 5192 5248 5305 768 5361 5418 5474 5531 5587 5644 6700 5757 5813 5870 769 5926 5983 6039 6096 6152 6209 .6265 6321 6378 6434 770 6491 6547 6604 6660 6716 6773 6829 6885 6942 6998 771 7054 7111 7167 7233 7280 7336 7392 7449 7505 7561 772 7617 7674 7730 7786 7842 7898 7955 8011 8067 8123 773 8179 8236 8292 8348 8404 8460 8516 8573 8629 8655 774 8741 8797 8853 8909 8965 56 9526 9021 9077 9134 9190 9246 775 9302 9358 9414 9470 9582 9638 9694 9750 9806 776 9862 9918 0974 ..30 ..86 .141 .197 .253 .309 .365 777 890421 0477 0533 0589 0645 0700 0756 0812 0868 0924 778 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 779 1537 1593 1649 1705 1760 1816 1872 1928 1983 2039 780 2095 2150 2208 2262 2317 2373 2429 2484 2540 2595 781 2651 2707 2762 2818 2873 2929 2985 3040 3096 3151 782 3207 3262 3318 3373 3429 3484 3540 3595 3651 3706 783 3762 3817 3873 3928 3984 4039 4094 4150 4205 4261 784 4316 4371 4427 4482 4538 4593 4648 4704 4759 4814 785 4870 4925 4980 5036 5091 5146 5201 5257 5312 5367 786 5423 5478 5533 6588 5644 5699 5754 5809 5864 5920 787 5975 6030 6085 6140 6195 6251 6306 6361 6416 6471 788 6526 6581 6636 6692 6747 6802 6857 6912 6967 7022 789 7077 7132 7187 7242 7297 7352 7407 7462 7517 7572 790 7627 7683 7737 7792 7847 7902 7957 8012 8067 8122 791 8176 8231 8286 8341 8396 8451 8506 8561 8615 8670 792 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 793 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 794 9821 9875 9930 9985 ..39 55 0586 ..94 .149 .203 .258 .312 795 900367 0422 0476 0531 0640 0695 0749 0804 0859 796 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 797 1458 1513 1567 1622 1676 1736 1785 1840 18*4 1948 798 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 799 2547 2601 2655 2710 2764 2818 2873 2927 2981 3036 OF NUMBERS 17 N. 1 2 3 4 5 6 7 8 9 800 903090 3144 3199 3253 3307 3361 3416 3470 3524 3578 801 3633 3687 3741 3795 3849 3904 3958 4012 4066 4120 802 4174 4229 4283 4337 4391 4445 4499 4553 4607 4601 803 4716 4770 4824 4878 4932 4986 5040 5094 5148 5202 804 5256 5310 5364 5418 5472 54 6012 5526 5580 5634 5688 5742 805 6796 5850 5904 5958 6086 6119 6173 6227 6281 80-3 6335 6389 6443 6497 6551 6604 6658 6712 6766 6820 807 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 808 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 809 7949 8002 8058 8110 8163 8217 8270 8324 8378 8431 810 8485 8539 8592 8646 8699 8753 8807 8860 8914 8967 811 9021 9074 9128 9181 9235 9289 9342 9396 9449 9503 812 9558 9810 9663 9716 9770 9823 9877 9930 9984 ..37 813 910091 0144 0197 0251 0304 0358 0411 0464 0518 0571 814 0824 0878 0731 0784 0838 0891 0944 0998 1051 1104 815 1158 1211 1264 1317 1371 1424 1477 1530 1584 1637 816 1690 1743 1797 1850 1903 1956 2009 2063 2115 2169 817 2222 2275 2323 2381 2435 2488 2541 2594 2645 2700 818 2753 2808 2859 2913 2966 3019 3072 3125 3178 3231 819 3284 3337 3390 3443 3496 3549 3602 3655 3708 3761 820 3814 3867 3920 3973 4026 4079 4132 4184 4237 4290 821 4343 4396 4449 4502 4555 4608 4660 4713 4766 4819 822 4872 4925 4977 5030 5083 5136 5189 5241 5594 5347 823 5400 5453 5505 5558 5611 5664 5716 5769 5822 5875 824 5927 5980 6033 6085 6138 6191 6243 6296 6349 6401 825 6454 6507 6559 6612 6664 6717 6770 6822 6875 6927 826 6980 7033 7085 7138 7190 7243 7295 7348 7400 7453 827 7508 7558 7611 7663 7716 7768 7820 7873 7925 7978 828 8030 8083 8185 8188 8240 8293 8345 8397 8450 8502 829 8555 8607 8659 8712 8764 8816 8869 8921 8973 9026 830 9078 9130 9183 9235 9287 9340 9392 9444 9496 9549 831 9601 9653 9708 9758 9810 9862 9914 9967 . .19 ..71 >832 920123 0176 0228 0280 0332 0384 0436 0489 0541 0593 833 0645 0697 0749 0801 0853 0903 0958 1010 1062 1114 834 1166 1218 1270 1322 1374 1426 1478 1530 1582 1634 835 1686 1738 1790 1842 1894 1946 1998 2050 2102 2154 836 2208 2258 2310 2362 2414 2466 2518 2570 2622 2674 837 2725 2777 2829 2881 2933 2985 3037 3089 3140 3192 838 • 3244 3296 3348 3399 3451 3503 3555 3607 3658 3710 839 3762 3S14 3865 3917 3969 4021 4072 4124 4147 4228 840 4279 4331 4383 4434 4486 4538 4589 4641 4693 4744 841 4796 4848 4899 4951 5003 5054 5106 5157 5209 6261 842 5312 5364 5415 5467 5518 5570 5621 5073 5725 5776 843 5828 5874 5931 5982 6034 6085 6137 6188 6240 6291 844 6342 6394 6445 6497 6548 52 7082 6600 6651 6702 6754 6805 845 6857 6908 6959 7011 7114 7165 7216 7268 7319 846 7370 7422 7473 7524 7576 7627 7678 7730 7783 7832 847 7883 7935 7986 8037 8038 8140 8191 8242 8293 8345 848 8396 8447 8498 8549 8601 8652 8703 8754 8805 8857 849 8908 8959 9010 9031 9112 9163 9216 9266 9317 9368 18 LOGARITHMS N. I 2 3 1 4 5 6 7 8 9 850 929419 9473 9521 9572 9623 9674 9725 9776 9827 9879 851 9930 9981 I . .32 ..83 .134 .185 .236 .287 .338 .389 852 930440 0491 0542 0592 0643 0694 0745 0796 0847 0898 853 0949 1000 1051 1102 1153 1204 1254 1305 1356 1407 854 1458 1509 1560 1610 1661 51 2169 1712 1763 1814 1865 1915 855 1966 2017 2068 2118 2220 2271 2322 2372 2423 856 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 857 2981 3031 3082 3133 3183 3234 3285 3335 3386 3437 853 3487 3538 3589 3639 3690 3740 3791 3841 3892 3943 859 3993 4044 4094 4145 4195 4246 4269 4347 4397 4448 860 4498 4549 4599 4650 4700 4751 4801 4852 4902 4953 861 5003 5054 5104 5154 5205 5255 5306 5356 5406 5457 862 5507 5558 5603 5658 5709 5759 5809 5380 5910 5960 863 6011 6051 6111 6162 6212 6262 6313 6363 6413 6463 864 6514 6564 6614 6665 6715 6765 6815 6865 6916 6966 865 7016 7056 7117 7167 7217 7267 7317 7367 7418 7468 866 7518 7568 7618 7668 7718 7769 7819 7869 7919 7969 867 8019 8069 8119 8169 8219 8269 8320 8370 8420 8470 868 8520 8570 8620 8670 8720 8770 8820 8870 8919 8970 869 9020 9070 9120 9170 9220 9270 9320 9369 9419 9469 870 9519 9569 9616 9669 9719 9769 9819 9869 9918 9968 871 940018 0058 0118 0168 0218 0267 0317 0367 0417 0467 872 0516 0566 0516 0566 0716 0765 0815 0865 0915 0964 873 1014 1064 1114 1163 1213 1263 1313 1362 1412 1462 874 1511 1561 1611 1660 1710 1760 1809 1859 1909 1958 875 2008 2058 2107 2157 2207 2256 2308 2355 2405 2455 876 2504 2554 2603 2633 2702 2752 2801 2851 2901 2950 877 3000 3049 3099 3148 3198 3247 3297 3346 3396 3445 878 3495 3544 3593 3643 3692 3742 3791 3841 3890 3939 879 3989 4038 40S8 4137 4186 4236 4285 4335 4384 4433 880 4483 4532 4581 4631 4680 4729 4779 4828 4877 4927 881 4976 5025 5074 5124 5173 5222 5272 5321 5370 5419 882 5469 5518 5567 5616 5665 5715 5764 5813 5862 5912 833 5961 6010 6059 6103 6157 6207 6256 6305 6354 6403 884 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 885 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 886 7434 7483 7532 7581 7630 7679 7728 7777 7826 7875 887 7924 7973 8022 8070 8119 8168 8217 8266 8315 8365 888 8413 8462 8511 8560 8609 8657 8708 3 755 8804 3853 889 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 890 9390 9439 9483 9536 9585 9634 9683 9731 9780 9829 891 98 78 9926 9975 ..24 ..73 .121 .170 .219 .267 .316 892 950365 0414 0462 0511 05G0 0508 0657 0/05 0754 6803 893 0351 0900 0949 0997 1046 1095 1143 1192 1240 1239 894 1338 1386 1435 1483 1532 48. 2017 1580 1629 1677 1726 17/5 895 1823 1872 1920 1969 2086 2114 2163 2211 2260 896 2303 2356 2405 2453 2502 2550 2599 2647 5696 2744 897 2792 2841 2889 2938 2986 3034 3033 3131 3180 3228 898 3276 3325 3373 3421 3470 3518 3666 3615 3663 3711 899 i 3760 3803 3856 3905 3953 4001 40 i9 4098 4146 4194 OF NUMBERS. 19 N. 4291 2 3 4 4435 5 6 7 8 9 900 954243 4339 4387 4484 4532 4580 4628 4677 901 4725 4773 4821 4869 4918 4966 5014 6062 5110 5158 902 520 1 5255 5303 5351 5399 5447 5495 5543 5592 5840 903 5688 5736 5784 5832 5880 5928 5976 6024 6072 6120 904 6168 6216 6265 6313 6361 48 6840 6409 6457 6505 6553 6601 905 6649 6697 6745 6793 6888 6936 6984 7032 7080 906 7128 7176 7224 7272 7320 7368 7416 7464 7512 7559 907 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 903 8086 8134 8181 8229 8277 8325 8373 8421 8468 8516 909 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 910 9041 9089 9137 9185 9232 9280 9328 9375 9423 9474 911 9518 9586 9614 9661 9709 9757 9804 9852 9900 9947 912 9095 ..42 ..90 .138 .185 .233 .280 .328 .376 .423 913 960471 0518 0566 0613 0861 0709 0756 0804 0851 0899 914 0946 0994 1041 1089 1136 1184 1231 1279 1326 1374 915 1421 1469 1516 1563 1611 1658 1706 1753 1801 1848 916 1895 1943 1990 2038 2035 2132 2180 2227 2275 2322 917 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 918 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 919 3316 3363 3410 3457 3504 3552 3599 3646 3693 3741 920 3788 3835 3882 3929 3977 4024 4071 4118 4165 4212 921 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 922 4731 4778 4825 4872 4919 4966 5013 5081 5108 5155 923 5202 5249 5296 5343 5390 5437 5484 5531 5578 5625 924 5672 5719 5766 5813 5860 5907 5954 6001 6048 6095 925 6142 6189 6236 6283 6329 6376 6423 6470 6517 6564 926 6611 6658 6705 6752 6799 6845 6892 6939 6986 7033 927 7080 7127 7173 7220 7267 7314 7361 7403 7454 7501 928 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 929 8016 8062 8109 8156 8203 8249 8296 8343 8390 8436 930 8483 8530 8576 8823 8670 8716 8763 8810 8858 8903 931 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 . 932 9416 9463 9509 9556 9602 9649 9695 9742 9789 9835 933 9882 9928 9975 ..21 ..63 .114 .161 .207 .254 .300 934 970347 0393 0440 0486 0533 0579 0626 0672 0719 0765 935 0812 0358 0904 0951 0997 1044 1090 1137 1183 1229 936 1276 1322 1369 1415 1461 1508 1554 1601 1647 1693 937 938 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 2203 2249 2295 2342 2388 2434 2481 2527 2573 2619 939 2666 2712 2758 '2804 2851 2397 2943 2989 3035 3082 940 3128 3174 3220 3266 3313 3359 3405 3451 3497 3543 941 3590 3836 3682 3728 3774 3820 3866 3913 3959 4005 942 4051 4097 4143 4189 4235 4281 4327 4374 4420 4466 943 4512 4558 4604 4650 4898 4742 4788 4834 4380 4926 944 4972 5018 5084 5110 5156 46 5616 5202 5248 5294 5340 5386 945 5432 5478 5524 5570 5662 5707 5753 5799 5845 946 5891 5937 5983 6029 6075 6121 6167 6212 6258 6304 947 6350 6396 6442 6488 6533 6579 6925 6671 6717 6763 948 6803 6854 6900 6946 6992 7037 7033 7129 7175 7220 949 7266 7312 7358 7403 7449 7495 7541 7586 7632 7678 19 20 LOGARITHMS N. 1 2 3 4 5 6 7 8043 8 9 950 977724 7769 7815 7861 7906 7952 7998 8089 8135 951 8181 8226 8272 8317 £363 8409 8454 8500 8546 8591 952 8637 8683 8728 8774 8819 8865 8911 8956 9002 9047 953 9093 9138 9184 9230 9275 9321 9366 9412 9457 9503 954 9548 9594 9639 9685 9730 46 0185 9776 9821 9867 9912 9968 955 980003 0049 0094 0140 0231 0276 0322 0367 0412 956 0458 0503 0549 0594 0340 0685 0730 0776 0321 0867 957 0912 0957 1003 1048 1093 1139 1184 1229 1275 1320 958 1366 1411 1456 1501 1547 1592 1637 1683 1728 1773 959 1819 1864 1909 1954 2000 2045 2090 2135 2181 2226 960 2271 2316 2362 2407 2452 2497 2543 2588 2633 2678 961 2723 2769 2814 2859 2904 2949 2994 3040 3085 3130 962 3175 3220 3265 3310 3356 3401 3446 3491 3536 3581 963 3626 3671 3716 3762 3807 3852 3897 3942 3987 4032 964 4077 4122 4167 4212 4257 43^2 4347 4392 4437 4482 965 4527 4572 4617 4662 4707 4752 4797 4842 4887 4932 966 4977 5022 5087 5112 5157 5202 5247 5292 5337 5382 96; 5426 5471 5516 5561 5606 5651 5699 5741 5786 5830 968 5875 5920 5965 6010 6055 6100 6144 6189 6234 6279 969 6324 6369 6413 6458 6503 6548 6593 6637 6682 6727 970 6772 6817 6861 6908 6951 6996 7040 7035 7130 7175 971 7219 7264 7309 7353 7398 7443 7488 7532 7577 7622 972 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 973 8113 8157 8202 8247 8291 8336 8381 8425 8470 8514 974 8559 8604 8648 8693 8737 8782 8826 8871 8916 8960 975 9005 9049 9093 9138 9183 9227 9272 9316 9361 9405 976 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 977 9895 9939 9983 ..28 ..72 .117 .161 .206 .250 .294 978 990339 0383 0428 0472 0516 0561 0605 0650 0694 0738 979 0783 0827 0871 0916 0960 1004 1049 1093 1137 1182 980 1226 1270 1315 1359 1403 1448 1492 1536 1580 1625 981 1669 1713 1758 1802 1846 1890 1935 1979 2023 2067 982 2111 2156 2200 2244 2288 2333 2377 2421 2465 2509 983 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 984 2995 3039 3083 3127 3172 3216 3260 3304 3348 3392 985 3436 3480 3524 3568 3613 3657 3701 3745 3789 3833 986 3877 3921 3965 4009 4053 4097 4141 4185 4229 4273 987 4317 4361 4405 4449 4493 4537 4581 4625 4669 4713 988 4757 4801 4845 4886 4933 4977 5021 5065 5108 5152 989 5196 5240 5284 5328 5372 5416 5460 5504 5547 5591 : 990 5635 5379 5723 5767 5811 5854 5898 5942 5986 6030 : 991 6074 6117 6161 6205 6249 6293 6337 6380 6424 6468 I 992 6512 6555 6599 6643 6687 6731 6774 6818 6862 6906 993 6949 6993 7037 7080 7124 7168 7212 7255 7299 7343 994 7386 7430 7474 7517 7561 44 7998 7605 7648 7692 7736 7779 995 7823 7867 7910 7954 8041 8085 8129 8172 8216 996 8259 8303 8347 8390 8434 8477 8521 8564 8608 8652 997 8695 8739 8792 8826 8869 8913 8956 9000 9043 9087 998 9131 9174 9218 9261 9305 9348 9392 9435 9479 9522 999 9565 9609 9652 9698 9739 9783 | 9826 9870 9913 9957 r -=? TABLE If. Log. Sines and Tangems. (0°) Natural Sines 21 Sine. D 10" Cosine. D.IO" Tang. D.IO" Coiang. N.sine N. cos. 9.OJ0000 10.000000 0.000000 Infinite. 00001 100000 60 1 6. 463726 000000 6.463726 13.536274 0002'd 100000 59 2 764766 039000 764756 235244 00058 iooooo 58 3 940847 009000 940847 059153 00087 100000 57 4 7.065786 000)0 J 7.065786 12.934214 00116 100000 56 5 162696 000003 162696 837304 00145 100000 55 6 241877 9.999999 241878 758122 00175 100000 54 7 308824 999999 308825 691175 00204 100000 53 366816 999999 366817 633183 00233 100000 52 9 417968 999999 417970 582030 00262 100000 51 10 463725 999998 463727 536273 00291 100000 50 11 7.505118 9.999998 7.505120 12.494880 00320 99999 49 1-2 542905 999997 542909 457091 00349 99999 48 13 577668 999997 577672 422328 00378 99999 47 14 609853 999996 609857 390143 00407 99999 46 15 639816 999996 639820 360180 00436 99999 45 16 667845 999995 667849 332151 00465 99999 44 17 694173 999995 694179 305821 00495 99999 43 18 718997 999994 719003 280997 00524 99999 42 19 742477 999993 742484 257516 00553 99998 41 20 764754 999993 764761 235239 00582 99998 40 21 7.785943 9.999992 7.785951 12.214049 00611 99998 39 22 806146 999991 808155 193845 00640 99998 38 23 825451 999990 825460 174540 00669 99998 37 24 843934 999989 843944 156056 00698 99998 36 25 861663 999988 861674 138326 00727 99997 35 26 878695 999988 878708 121292 00756 99997 34 27 895085 999987 895099 104901 00785 99997 33 28 910879 999986 910894 089106 00814 99997 32 29 926119 999985 926134 073866 00844 99996 31 30 940842 999983 940858 059142 00873 99996 30 31 7.955082 2298 2227 2161 2098 2039 1983 1930 1880 1832 1787 1744 1703 1664 1626 1591 1557 1524 1492 1462 1433 1405 1379 1353 1328 1304 1281 1259 1237 1216 9.999982 7.955100 2298 2227 2161 2098 2039 1983 1930 1880 1833 1787 1744 1703 1664 1627 1591 1557 1524 1493 1463 1434 1406 1379 1353 1328 1304 1281 1259 1238 1217 12 . 044900 00902 99996 29 32 968870 999981 0.2 0.2 0.2 968889 031111 00931 99996 28 ,33 982233 999980 982253 017747 00960 99995" 27 34 995198 999979 995219 004781 00989 99995 26 35 8.007787 999977 0-2 0-2 0-2 8.007809 11.992191 01018 99995 25 36 020021 999976 020045 979955 01047 99995 24 37 031919 999975 031945 968055 01076 99994 23 38 043501 999973 02 0-2 043527 956473 01105 99994 22 39 054781 999972 054809 945191 01134 99994 21 40 065776 999971 0'2 0'2 065806 934194 01164 99993 20 41 8.076500 9.999969 8.076531 11.923469 01193 99993 19 42 086965 999968 2 0*2 0'2 0'3 0'3 o's 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.4 0.4 0.4 0.4 086997 913003 01222 99993 18 43 097183 999966 097217 902783 01251 99992 17 44 107167 999964 107202 892797 01280 99992 16 45 116926 999963 116963 883037 01309 99991 15 46 126471 999961 126510 873490 01338 99991 14 47 135810 999959 135851 864149 01367 99991 13 48 144953 999958 144996 855004 01396 99990 12 49 153907 999956 153952 846048 01425 99990 11 50 162681 999954 162727 837273 01454 99989 10 51 8.171280 9.999952 3.171328 11.828672 01483 99989 9 52 179713 999950 179763 820237 01513 99989 8 53 187985 999948 188036 811964 01542 99988 7 54 196102 999946 196156 803844 01571 99988 6 55 204070 999944 204126 795874 101600 99987 5 56 211895 999942 211953 788047 j 01629 99987 4 57 58 219581 227134 999940 999938 219641 227195 780359 101658 772805 |i 01687 99986 99986 3 2 59 234557 999936 234621 765379 !| 01716 99985 1 60 241855 999934 241921 758079 j 01745 N. cos. 99985 N. sine- ~ 7 Cosine. Sine. Cot an?. Tan jr. 89 Degrees. ; 22 Lo-r. Sines and Ta 0°) TABLE |D 10' 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1? 18 19 20 21 22' 23 24 25 26 2? 28 29 30 31 32 33 34 35 36 37 as 39 40 41 42 43 44 45 40 47 48 49 59 51 52 53 54 55 53 57 58 59 60 3.241855 11Qp 249033 }{^2 256094 {}'' 263042 }}» 269881 [{$" 276614 \\jt 283243 }££ *° 3 ' ia 1079 296207 Jx-fi 302546 ^ 308794 Jnoi $.314954 {xf* 321027 q Q « 327016 gg° 332924 g*° 338753 q-q 344504 HI 350181 ™ 355783 qS 361315 qTq 366777 goo 5.372171 o«« 377499 o„ 382762 2' ' 387962 o~' 393101 g™ 398179 007 4031991 qJ 7 408161 ! o7o 4130381 o,,t; 417919 jJJn $.422717 °g° 427462 ^ 432156 ,~ 436800! «^ UlSt I 7 ^ 4o0440 ?42 454893 '™ 459301 ! Lfrj 463665 i^n $.467985 ifi 472263 ! ^ 476498 '$ 480693 *£ 484848 ££ 488963 ^°q 493040 ^ 497078 ^ 501080 }*' 505045 w» $.508974 ™? 512867 £** 516726 °~ 520551 ™ 7 524343 ^ 528102 Jo? 531828 °f; 535523 °|° 539186 J?. 1 * 542819 UD Cosine. D.10' 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0-4 0-4 0-5 05 0-5 0.5 0.5 0-5 05 0.5 0.5 0-5 0.5 0.5 0.5 0-5 06 0-6 0.6 0-6 0-6 0.6 0.6 0.6 0-6 0-6 0-6 0-6 0-6 0.6 0.6 0-7 07 0.7 0.7 0.7 0.7 0.7 0.7 $.241921 249102 256165 263115 269956 276691 283323 289856 298292 302634 308884 $.315046 321122 327114 333025 333856 344610 35U289 355895 361430 366895 t. 372292 377622 382889 388092 393234 398315 403338 408304 413213 418088 ;. 422869 427618 432315 436962 441560 446110 450613 455070 459481 463849 .468172 472454 476693 480892 485050 489170 493250 497293 501298 505267 .509200 513098 516981 520790 . 524586 528349 532089 535779 539447 543084 "Couing-. D 10' 1197 1177 1158 1140 1122 1105 1089 1073 1057 1042 1027 1013 999 985 972 959 946 934 922 911 899 888 879 867 857 847 837 828 818 809 800 791 783 774 766 758 750 743 735 728 720 713 707 700 693 686 680 674 668 661 655 650 644 638 633 627 622 616 611 606 G 'N. sine. X. cos.l 11 11 11.758079 750898 743835 736885 730044 723309 716677 710144 703708 697366 691116 684954 678878 672886 666975 661144 655390 649711 644105 638570 633105 627708 622378 617111 611908 606766 601685 596662 591696 586787 581932 11.577131 572382 567685 563038 558440 553890 549387 544930 540519 536151 11-531828 527546 523307 519108 514950 510830 506750 502707 498702 494733 11.490800 486902 483039 479210 475414 471651 467920 464221 460553 ; 456916 , Tana:. I 0174 01774 0180; 01832 01862 01891 01920 0194! 01978 02007 02036 02065 02094 02123 02152 02181 02211 02240 02269 02298 0232 02356 02385 02414 02443 02472 02501 02530 02560 02589 02618 0264; 02676 02705 02734 02763 02792 02821 02850 02879 02908 02938 02967 02996 03025 03054 03083 03112 03141 03170 03199 03228 03257 ! 03286 '< 03316 ■ 03345 : 03374 i 03403 : 03432 03461 , 03490 99385; 60 )9984 i 59 99934 58 99983! 57 99983 1 56 99982 55 99982 54 99^81 63 99980 52 99980 51 99979' 50 Jy979j 48 y9978| 48 99977 47 ); 977 46 99976 45 99976 44 ^9975 43 99974^ 42 99974! 41 99973 40 b972 39 99972 38 99971 37 99970 36 89969 35 99969 34 99968' 33 99967| 32 99966! 31 ;9966 99985 99984 99963 99963 99962 99961 99960 99959 99959 99958 99957 99956 99955 99954 99953 99952 99952 99951 99950 99949 99948 99947 99946 99945 99944 99943 99942 99941 99940 999b9 X. cos. X.sine. ' 83 Degrees. TARLK II. Log. Sines and Tangents. (-2°) Natural Sines. 23 1 2 3 4 5 6 7 S 9 10 11 12 13 14 15 16 17 18 IS 20 •21 2^ 23 21 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 53 54 55 56 57 58 59 GO 8.542819 546422 549995 553539 557054 560540 563999 567431 570836 574214 577566 8.580892 584193 587469 590721 593948 597152 600332 603489 606623 609734 8.612823 615891 618937 621962 624965 627948 630911 633854 636776 639680 8.642563 645428 648274 651102 653911 656702 659475 662230 664968 667689 8.670393 673080 675751 678405 681043 683665 686272 688863 691438 693998 8.696543 699073 701589 704090 706577 709049 711507 713952 716383 718800 Cosine. D. 10 600 595 591 586 581 576 572 567 563 559 554 550 546 542 538 534 530 526 522 519 515 511 508 504 501 497 494 490 487 484 481 477 474 471 468 465 462 459 456 453 451 448 445 442 440 437 434 432 429 427 424 422 419 417 414 412 410 407 405 403 9.999735 999731 999726 999722 999717 999713 999708 999704 999899 999694 999689 9.999685 999680 999675 999670 999665 999660 999655 999650 999645 999640 999635 999629 999324 999619 999614 999608 999603 999597 999592 999586 ^.999581 999575 999570 999564 999558 999553 999547 999541 999535 999529 999524 999518 999512 999506 999500 999493 999487 999481 999475 999469 999463 939456 999450 999443 999437 999431 999424 999418 999411 999404 D. 10 y 0.7 0.7 0-7 0-8 0-8 0-8 0.8 0.8 0-8 0-8 0.8 0-8 0-8 0.8 0.8 0.8 0.8 0.8 0.8 0.8 0.9 Tansr. 0.9 0.9 0-9 0-9 0-9 0-9 0-9 0-9 0.9 0-9 0.9 0.9 0.9 0.9 1 1 1 1-0 1.0 1-0 1-0 1.0 1.0 1.0 1.0 1.0 1.0 1.1 1 1 1 1 1 1 8.543084 546691 550268 653817 557336 560828 564291 567727 571137 574520 577877 8.581208 584514 587795 591051 594283 597492 600677 603839 608978 610094 8.613189 616262 619313 622343 625352 628340 631308 634256 637184 640093 8.642982 645853 648704 651537 654352 657149 659928 662689 665433 668160 8.670870 673563 676239 678900 681544 684172 6 6784 689381 691963 694529 .697081 699617 702139 704246 707140 709618 702083 714534 716972 719396 "Coiaiisr. D. W 602 59;} 591 587 582 577 573 568 564 559 555 551 547 543 539 535 531 527 523 519 516 512 508 505 501 498 495 491 488 485 482 478 475 472 469 466 463 460 457 454 453 449 446 443 442 438 435 433 430 428 425 423 420 418 415 413 411 408 406 404 Cotang. | IN. sine 11 11.456916 453309 449732 446183 442664 439172 435709 432273 428863 425480 422123 ,418792 415486 412205 408949 405717 402508 399323 396161 393022 389906 11.386811 383738 380687 , 377657 I 374648 371660 368692 365744 362816 359907 11.357018 354147 351296 348463 ! 345648 j 342851 i 340072 337311 334567 331840 11.329130 326437 323761 321100 318456 315828 313216 310819 308037 305471 11.302919 300383 j 297861 j 295354 | 292860 ; 290382 i 287917 ! 285465 ! 283028 280604 03490 03519 03548 03577 03606 03635 ,03664 03693 03723 03752 03781 03810 03839 03868 03897 03926 03955 03984 04013 04042 04071 04100 03129 04159 04188 04217 04246 04275 04304 04333 N. cos 99.939 99938 99937 99936 99935 99934 99933 99932 99931 99930 99929 99927 99926 99925 99924 99923 99922 99921 99919 99918 99917 99916 99915 99913 99912 99911 99910 99909 99907 99908 04362J99905 04391 04420 04449 04478 04507 04536 04565-99896 04594 04623 04653 04682 04711 04740 04769 04798 04827 04856 04885 04914 04943 04972 05001 05030 05059 05088 05117 05146 05175 05205 05234 N. cos. N.siire 99904 99902 99901 99900 99898 9989' 99894 99893 99892 99890 99889 99888 99886 99885 99883 99882 99881 99879 99878 99876 99875 99873 99872 99870 99869 99867 99866 99864 99863 60 59 58 57 56 55 54 58 52 51 5C 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 80 29 28 27 28 25 24 23 22 21 20 19 18 17 16 15 14 13 12 10 9 8 7 6 5 4 3 2 1 87 Degrees. 24 Log. Sines and Tangents. (3°; Natural Sines. TABLE II. 1 2 3 4 5 6 7 8 B 10 11 12 13 14 15 16 17 18 & 20 21 22 23 24 25 26 27 38 29 30 31 32 33 34 35 30 3? 38 39 40 41 42 43 44 45 46 4? 48 49 50 51 52 53 54 55 56 57 58 59 60 Shit-. i. 718809 721204 723595 725972 728337 730888 733027 735354 737667 739969 742259 1.744536 746802 749055 751297 753528 755747 757955 760151 762337 764511 ;. 766675 768828 770970 773101 775223 777333 779434 781524 783605 785675 1.787736 789787 791828 793859 795881 797894 799897 801892 803876 805852 i. 8078 19 809777 811726 813667 815599 817522 819436 821343 823240 825130 1.827011 828884 830749 832607 834456 836297 838130 839956 841774 843585 Cosine. 401 398 396 394 392 390 388 386 384 382 380 378 376 374 372 370 368 366 364 362 361 359 357 355 353 352 350 348 347 345 343 342 340 339 337 335 334 332 331 329 328 326 325 323 322 320 319 318 316 315 313 312 311 309 308 307 306 304 303 302 Cosine. .999404 999398 999391 999384 999378 999371 999364 999357 999350 999343 999336 .999329 999322 999315 999308 999301 999294 999286 999279 999272 999265 .999257 999250 999242 999235 999227 999220 999212 999205 999197 999189 .999181 999174 999166 999158 999150 999142 999134 999126 999118 999110 .999102 999094 999086 999077 999069 999061 999053 999044 999036 999027 .999019 999010 999002 998993 998984 998976 998967 998958 998950 998941 Sin«\ D. 10' Cotang. |(N.sine .719396 721806 724204 726588 728959 731317 733663 735996 738317 740S26 742922 .745207 747479 749740 751989 754227 756453 758668 760872 763065 765246 .767417 769578 771727 773866 775995 778114 780222 782320 784408 786486 .788554 790613 792662 794701 796731 798752 800763 802765 804858 806742 .808717 810683 812641 814589 816529 818461 820384 822298 824205 826103 .827992 829874 831748 833613 835471 837321 839163 840998 842825 844644 Cotang. 402 399 397 395 393 391 389 387 385 383 381 379 377 375 373 371 369 367 365 364 362 360 358 356 355 353 351 350 348 346 345 343 341 340 338 337 335 334 332 331 329 328 326 325 323 322 320 319 318 316 315 314 312 311 310 308 307 306 304 303 11.280604 278194 275796 273412 271041 268683 266337 264004 261683 259374 257078 11.254793 252521 250260 248011 245773 243547 241332 239128 236935 234754 11.232583 230422 228273 226134 224005 221886 219778 217680 215592 213514 11.211446 209387 207338 205299 203269 201248 199237 197235 195242 193258 191283 189317 187359 185411 183471 181539 179616 177702 175795 173897 11.172008 170126 168252 166387 164529 162679 160837 159002 157175 155356 05234 05263 05292 05321 05350 05379 05408 05437 05466 05495 05524 05553 05582 05611 05640 05669 05698 05727 05756 05785 05814 05844 05873 05902 05931 05960 05989 06018 06047 06076 06105 06134 06163 06192 08221 06250 06279 06308 06337 06366 06395 08424 06453 06482 06511 08540 08569 06598 0662' ,06656 06685 06714 06743 06773 06802 1 06831 1 06860 ! 06889 1 06918 | 0694 ! 06976 Tang. II N. cos. X.sine. N. cos 99863 99861 99860 99858 99857 99855 99854 99852 99851 99849 99847 99846 99844 99842 99841 99839 99838 99836 99834 99833 99831 99829 99827 99826 99824 99822 99821 99819 99817 99815 99813 99812 99810 99808 99806 99804 99803 99801 99799 99797 99795 99793 99792 99790 99788 99786 99784 99782 99780 99778 99776 99774 99772 99770 99768 99766 99764 99762 99760 99758 99756 86 Degrees. TABLE H. Log. Sines and Tangents. (4°) Naiu.-al Sines. Sino. 8.843585 845387 847183 848971 850751 852525 854291 856049 857801 859546 861283 8.863014 864738 866455 868165 869868 871565 873255 874938 876615 878285 8.879949 881607 883258 884903 886542 888174 889801 891421 893035 894643 8.896246 897842 899432 901017 902596 904169 905736 907297 908853 910404 8.911949 913488 915022 916550 918073 919591 921103 922610 924112 925609 8.927100 928587 930088 931544 933015 934481 935942 937398 938850 94u296 Cosine 1). 10" 287 286 285 284 283 282 281 279 279 277 276 275 274 273 272 271 270 269 268 267 266 265 264 263 262 261 260 259 258 257 257 256 255 254 253 252 251 250 249 249 248 247 248 245 244 243 243 242 241 Cosine. 1.998941 998932 998923 998914 998905 998896 998887 998878 998869 998860 993851 1.998841 998832 998823 998813 993804 998795 998785 998776 998766 998757 1.998747 998738 998728 998718 998708 998699 998689 998679 998669 998659 1.998649 998639 998629 998619 998609 998599 998589 998578 998568 998558 .998548 998537 998527 998516 998508 998495 998485 998474 998464 998453 .998442 998431 998421 998410 998399 f 98388 998377 998366 998355 998344 Sine. I). 10" 1.5 1.5 1.5 1.5 1-5 1-5 1.5 1.5 1-5 1-5 1.5 1.5 1-5 1.6 1 1 1 1 1 1 1 J 1 1 1.6 1-6 1.6 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 17 1.7 1.7 1.7 1.7 1.8 1.8 1.8 1.8 1.8 1.8 1 1 8 8 1.8 Tang. >. 844644 846455 848260 850057 851846 853628 855403 857171 858932 860686 862433 5.864173 865906 867 632 869351 871064 872770 874469 876162 877849 879529 i. 881202 882869 884530 886185 887833 889476 891112 892742 894366 895984 i. 897596 899203 900803 902398 903987 905570 907147 908719 910285 911846 i. 91 3401 914951 916495 918034 919568 921096 922619 924136 925649 927156 !. 928658 930155 931647 933134 934616 936093 937565 939032 940494 941952 Cotanjr. D. 10/ 302 301 299 298 29/ 29 > 295 293 292 291 290 289 288 287 285 284 283 282 281 280 279 278 277 276 275 274 273 272 271 270 269 268 267 266 265 264 263 262 261 260 259 258 257 256 256 255 254 253 252 251 250 249 24y 248 247 246 245 244 244 243 Cotang. : N. sine. X. cos 11.155356 153545 151740 149943 148164 146372 144597 142829 141068 139314 137567 11.135827 134094 132368 130649 128936 127230 125531 123838 122151 120471 11.118798 117131 115470 113815 112167 110524 108888 1 07258 105634 104016 11.102404 100797 099197 097602 096013 094430 092853 091281 089715 088154 11.086599 085049 083505 081966 080432 078904 077381 075864 074351 072844 11.071342 089845 068353 066866 065384 063907 062435 060968 059506 058048 Tang-. 06976 07005 07034 07083 07092 07121 07150 07179 07208 07237 07266 99756 99754 99752 99750 99748 99746 99744 99742 99740 99738! 51 99736 50 07295 99734 i 1 08339 99652 •108368 99649 I 08397 99647 'i 08426 99644 ; 08455 99642 |: 08484 99689 •0851399637 1 08542 99635 :' 0857 199632 1 108600 99630 II 08629 9y627 1 108658 99625 08687 99622 108716 99619 49 07324 99731 07353 99729 07382 99727 0741199725 07440 99723 07469 99721 07498 99719 07527 99716 07556 99714 07585 99712 07614 99710 07643 99708 j 07672 99705 ! 107701 99703 ! i 07730 99701J 34 ! ; 07759 99699 33 1 107788 99696:82 !i 07817 99694 31 1 1 07846 99692 30 !! 07875 996891 29 i 1 07904 99387, 28 1 107933 99885 27 ; 1 07962 99683 10799199680 ! 108020 99878 ,,08049 99676 i! 08078 99873 J 08107 4*9671 108136 99668 1108165 99666! 19 ! | 08194 99864' 18 ;i 08223 99661 1 17 j J08252 99659 16 !; 08281 99657| 15 108310 99654 14 .\. cus. N.sine. ' 85. Degrees. 26 Log. Sines and Tangents. (5°) Natural Sines. —1 1 tabu: i?. D. 10" 3.940290 941 738 943174 944693 946034 947453 948874 950287 951693 953100 954499 3.955894 957284 958370 960052 961429 962801 964170 965534 968893 968249 3.969800 970947 972289 973628 974932 976293 977619 978941 980259 981573 8.982S33 984189 985491 988789 938083 989374 990360 991943 993222 994497 3.995768 997036 998299 999580 3.009816 002039 003318 004563 005805 097044 9.003278 009510 010737 011982 013182 014400 015613 016824 018031 019235 Cosine. 240 239 239 238 237 236 235 235 234 233 232 232 231 230 229 229 228 227 227 226 225 224 224 223 222 222 221 220 220 219 218 218 217 216 216 215 214 214 213 212 212 211 211 210 209 209 203 208 207 203 203 205 205 204 203 203 2U2 202 201 201 9. 993344 993333 998322 993311 998300 998289 998277 998266 998255 998243 998232 .998220 998209 998197 998186 998174 998163 998151 998139 998128 998116 .998104 998092 998030 998088 998053 998044 998032 998020 998008 997998 .997984 997972 997959 997947 997935 997922 997910 997897 997885 997872 .997860 997847 997835 997822 997809 997 797 997784 997771 997758 997745 .997732 997719 997706 997693 997680 997667 997654 997641 997628 997614 Sine. D. 10' 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.9 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.1 2.2 2.2 2.2 2.2 2.2 2.2 .941952 943404 944852 948295 947734 949168 950597 952021 953441 954856 956267 .957674 959075 960473 981866 983255 984639 966019 967394 968766 970133 .971496 972855 974209 975560 976908 978248 979586 980921 982251 983577 .984899 986217 987532 988842 990149 991451 992750 994045 995337 996624 .997908 999188 .000465 001738 003007 004272 005534 008792 008047 009298 •010546 011790 013031 014268 015502 016732 017959 019183 020403 021620 Cotang. 0. 10"| Coiang. N. sine. 242 241 240 240 239 238 237 237 236 235 234 234 233 232 231 231 230 229 229 228 227 226 22S 225 224 224 223 222 222 221 220 220 219 218 218 217 216 216 215 215 214 213 213 212 211 211 210 210 209 208 208 207 207 206 206 205 204 204 203 203 ill. 058048 05i>596 055148 053705 052266 050832 049403 047979 046559 045144 043733 11.042326 040925 039527 038134 036745 035361 033981 032606 031234 029867 11.028504 027145 025791 024440 023094 021752 020414 019079 017749 016423 11.015101 013783 012468 011158 009851 008549 007250 005955 004663 003376 11.002092 000812 t0. 999535 998262 996998 995728 994466 993208 991953 990702 10.989454 988210 686969 985732 984498 983268 983041 980817 979597 978380 0871699619 08745 99617 i 087 74 108808 ! 08831 108860 99614 99612 99609 99607 0888999604 : 108918 jl 08947 ! J 08976 09006 ! ! 09034 1 1 09063 ; 109092 '109121 99802 99599 99596 99594 99591 99588 99586 99583 I 09179 i ! 09208 i 109237 '09266 0915099580 99578 99575 99572 99570 i 09295199567 i 09324 89564 : 09353 j 09382|99559 10941 1199556 ! 09440J99553 1 09469199551 09498J99548 09527:99545 59 5H 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 33 37 36 o5 34 33 32 Tana. I:09556|99542l31 | i 09585199540 09614)99537 ! 109642199534 '0967199531 ; |09700i99528 '09729 99626 !' 03758 99523 j 09787 99520 ! | 09816 99517 1 1 09846(99614 ji 09874 i i 09903:99508 ; 109932 9950'; l| 09961 (99503 03990 99500 10019J99497 | 10048 99494 ii 10077 ! 110103 99488 | 10135 99485 I 10164 99482 I 10192)99479 '110221 99470 1025099473 10279 99470 I 10308 99467 110337 99464 10366 99401 1 10398 99458 110424 99455 10453 99452 N. cos. NVir>f. S4 Degree*. TABLE II. Log. Sines and Tangents. (6°) Natural Sines. Cotang. j N. sine. N. cos! Tang. iD. 10'' 021620 022834 024044 025251 026455 027655 028852 030046 031237 032425 033609 ,034791 035969 037144 038316 039485 040651 041813 .042973 044130 045284 , 046434 047582 048727 049869 051008 052144 053277 054407 055535 056659 ,057781 058900 060016 061130 062240 083348 064453 065556 066655 067752 ,068846 069038 071027 072113 073197 074278 075356 076432 077505 078576 .079644 080710 081773 082833 083891 084947 086000 087050 088098 089144 Cotang. 202 202 201 201 200 199 199 198 198 197 197 196 196 195 195 194 194 193 193 192 192 191 191 190 190 189 189 188 188 187 187 186 186 185 185 185 184 184 183 183 182 182 181 181 181 180 180 179 179 178 178 178 177 177 176 176 175 175 175 174 10.9783801, 977166 j 975956 97474911 973545 | j 972345 1 1 971148 I; 969954;! 968763 !| 967575 1 1 966391 1 ! 10.965209 ' 964031 962856 961684'| 960515 959349 958187 957027 955870 954716 10-953566 952418 951273 950131 94.8992 947856 946723 945593 944465 943341 10.942219 941100 939984 938870 937760 936652 935547 934444 933345 932248 10-931154 930062 928973 927887 926803 925722 924644 923568 922495 921424 10-920356 919290 918227 917167 916109 915053 914000 912950 911902 ! 910856 I Tang! 0453 0482 0511 0540 0569 059; 0626 0655 0684 0713 0742 0771 0800 0829 0858 0887 0916 0945 0973 1002 1031 1060 1089 1118 1147 1176 1205 1234 1263 1291 1320 1349 1378 1407 1436 1465 1494 1523 1552 1580 1638 1667 1696 1725 1754 1783 1812 1840 1869 1898 192' 1956 1985 2014 2043 2071 2100 2129 2158 218 99452 >9449 99446 99443 99440 99437 99434 99431 9y428 99424 99421 99418 99415 99412 99409 99406 99402 99399 99396 99393 99390 99386 99383 99380 99377 H9374 99370 99367 99364 99360 99357 99354 99351 99347 99344 99341 99337 99334 99331 99327 160999324 99320 99317 99314 99310 99307 99303 99300 99297 99293 99290 99286 99283 99279 99276 99272 99269 992o5 99262 99258 99266 N. ens. N.sine 83 Degrees. 28 Log. Sines and Tangents. (7°) Natural Sines. TABLE II. Sine. 9.035894 086922 087947 038970 089990 091008 092024 093037 094047 095056 096082 9.097065 098066 099065 100082 101056 102048 103037 104025 105010 105992 9.106973 107951 108927 109901 110873 111842 112809 113774 114737 115698 9.116656 117613 118567 119519 120469 121417 122362 123306 124248 125187 9.126125 127060 127993 128925 129854 130781 131706 132630 133551 134470 9.135387 136303 137216 138128 139037 139944 140850 141754 142655 143555 Cosine D. 10' 171 171 170 170 170 169 169 168 168 168 167 167 166 166 166 165 165 164 164 164 163 163 163 162 162 162 161 161 160 160 160 109 159 159 158 158 158 157 157 157 156 156 156 155 155 154 154 154 153 153 153 152 152 152 152 151 151 151 150 150 Cosine. D. lu' .996751 996735 996720 996704 996688 996673 996657 996641 996625 996610 996594 .996578 996562 996546 996530 996514 996482 996465 996449 996433 .996417 996400 996384 996368 906351 996335 996318 996302 996285 996269 .996252 996235 996219 996202 996185 996168 996151 996134 996117 996100 .996083 996066 996049 996032 996015 995998 995980 995963 995946 995928 .995911 995894 995878 995859 995841 995823 995808 995788 995771 995753 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.6 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.7 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.8 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 2.9 Sine. Taiiy;. .089144 090187 091228 092266 093302 094336 095367 096395 097422 098446 099468 , 100487 101504 102519 103532 104542 105550 106556 107559 108560 109559 .110556 111551 112543 113533 114521 115507 116491 117472 118462 119429 . 120404 121377 122348 123317 124284 125249 126211 127172 128130 129087 .130041 130994 131944 132893 133839 134784 135726 136667 137605 138542 . 139476 140409 141340 142269 143196 144121 145044 145966 146885 147803 Cotang. D. ju' 174 173 173 173 172 172 171 171 171 170 170 169 169 169 168 168 168 167 167 166 166 166 165 166 165 164 164 164 163 163 162 162 162 161 161 161 160 160 160 159 159 159 158 158 158 157 157 157 166 156 156 155 155 155 154 154 154 153 153 153 uiaag. kN. sine. N. cos 10.910356 909813 908772 907734 906698 905664 904633 903605 902578 901554 900532 10.899513 898496 897481 896468 895458 894450 893444 892441 891440 890441 10.889444 888449 887457 886467 885479 884493 883509 882528 881548 880571 10.879596 878623 877652 876683 875716 874761 873789 872828 871870 870913 10.869959 869006 868056 867107 866161 865216 864274 863333 862395 861458 10.860524 859591 858660 857731 856804 855879 854956 854034 853115 852197 12187 12216 12245 12274 12302 12331 12360 12389 12418 12447 12476 12504 12533 12562 12591 12620 12649 12678 12708 12735 12764 12793 12822 12851 12880 12908 12937 12966 12995 13024 13053 13081 13110 13139 13168 13197 13226 13254 13283 13312 13341 13370 99255 99251 99248 99244 99240 99237 99233 99230 9226 99222 99219 99215 99211 99208 99204 99200 99197 99193 99189 99180 99182 99178 99175 99171 99167 99163 99160 99156 99152 99148 99144 99141 99137 99133 99129 99125, 99122" 99118 99114 99110 99106 )9102 13899 f99098 13427 13456 13485 13514 13543 13572 13600 13629 13658 13687 13716 13744 13773 13802 13831 13860 13889 1S917 TllllL 99094 99091 99087 99083 99079 99076 99071 99067 99063 99059 99055 )9051 J9047 59043 99039 )9035 J9031 J9027 60 59 58 57 56 55 64 53 62 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 3D 29 28 27 26 25 24 •,'3 22 21 20 19 18 17 16 15 14 13 12 11 10 9 S 7 6 5 4 3 2 1 82 Degrees. 30 Log. Sines and Tangents. (9?) Natural Sines. TABLE II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 V 18 19 20 •2! 22 23 24 25 26 27 28 29 30 31 32 53 34 35 38 39 49 41 42 43 44 45 49 47 4o 48 50 51 52 53 54 55 56 57 58 59 60 Sine. .194332 195129 195925 196/19 197511 198302 199091 199879 200366 201451 202234 .203017 203797 204577 205354 208131 203906 207679 203452 209222 209992 .210760 211526 212291 213055 213818 214579 215338 216097 216854 217009 .218363 219116 219868 220318 221307 222115 222861 223600 224349 225092 ,225833 226573 227311 228048 228784 229518 230252 230984 231714 232444 .233172 233899 234625 235349 236073 236795 237515 238235 238953 239070 Cosine. I). 10' Cosine. D. 10 133 133 132 132 132 132 131 131 131 131 130 130 130 130 129 129 129 129 128 128 128 128 127 127 127 127 127 126 126 126 126 125 125 125 125 125 124 124 124 124 123 123 123 123 123 122 122 122 122 122 121 121 121 121 120 120 120 120 120 119 '.994620 994600 994580 994560 994540 994519 994499 994479 994459 994438 991418 '.994397 99437/ 994357 994336 994316 994295 994274 994254 994233 994212 .994191 994171 994150 994129 99410c 994087 994066 994045 994024 994003 .993981 993960 993939 993918 993896 9938/5 99385 i 993832 993811 993789 .993768 993746 993725 993703 993681 993660 993638 993616 993594 993572 .993550 994528 993506 993484 993462 993440 993418 -993396 993374 993351 Sine. 3.3 3.3 3.3 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.4 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.7 3.7 3.7 3.7 3.7 3.7 3.7 3.7 3.7 3.7 3.7 Tun g. .199713 20J529 201345 202159 202971 203782 204592 205400 206207 207013 20/817 .208619 209420 210220 211018 211815 212611 213405 214198 214989 215780 .216568 217356 218142 218926 219710 220492 221272 222052 222830 223606 ^.224382 225156 225929 226700 227471 228239 229007 229773 230539 231302 .232065 232826 233586 234345 235103 235859 236614 237368 238120 238872 .239622 240371 241118 241865 242610 243354 244097 244839 245579 246319 Cotang. 1). 10' 136 136 136 135 135 135 135 134 134 134 134 133 133 133 133 133 132 132 132 132 131 131 131 131 130 130 130 130 130 129 129 129 129 129 128 128 128 128 127 127 127 127 127 126 126 126 126 126 125 125 125 125 125 124 124 124 124 124 123 123 Cotang. i IN. sine. N. cos.l 10.800287 799471 798655 797841 797029 796218 795403 794600 793793 792987 792183 10.791381 790580 789780 788982 788185 787389 786595 785802 735011 784220 10.783432 782644 781858 781074 780290 779503 778728 777948 777170 776394 10.775618 774844 774071 773300 772529 771761 770993 770227 769461 768698 10.767935 767174 766414 765655 764897 764141 763386 762632 761880 761128 10.760378 759629 758882 758135 757390 756646 755903 755161 754421 753681 15643 15672 15701 15730 1575b 15787 15816 15845 15873 15902 15931 15959 159P8 16017 16046 16074 16103 16132 16160 j 16189 16218 16246 769 93764 98760 98755 93/51 98746 93741 98737 98732 98728 98723 98718 98714 98709 98704 98700 ^8695 98690 98686 98681 98676 98671 16275 93667 16304193662 16333J98U57 163;il98C,52 1639098648 16419 98643 16447 198638 || 16476 98633 j! 16505 98629 i! 16533198624 !ll6562:98(il9 ij 16591 198614 1662098609 i! 16648:98604 1667 7 198600 Tani j! 16708 98595 116734^8690 1 16763^8585 j| 16792 98580 ij 16820,98575 I ! 16849198570 16878J98566 1690698661 1693698686 16964 9:-; 551 I 16992 98646 li 17021 98541 17050 98536 ill7078|98531 I! 17107 98626 17136 98521 1716498516 17193 98511 17222 98506 17250 98501 17279 98496 17308 98491 17336198486 17365J98481 V cos.l N. sine, 80 Dsgraas. TABLE II. Log. Sines and Tangents. (10°) Natural Sines. 31 9 i Sine. 239670 240385 241101 241814 342526 243037 243947 244656 245363 246069 246775 1.247478 248181 248883 249583 250282 250980 251677 252373 253037 253761 ► .254453 255144 255834 256523 257211 257898 258583 259268 259951 200633 '.261314 261994 262673 263351 264Q27 264703 265377 266051 266723 267395 L268065 268 734 269402 270039 270/35 271400 272064 272726 273388 274049 .274708 275367 276024 276681 277337 277991 278644 279297 279948 280599 Cosine. D. lo"| Cosine, 1.993351 993329 993307 993285 993262 993240 993217 993195 993172 993149 993127 '.993104 993081 993059 993036 993013 992990 992967 992944 992921 992898 •992875 992852 993829 992806 992783 992759 992736 992713 992690 992666 .992643 992619 992596 992572 992549 992525 992501 992478 992454 992430 .992406 992382 992359 992335 992311 992287 992263 992239 992214 992190 .992166 992142 992117 992093 992069 992044 992020 991996 991971 991947 119 119 119 119 118 118 118 118 118 117 117 117 117 117 116 116 116 116 116 116 115 115 115 115 115 114 114 114 114 114 113 113 113 113 113 113 112 112 112 112 112 111 111 111 111 111 111 110 110 110 110 110 110 109 109 109 109 109 109 108 Sine. D. W 3.7 3.7 3.7 3.7 3.7 3.7 3.8 3.8 3.8 3.8 3.8 3.8 3.8 3.8 3 3 3 3 3 3 3, 3.8 3.8 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 3.9 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.1 4.1 Tan' ,246319 247057 247794 248530 249264 249998 250730 251461 252191 252920 253648 .254374 255100 255824 256547 257269 257990 258710 259429 260146 260863 .261578 262292 263005 263717 264428 265138 265847 266555 267261 267967 .268671 269375 270077 270779 271479 272178 272876 273573 274269 274964 .275658 276351 277043 277734 278424 279113 279801 280488 281174 281858 .282542 283225 283907 284588 285268 285947 286624 287301 287977 288652 D. 10"| Cotang. 123 123 123 122 122 122 122 122 121 121 121 121 121 120 120 120 120 120 120 119 119 119 119 119 118 118 118 118 118 118 117 117 117 117 117 116 116 116 116 116 116 115 115 115 115 115 115 114 114 114 114 114 114 113 113 113 113 113 113 112 Cotang. Degrees. IN.sine.lN. cos, 10.753681 75v943 75^205 751470 750736 750002 749270 748539 747809 747080 746352 10.745626 744900 744176 743453 742731 742010 741290 740571 739854 739137 10.738422 737708 736995 736283 735572 734862 734153 733445 732739 732033 10.731329 730625 729923 729221 728521 jj 727822 727124 726427 725731 725036 724342 723649 722957 722266 721576 720887 720199 719512 718826 718142 10.717458 716775 716093 715412 714732 714053 713376 712699 712023 711348 .7365198481 10 Tang. 98476 98471 98466 98461 98455 98450 98445 98440 98435 98430 98425 98420 98414 98409 98404 98399 983y4 98389 98383 98378 98373 98368 98362 98357 98S52 98347 98341 98336 98331 98325 18252.98320 18281 '983 J 5 18309 ! 983l0 IS338:9ho04 18367198299 1S395I98294 1842498288 18452 98283 17393 17422 17451 17479 17508 17537 17565 17594 17623 17651 17680 17708 17737 17766 17794 17823 17852 17880 17909 17937 17966 17995 18023 18052 18081 18109 18138 18166 18195 18224 18481 18509 18538 18567 18595 18624 18652 18681 98277 98272 98267 98261 98256 98250 98245 98240 18710 98234 18738;98229 18 767 198223 18795J98218 1882498212 1885298207 1888198201 1891098196 18938 98190 18967;98185 18995i'98179 19024I98174 19052198168 19081 [98163 N. cos. N.s-ine. > and Tangents. (11°) Natural Bines. TABLE II. Sin. ID. 10 1 2 3 4 5 6 7 4 9 10 11 12 13 14 15 16 r 18 19 20 21 22 23 24 20 2G 2; 24 •J') 30 31 32 33 34 30 36 3^ 38 39 40 41 42 43 41 45 46 4? 48 49 50 51 52 53 54 55 56 57 53 59 60 .280599 281248 281897 282544 283190 283836 284480 285124 285766 286408 287048 .287687 288326 288964 289600 290236 290870 291504 292137 292768 293399 .294029 294658 295286 295913 296539 29/164 297788 298412 299034 299655 .300276 300895 301514 302132 302748 303364 303979 304593 305207 305819 .306430 307041 307650 308259 308867 309474 310080 310685 311289 311893 .312495 313097 313698 314297 314897 315495 316092 316689 317284 317879 ('(.•sine 103 103 108 108 108 107 107 107 107 107 107 106 106 106 106 106 106 105 105 105 105 105 105 104 104 104 104 104 104 104 103 103 103 103 103 103 102 102 102 102 102 102 102 101 101 101 101 101 101 100 100 100 100 100 100 100 100 99 99 99 ID. i, .991947 991922 991897 991873 991848 991823 991799 991774 991749 991724 991699 .991674 991649 991824 991599 991574 991549 991524 991498 991473 991448 .991422 991397 991372 991346 991321 991295 991270 991244 991218 991193 .991167 991141 991115 991090 991064 991038 991012 990986 990960 990934 990908 990842 990855 990829 990803 990777 990750 990724 690697 990671 .990644 990618 990591 990565 990538 990511 990485 990458 990431 990404 Sine. 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.3 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.5 4.5 4.5 4.5 ,288652 24932 289999 290671 291342 292013 292682 293350 294017 294684 295349 .298013 298677 297339 298001 298662 299322 299980 300638 301295 301951 .302607 303261 303914' 304567 305218 305869 306519 307168 307815 308463 .309109 309754 310398 311042 311685 312327 312967 313608 314247 314885 .315523 316159 316795 317430 318064 318697 319329 319961 320592 321222 .321851 322479 323106 323733 324358 324983 325607 326231 326853 327475 D. IK) 112 112 112 112 112 111 111 111 111 111 111 111 110 110 110 110 110 110 109 109 109 109 109 109 109 108 108 108 108 108 108 107 107 107 107 107 107 107 106 106 106 106 106 106 108 105 105 105 105 105 105 105 104 104 104 104 104 104 104 104 Cotan<>;. 10 10.711344 71. 10 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4 4 4 4 4 4 4 4 4.6 4.6 4.6 4.6 4.6 4.6 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.7 4.8 4.8 4.8 4.8 4.8 4.8 4.8 4.8 4.8 4.8 4.8 4.9 4.9 4.9 9.327474 328095 328715 329334 329953 330570 331187 331803 332418 333033 333646 334259 334871 335482 336093 336702 337311 337919 338527 339133 339739 340344 340948 341552 342155 342757 343358 343958 344558 345157 345755 9.346353 346949 347545 348141 348735 349329 349922 350514 351106 351697 352287 352876 353465 354053 354640 355227 355813 356398 356982 357566 9.358149 358731 359313 359893 360474 361053 361632 362210 362787 363364 Cotang. 0. lu 103 103 103 103 103 103 103 102 102 102 102 102 102 102 102 101 101 101 101 101 101 101 101 100 100 100 100 100 100 100 100 99. 99. 99, 99, 99, Cocang. 10.672526 671905 671285 670666 670047 689430 668813 668197 667582 666967 666354 10.665741 665129 664518 663907 663298 662689 662081 661473 660867 660261 10.659656 659052 658448 657845 657243 656642 656042 655442 654843 654245 10.653647 653051 652455 651859 651265 650671 650078 649486 648894 648303 10.647713 647124 646535 645947 645360 644773 644187 643602 643018 642434 10.641851 641269 640687 640107 639526 638947 638368 637790 637213 636636 N. sine. iN . cos. 20791 97815 20820 97809 20848 97803 20877 97797 20905 97791 20933 97784 20962 97778 20990 97772 21019 97766 21047 97760 21076 97754 21104 9774- 21132 97742 2116197735 21189 97729 21218 97723 21246 97717 21275 97711 21303 97705 21331 97698 21360 97692 21388 97686 21417 97689 21445 97673 21474 97667 21502 97661 2153097655 21559 97648 ii 21587 97642 21616 97636 21644 97630 21672 97623 21701 ;976l7 21729 97611 j| 21758 97604 21786 97598 II 21814 97592 i!21843:97585 1 21871 197579 !21899 ! 97573 1 121928 197566 1121956 97560 ! |2198597553 1 122013197547 ,'22041:97541 122070:97534 122098 97528 | ! 22126 97521 i| 22155 97515 I' 22183 '97508 122212 97502 1 22240|97496 i 22268 97489 22297197483 22325197476 22353|97470 2238297463 22410197457 22438 97450 22467 22495 Tang. N. cos. N.sine. 97444 97437 77 Degrees. Log. Sines and Tangents. (13°) Natural Sines. TABLE II. Sine. D. 1U"| Cosine. 1 2 3 4 5 6 7 8 9 10 11 1-2 13 14 15 lb 1? IS 19 20 21 ■>■: 23 24 25 26 27 28 29 30 31 32 33 34 35 3(3 37 38 39 40 41 42 43 44 45 46 47 48 49 50 61 62 53 54 55 56 67 58 59 GO .352088 352635 353181 353726 354271 354815 355358 355901 356443 356984 357524 .358064 358603 359141 359578 360215 360752 361287 361822 362356 3W889 .363422 363954 364485 365016 365546 366075 366604 367131 367659 368185 .368711 369236 369761 370285 370808 371330 371852 372373 372894 373414 . 373933 374452 3749/0 375487 376003 376519 377035 377549 378063 378577 .379089 379601 380113 380624 381134 381643 382152 382661 383168 383675 Cosine. 91.1 91.0 90.9 90.8 90.7 90.5 90.4 90.3 90.2 90.1 89.9 89.8 89.7 89.6 89.5 89.3 89.2 89.1 89.0 88.9 88.8 88.7 88.5 88.4 88.3 88.2 88.1 88.0 87.9 87.7 87.6 87.5 87.4 87.3 87.2 87.1 87.0 86.9 86.7 86.6 86.5 86.4 86.3 86.2- 86.1 86.0 85.9 85.8 85.7 85.6 85.4 85.3 §5.2 85.1 85.0 84.9 84.8 84.7 84.6 84.5 .988724 938695 988666 988636 988607 988578 988548 988519 988489 988460 988430 .988401 988371 988342 988312 988282 988252 988223 988193 988163 988133 .988103 988073 988043 988013 987983 987953 987922 987892 987862 987832 .987801 987771 987740 987710 987679 987649 987618 987588 987557 987526 .987496 987465 987434 987403 987372 987341 987310 987279 987248 987217 '.987186 987155 987124 987092 987061 987030 986998 986967 986936 986904 Sine. ). 10' 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 4.9 5.0 5.0 5.0 5.0 5.0 5.0 6.0 5.0 6.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.1 6.1 5.1 5.1 5.1 1 1 1 2 2 5.2 5.2 5.2 5.2 5.2 5.2 6.2 5.2 5.2 5.2 5.2 5.2 6.2 6.2 I'm;. 9.363364 363940 364515 365090 365664 366237 366810 367382 367953 368524 369094 9.369663 370232 370799 371367 371933 372499 373084 373629 374193 374756 375319 375881 376442 377003 377563 378122 378681 379239 379797 380354 9.380910 381466 382020 382575 383129 383682 384234 384786 385337 385888 386438 386987 387536 388084 388631 389178 389724 390270 390815 391360 9.391903 392447 392989 393531 394073 394614 395154 395694 396233 396771 Cotam D. 10"| (Jotanj*. i N.sine -N. cos, 10 10 10 10 10 10 636636 636060 635485 634910 634336 633763 633190 632618 632047 631476 630906 630337 629768 629201 628633 628067 627501 626936 626371 625807 625244 624681 624119 623558 622997 622437 621878 621319 620761 620203 619646 ,619090 618534 617980 617425 616871 616318 615766 615214 614663 614112 ,613562 613013 612464 611916 611369 610822 610276 609730 60J185 603640 .608097 607553 607011 606469 605927 605386 604846 604306 603767 603229 TangT i 22495 : 22523 | 22552 ! 22580 | 22608 1 22637 ! 22665 | 22693 ! 22722 22750 97378 22778 22807 22835 22977 23005 23033 23062 97437 97430 97424 97417 97411 97404 97398 97391 97384 9737 M 97365 97358 22863'97351 22892197345 22920 973L-8 22948 97331 97325 97318 97311 97304 23090 97298 23118 23146 23175 23203 23231 23260 23288 23316 23345 23373 23401 23429 23458 23486 23514 23542 23571 23599 23627 23656 23684 23712 23740 97141 23769 23797 23825 23938 97291 97*84 97278 97271 97264 9/257 97251 97244 97237 97230 97223 97217 97210 97203 97 96 97189 ;>7182 97176 97169 97162 97155 97148 97134 97127 97120 23853 97113 23882 97103 23910 97100 •J7093 23966 9 70^6 23995 97079 24023 970/2 24051 97065 24079 97053 24108 97051 2413b 97044 24164 97037 2419-2 97030 n. cos. |n. \Q Degrees. TABLE II. Log. Sines and Tangents. (14°) Natural Sines. 35 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 SO 31 32 33 34 35 86 3? 3d 30 40 41 42 40 44 45 46 47 43 40 50 51 52' 53 54 55 50 57 58 50 60 9. .333675 384182 384687 385102 385697 380201 386704 38720/ 387709 388210 388711 .389211 389711 390210 390708 391206 391703 392199 392695 393191 393685 .394179 394673 395166 395658 396150 396641 397132 397621 398111 398000 .399088 399575 400062 400549 401035 401520 402005 402489 402972 403455 403938 404420 404901 405382 405862 406341 406820 407299 407777 408254 408731 409207 409682 410157 410632 411100 411579 412052 412524 412996 Cosine. L>. lu' 84.4 84.3 84.2 81.1 81.0 83.9 83.8 83.7 S3. 6 83.5 33.4 83.3 83.2 83.1 83.0 82.8 82.7 82.6 82.5 82.4 82.3 82.2 82.1 82.0 81.9 81.8 81.7 81.7 81,6 81.5 81.4 81.3 81.2 81.1 81.0 80.9 80.8 80.7 80.6 80.5 80.4 80.3 80.2 80.1 80.0 79.9 79.8 79.7 7y.6 79.5 79.4 79.4 79.3 79.2 79.1 79.0 78.9 78.8 78.7 78.6 9. .986904 986873 986841 986809 986778 936746 986714 986683 986651 980619 986587 .986555 986523 986491 986459 986427 986395 986363 986331 986299 986266 .986234 986202 986169 986137 986104 986072 986039 986007 985974 985942 985909 985876 985843 985811 985778 985745 985712 985679 985646 985613 985580 985547 985514 985480 985447 985414 985380 985347 985314 985280 985247 985213 985180 985146 985113 985079 985045 985011 984978 984944 D. lu ; Sim 5.2 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.4 5.4 5.4 5.4 5.4 6.4 6.4 5.4 5.4 5.4 5.4 5.4 5.4 5.4 5.5 5.5 5.5 5.5 5.5 5.5 5.6 5.6 5.5 5.5 5.5 5.5 5.5 5.6 5.5 5.6 5.6 5.6 5.6 Taas 5.6 5.6 5.6 5.6 9.396771 397309 397846 398383 398919 399455 399990 400524 401058 401591 402124 9.402656 403187 403718 404249 404778 405308 405836 406364 406892 407419 9.407945 408471 408997 409521 410045 410569 411092 411615 412137 412658 413179 413699 414219 414738 415257 415775 416293 416810 417326 417842 9.418358 418873 419387 419901 420415 420927 421440 421952 422463 422974 9.423484 423993 424503 425011 425519 426027 426534 427041 427547 428052 Co tang. D. 10'1 89.6 89.6 89.5 89.4 89.3 89.2 8.9.1 89.0 88.9 88.8 88.7 88.6 88.5 88.4 88.3 88.2 88". 1 88.0 87.9 87.8 87.7 87.6 87.5 87.4 87.4 87.3 87.2 87.1 87.0 86.9 86.8 86.7 86.6 86.5 86.4 86.4 86.3" 86.2 86.1 86.0 85.9 85.8 85.7 85.6 85.5 85.5 85.4 85.3 85.2 85.1 85.0 84.9 84.8 84.8 84.7 84.6 84.5 84.4 84.3 84.3 Cotang. X. sin-\ N. cos. 10.603229! 124192 602691 1 1 24220 602154 ; 24249 601617 1 1 24277 601081!! 24305 030 97023 97015 97003 97001 600545 6000101 59J476 | 598942 i 598409 ! 597876 ! 10.597344! 596813 ! 596282 ! 595751 I 595222 : 594692 ! 504164 i 5936361 • 593108 | 592581 10.592055 24333:90994 24302 24390 24418 24446 96987 96980 96973 96906 24474^6959 »£ 591529 | 24813 24503 24531 24559 24587 24615 24644 24672 24700 24728 24756 24784 691003 590479 ! 589955 j 589431 ! 588908 588385 687863 587342 10.586821 586301 585781 585262 584743 584225 583707 583190 582674 582158 10.581642 581127 680613 680099 679585 579073 ! 678560 ! 578048 ! 577537 I 577026 I 10.576516 576007 575497 24841 24869 24897 96851 24925 i 24954 ■' ll 24982 25010 25038 25006 25094 25122 25151 25179 25207 25235 25263 25291 25320 25348 25376 25404 25432 25460 25488 25516 25545 25573 25601 25629 25657 j 25685 574989! 25713 574481! 25741 573973! 25766 673466! 25798 572959 ! 25826 572453 1 1 25854 5 71948 ; 1 25882 "Tang. | IN 96952 96945 96937 96930 96923 96916 96909 96902 96894 96887 96880 96873 96866 96858 96844 96837 96829 96822 96815 96807 96800 96793 96786 96778 96771 96764 96756 96749 96742 96734 96727 96719 96712 96705 96697 96690 96682 96675 96667 96660 96653 96645 96638 96630 96623 96615 96608 96600 96593 N.nlne. 75 Degrees. 20 26 Log. Sines and Tangents. (15°) Natural Sines. TABLE II. bine. 9.412993 413467 413938 414408 414878 415347 415815 416283 416751- 417217 417684 9.418150 418615 419079 419544 420007 420470 420933 421395 421857. 422318 9.422778 423238 4236*7 424158 424615 425073 425530 425987 426443 426899 9.427354 427809 428263 428717 429170 429823 430075 430527 430978 431429 9.431879 432329 432778 433226 433675 434122 434569 435016 435462 435908 436353 436798 437242 437686 438129 438572 439014 439456 439897 440338 D. 1U"| 78.5 78.4 78.3 78.3 re.s 78-1 78.0 a .a 77.8 77.7 77.6 77.5 77.4 77.3 77.3 77.2 77.1 77.0 76.9 76.8 76.7 76.7 76.6 76.5 76.4 76.3 76.2 76.1 76.0 76.0 75.9 75.8 75.7 75.6 76 75 75 75 75 75 75 74.9 74.9 74.8 74 74 74 74 74 74 74 74.1 74.0 74.0 73.9 73.8 73.7 73.6 73.6 73.5 Cosine, j .984944 984910 984876 984842 984808 984774 984740 984706 984672 984637 984603 .984569 984535 984500 981466 984432 984397 984363 984328 984294 984259 .984224 984190 984155 984120 984085 984050 984015 983981 983946 983911 .983875 983840 983805 983770 983735 983700 983664 983629 983594 983558 .983523 983487 983452 983416 983381 983345 983309 983273 983238 983202 .983166 983130 983094 983058 983022 982986 982950 982914 982878 982842 D. 10' 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.7 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.8 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 5.9 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 6.0 9.428052 428557 429082 429566 430070 430573 431075 431577 432079 432580 433080 433580 434080 434579 435078 435576 436073 436570 437067 437563 438059 9.438554 439048; 439543 440036 440529 441022 441514 442008 442497 442988 9.443479 443968 444458 444947 445435 445923 446411 446898 447384 447870 .448356 448841 449326 449810 450294 450777 451260 451743 452225 452706 .453187 453668 454148 454628 455107 455586 456064 456542 457019 457496 Co tang. D. 10" N. sin 10.571948 571443 570938 570434 569930 569427 538925 568423 567921 567420 566920 10.566420 565920 565421 584922 564424 563927 563430 562933 562437 561941 10.561446 580952 560457 559964 559471 558978 558486 557994 557503 557012 10.556521 556032 555542 555053 554565 554077 553589 553102 552616 552130 10.551644 551159 550674 550190 549706 549223 548740 548257 547775 547294 10.546813 546332 545852 545372 544893 544414 543936 543458 542981 542504 25882 25910 2593- 2596o 25994 26022 26050 26079 26107 26135 26163 26191 26219 |26247 i 26275 | 26303 | 26331 j 26359 ! 26387 26415 26443 26471 26500 26528 26556 26584 26612 26640 26668 26696 26724 I 26752 26780 ; 26808 26836 ! 26864 | 26892 26920 26948 26976 27004 27032 27060 27088 27116 27144 27172 27200 27228 27256 27 27312 27340 27368 27396 27424 27452 274S0 27508 27536 27564 96593 96585 96578 96570 96562 96555 96547 96540 96532 96524 96517 96509 96502 96494 96486 98479 96471 96463 96456 96448 96440 96433 96425 96417 96410 9ti402 96394 96386 96379 96371 96363 96355 96347 96340 96332 96324 96316 96308 96301 96293 96285 96277 96269 96261 96253 96246 96238 96230 96222 96214 284 96206 96198 96190 96182 96174 96166 96158 96150 96142 96134 96126 Tang. | N. coP.jN.gme, 74 Degrees. TABLE II. Lou'. Sinefl and Tangents. (16°) Natural Sines. 37 3.410338 1 440778 .» 441218 fl 441058 4 44-20i)0 5 442535 442973 7- 443410 8 443847 9 444284 10 444720 11 9.445155 1-2 445590 13 446025 14 446459 16 446893 16 447326 17 447759 18 448191 19 448623 •20 449054 21 9.449485 22 449915 2:; 450345 24 450775 25 451204 26 451632 •27 452060 28 452488 29 452915 80 453342 81 9.453768 32 454194 83 454619 34 455044 85 455469 80 455893 87 456316 88 456739 89 457162 40 457584 41 9.458006 42 458427 48 458848 44 459268 45 459688 40 460108 47 460527 48 460946 49 461364 50 461782 51 9.462199 52 462616 58 463032 54 ' 463448 55 463864 50 464279 57 464694 58 465108 5!) 466522 00 465935 Cosine. I), lo' 73.4 73.3 73.2 73-1 73.1 73.0 72.9 72.8 72.7 72.7 72.6 72.5 72.4 72.3 72.3 72.2 72.1 72.0 72.0 71.9 71.8 71.7 71.6 71.6 71.5 71.4 71.3 71.3 71.2 71.1 71.0 71.0 70.9 70.8 70.7 70.7 70.6 70.5 70.4 70.4 70.3 70.2 70.1 70.1 70.0 69.9 69.8 69.8 69.7 69.6 69.5 69.5 69.4 69.3 69.3 69.2 69.1 69.0 69.0 68.9 .982842 982805 982709 982733 982690 982660 982624 982587 982551 982514 982477 .982441 982404 982367 982331 982294 982257 982220 982183 982146 982109 .982072 982035 981998 981961 981924 981886 981849 981812 981774 981737 .981699 981662 981625 981587 981549 981512 981474 981436 981399 981361 .981323 981285 981247 981209 981171 981133 981095 981057 981019 989^81 .980942 980904 980866 980827 980789 980750 980712 980673 980635 980596 Sine. D. 10" 6.0 0.0 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.1 6.2 6.2 6.2 6.2 6.2 6.2 6.2 9.457496 6.3 6.3 6.3 6.3 6.3 6.3 6.3 6.3 6.3 6.3 6.3 6, 6 6 6 6 (i 3 8 3 3 4 4 4 4 4 4 4 4 4 4 6.4 6.4 6.4 6.4 45 45 4589 4594 4598 458449 459400 73 49 i25 00 75 460349 460823 461297 461770 462242 9 462714 463186 463658 464129 464699 465069 465539 466008 466476 466945 467413 467880 468347 468814 469280 469746 470211 470676 471141 471605 9 472068 '472532 472995 473457 473919 474381 474842 475803 475763 476223 476683 477142 477601 478059 478517 478975 479432 479889 480345 480801 481257 481712 482167 482621 483075 483529 483982 484435 484887 485339 9. Cotang. D. 10" 79.4 79.3 79.3 79 2 79.1 79.0 79.0 78.9 78.8 78.9 78.7 78.6 78-5 78.5 78.4 78.3 78.3 78.2 78.1 78.0 78.0 77-9 77-8 77.8 77.7 77-6 77-5 77.5 77-4 77-3 77-3 77.2 77.1 77.1 7770 76.9 76.9 76.8 76.7 76.7 76.6 76.5 76.5 76.4 76.3 76.3 76.2 76.1 76.1 76.0 75.9 75.9 75.8 75.7 75.7 75.6 75.5 75.5 75.4 75.3 Cotain N. sine. X. cos. 10.5425041 542027 ; 541551 I 541075 ! 640600 540125 539651 639177 638703 538230 537758 10.537286 536814 636342 635871 535401 534931 534461 533992 633524 533055 10.532587 532120 531653 531186 630720 530254 629789 629324 528859 628395 10.527932 527468 527005 626543 526081 525619 525158 524697 524237 523777 10.523317 522858 522399 521941 521483 621025 520568 520111 519655 519199 10.518743 518288 517833 517379 516925 516471 516018 515565 515113 514661 27564 27592 27620 27648 27676 27704 27731 27759 27787 27815 27843 27871 27899 27927 27955 27988 28011 28039 28067 28095 28123 28150 28178 28206 28234 28262 28290 28318 28346 28374 28402 28429 28457 28485 28513 28541 28569 28597 28625 28652 28680 28708 28736 28764 28792 28820 28847 28875 28903 28931 28959 28987 29015 29042 29070 29098 29126 29154 29182 29209 29247 Tang. Il N. cos. N.sine. ' 96126 96118 96110 96102 96094 96086 96078 96070 96062 96054 96046 96037 96029 96021 96013 96005 y5997 95989 95981 95972 95964 95956 95948 95940 95931 95923 95915 95907 95898 95890 95882 95874 95865 95857 95849 95841 95832 95824 95816 y5807 95799 95791 95782 95774 95766 95767 95749 95740 95732 95724 95715 95707 95698 95690 95681 95673 95664 95656 95647 95639 95630 73 Degrees. 38 Log. Sines and Tangents. (17°) Natural Sines. TABLE II. Sine. |D. 10" 9 10 1! 12 13 14 16 10 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 3-2 33 34 86 86 37 38 39 40 41 42 43 44 45 40 47 48 40 50 51 5-2 53 54 55 56 57 58 59 00 9.465935 406348 466761 467173 467585 467996 468407 468817 469227 469637 470046 9.470455 470863 471271 471679 472036 472492 472898 473304 473710 474115 9.474519 474923 475327 475730 476133 476536 476938 477340 477741 478142 9.478542 478942 479342 479741 480140 480539 480937 481334 481731 482128 9.482525 482921 483316 483712 484107 484501 484895 485289 485682 486075 9.486467 486860 487251 487643 488034 488424 488814 489204 489593 489982 Cosine. 68.8 68.8 68.7 68.6 68.5 68.5 68.4 68.3 68.3 68.2 68.1 68.0 68.0 67.9 67.8 67.8 67.7 67.6 67.6 67.5 67.4 67.4 67.3 67.2 67.2 67.1 67.0 66.9 66.9 66.8 66.7 66.7 66.6 66.5 66.5 66.4 66.3 66.3 66.2 66.1 66.1 66.0 65.9 65.9 65.8 65.7 65.7 65.6 65.5 65.5 65.4 66.3 65.3 65.2 65.1 65.1 65.0 65.0 64.9 64.8 Cosine. 9.980596 980558 980519 980480 980442 980403 980364 980325 980288 980247 980208 ). 980169- 980130 980091 980052 980012 979973 979934 979895 979855 979816 >. 979776 979737 979697 979658 979618 979579 979539 979499 979459 979420 9.979380 979340 979300 979260 979220 979180 979140 979100 979059 979019 9.978979 978939 978898 978858 978817 978777 978736 978696 978655 978615 | 978574 978533 978493 978452 978411 978370 978329 978288 978247 978208 "sTneT D. 10" 6.4 6.4 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.5 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.6 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6.7 Tain 6.7 6.7 6.7 6.8 6.8 6.8 6.8 6.8 6.8 6.8 6.8 6.8 6.8 6.8 6.8 9.485339 485791 486242 486693 487143 487593 488043 488492 488941 489390 489838 9.490286 490733 491180 491627 492073 492519 492965 493410 493854 494299 9.494743 495186 495630 496073 496515 496957 497399 497841 468282 498722 9.499163 499603 500042 500481 500920 501359 501797 602235 502672 503109 9.503546 503982 504418 504854 505289 505724 508159 506593 507027 507460 '.507893 603326 508759 509191 609622 510054 510485 610916 511346 511776 Cotang. D. 10' 75.3 75.2 75.1 75.1 75.0 74.9 74.9 74.8 74.7 74.7 74.6 74.6 74.5 74.4 74.4 74.3 74.3 74.2 74.1 74.0 74.0 74.0 73.9 73.8 73.7 73.7 73.6 73.6 73.5 73.4 73.4 73.3 73.3 73.2 73.1 73.1 73.0 73.0 72.9 72.8 72.8 72.7 72.7 72.6 72.5 72.5 72.4 72.4 72.3 72.2 72.2 72.1 72.1 72.0 71.9 71.9 71.8 71.8 71.7 71.6 Cotanu - . i N. sine. IN. cos. 10.514661 J I 29237 514209 129265 613758 I 29293 513307 jj 29321 512857! '29348 612407 | 29376 611957;; 29404 511508 129432 5110591! 29460 510610 |29487 95622 95613 95605 95596 95588 95579 95571 95562 95554 5101621129515195645 10.509714 129543 609267 608820 508373 507927 507481 507035 503590 508146 605701 10.505257 504814 504370 503927 503485 503043 502601 502159 601718 501278 10.500837 600397 499958 499519 499080 498641 498203 497765 497328 496891 10.496454 496018 495582 495146 494711 494276 493841 493407 492973 492540!! 10.492107 491674 491241 490809 490378 489946 489515 489084 488654 488224 ! 29571 ! 29599 | 29626 ! 29654 i 29682 129710 | 29737 I 29765 1 29793 29821 95630 I 60 29849-95441 | 29876 j 29904 j 29932 ! 29960 12998/ (30015 j 30043 1300/1 30098 I 30126 130154 130182 I 30209 30237 30265 30292 30320 30348 30376 30403 30431 30459 30486 30514 30542 30570 3059/ 30625 30653 30680 30708 30736 3076 a 30791 30819 30846 30374 30902 95536 95528 95519 95511 95502 95493 95485 95476 95467 95459 95450 95433 95424 95415 95407 95398 95389 95380 95372 95363 95354 95345 95337 95328 95319 95310 95301 95293 95284 95275 95266 95257 95248 95240 95231 95222 95213 95204 95195 95186 95177 95168 95159 95150 95142 95133 95124 95115 95106 Tan<: N. cos. N.si 59 58 57 56 55 54 53 52 61 50 49 48 47 48 45 44 43 42 41 40 39 38 87 36 35 34 33 82 81 30 29 28 27 26 26 •24 23 22 21 20 19 18 17 16 16 14 13 12 11 10 9 8 7 6 5 4 8 2 1 7 9 Degrees. TABLE IT. jog. Sines and Tangents. (18°) Natural Sines. 39 JSllK'. 9.489982 490371 490759 491147 491535 491922 492308 492695 493081 493466 493851 9.494236 494621 495005 495388 495772 496154 496537 496919 497301 497682 9.498064 498444 498825 499204 499584 499963 500342 500721 501099 501476 9.501854 502231 502607 502984 503360 503735 504110 504485 504860 505234 9.505608 505981 506354 506727 507099 507471 507843 508214 508585 508956 9.509326 509696 510065 610434 510803 511172 511540 511907 512275 512642 D. 10' Cosine. 64.8 64.8 64.7 64.6 64.6 64.5 64.4 64.4 64.3 64.2 64.2 64.1 64.1 64.0 63.9 63.9 63.8 63.7 63.7 63.6 63.6 63.5 63.4 63.4 63.3 63.2 63.2 63.1 63.1 63.0 62.9 62.9 62.8 62.8 62.7 62.6 62.6 62.5 62.5 62.4 62.3 62.3 62.2 62.2 62.1 62.0 62.0 61.9 61.9 61.8 61.8 61.7 61.6 61.6 61.5 61.5 61.4 61.3 61.3 61.2 Cosine. .978206 978165 978124 978083 978042 978001 977959 977918 977877 977835 977794 .977752 977711 977669 977628 977586 977544 977503 977461 977419 977377 9.977335 977293 977251 977209 977167 977125 977083 977041 976999 976957 9.976914 976872 976830 976787 976745 976702 976660 976617 976574 976532 976489 976446 976404 976361 976318 976275 976232 976189 976146 976103 976060 976017 975974 975930 975887 975844 975800 975757 975714 975670 D. 10" Sine. 6.8 6.8 6.8 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 6.9 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.0 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.1 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 7.2 Tang. 9.511776 512206 512635 513064 513493 513921 514349 514777 515204 515631 516057 9.516484 516910 517335 517761 518185 518610 519034 519458 519882 I 520305 520728 521151 621573 621995 522417 522838 523259 523680 624100 624520 524939 525359 525778 526197 626615 627033 527451 627868 528285 528702 629119 529535 529950 530366 530781 531196 531611 532025 532439 532853 533266 533679 534092 534504 534916 635328 535739 536150 536561 636972 D. 10" Cotang. Cotang. IN.sine.iN. 10 10-47927S 10 10 10 488224 ! 487794 I 487365 j 486936 : 486507 ! 486079 ! 485651 | 485223 | 484796 484369 j 483943 I 483516 | 483090 j 482665 I 482239 I 481815 j 481390 | 480966 480542 J 480118 479695 479272 478849 478427 478005 47?583 477162 476741 476320 475900 475480 .476061 474641 474222 473803 473385 472967 472549 472132 471715 471298 .470881 470465 470050 469634 469219 468804 468389 467975 467561 467147 .466734 466321 465908 465496 465084 464672 464261 463850 463439 463028 Tang. 30929 30957 30985 31012 31040 31068 31095 31123 31151 31178 31206 31233 31261 31289 31316 31344 31399 31427 31454 31482 31510 31537 31565 31593 31620 31648 31675 31703 95097 95088 95079 95070 95061 95052 95043 95033 95024 95015 95006 94997 94988 94979 94970 94961 31372i94952 94943 94933 94924 94915 94906 94897 94888 94878 94869 94860 94851 94842 31730;94832 31758194823 31786194814 31813J94805 31841194795 31868|94786 31896194777 31923194768 3 1951 194758 31979|94749 32006194740 32034;94730 32061 194721 3208994712 32116194702 32144(94693 32171194684 32199J94674 32227 94665 32250 ! 94656 32282J94646 3230994637 32337 32364 32392 32419 32447 32474 32502 32529 32557 94627 94618 94609 94599 94590 94580 94571 94561 94552 N. cos.jN.sine, 71 Degrees. 43 Log. Sines and Tangents. (19°) Natural Sines. TABLE II. 9.512042 1 5i300y 2 513375 3 513741 4 514107 5 514472 6 514837 7 515202 8 515500 9 515930 10 510294 11 9.510057 12 517020 13 517382 14 517745 15 518107 16 518468 17 518829 18 519190 19 519551 20 519011 21 9.520271 22 520031 23 520990 24 521349 25 521707 26 522000 27 522424 28 522781 29 523138 31) 523495 31 9.523852 32 524208 33 524504 31 524920 35 525275 36 525680 37 525984 38 526339 39 520093 40 527040 41 9.527400 42 527753 43 528105 44 528458 45 528810 46 529101 47 529513 48 529804 4. iu J Cotang. ||N. sine.|N. cos.| 68.4 68.3 68.3 68.2 68.2 68.1 68.1 68.0 68.0 67.9 67.9 67.8 67.8 67.7 67.7 67.6 67.6 67.5 67.5 67.4 67.4 67.3 67.3 67.2 67.2 67.1 67.1 67.0 67.0 66.9 66.9 66.8 66.8 66.7 66.7 66.6 66.6 5 10.463028 3255 66, 00. 00. 00. GO. 66. 00. 66. 66.2 66.1 66.1 66.0 66.0 65.9 65.9 65.9 65.8 65.8 65.7 65.7 65.6 65.6 65.5 402018 :i258; 402208 401798 32039 401389 32667 400980 i 32694 400571 I 32722 400163 I 32749 459755 32777 459347; 32804 458939: 32832 10.458532 32859 458125 I 32887 457719 132914 457312 32942 450900 132909 450501 32997 456095 : 33024 455690 133051 455285! 33079 454881 33100 10.454476 33134 454072 j 33161 453669; 133189 4532651:33216 452862 !j 33244 452400 j 33271 452057 ! ' 33298 451055 33320 451253 33353 450851 i| 33381 10.450450 ! 33408 450049 133436 449648 33463 449248 33490 448848 33518 418448 ' 33545 448048 33573 447049 33000 447250 133627 446851 10.446452 440054 33710J94147 94552 94542 94533 94523 94514 94504 94495 94485 94476 94466 94457 94447 94438 94428 94418 94409 94399 94390 94380 94370 94361 94351 94342 94332 94322 94313 94303 94293 94284 94274 94264 94254 94245 94235 94225 94215 94206 94196 94186 94176 33655 94167 33682 94157 445656 33737 445259 33764 444861 33 792 444404 33819 444067 '133846 443671 443275 442879 10.442483 442087 441692 441298 440903 440509 33874 33901 33929 33950 94058 33983 34011 34038 34065 34093 440115 34120 439721 : 439327 34175 438934 34202 Tang. N. cos. X.sine 94137 94127 94118 94108 94098 94088 94078 94068 94049 94039 94029 94019 94009 93999 34147 93989 93979 93969 00 59 58 57 ^ 66 54 53 52 51 50 49 48 47 40 45 44 43 42 41 40 39 38 37 36 85 84 83 82 81 30 20 28 27 20 25 24 23 22 21 •JO 10 18 17 10 15 14 13 12 11 10 8 7 6 5 4 3 2 1 70 Degrees. TABLE II. Log. Sines and Tangents. (2C°) Natural Sines. 11 534399 534745 53509-2 635438 535783 536129 536474 536818 537163 53750/ 9.537851 538194 538538 538880 539223 539565 539907 540249 540590 540931 9.541272 541613 541953 542293 542632 542971 543310 543649 543987 544325 9.544663 545000 545338 545674 546011 546347 546683 547019 547354 547689 9.548024 548359 548693 549027 549360 549693 550026 550359 550692 551024 9.551356 551687 552018 552349 552680 553010 553341 553(570 554000 554329 Cosine. D. 10" Cosine. D. lu 57.8 57.7 57.7 57-7 57 57 5 V 57 57 57.3 57.3 57.2 57.2 57.1 57.1 57.0 57.0 56.9 56.9 56.8 56.8 56.7 56.7 56.6 56.6 56.5 56.5 56.4 56.4 56.3 56.3 56.2 56.2 56.1 56.1 56.0 56.0 55.9 55.9 55.8 55.8 55.7 55.7 55.6 55.6 55.5 55.5 55.4 55.4 55.3 55.3 55.2 55.2 55.2 55.1 55.1 55.0 55.0 54.9 54.9 9.972986 972940 972894 972848 972802 972755 972 70 J 9J 2663 972617 972570 972524 9.972478 972431 9*2385 972338 972291 972245 972198 972151 972105 972058 9.972011 971964 971917 971870 971823 971776 971729 971682 971635 971588 9.971540 971493 971446 971398 971351 971303 971256 971208 971161 971113 9.971066 971018 970970 970922 970874 970827 970779 970731 970683 970635 9.970586 970538 970490 970442 970394 970345 970297 970249 970200 970152 Sine. 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.8 7.9 7.9 7.9 7.9 7.9 7.9 7.9 7.9 7.9 7.9 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.1 8.1 8.1 8.1 Tang. 9.561066 561459 561851 562244 562636 563028 563419 563811 564202 564592 564983 9.565373 565763 566153 566542 566932 667320 567709 568098 568486 568873 9.569261 569648 570035 570422 570809 571195 571581 571967 572352 572738 9.573123 573507 573892 574276 574660 575044 575427 575810 676193 576576 9.576958 577341 577723 578104 578486 578867 579248 579629 580009 580389 9.580769 581149 581528 581907 582286 582665 683043 583422 583800 684177 Cotam D. 10 65.5 65.4 65.4 65 3 65.3 65.3 65.2 65.2 65.1 65.1 65.0 65.0 64.9 64.9 64.9 64 8 64.8 64.7 64.7 64.6 64.6 64.5 64.5 64.5 64.4 64.4 64.3 64.3 64.2 64.2 64.2 64.1 64.1 64.0 64.0 63.9 63.9 63.9 63.8 63.8 63.7 63.7 63.6 63.6 63.6 63.5 63.5 63.4 63.4 63.4 63.3 63.3 63.2 63.2 63.2 63.1 63.1 63.0 63.0 62.9 Co tang. 10.438934 438541 438149 437756 437364 436972 436581 436189 435798 435408 435017 10.434627 434237 433847 433458 433068 432680 432291 431902 431514 431127 10.430739 430352 429965 429578 42919*1 428805 428419 428033 427648 427262 10.426877 426493 426108 425724 425340 424956 424573 424190 423807 423424 10.423041 422659 422277 421896 421514 421133 420752 420371 419991 419611 10.419231 418851 418472 418093 417714 417335 416957 416578 416200 415823 Tang. I N. sine. 1 34202 ! 34229 ! 34257 ! 34284 3431 ! 34339 i 34366 ! 34393 i 34421 ! 34448 34475 34503 34530 3455 34584 34612 34639 34666 34694 34721 34748 34775 34803 34830 34857 34884 34912 34939 34966 34993 35021 35048 35075 35102 35130 35157 35184 35211 35239 35266 35293 35320 35347 35375 35402 35429 35456 35484 35511 35538 35565 35592 35619 35647 35674 35701 35728 35755 35782 35810 35837 N.coa. 93969 93959 93949 93939 93929 93919 93909 93899 93889 93879 93869 93859 93849 93839 93829 93819 93809 93799 93789 93779 93769 93759 93748 93738 93728 93718 93708 93698 93688 93677 93667 93657 93647 93637 93626 93616 93606 93596 93585 93575 93565 93555 93544 93534 93524 93514 93503 93493 93483 93472 93462 93452 93441 93431 93420 93410 93400 93389 93379 93368 93358 N. cos. N.sine. 69 Degrees. 4:2 Log. Sines and Tangents. (21°) Natural Sines. TABLE II. 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 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Sine. 3.554329 554658 554987 555315 555643 555971 556299 556626 556953 557280 557608 9.557932 558258 558583 558909 559234 55y558 559883 560207 560531 560855 9.561178 561501 561824 562146 582468 562790 563112 563433 563755 564075 9.564396 564716 565036 565356 565676 565995 566314 566632 566951 567269 9.567587 567904 568222 568539 568856 569172 569488 569804 570120 570435 9.570751 571066 571380 571695 572009 572323 572636 572950 573263 573575 Cosine. D. 10"[ Cosine. 54.8 54.8 54.7 54.7 54.6 54.6 54.5 54.5 54.4 54.4 54.3 54.3 54.3 54.2 54.2 54.1 54.1 54.0 54.0 53.9 53.9 53.8 53.8 53.7 53.7- 53.6 53.6 53.6 53.5 53.5 53.4 53.4 53 53 53 53 53 53.1 53.1 53.0 53.0 52.9 52.9 52.8 52.8 52.8 52.7 52.7 52.6 52.6 52.5 52.5 52.4 52.4 52.3 52.3 52.3 52.2 52.2 52.1 9.970152 970103 970055 970006 969957 969909 939860 969811 989762 969714 969665 3.969616 969567 969518 969469 969420 969370 969321 969272 969223 989173 3.969124 969075 969025 968976 968926 968877 968827 968777 968728 968678 >. 968628 968578 968528 968479 968429 968379 988329 968278 968228 968178 9.988128 9680 ?8 968027 967977 967927 967876 937826 967775 967725 967674 9.967624 987573 957522 967471 987421 967370 967319 987268 987217 96,166 D. 10"i Tang. 8.1 8.1 8.1 8.1 8.1 8.1 8.1 8.1 8.1 8.1 8.1 8.2 8.2 8 S 8 8 8 8 8 8 8 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.3 8.4 8.4 8.4 8.4 8.4 8.4 8.4 8.4 8.4 8,4 8.4 8.4 8.4 8.4 8.5 8.5 8.5 8.5 8.5 8.5 8.5 9.584177 584555 584932 585309 585888 588062 586439 586815 587190 587566 587941 9.588316 588891 589066 589440 589814 590188 590562 590935 591308 591681 9.592054 592426 592798 693170 593542 593914 594286 594656 595027 595398 9.595768 596138 596508 596878 597247 597616 597985 598354 598722 599091 599459 599827 600194 600562 600929 601296 601662 602029 602395 602761 9.603127 603493 603858 604223 604588 604953 605317 605682 ( 08046 608410 0. 10' Sine. Cotang. 62.9 62.9 62.8 62.8 62.7 62.7 62.7 62.6 62.6 62.5 62.5 62.5 62.4 62.4 62.3 62.3 62.3 62.2 62.2 62.2 62.1 62.1 62.0 62.0 61.9 61.9 61.8 61.8 61.8 61.7 61.7 61.7 61.6 61.6 61.6 61.5 61.5 61.6 61.4 61.4 61.3 61.3 61.3 61.2 61.2 61.1 61.1 61.1 61.0 61.0 61.0 60.9 60.9 60.9 60.8 60.8 60.7 60.7 60.7 60.6 Cotam 10.415823 415445 415038 414691 414314 413938 413561 413185 412810 412434 412059 10.411684 411309 410934 410560 410186 409812 409438 409065 408692 408319 10.407946 407574 407202 406829 406458 406086 405715 405344 404973 404602 10.404232 403862 403492 403122 402753 402384 402015 401646 401278 400909 10.400541 400173 399806 ! 399438 ! 399071 | 398704 ' 398338 ' 397971 397605 397239 J 10.396873 | 396507|j 396142!| 395777! j 395412 ; 395047 394683 394318 ! i 393954 ' 393590 :| 35837 35864 X. cos. 93358 93348 35891 93337 35918 35945 3597^ 36000 36027 36054 36081 36108 36135 36162 36190 36217 93211 36244 36271 36298 36325 93327 93316 93308 93295 93285 93274 93264 93253 93243 93232 93222 93201 93190 93180 3169 36352 93159 36379 36406 36434 36461 36488 36515 36542 36569 3659t 36623 36650 3667 36704 36731 36758 36785 36812 36839 3686 ! 36894 36921 1 36948 ! 36975 | 37002 J37029 37056 92881 1 37083 37110 i 37137 37164 137191 37218 37245 37272 37299 37326 37353 37380 37407 37434 37461 93148 93137 93127 y3116 93108 I 36 93095 I 35 93084 93074 93063 93052 93042 93031 93020 93010 92999 92988 92978 92967 92956 92945 92935 92926 92913 92902 92892 92870 92859 92849 92838 32827 92816 92805 92794 92784 92773 92762 92751 2740 92729 92718 Tang. IN. cos. A. sine 68 Decrees TABLE II. Log. Sines and Tangents. (22°) Natural Sines. 43 D. 10" Cosi .573575 573888 5 74200: 574512 574824 575136 575447 575/58 576059 570379 570089 .576999 577309 577018 577927 578230 578545 578853 579102 579470 579777 .580035 580392 580699 581005 581312 581618 681924 582229 fc.,, 582535 ~' 582840 .583145 583449 583754 584058 584301 584065 584968 585272 585574 585877 .586179 586482 586783 587085 587386 587688 587989 588289 588590 588890 .589190 589489 589789 590088 590387 590686 590984 591282 591580 591878 Cosine. ,967166 967115 967064 907013 966961 966910 906859 96880S 906756 966705 966653 .966602 966550 966499 966447 966395 966344 966292 966240 966188 966136 9.966085 966033 965981 965928 965876 965824 965772 965720 965668 965615 9.965563 965511 965458 965403 965353 965301 965248 965195 965143 965090 965037 964984 964931 964879 964826 964773 964719 964666 964613 964560 .964507 964454 964400 964347 964294 964240 964187 964133 964080 9J4026 Sme. D. 10' 8.5 8.5 8.5 8.5 8.5 8.5 8.5 8.5 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.6 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.7 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.8 8.9 8.9 8.9 8.9 8.9 8.9 8.9 8.9 8.9 8.9 8.9 8.9 Tans 606410 006773 607137 607500 607863 608225 608588 008950 609312 609674 610030 010397 610759 611120 611480 611841 612201 612561 612921 613281 613641 614000 614359 614718 615077 615435 615793 616151 616509 616867 617224 9.617582 617939 618295 618652 619008 619364 619721 620076 620432 620787 621142 621497 621852 622207 622561 622915 623269 623623 623976 624330 624683 625036 625388 625741 626093 626445 626797 627149 627501 627852 Cotang. D. 10" 60.6 60.6 60.5 60.5 60.4 60.4 00.4 60.3 60.3 60.3 60.2 60.2 60.2 60.1 60.1 60.1 60.0 60.0 60.0 59.9 59.9 59.8 59.8 59.8 59.7 59.7 59.7 59.6 59.6 59.6 59.5 59.5 59.5 59.4 59.4 59.4 59.3 59.3 59.3 59.2 59.2 59.2 59.1 59.1 59.0 59.0 59.0 58.9 58.9 58.9 58.8 58.8 58.8 58.7 58.7 58.7 58.6 58.6 58.6 58.5 Cotang. j N . sine.l N. cos . 10.393590 S93227 392863 392500 392137 391775 391412 39105t) 390888 390326 389964 10.389603 389241 388880 388520 388159 387799 387439 387079 386719 386359 10-386000 385641 385282 384923 384565 384207 383849 383491 383133 382776 10-382418 382061 381705 381348 380992 380636 380279 379924 379568 379213 10-378858 378503 378148 377793 377439 377085 376731 376377 376024 375670 10-375317 374964 374612 374259 373907 373555 373203 372851 372499 372148 37461192718 ! 37488 ! 37515 :; 37542 ! 37569 1137595 1 1 37622 37649 37676 37703 37730 37757 37784 37811 92707 92697 92686 92675 92664 92653 92642 92631 92620 92609 92598 92587 92576 37838 92565 37865 37892 37919 37946 37973 37999 38026 38053 38080 38107 38134 38161 38188 38215 38241 38268 38295 38322 38349 38376 38403 92554 92543 92532 92521 92510 92499 92488 92477 92466 92455 92444 92432 92421 92410 92399 92388 92377 92366 92355 92343 92332 3843092321 38456 38483 38510 38537 38564 3.8591 38617 38644 38671 38698 38725 38752 38778 38805 38832 38859 38886|92130 38912 92119 3893992107 J 92096 92310 92299 92287 92276 92265 92254 92243 92231 92220 92209 92198 92186 92175 92164 92152 92141 38966 38993 39020 39046 92085 92073 92062 39073|92050 Tang. | N. cos.) N.sine, 67 Degrees. 44 Log. Sines and Tangents. (23°) Natural Sines. TABLE II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1? IS 1!) 20 21 22 23 24 25 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 40 50 51 52 53 54 55 56 57 58 50 60 9.591878 592176 592473 592770 593087 593363 593659 593955 594251 594547 594842 9.595137 595432 595727 595021 596315 598609 596903 597196 597490 597783 598075 598368 598660 598952 599244 599536 599827 600118 600409 600700 600990 601280 601570 601860 602150 602439 602728 603017 603305 603594 9.603882 604170 604457 604745 605032 605319 605606 605892 603179 603465 .608751 607036 607322 607607 607892 608177 608461 608745 609029 609313 Cosine, i D. 10' 49.8 49.5 49.5 49.5 49.4 49.4 49.3 49.3 49.3 49.2 49.2 49.1 49.1 49.1 49.0 49.0 48.9 48.9 48.9 48.8 48.8 '48.7 48.7 48.7 48.6 48.6 48.5 48.5 48.5 48.4 48.4 48.4 48.3 48.3 48.2 48.2 48.2 48.1 48.1 48.1 48.0 48.0 47.9 47.9 47.9 47 8 47.8 47.8 47.7 47.7 47-6 47-6 47-6 47-5 47-5 47-4 47.4 47.4 47.3 47.3 Cosine. 1.964026 963972 963919 963865 963811 963757 983704 963650 963596 963542 963488 '.963434 963379 963325 963271 963217 963163 963108 963054 962999 962945 .962890 962836 962781 962727 962672 962617 962562 962508 962453 962398 .962343 962288 962233 962178 962123 962067 962012 961957 961902 961846 .961791 961735 961680 961624 961569 961513 961458 961402 961346 961290 .961235 961179 961123 961087 961011 960955 960899 960843 960786 930730 Sine. 8.9 8.9 8.9 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9.1 9 2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.2 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.3 9.4 9.4 T;ti! 45 44 43 42 41 40 39 37 06 35 34 33 32 31 30 29 •28 27 20 25 24 23 22 21 m 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 65 Degrees. 46 Log. Sines and Tangents. (25°) Natural Fines. TABLE II. Sine. ID. 10" 9. 625948 1 626219 2 625490 3 636750 4 627030 5 627300 6 627570 7 627840 8 628109 9 6283 ?8 If) 628647 11 9.628913 12 629185 13 629453 14 629721 15 620989 lii 630257 17 630524 18 630792 19 631059 20 631326 21 9.631593 22 631859 33 632125 2-i 632392 25 632658 36 632923 27 633189 38 633454 39 633719 30 633984 31 9.634249 32 634514 33 634778 34 635042 33 635305 36 635570 37 635834 38 63609? 39 636360 40 636623 •it 9.635886 43 637148 43 637411 44 637673 45 637935 46 638197 47 638458 48 638720 49 638981 60 639242 51 9.639503 5-2 639764 53 640024 54 640284 55 640544 56 640304 5? 641064 58 641324 59 641584 60 641842 Cosine. 45.1 45.1 45.1 45.0 45.0 45.0 44.9 44.9 44.9 44.8 44.8 44.7 44.7 44.7 44.6 44.6 44.6 44.6 44.5 44.5 44.5 44.4 44.4 44.4 44.3 44.3 44.3 44.2 44.2 44.2 44.1 44.1 44.0 44.0 44.0 43.9 43.9 43.9 43.8 43.8 43.8 43.7 43.7 43.7 43.7 43.6 43.6 43.6 43.5 43.5 43.5 43.4 43.4 43.4 43.3 43.3 43.3 43.2 43.2 43.2 9.8 9.8 9.8 9.8 9.8 9.8 9.9 9.9 Cos ine. ?D. 10" 9.957276 957217 957158 957099 957040 956981 956921 956862 956803 956744 956684 9.956625 956566 958598 956447 958387 958327 955268 956208 956148 956089 958029 955969 955909 955849 955789 955729 955S69 95560!) 955548 955488 955428 955368 955307 955247 955186 955126 955055 955005 954944 954883 9548-23 954762 954701 954640 954579 954518 954457 954396 954335 954274 954213 954153 954090 954029 953968 953906 953845 953783 953722 953660 9 9 9 9 9 9 9 9 9.9 9.9 9.9 9.9 9.9 9.9 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 10.1 10 10 Tang. i 1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.1 10.3 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.2 10.3 1.868673 669002 669332 669661 669991 670320 670849 670977 671308 671634 671983 '.672291 672619 672947 673274 673602 673929 674257 674584 674910 675237 1.675564 675890 676216 676543 676859 677194 677520 677846 678171 678496 '•678821 679146 679471 679795 680120 680444 680768 681092 681416 681740 1.682083 682387 682710 683033 683356 683679 684001 684324 684646 684968 1.685290 685612 685934 686255 686577 686898 687219 687540 687861 688182 Cotang. D. 10' Ootanjj. ; N .sine. a. cos 10.331327! 330998 ! 330568 330339 | 330009 339680 329351 329023 328694 328366 328037 10.327709 327381 327053 326726 326398 326071 ! 325743 j 325416 | 325090 j 324763 i 10.324436! 324110 ' 323784; 323457 ! 323131 | 322806! 322480 333154 321829; 321504| 10.3211791 320854 | 320529 i 320205 | 319880 | 319556 j 319232 I 318908 318584! 318260 i 10.317937! 317613! 317290 | 316967] 316644 316321 315999 315676 315354 315032 | ! 10.314710 j 314388!! 314036 | ! 313745 ! 313423 | i 313102,! 312781 || 312460!j 312139!| 311818 | 42262 42288 42315 42341 4236', 42894 42420 42446 42473 42499 42525 42552 42578 42604 42631 4265; 42683 42709 4273b 42762 42788 42815 42841 42867 42894 42920 42946 42972 42999 43036 43051 43077 43104 43180 43156 43182 43209 43235 43261 43287 43313 43340 43366 43392 43418 43445 43471 43497 43523 43549 43575 43602 43536 43654 43680 43705 4373o 43755 43785 43811 43837 Tang. II N. cos. N.f 90331 yoeis 90606 90594 90582 90569 90557 90545 90532 90520 90507 90495 90483 90470 90458 90446 90433 90421 90408 90396 90383 90371 90358 90346 90334 90321 90^0y 90296 90284 90271 90-259 90246 90233 90221 90208 90196 90183 90171 90158 90146 90183 90120 90108 900i;5 90082 900/0 90057 90045 90032 90019 90007 89994 89981 89968 89956 89943 89930 89918- 89905 89892 89879 64 De grees. ■CAliLT? IT. Sina. Log. Sines and Tangents. (-2CP) Natural Sinea D. 10"i Cosine 1 o 3 4 5 6 8 9 10 11 12 13 14 15 16 r, 18 19 20 21 23 24 25 26 27 u8 29 30 31 82 33 34 35 36 3T 38 39 40 41 42 43 44 45 46 47 48 49 53 51 62 53 51 55 66 57 68 59 30 ). 04 1842 042101 642360 642618 642877 643135 643393 643650 643908 644165 644423 ). 644680 644936 645193 645450 645703 645962 646218 646474 646729 648984 ). 647240 647494 647749 648004 648258 648512 648766 649020 649274 649527 1.649781 650034 650287 650539 650/92 651044 651297 651549 651800 652052 .652304 652555 652806 653057 653308 653558 653808 654059 654309 654558 .654808 655058 65530 1 655556 655805 656054 655392 ':■-.-■ 5 jl 650793 05/017 L). lo"' 43.1 43.1 43.1 43.0 43.0 43.0 43.0 42.9 42.9 42.9 42.8 42.8 42.8 42.7 42.7 42.7 42.6 42.6 42.6 42.5 42.5 42.5 42.4 42.4 42.4 42.4 42.3 42.3 42.3 42.2 42.2 42.2 42.2 42.1 42.1 42.1 42.0 42.0 42.0 41.9 41.9 41.9 41.8 41.8 41.8 41.8 41.7 41.7 41.7 41.6 41.6 41.6 41.6 41.5 41.5 41.5 41.4 41.4 41.4 41.3 9.95 953660 953539 953537 953475 953413 953352 953290 953228 953166 953104 953042 952980 952918 952855 952793 952731 952669 952608 952544 952481 952419 .952356 952294 952231 952168 952108 952043 951980 951917 951854 951791 .951728 951665 951602 951539 951476 951412 951349 951286 951222 951159 •951096 951032 950968 950905 959841 950778 950714 950650 950586 950522 .950458 95U394 950380 950366 950202 950138 950074 950010 949945 949881 Sine. 10.8 10.3 10.3 10.3 10.3 10 li) 10 10 10 10 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.4 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.5 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.6 10.7 10.7 10.7 10.7 10.7 10.7 10.7 10.7 10.7 10 Tang. 688182 688502 688823 689143 689463 689783 690103 690423 690742 691062 691381 691700 692019 692338 692656 692975 693293 693612 693930 694248 694566 694883 695201 695518 695836 696153 696470 696787 697103 697420 697736 698053 698369 698685 699001 699316 699632 699947 700263 7005/8 700893 01208 701523 701837 702152 702466 702180 703095 703409 703/23 704038 ). 704350 704663 704977 705290 705608 705916 706228 703541 706854 707166 * ofcang. 9.7 P. 10 ' 53.4 53.4 53.4 53.3 53.3 53.3 53.3 53.3 53.2 53.2 53.2 53.1 53.1 53.1 53.1 63.1 53.0 53.0 53.0 53. 52.9 52.9 52.9 52.9 52.9 52.8 52.8 52.8 62.8 52.7 62.7 52.7 62.7 52.6 52.6 52.6 52.6 52.6 52.5 52.5 62.5 52.4 52.4 52.4 52.4 52.4 52.3 52.3 52.3 52.3 52.2 52.2 52.2 52.2 52.2 52.1 52.1 52.1 52.1 52.1 Cotang. N. siue 10.311818 311498 311177: 310857 i 8105371 310217 ; 309897 309577 ! 309258 ! 308938 I 308619 | 10.308300! 307981 ! 307662 j 307344 | 307025 i 306707 | 306388 ! 306070 | 305752 I 305434 j 10.305117 | 304799 304482 | 304164 303847 303530 303213 302897 302580 302264 10-301947 301631 301315 300999 300684 300368 300053 299737 299422 299107 10-298792 298477 298163 297848 297534 297220 296905 296591 296277 285964 .295650 295337 295023 294710 294397 294084 293772 293459 293146 292834 10 Tang. ! 43837 43863 i 43889 | 43916 ! 43942 1 43968 I 43994 i 44020 ! 44046 : 44072 ! 44098 '44124 ; 44151 ! 44177 ; 44203 44229 44255 44281 44307 44333 44359 44385 44411 44437 44464 44490 44516 44542 44568 44594 44620 44646 44672 44698 44724 N. cos. 89879 89867 89854 89841 89828 89816 89803 89790 89777 89764 89752 89739 89726 89713 89709 89687 89674 89662 89649 89636 89623 89610 89597 89584 8957 1 89558 89545 89532 89519 89508 89493 89480 89467 89454 89441 4475089428 44776 89415 4480289402 44828 89389 44854 44880 44906 44932 44958 44984 45010 45036 45062 45088 45114 45140 45166 45192 45218 15243 45269 45295 45321 45347 45373 45599 89376 89363 89350 89337 89324 89311 89298 89285 89272 89259 89245 89232 89219 89206 B9193 89180 89167 89153 8yl40 89127 89114 89101 :\. cos. N.sine 60 69 58 57 56 55 54 63 52 51 50 49 4H 47 46 45 44 48 43 41 40 89 58 37 86 85 84 83 32 81 29 28 27 26 26 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 8 2 1 G3 Degrees. 48 Log. Slues and Tangents. (27°) Natural Sines. TABLE II. Sine. D. io 9.657047 1 657295 2 657542 3 657790 4 658037 6 658284 6 658531 7 658778 8 659025 9 659271 10 659517 11 9.659763 12 6601)09 13 660255 14 660501 16 660 743 16 660991 1? 661238 is 661481 19 661726 80 661970 21 9.662214 22 662459 23 662703 24 662946 25 663190 26 663433 27 663677 28 663920 29 664163 30 664406 31 9.664648 32 664891 33 665133 34 665375 35 665617 36 665859 37 666100 CS 666342 39 666583 40 666824 41 9.667055 42 667305 43 667546 44 667786 45 668027 46 668267 47 668506 4S 668746 49 668986 54 659225 51 9.669464 52 ' 669703 53 669942 54 670181 55 670419 56 670658 57 670896 58 671134 59 671372 60 671609 | Cosine. 41.3 41.3 41.2 41.2 41.2 41.2 41.1 41.1 41.1 41.0 41.0 41.0 40.9 40.9 40.9 40.9 40.8 40.8 40.8 40.7 40.7 40.7 40.7 40.6 40.6 40.6 40.5 40.5 40.5 40.5 40.4 40.4 40.4 40.3 40.3 40.3 40.2 40.2 40.2 40.2 40.1 40.1 40.1 40.1 40.0 40 40.0 39.9 39.9 39.9 39.9 39.8 39.8 39.8 39.7 39.7 39.7 39.7 39.6 39.6 9. 949881 949816 949752 949688 949623 949558 949494 949429 949364 949300 949235 949170 949105 949040 948975 948910 948845 948780 948715 948650 948584 948519 948454 948388 948323 948257 948192 948126 948060 947995 947929 947863 947797 947731 947665 947600 947533 947467 947401 947335 947269 947203 947 \36 947070 947004 946937 946871 946804 946738 946671 946604 946538 946471 946404 946337 946270 946203 946136 946069 946002 945935 10.7 10.7 10.7 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.8 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 10.9 11.0 11.0 11.0 11.0 11.0 11.0 11.0 11.0 11.0 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11. 11.2 11.2 11.2 11.2 11.2 D. n» .707166 707478 707790 708102 708414 708726 709037 709349 709660 709971 710282 .710593 710904 711215 711525 711836 712146 712456 712766 713076 713388 .713696 714005 714314 714624 714933 715242 715551 715860 if . 716168 \l\ A . 716477 & * .716785 !&| A . 717093 I "•* 717401 I b \-% 717709!° \'% 718017 «•» 718325 "-J 718633 j" -| 718940 « [l 719248 "J o 719555 .719862 720169 720476 52.0 52.0 52.0 52.0 51.9 51.9 51.9 51.9 51.9 51.8 51.8 51.8 51.8 51.8 51.7 51.7 51.7 51.7 51.6 51.6 51.6 51.6 51.6 51.5 51.5 51.5 51.5 51.4 51.2 51.2 51.2 51.1 7204 /o--r 720783 J}'} 721089 I °}'J 7213961°}-; 721702 J}'* 722009 °*" 722315 ■?{•" 722621 $H .722927 723232 723538 723844 724149 724454 724759 725065 725369 725874 51.0 51.0 50.9 50.9 50.9 50.9 50.9 50.8 50.8 50.8 Cotang. (Jotang. N. sine. N. cos. 10.292834 292522 292210 291898 291586 291274 290963 290651 290340 290029 289718 10.289407 289096 288785 288475 288164 287854 287544 287234 286924 286614 10.286304 285995 285686 285376 285067 284758 284449 284140 283832 283523 10.283215 282907 282599 282291 281983 281675 281367 281060 280752 280445 10.280138 279831 279524 .279217 278911 278604 278298 277991 277685 277379 10.277073 276768 276462 276156 275851 275546 275241 274935 274631 274326 45399 89101 45425 89087 I 4545189074 45477 89061 J '45503 89048 .i 45529 89035 t 45554 89021 .1145580 89008 > 4560688995 1:45632 88981 1145658 88968 45684 88955 45710 88942 45736 88928 J 45762 88915 1 45787 88902 1145813 88888 ■ji 45839 88875 145865 88862 4589188848 45917 88835 45942 88822 45968 88808 45994 8S795 46020 88782 46046 88768 46072 88755 46097 88741 46123 88728 46149 88715 46175 88701 46201 88688 46226 88674 46252 88661 46278 88647 46304 88634 46330 88620 46365 88607' 46381 88593 46407 88580 46433 88566 46458 88553 46484 88539 46510 88526 46536 88512 46561 88499 46587 88485 46613 88472 46639 88458 46664 88445 46696 88431 46716 88417 46742 88404 46767 88390 46793 88377 46819 88363 46844 88349 46870 88336 46896 88322 46921 88308 46947 8829f Tang. ! N. cos. 62 Degrees. TABLE II. Log. Sines and Tangents. (2S°) Natural Sines. 49 N. sine. IN. cos Siue. • .671639 671847 672084 67-2321 672558 672795 673032 673268 673505 673741 673977 '.674213 674448 674684 674919 675155 675390 675624 675859 676094 676328 '.676562 676796 677030 677264 677498 677731 677964 678197 678430 678663 .678895 679128 679360 679592 679824 680356 680288 680519 680750 680982 .681213 681443 681674 681905 682135 682365 682595 682825 683055 683284 .683514 683743 683972 684201 684430 684658 684887 685115 685348 685571 D. 10' 39.6 39.5 39.5 39 39 39 39 39 39 39 39 39 39 39.2 39.2 39.2 39.1 39.1 39.1 39.1 39.0 39.0 39.0 39.0 38.9 38.9 38.9 38.8 38.8 38.8 38.8 38.7 38 88 38 38 38 as 38 38.5 38.5 38.5 38.4 38.4 38.4 38.4 38.3 38.3 38.3 38.3 38.2 38.2 38.2 38.2 38.1 38.1 38.1 38.0 38.0 38.0 Cosine. .945935 945868 945809 945733 945666 945598 945531 945464 945396 • 945328 945261 .945193 945125 945058 944990 944922 944854 944786 944718 944650 944582 .944514 944446 944377 944309 944241 944172 944104 944036 943967 943899 .943830 943761 943693 943624 943555 943486 943417 943348 943279 943210 .943141 943072 943003 942934 942864 942795 942726 942656 942587 942517 .942448 942378 942308 942239 942169 942099 942029 941959 941889 941819 Sine. D. 10' Tang. 725674 725979 726284 726588 726892 727197 727501 727805 728109 728412 728716 72902C 729323 729626 729929 730233 730535 730838 731141 731444 731746 9.732048 732351 732653 732955 733257 733558 733860 734162 734463 734764 9.735066 735367 735668 735969 736269 736570 736871 737171 737471 737771 738071 738371 738671 738971 739271 739570 739870 740169 740468 740767 9.741066 741365 741664 741962 742261 742559 742858 743156 743454 743752 Cotang. D. 10" 50.8 50.8 50 7 50.7 50.7 50.7 50.7 50.6 50.6 50.6 50.6 50.6 50-5 50.5 50.5 50.5 50.5 50 50 50 50 50 50 50 50 50 50 50 50-2 50-2 50-2 50.2 50.2 50.1 50.1 50.1 50.1 50.1 50.0 50.0 50.0 50.0 50.0 49.9 49.9 49.9 49.9 49.9 49.9 49.8 49.8 49.8 49.8 49.8 49.7 49.7 49.7 49.7 49.7 49.7 Co tan; 10.274326 274021 273716 273412 273108 272803 272499 272195 271891 271588 271284 10.270980 270677 270374 270071 269767 269465 269162 268859 268556 268254 10.267952 267649 267347 267045 266743 266442 266140 265838 265537 265236 10.264934 264633 264332 264031 263731 263430 263129 262829 262529 262229 10.261929 261629 261329 261029 260729 260430 260130 259831 259532 259233 10.258934 258635 258336 258038 257739 257441 257142 258844 256546 256248 ; ! 46947 88295 46973 88281 4699988267 4702488254 4705088240 47076 88226 47101 88213 47127 88199 ,.47153 88185 "47178 88172 4720488158 47229 47255 47281 47306 47332 47358 47383 47409 47434 4746088020 47486 88006 47511 87993 47537 87979 47562 87965 47588 87951 47614 87937 88144 88130 88117 88103 88089 88075 88062 88048 88034 47639 47665 47690 i 47716 : ! 47741 47767 47793 47818 87923 87909 87896 87882 87868 87854 87840 87826 47844J87812 47869 87798 47895 87784 47920 87770 | 47946 ij 47971 47997 .48022 ;' 48048 I 48073 ;! 48099 87756 87743 87729 87715 87701 87687 87673 148124 87659 I 48150 87645 (48175187631 48201 87617 ! 48226 87603 : 48252 | 48277 ! 48303 : 48328 I 48354 148379 ! 48405 ! 48430 1 48456 48481 Tans. ! N. cos. N.sine 87589 87575 87561 87546 87532 87518 87504 87490 87476 87462 61 Degrees. 50 Log. Sines and Tangents. (29°) Natural Sines. TABLE II. Sine. 9.685571 1 685799 2 686027 3 686254 4 686482 5 686709 6 686936 7 687163 8 687389 9 687616 10 687843 11 ). 688059 12 688295 13 688521 14 688747 15 688972 16 689198 17 689423 18 689648 19 689873 20 690098 21 3.690323 22 690j48 23 690772 24 690996 25 691220 36 691444 27 691668 28 691892 29 692115 30 692339 31 9.692562 32 692785 33 693008 34 693231 35 693453 36 693676 37 693898 38 694120 39 694342 40 694564 41 9.694786 42 695007 43 695229 44 695450 45 695671 46 695892 47 696113 48 696334 49 696554 50 696775 51 9.696995 52 697215 53 697435 54 697654 55 697874 56 698094 57 698313 68 698532 69 698751 60 698970 D. 10"| Cosine. [ D. lu" 38.0 37.9 37.9 37.9 37.9 37.8 37.8 37.8 37.8 37.7 37.7 37.7 37.7 37.6 37.6 37.6 37.6 37.5 37.5 37.6 37.5 37.4 37.4 37.4 37.4 37.3 37.3 37.3 37.3 37.5 37.2 37.2 37.1 37.1 37.1 37.1 37.0 37.0 37.0 37.0 36.9 36.9 36.9 36.9 36.8 36.8 36.8 36.8 36.7 36.7 36.7 36.7 36.6 36.6 36.6 36.6 36.5 36.5 36.5 36.5 Cosine. .941819 941749 941679 941609 941539 941469 941398 941328 941258 941187 941117 .941046 940975 940905 940834 940763 940693 940622 940551 940480 940409 .940338 940267 940196 940125 940054 939982 939911 939840 939768 939697 .939625 939554 939482 939410 ,939339 939267 939195 939123 939052 938980 .938908 938836 938763 938691 938619 938547 938475 938402 938330 938258 1.938185 938113 938040 937967 937895 937822 937749 937676 937604 937531 Sine. 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.7 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.9 11.9 11.9 11.9 11.9 11.9 11.9 11,9 11.9 11.9 11.9 11.9 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.0 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 12.1 Tang. .743752 744050 744348 744645 744943 745240 745538 745835 746132 746429 746726 . 747023 747319 747616 747913 748209 748505 748801 749097 749393 749689 .749985 750281 750576 750872 751167 751462 751757 752052 752347 752642 .752937 753231 753526 753820 754115 754409 754703 754997 755291 755585 1.755878 756172 756465 756759 757052 757345 757638 757931 758224 758517 1.758810 759102 769395 759687 759979 760272 760564 760856 761148 761439 Co tang. D. 10" 49.6 49.6 49.6 49.6 49.6 49.6 49.5 49.5 49.5 49.5 49.5 49.4 49.4 49.4 49.4 49.4 49.3 49.3 49.3 49.3 49.3 49.3 49.2 49.2 49.2 49.2 49.2 49.2 49.1 49.1 49.1 49.1 49.1 49.1 49.0 49.0 49.0 49.0 49.0 49.0 48.9 48.9 48.9 48.9 48.9 48.9 48.8 48.8 48.8 48.8 48.8 48.8 48.7 48.7 48.7 48.7 48.7 48.7 48.6 48.6 Cotang. 10.256248 255950 255652 255355 255057 254760 254462 254165 253868 253571 253274 10.252977 252681 252384 252087 251791 251496 251199 250903 250607 250311 10.250015 249719 249424 249128 248833 248538 248243 247948 247653 247358 10.247063 246769 246474 246180 245885 545591 245297 245003 244709 244415 10.244122 243828 243535 243241 242948 242655 242362 242069 241776 241483 10.241190 240898 240605 240313 240021 239728 239436 239144 238852 238561 N. sine. X. cos. 87462 87448 87434 87420 87406 87391 87377 87107 .- 87093 ""' 87079 87064 87050 87036 87021 87007 86993 49344J86978 49S69|86964 49394*86949 4941986935 49445 86921 i 149470186906 4949586892 4952186878 49546 ! 86863 4957186849 49596 J86834 49622J86820 49647 86805 49672 49697 49723 49748 86791 86777 86762 86748 49798J86719 49824 86704 49849 86690 Tang. 49874 49899 49924 49950 86675 86661 86646 86632 499-15:86617 5000U|86603 I N. cos. L\. Pint- 60 69 58 57 56 55 54 53 52 61 50 49 4b 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 39 28 ■21 26 25 24 23 22 21 30 19 IS 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 60 Degrees. Log. Sines aril Tangents, (30°) Natural Sines. 51 s 4 5 6 7 8 9 10 11 12 13 14 16 16 1? 18 19 •Jo ■21 J2 23 24 25 26 27 28 25 31 31 32 33 34 36 36 37 3^ 30 •10 -41 4-2 43 44 45 46 4? 48 49 50 51 52 53 54 55 56 57 68 50 60 iD. 10' 9.698970 „, 699189 ™ 699497 I ^ 699626 i ™ 699844 : %* 700082 *■ 700280 \™ ■ 700498 \%°- 700716 1 5' 700933 S 701151 i^ .701388 ^ 701585 il 701802 ^ 702019 ™' 702236 H 702452 ™- 702669 %l 702885 ^ 703101 tl 703317 * 9.703533 Jg" 703749 JJ' 703984 g? 704179 *?' 704395 Jg' 704810 | r!?' 704825 |*?" 705040 I g 705254 | J 705469 I J 9.705683 *? 705898 j?? 706112 £ 706326 JJ' 705539 J5- 706753 J? r f 0596 7 r^S 707180 ;;?■ 707393 *>' 707603 £?■ "07819 Jg 35 35 35 36 35, 35 35. 35. 35. 9.7 708032 708245 708458 703670 703882 709094 709305 709518 709730 9.709941 710153 710J64 710575 710785 71096 7 711208 g 711419 jj? 711629 t? 711839 do Cosine. 35 35. 35 35 9.937531 937458 937385 937312 937238 937165 937092 937019 936946 936872 936799 9.936725 936652 936578 936505 938431 936357 936284 936210 936136 936062 .935988 935914 935840 935766 935692 935618 935543 935459 935395 935320 9.935246 935171 935097 935022 934948 934873 934798 934723 934649 934574 9.934499 934424 934349 934274 934199 934123 934048 933973 933898 933822 .933747 933671 933596 933520 933445 933369 933293 933217 933141 933056 D. 10' 12.1 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.2 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.3 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.4 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.6 12.6 12.6 12.6 12.6 12.6 12.6 12.6 12.6 12.6 12.6 Tan ^. 1.761439 761731 762023 762314 762603 762897 763188 763479 763770 784081 764352 '.764643 764933 765224 765514 765805 766095 766385 768675 766965 767255 . 767545 767834 768124 768413 768703 768992 769281 769570 769380 770148 .770437 770726 771015 771303 771592 771880 772168 772457 772745 773033 .773321 773608 773896 774184 774471 774759 775046 775333 775621 775908 .776195 776482 776769 777055 777342 777628 777915 778201 778487 778774 D. 10' Cotanjr. 48.6 48.6 48.6 48.6 48.5 48.5 48.5 48.5 48.5 48.5 48.4 48.4 48.4 48.4 48.4 48.4 48.4 48.3 48.3 48.3 48.3 48.3 48.3 48.2 48.2 48.2 48,2 48.2 48.2 48.1 48.1 48.1 48.1 48.1 48.1 48.1 48.0 48.0 48.0 48.0 48.0 48.0 47.9 47.9 47.9 47.9 47.9 47.9 47.9 47.8 47.8 47.8 47.8 47.8 47.8 47.8 47.7 47.7 47.7 47.7 N. sin 10.235357 : 50277 235037 jj 50302 234776 |! 50327 234488! '50352 50377 50103 50428 50453 50478 50503 50528 50553 50578 234195 233905 233615 233325 233035 232745 10.232455 232166 231876 231587 II 50503 231297 !| 50628 231008 |50854 230719 ij 50679 230430; 50704 230140 150729 229852 j 50754 10-229563j!50779 229274 'I 50304 228985 '50820 228697 1150854 228408 50879 228120 50904 227832 50929 227543 50954 227255 j 50979 226967! 51004 10-226679 I j 51029 226392: 1 51054 226104|; 51079 225816 | j 51104 225529 511589 85941 225241 ! 1 51 154 35926 224954 151179 85911 X. cos 36603 86588 86573 36559 36544 36530 36515 86501 86486 36471 38457 38442 88427 86413 36398 86384 86369 33354 86340 86325 36310 38295 86281 85266 88251 86237 86222 86207 86192 86178 86163 83148 86133 86119 86104 86089 86074 86059 86045 86030 86015 88000 85985 35970 85955 224667 224379 | 224092 i 10-223805' 223518 i 223231 i 222945 I 222658 I 222372 | 222085 | 221799 | 221512 221226 I Tang. 51204 51229 51254 51279 51304 51329 51354 51379 51404 51429 51454 51479 fl504 85898 85881 85868 85851 85836 85821 85808 85792 85777 85762 85747 85732 85717 N. cos. N.sine. Degrees. 21 Log. Sines and Tangents. (31 c ) Natural Sines. ) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 16 17 18 19 20 | 21 9 22 23 24 25 36 27 28 29 30 j Sine. .711839 712050 712260 712409 712679 712889 713098 713308 713517 713726 713935 .714144 714352 714561 714769 714978 715186 715394 715602 715809 716017 .716224 716432 716639 716846 717053 717259 717466 717673 717879 718085 .718291 718497 718703 718909 719114 719320 719525 719730 719935 720140 .720345 720549 720754 720958 721162 721366 721510 721774 721978 722181 .722385 722588 722791 722994 723197 723400 723603 723805 724007 724210 Cosine. lD. 10" 35.0 35.0 35.0 34.9 34.9 34.9 34.9 34.9 34.8 34.8 34.8 34.8 34.7 34.7 34.7 34.7 34.7 34.6 34.6 34.6 34.6 34.5 34.5 34.5 34.5 34.5 34.4 34.4 34.4 34.4 34.3 34.3 34.3 34.3 34 34 34 34 34 34 34 34 34 34.0 34.0 34 34.0 34.0 33.9 33.9 33.9 33.9 33.9 33.8 33-8 33.8 33.8 33.7 33.7 33.7 Uosme. .933036 932999 932914 932838 932762 932685 932609 932533 932457 932380 932304 .932228 932151 932075 931998 931921 931845 931768 931691 931614 931537 .931460 931383 931306 931229 931152 931075 930998 930921 930843 930766 1.930688 930611 930533 930456 930378 930300 930223 930145 930067 929989 ). 9299 11 929833 929755 929677 929599 929521 929442 929364 929286 929207 ). 929129 929050 928972 928893 928815 928736 928657 928578 928499 928420 Sine. - D. 10' 12.6 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.7 12.8 12.8 12.8 12.8 12.8 12.8 12.8 12.8 12.8 12.8 12.8 12.9 12.9 12.9 12.9 12.9 12.9 12.9 12.9 12.9 12.9 12.9 12.9 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.0 13.1 13.1 13.1 13.1 13.1 13.1 13.1 13.1 13.1 13.1 13.1 13.1 Tang. 778774 779030 779346 779332 779918 780203 780489 780775 781060 781346 781631 781916 782201 782480 782771 783056 783341 783626 783910 784195 784479 784764 785048 785332 785616 785900 786184 786468 786752 787036 787319 787603 787886 788170 788453 788736 789019 789302 789585 789868 790151 790433 790716 790999 791281 791563 791846 792128 792410 792692 792974 ,793256 793538 793819 794101 794383 794664 794945 795227 795508 795789 D. 10' (Jotam 47.7 47.7 47.6 47.6 47.6 47.6 47.6 47.6 47.6 47.5 47.5 47.5 47.5 47.5 47.5 47.5 47.5 47.4 47.4 47.4 47.4 47.4 47.4 47.3 47.3 47.3 47.3 47.3 47.3 47.3 47.2 47.2 47.2 47.2 47.2. 47.2 47.2 47.1 47.1 47.1 47.1 47.1 47.1 47.1 47.1 47.0 47.0 47.0 47.0 47.0 47.0 47.0 46.9 46.9 46.9 46.9 46.9 46.9 46.9 46.8 10 Cotang. Degrees. 10. 10 10 10 10 221226 220940 220654 220368 220082 219797 219511 219225 218940 218654 218369 218084 217799 217514 217229 216944 216659 216374 216090 215805 215521 215236 214952 214668 214384 214100 213816 213532 213248 212964 212681 212397 212114 211830 211547 211264 210981 210698 210415 210132 209849 209567 209284 209001 208719 208437 208154 207872 207590 207308 207026 .206744 206462 206181 205899 205617 205336 205055 204773 204492 204211 N.sine.jN. co.s. 51504185717 51529(85702 51554|85687 51579185672 51604)85657 51628185642 85627 j 51653 I 51678 ' 51703 51728 I 51753 151778 51803 85612 85597 85582 85567 85551 85536 51828185521 51852 51877 51902 51927 85506 85491 85476 85461 51952185446 51977185431 52002185416 52026185401 52051 185385 52076 52101 52126 52151 85370 85355 85340 85325 52175185310 52200185294 52225J85279 5225085264 52275i85249 52299J85234 52324 52349 52374 52399 52423 52448 52473 85218 85203 85188 85173 85157 85142 65127 52498 85112 52522 52547 52572 52597 52621 52646 52671 52696 52720 52745 52770 152794 i 52819 i 52844 1 52869 1 52893 | 52918 1 52943 |52967 I 52992 85096 85081 85066 85051 85035 85020 85005 84989 84974 84959 84943 84928 84913 84897 84882 84866 84851 84836 84820 84805 Tang. '' N. cos.JN.aine, ' TABLE II. Log. Sines and Tangents. (32°) Natural Sines. 53 l 2 S 4 6 6 7 8 9 10 11 12 13 14 16 16 17 18 19 20 •21 2-2 23 24 25 26 27 28 29 30 31 32 33 84 35 86 37 38 40 41 42 43 44 45 46 47 4S 40 50 51 52 53 54 55 50 57 68 69 60 Sine. .724210 724412 724614 724816 725017 725219 725420 725S22 725823 726024 726225 .726426 726626 726827 727027 727228 727428 727628 727828 728027 728227 .728427 728626 728825 729024 729223 729422 729621 729820 730018 730216 . 730415 730613 730811 731009 731208 731404 731602 731799 731996 732193 . 732390 732587 732784 732980 733177 733373 733569 733765 733961 734157 ,734353 734549 734744 734939 735135 735330 735525 735719 735914 736109 Cosine. D. 10" 33.7 33.7 33.6 33.6 33.6 33.6 33.5 33.5 33.5 33.5 33.5 33.4 33.4 33.4 33.4 33.4 33.3 33.3 33.3 33.3 33.3 33.2 33 33 33 33 33 33 33 33.0 33.0 33.0 33.0 33.0 32.9 32.9 32.9 32.9 32.9 32.8 32.8 32.8 32.8 32.8 32.7 32.7 32.7 32.7 32.7 32.6 32.6 32.6 32.6 32.5 32.5 32.5 32.5 32.5 32.4 32.4 Cosine. |D. 10' 9.928420 928342 928263 928183 928104 928025 927946 927867 927787 927708 927629 9.927549 927470 927390 927310 927231 927151 927071 926991 92691 1 926831 9.926751 926671 926591 926511 926431 926351 926270 926190 926110 926029 9.925949 925868 925788 925707 925626 925545 925465 925384 925303 925222 9.925141 925060 924979 924897 924816 924735 924654 924572 924491 924409 9.924328 924246 924164 924083 924001 923919 923837 923755 923673 923591 "Sine. - " 13.2 13.2 13.2 13.2 13.2 13.2 13.2 13 13 13 13 13 13 13.3 13.3 13.3 13.3 13.3 13.3 13.3 13.3 13.3 13.3 13.3 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.5 13.5 13.5 13.5 13.5 13.5 13.5 13.5 13.5 13.5 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.6 13.7 13.7 Tani?. 9.795789 796070 796351 796632 796913 797194 797475 797755 798036 798316 798596 9.798877 799157 799437 799717 799997 800277 800557 800836 801116 801396 9.801675 801955 802234 802513 802792 803072 803351 803630 803908 804187 9.804466 804745 805023 805302 805580 805859 806137 806415 806693 806971 9.807249 807527 807805 808083 808361 808638 808916 809193 809471 809748 9.810025 810302 810580 810857 811134 811410 811687 811964 ■812241 812517 "Cotang. D. 10' 46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.7 46.7 46.7 46.7 46.7 46.7 46.7 46.6 46.6 46.6 46.6 46.6 46.6 46.6 46.6 46.5 46.5 46.5 46.5 46.5 46.5 46.5 46.5 46.4 46.4 46.4 46.4 46.4 46.4 46.4 46.3 46.3 46.3 46.3 46.3 46.3 46.3 46.3 46.2 46.2 46.2 46.2 46.2 46.2 46.2 46.2 46.2 46.1 46.1 46.1 46.1 46.1 Co tail'. 10.204211 203930 203649 203368 203087 202806 202525 202245 201964 201684 201404 10.201123 200843 200563 200283 200003 199723 199443 199164 198884 198604 10.198325 198045 197766 197487 197208 196928 196649 196370 196092 195813 10.195534 195255 194977 194698 194420 194141 193863 193585 193307 193029 10.192751 192473 192195 191917 191639 191362 191084 190807 190529 190252 10.189975 189420 189143 188866 188590 188313 188036 187759 187483 N. s;iH\[_\. COS. 84805 84789 84774 84759 84743 84728 84712 84697 84681 84666 84650 84635 84619 } 152992 53017 53041 53036 53091 53115 53140 53164 53189 53214 53238 53263 53288 Tang. 53312 84604 ! 53337 84588 j 53361 84573 53386J84557 S53411J84542 j 53435 84526 5346084511 ! 53484 84495 ' 53509 84480 ! 53534 84464 1 53558 84448 53583 [84433 53007:84417 5363284402 5365684386 53681 184370 ; 53705 ! 84355 53730J84339 53754184324 I 53779184308 j 53804184292 |53828|84277 | 53853J84261 I 53877 [84245 153902 '84230 i 53926J84214 163951 '84198 J53975J84182 ! 54000 84167 { 54024 ! 84151 ! 54049 ! 84135 I54073J84120 154097184104 1 54122^84088 | 54146J84072 5417184057 54195:84041 54220J84025 5424484009 54269 83994 54293 183978 54317183962 54342 '83946 54366J83930 54S91 83915 5441583899 54440 83883 54464 83867 N. cos. N.sine. 57 Degrees. 54 Log. Sines and Tangents. (33°) Natural Sines. TABLE II. ■me. {P. 10 1 2 3 4 5 6 7 8 9 id ii 12 13 14 16 16 17 18 1!) w 21 22 23 24 25 26 21 28 2!) 30 31 32 33 34 35 36 37 38 89 40 41 42 4:1 44 45 46 47 48 49 50 51 52 53 64 55 54 57 58 54 60 K 736109 736303 736498 736692 736886 737080 737274 737467 737661 737855 738018 '.738241 738434 738027 738820 739013 739208 739398 7395J0 739783 739975 .740167 740359 740550 740742 740934 741125 741316 741508 741699 741889 .742080 742271 742462 742652 742342 743033 743223 743413 743602 743792 . 743982 744171 744361 744550 744739 744928 745117 745305 745494 745I;83 .745871 746059 746248 746436 746624 746812 746999 747187 747374 747562 Cosine. 32.4 32.4 32.4 32.3 32.3 32.3 32.3 32.3 32.2 32.2 32. 2 32.2 32.2 32.1 32.1 32.1 32.1 32.1 32.0 32.0 32.0 32.0 32.0 31.9 31.9 31.9 31.9 31.9 31.8 31.8 31.8 31.8 31.8 31.7 31.7 31.7 31.7 31.7 31.6 31.6 31.6 31.6 31.6 31.5 31.5 31.5 31.5 31.6 31.4 31.4 31.4 31.4 31.4 31.3 31.3 31.3 31.3 31.3 31.2 31.2 .923591 923509 923427 923345 923263 923181 923098 923016 922933 922851 922768 .922086 922603 922520 922438 922355 922272 922189 922105 922023 921940 .921857 921774 921691 921607 921524 921441 921357 921274 921190 921107 .921023 920939 920856 920772 920688 920304 920520 920436 920352 921)268 .920184 920J99 920015 919931 9lyS46 919762 919677 919593 919508 919424 .919339 919254 919169 919085 919000 918915 918830 918745 918659 9)8574 Sim-. L>. lo' 13.7 13.7 13.7 13.7 13.7 13.7 13.7 13.7 13.7 13.7 13.8 13.8 13.8 13.8 13.8 13.8 13.8 13.8 13.8 13.8 13.8 13.9 13.9 13.9 13.9 13.9 13.9 13.9 13.9 13.9 13.9 13.9 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.1 14.2 14.2 14.2 14.2 Tang. ,812517 812794 813070 813347 813623 813899 814175 814452 814728 815004 815279 815555 815831 816107 816382 816658 816933 817209 817484 817759 818035 818310 818585 818860 819135 819410 819684 819959 820234 820508 820783 9.821057 821332 821606 821880 822154 822429 822703 822977 823250 823524 823798 824072 824345 824619 824893 825166 825439 825713 825986 826259 9.826532 826805 827078 827351 827624 827897 828170 828442 828715 828987 Cotang. D. 10 46.1 46.1 46 1 46.0 46.0 46.0 46.0 46.0 46.0 46.0 46.0 45.9 45.9 45.9 45.9 45.9 45. 9H 45.9 45.9 45.9 45.8 45.8 45.8 45.8 45.8 45.8 45.8 45.8 45.8 45.7 45.7 45.7 45.7 45.7 45.7 45.7 45.7 45.7 45.6 45.6 45.6 45.6 45.6 45.6 45.6 45.6 45.6 45.5 45.5 45.5 45.5 45.6 45.5 45.5 45.5 45.5 45.4 45.4 45.4 45.4 Cotang 10 10 10 10 183893 | 1836181 183342 183067 182791 1825161 182241 ! 181965 j .181696! 181415! 181140;! 180865!; 180590 ! ! 180316 i' 180041 j 179766 h 179492 j! 1792171; ,178943 ij 178668 i ! 178394 178120 177846 j 177571 177297 ' 177023 176750 176476 ■ 176202 175928 i 1 175655H 175381 | j 175107! 174834 ! 174561 II 174287 174014 173741 173458 173195 172922 172649 172376 172103 171830 171558 171285 171013 54805:83645 ,54829 83629 ! ! 54854 83613 |j54878!83597 '54902183581 ! 54927183565 1 54951 ;83549 54975 83533 I54999J83517 ! 55024:83501 55048:83485 55072 !83469 5509i ! 83453 55121183437 ;55145 ! 83421 55169 ! 83405 55194 i 83389 55218|83373 55242 83o56 j! 55266 83340 I 55291 !83324 ■j55315 ! 83308 55339 183292 55363 832 i 6 55388 83260 55412 83244 55436;83228 55460183212 55484'83195 55509;83179 55533:83163 55557!83147 55581 ! 83131 55605 ^831 15 55630 '83008 55654:83082 i.5678 ! 830.,0 55702 83050 55726:83034 55750'83017 55775183001 55799I82L85 55823 82669 82953 82936 82920 55847 55871 55895 Tarn 55919182904 N. cos.JN.sine. 56 Degrees. Lop;. Sines and Tangents. (34°) Natural Sines. 55 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1? is 19 20 21 22 23 24 25 36 2? 28 29 80 81 32 33 84 35 86 37 88 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 66 57 58 59 60 .747502 747749 747936 748123 748310 748497 748683 748870 749056 749243 749426 .749815 749801 749987 750172 750358 750543 750729 750914 751099 751284 .751469 751654 751839 752023 752208 752392 752576 752760 752944 753128 .753312 753495 753679 753862 754046 754229 754412 754595 754778 754960 755143 755326 755508 755690 755872 756054 756236 756418 756600 756782 .756963 757144 757326 757507 757688 757869 758050 758230 758411 J758591 Cosine. D. 10" 9.7 31 31 31 31 31 31 31 31.1 31.0 31.0 31.0 31.0 31.0 30.9 30.9 30.9 30.9 30.9 30.8 30.8 30.8 30.8 30.8 30.8 30.7 30.7 30.7 30.7 30.7 30.6 30.6 30.6 30.6 30.6 30.5 30.5 30.5 30 5 30.5 30.4. 30.4 30.4 30.4 30.4 30.2 Cosine. 1.918574 918489 918404 918318 918233 918147 918032 917976 917891 917805 917719 1.917634 917548 917462 917376 917290 917204 917118 917032 916946 916859 .916773 916687 916600 916514 916427 916341 916254 916167 916081 915994 .915907 915820 915733 915646 915559 915472 915385 915297 915210 915123 .915035 914948 914860 914773 914685 914598 914510 914422 914334 914246 .914158 914070 913982 913894 913806 913718 913630 913541 913453 913365 ~SimT~ D. 10" 14.2 14.2 14.2 14.2 14.2 14.2 14.2 14.3 14.3 14.3 14.3 14.3 14.3 14.3 14.3 14.3 14.3 14.4 14.4 14.4 14.4 14.4 14.4 14.4 14.4 14.4 14.4 14.4 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.5 14.6 14.6 14.6 14.6 14.6 14.6 14.6 14.6 14.6 14.6 14.7 14.7 14.7 14.7 14.7 14.7 14.7 14.7 14.7 14.7 9. Tang. .828987 829260 829532 829805 830077 830349 830621 830893 831165 831437 831709 .831981 832253 832525 832796 833068 833339 833611 833882 834154 834425 834696 834967 835238 835509 835780 836051 836322 836593 836864 837134 ►.837405 837675 837946 838216 838487 838757 839027 839297 839568 839838 1.840108 840378 840647 840917 841187 841457 841726 841996 842266 842535 1.842805 843074 843343 843612 843882 844151 844420 844689 844958 845227 Cotang. D. 10" 45.4 45.4 45.4 45.4 Cotang. jN. sine X. cos. 45.4 45.3 45.3 45.3 45.3 45.3 45.3 45.3 45.3 45 45 45 45 45 45 45 45 45 45 45 45 45.1 45.1 45.1 45.1 45.1 45.1 45.1 45.1 45.1 45.1 45.0 45.0 45.0 45.0 45.0 45.0 45.0 45.0 45.0 44.9 44.9 44.9 44.9 44.9 44.9 44.9 44.9 44.9 44.8 44.8 44.8 44.8 44.8 10 10 10 10 10 171013 170740 170468 170195 169923 169651 169379 169107 168835 168563 168291 168019 167747 167475 167204 166932 166661 166389 166118 165846 165575 165304 165033 164762 164491 164220 163949 163678 163407 163136 162866 162595 162325 162054 161784 161513 161243 160973 160703 160432 160162 159892 159622 159353 159083 158813 158543 158274 158004 157734 157465 157195 156926 156657 156388 156118 155849 155580 155311 155042 154773 Ting- : 55919 | 55943 55968 ! 55992 I 56016 ! 56040 J 56064 56088 156112 | 56136 | 56160 56184 32904 8288' 82871 82855 82839 82822 82806 82790 82773 82757 82741 82724 56208 82708 56232 56256 56280 56305 56329 56353 56377 56401 56425 56449 56473 56497 56521 56545 56569 56593 56617 56641 56665 56689 56713 56736 56760 56784 56808 56832 56856 56880 56904 56928 56952 56976 57000 57024 57047 57071 57095 57119 57143 57167 57191 57215 57238 57262 57286 57310 57334 57358 82692 82675 82659 82643 82626 82610 82593 82577 82561 82544 82528 82511 82495 82478 82462 82446 82429 82413 82396 82380 82363 82347 82330 82314 82297 82281 82264 82248 82231 82214 82198 82181 82165 82148 82132 82115 82098 82082 82085 82048 82032 82015 81999 81982 81965 81949 81932 81915 X. cos. X sine, 55 Decrees. 5G Log. Sines and Tangents. (35°) Natural Sines. TABLE II. jsttne. 9.758591 758772 758952 759132 759312 759492 759672 759852 760031 760211 760390 9.760569 760748 760927 761106 761285 761464 761642 761821 761999 762177 9.762356 762534 762712 762889 763067 763245 763422 763600 763777 763954 9.764131 764308 764485 764662 764838 765015 765191 766367 765544 765720 9.765896 766072 766247 766423 766598 766774 766949 767124 767300 767475 9.767649 767824 767999 768173 768348 768522 768697 768871 769045 769219 Cosine. D. 10' 30.1 30.0 30.0 30 30. o 30.0 29.9 29.9 29.9 29.9 29.9 29.8 29.8 29.8 29.8 29.8 29.8 29.7 29.*? 29.7 29.7 29.7 29.6 29.6 29.6 29.6 29.6 29.6 29.5 29.5 29.5 29.5 29.5 29.4 29.4 29.4 29.4 29.4 29.4 29.3 29.3 29.3 29.3 29.3 29.3 29.2 29.2 29.2 29 20 ■29 ■29 29 ■29 29 29.0 29.0 29.0 29.0 29.0 Cosine. .913365 913276 913187 913099 913010 912922 912833 912744 912655 912566 912477 .912388 912299 912210 912121 912031 911942 911853 911763 911674 911584 .911495 911405 911315 911226 911136 911046 910956 910866 910776 910686 .91U596 910506 910415 910325 910235 910144 910054 909963 909873 909782 .909691 909601 909510 909419 909328 909237 909146 909055 908964 908873 1.908781 908690 9035-9 908507 908416 908324 908233 908141 903049 907958 Sine. U. 10" 4 4 4 4 4 4 4.8 4.8 4.8 4.8 4.8 4 4 4.9 4.9 4.9 4.9 4.9 4.9 4.9 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.2 5.2 5.2 5.2 5.2 5.2 5.2 5.2 5.2 5.2 5.3 5.3 5 . 3 5.3 6 . 3 Tanir. .845227 845496 845764 846033 846302 846570 846839 847107 847376 847644 847913 '.848181 848449 848717 848986 849254 849522 849790 850058 850325 850593 .850861 851129 851396 851664 851931 852199 852466 852733 853001 853268 .853535 853802 854069 854336 854603 854870 855137 855404 855671 855938 '.856204 856471 856737 857004 857270 857537 857803 858069 858336 858602 1.858868 859134 859400 859666 859932 860198 860464 860730 860995 861261 Cotang. 44.8 44.8 44.8 44.8 44.8 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.7 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.6 44.5 44.6 44.5 44.5 44.5 44.5 44.5 44.5 44.5 44.5 44.5 44.4 44.4 44.4 44.4 44.4 44.4 44.4 44.4 44.4 44.4 44.4 44.3 44.3 44.3 44.3 44.3 44.3 44.3 44.3 44.3 44.3 Cotang. I N. ?ine. N. cos. 10.154773 154504 154236 153987 153698 153430 153161 152893 152624 152356 152087 10.151819 151551 151283 151014 150746 150478 150210 149942 149675 149407 10-149139 148871 148604 148336 148089 147801 147534 147267 146999 146732 10-146465 146198 145931 145664 145397 145130 144863 144596 144329 144062 10-143796 143529 143263 142996 142730 142463 142197 141931 141664 141398 10-141132 140866 140600 140334 140068 139802 139536 139270 139005 138739 Tang. I 57358 j; 57381 I; 57405 I 57429 [ 57453 I I 57477 i 57501 i 57524 57548 57572 < 67596 81915 81899 81882 81865 81848 81832 81815 81798 81782 81765 81743 | 57643 81714 |57667 81698 57691 181681 57715J81664 67738181647 57762 81631 ,57786 81614 !57810 ! 81597 157833 81580 57857181563 5788181546 5790481530 5792881513 57952 l 81496 57976J81479 | 57999,81462 58023;81445 I 58047181428 158070 81412 58094 58118 58141 58165 58189 58212 58236 58260 81395 81378 81361 81344 81327 81310 81293 HI 276 58283 81259 58307 58330 58354 58378 58401 58425 58449 58472 58496 58519 58543 I 58567 1 58590 I 58614 81242 81225 81208 81191 81174 81157 81140 81123 81106 81039 81072 81055 81038 81021 58637 \s 1004 ! 58661 ! 58684 j 58708 158731 ! 58755 i 58 779 80987 80370 80953 30036 80019 809U2 N. C06. -'•..? 54 Degrees. TABLE II. Log. Sines and Tangents. (36°) Natural Sines. 57 9.769219 1 769393 2 769566 3 769740 4 769913 5 770087 6 770260 7 770433 6 770606 9 770779 10 770952 11 9.771125 13 771298 13 771470 11 771643 16 771815 16 771987 17 772159 18 772331 19 772503 30 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 31 775070 35 775240 36 775410 :)7 775580 38 775750 39 775920 40 776090 41 9.776259 42 776429 43 776598 44 776768 45 776937 46 777108 47 777275 48 777444 49 777613 50 777781 51 9.777950 52 778119 53 778287 54 778455 55 778824 56 778792 57 778960 58 779128 69 779295 60 779483 Cosine. D. 10" Cosine. 29.0 28.9 28.9 28.9 28.9 28.9 28.8 28.8 28.8 28.8 28.8 28.8 28.7 28.7 28.7 28.7 28.7 28.7 28.6 28.6 28.6 28.6 28.6 28.6 28.5 28.5 28.5 28.5 28.5 28.5 28.4 28.4 28.4 28.4 28.4 28.4 28.3 28.3 28.3 28.3 28.3 28 28 28 28 28 28 28 28 28 28 28 28.1 28.0 28.0 28.0 28.0 28.0 28.0 27.9 .907958 907866 907774 907682 907590 907498 907406 907314 907222 907129 907037 .908945 906852 906760 906667 906575 906482 906389 906296 906204 906111 .906018 905925 905832 905739 905645 905552 905459 905366 905272 905179 .905085 904992 904898 904804 904711 904617 904523 904429 904335 904241 .904147 904053 903959 903864 903770 903676 903581 903487 903392 903298 .903202 903108 903014 902919 902824 902729 902634 902539 902444 902349 Sine. D. 10" Tang. D. 10 15.3 15.3 15.3 15.3 15.3 15.3 15.3 15.4 15.4 15.4 15 15 15 15 15 15 15.4 15.5 15.5 15.5 15.5 15.5 15.5 15.5 15.5 15.5 15.5 15.5 15.6 15.6 15.6 15.6 15.6 15.6 15.6 15.6 15.6 15.6 15.7 15.7 15.7 15.7 15.7 15.7 15.7 15.7 15.7 15.7 15.7 15.8 15.8 15.8 15.8 15.8 15.8 15.8 15.8 15.8 15.9 15.9 9. .861261 861527 861792 862058 862323 862589 862854 863119 863385 863650 863915 .864180 864445 864710 864975 865240 865505 865770 866035 866300 866564 .866829 867094 867358 867623 867887 868152 868416. 868680 868945 869209 .869473 869737 870001 870265 870529 870793 871057 871321 871585 871849 872112 872376 872640 872903 873167 873430 873694 873957 874220 874484 874747 875010 875273 875536 875800 876063 876326 876589 876851 877114 Co tan c 44.3 44.3 44 2 44.2 44.2 44.2 44.2 44.2 44.2 44.2 44.2 44 44 44 44 44 44 44 44 44 44 44 44.1 44.1 44.1 44.1 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 44.0 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.9 43.8 43.8 43.8 43.8 43.8 43.8 43.8 Cotang. 10.138739 138473 138208 137942 137677 137411 137146 136881 136615 136350 136085 10.135820 135555 135290 135025 134760 134495 134230 133965 133700 133436 10.133171 132906 132642 132377 132113 131848 131584 131320 131055 130791 10.130527 130263 129999 129735 129471 129207 128943 128679 128415 128151 10.127888 127624 127360 127097 126833 126570 126306 126043 125780 125516 125253 124990 124727 124464 124200 123937 123674 123411 123149 122886 Tang. N. sine. N. cos 58779 80902 58802 80885 58826J80867 5884980850 5887380833 58896,80816 58920 80799 58943 80782 58967 80765 58990 80748 59014180730 59037180713 5906180696 5908480679 59108|80662 59131)80644 5915480627 5917880610 5920180593 59225 80576 59248 80558 5927280541 59295 ! 80524 59318!80507 59342 80489 59366 180472 59389)80455 59412J80438 59436:80422 59459|80403 59482:80386 59606180368 69529|80361 59552J80334 59576 80316 59599!80299 59622 J80282 59646 J80264 59669 [80247 59693 80230 5971680212 59739 80195 59763J80178 59786|80160 59809|80143 59832 8012 10 59856 59879 59902 59926 59949 80108 80091 80073 80056 80038 69972 J80021 | 59995 80003 60019179986 ! 60042|79968 I 60065 : 60089 j 60112 60135 160158 160182 ! N. cos* N.sine 79951 ?9934 ,"9916 ?9899 J 9881 79864 53 Degrees. 58 Log. Sines and Tangents. (37°) Natural Sines. TABLE II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Ifl 20 21 22 23 24 25 26 2 7 28 20 39 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 5L 52 53 54 55 56 57 58 50 00 Sine. 1.779463 779631 779798 779966 780133 780300 780467 780534 780801 780968 781134 .781301 781468 781634 781800 781966 782132 782298 782464 782630 782796 .782961 783127 783282 783458 783623 783788 783053 784118 784282 784447 .784612 784776 784941 785105 785269 785433 785597 785761 785925 786089 . 786252 786416 786579 786742 786905 787069 787232 787395 787557 787720 .787883 788045 788208 788370 788532 788694 788856 789018 789180 789342 Cosine. D. 10' 27.9 27.9 27.9 27.9 27.9 27.8 27.8 27.8 27.8 27.8 27.8 27.7 27.7 27.7 27.7 27.7 27.7 27.6 27.6 27.6 27.6 27.6 27.6 27.5 27.5 27.5 27.5 27.5 27 27 27 27 27 27 27 27.3 27.3 27.3 27.3 27.3 27.3 27.2 27.2 27^2 27.2 27.2 27.2 27 27 27 27 27 27 27 27.0 27.0 27.0 27.0 27.0 27.0 Cosine. |D. 10' 1.902349 902253 902158 902063 901967 901872 901776 901681 901585 901490 901394 •.901298 901202 901106 901010 900914 900818 900722 900626 900529 900433 '.900337 900242 900144 900047 899951 899854 899757 899660 899584 899467 '.899370 899273 899176 899078 898981 898884 898787 898689 898592 898494 1.898397 898299 898202 898104 898006 897908 897810 897712 897614 897516 1.897418 897320 897222 897123 897025 896926 896828 896729 896631 895532 Sine. Tans. 9.877114 877377 877640 877903 878165 878428 878691 878953 879216 879478 879741 9.880003 880265 880528 880790 881052 881314 881576 881839 882101 882363 9.882625 882887 883148 883410 883672 883934 884196 884457 884719 884980 9.885242 885503 885765 886026 885288 886549 886810 887072 887333 887594 9.887855 888116 888377 888639 888900 889160 889421 889682 889943 890204 1.890465 890725 890986 891247 891507 891768 892028 892289 892549 892810 Cotang. D. 10" Cotang. 10.122886 122623 122360 122097 121835 121572 121309 121047 120784 120522 120259 10.119997 119735 119472 119210 118948 118686 118424 118161 117899 117637 10.117375 117113 116852 116590 116328 116056 115804 115543 115281 115020 10.114758 114497 114235 113974 113712 113451 113190 112928 112667 112405 10.112145 111884 111623 111361 111100 110840 110579 110318 110057 109796 10.109535 109275 109014 108753 108493 108232 10 i 9 72 107711 107451 107190 jN.sine 60182 60205 60228 60251 60274 60298 60321 60344 60367 60390 60414 60437 60460 60483 60506 60529 60553 60576 60599 60622 60645 60668 60691 60714 60738 60761 60784 6080 60830 60853 60876 60899 60922 60945 60968 60991 61015 61038 61061 61084 61107 61130 61153 61176 61199 61222 61245 i 61268 i 61291 161314 61337 161360 161383 ! 61406 ! 61429 161451 j 61474 161497 | 61520 61543 ; 6156b N. cos 79864 79846 79829 79811 79793 79776 79758 79741 79723 79706 79688 79671 79658 79635 79618 79600 79583 79565 79547 79530 79512 79494 79477 79459 79441 79424 79406 79388 79371 79353 79335 79318 79300 79282 79264 79247 79229 79211 79193 .9176 79158 79140 79122 79105 79087 79069 79051 ,9033 79016 ;8998 78980 78962 78944 78926 78908 78891 78873 78855 78837 78819 78801 Tang, h N. eo8.jy.8toe 52 Degrees. TABLE II. Log. Sines and Tangents. (08°) Natural Sines. 59 l 3 3 4 5 6 7 8 9 10 11 12 13 14 15 16 I? lb 19 20 •21 22 33 24 25 26 27 28 29 30 31 \9 32 33 34 35 30 37 38 39 40 41 42 43 44 45 45 47 48 49 50 51 5-2 53 54 55 56 5 7 58 59 60 789342 789504 789665 789827 789988 790149 790310 790471 790832 799793 790954 791115 791275 791436 791596 791757 791917 792077 792237 792397 792557 ,792716 792876 793035 793195 793354 793514 793673 793832 793991 794150 .794308 794467 794626 794784 794942' 795101 795259 795417 795575 795733 .795891 796049 790206 790364 796521 790679 796836 796993 797150 797307 .797464 797021 797777 797934 798091 798247 798403 79S560 798716 798872 D, 10' 26.9 26.9 26.9 26.9 26.9 26.9 26.8 26.8 26.8 26.8 28.8 28.8 26.7 26.7 26.7 26.7 26.7 26.7 26.6 26.6 28.6 26.6 26.6 26.6 28.5 26.5 28.5 26.5 26.5 26.5 26.4 26.4 26.4 26.4 26-4 26-4 26 20 20 20 20 20 20 25 20 20 20 20 26 26-1 26-1 26 20 26 20 20 26 26.0 20.0 20.0 Cosine. Cosine. .898532 898433 896335 896238 896137 896038 895939 895840 895741 895641 895542 1.895443 895343 895244 895145 895045 894945 894846 894746 894646 894546 '.894446 894346 894246 894146 894046 893946 893846 893745 893645 893544 1.893444 893343 893243 893142 893041 892940 892839 892739 892638 892536 1.892435 892334 892233 892132 892030 891929 891827 891726 891624 891523 >. 891421 891319 891217 891115 8910U 890911 890809 890707 890505 890503 I). io r; 10 10 10 10 10 16 16.5 16.5 16.5 16.5 16.5 16.6 16.6 16.6 16.6 16.6 16.6 16.6 16.6 16.6 16.6 16.7 16.7 16.7 16.7 16.7 16.7 16.7 16.7 16.7 16.7 16.8 16.8 16.8 16 10 10 16 16 16.8 16.8 16.9 16.9 16.9 16.9 16.9 16.9 16.9 16.9 16.9 17.0 17.0 17.0 17.. 17.0 17.0 17.0 17.0 17.0 17.0 Tani .£92310 893070 893331 893591 893851 894111 894371 894632 894892 895152 895412 .895672 895932 896192 896452 896712 896971 897231 897491 897751 898010 .898270 898530 898789 899049 899308 899563 899827 900086 900340 900605 .900864 901124 901383 901042 901901 902100 902419 902679 902938 903197 1.903455 903714 903973 904232 904491 904750 905008 905267 905526 905784 (.900043 900302 900500 906819 907077 907330 907594 907852 908111 908369 Cotang. D. 10'- 43.4 43.4 43.4 43.4 43.4 43 43 43 43 43 43 43 43 43 43 43 43.3 43.3 43.3 43.3 43.3 43.3 43.3 43.3 43.2 43.2 43.2 43.2 43.2 43.2 43.2 43.2 43 43 43 43 43 43 43 43 43 43 43 43 43 43 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.1 43.0 Cotang. 10.107190 106930 106609 100409 106149 105889 105629 105368 105108 104848 104588 10.104328 104088 103808 103548 103288 103029 102769 102509 102249 101990 10.101730 101470 101211 100951 100692 100432 100173 099914 099654 099395 10.099136 098876 098617 098358 098039 097840 097581 097321 097062 096803 10.090545 090280 096027 095708 095509 095250 094992 094733 094474 094216 10.093957 093698 093440 093181 092923 092664 092406 092148 091889 091031 N. sine.iN. cos. | 61560,78801 78783 78705 78747 78729 78711 78694 78670 78658 78640 78622 61589 ! 61612 I 61635 J 61658 ! 61081 ! 61704 I 61726 II 61749 61772 61795 61818 78604 61841 61864 61887 61909 61932 78586 78568 78550 78532 78514 61355I78495 61978 78478 78460 78442 78424 78405 78387 78369 78351 78333 78315 78297 78279 78261 8243 8225 8200 8188 78170 78152 78134 78116 78098 78079 78001 78043 78025 8007 77988 77970 77952 77934 77916 77897 77879 77801 77843 77824 77808 77788 77769 77751 77733 77715 Tang. |! N. coy. N.sine 62001 1 1 62024 ! 1 62046 J ; 02069 62092 62115 62138 62160 62183 62208 62229 62251 62274 62297 62320 62342 62365 62388 62411 62433 62456 62479 625U2 62524 62547 62570 62592 ij 62615 I i 62638 j 1 62660 ; J62683 1 1 62706 ! 1 62728 C2751 j 1 62774 j! 62796 'J62819 [62842 | 62864 | 62887 ! 1 62909 62935 51 Degrees. 60 Log. Sines and Tangents. (39°) Natural Sines. TABLE II. 10 1 i 12 13 14 15 16 1? 18 19 20 21 •2-2 23 ■21 26 26 •27 28 29 30 31 32 33 34 35 36 3? 38 39 40 41 42 43 44 15 46 47 48 49 50 51 52 53 54 65 56 57 58 59 60 9.798772 799028 799184 799339 799495 799651 799803 799962 800117 800272 800427 9.800582 800737 800892 801047 801201 801356 801511 801665 801819 801973 3.802128 802282 802436 802589 802743 802897 803050 803204 803357 803511 >. 803664 803817 803970 804123 804276 804428 804581 804734 804886 805039 '.805191 805343 805495 805547 805799 805951 805103 808254 805405 805557 9.808709 805860 807011 807163 807314 807465 807615 807766 807917' 803067 Cosine. D. 10 I Cosine. 26.0 26.0 26.0 25.9 25.9 25.9 25.9 25.9 25.9 25.8 25.8 25.8 25.8 25.8 25.8 25.8 25.7 25.7 25.7 25.7 25.7 25.7 25.6 25.6 25.6 25.6 25.6 25.6 25.6 25.5 55.5 25.5 25.5 25.5 25-5 25.4 25.4 25.4 25.4 25.4 25.4 25.4 25.3 25.3 25.3 25.3 25.3 25.3 25.3 25.2 25.2 25.2 25.2 25.2 25.2 25.2 25.1 25.1 25.1 25.1 9.890503 890400 890298 890195 890093 889990 889888 889783 889682 889579 889477 9.889374 889271 889168 889054 888961 888858 888755 888651 888548 888444 9.888341 888237 888134 888030 887926 887822 887718 887614 887510 887406 9.887302 887198 887093 888989 83.885 886780 886676 836571 886466 886362 9.888257 888152 886047 885942 885837 885732 885627 885522 885416 885311 9.885205 885100 884994 884839 884783 884677 884572 881466 884360 884254 Bine. D. 10 17.0 17.1 17.1 17.1 17 17 !7 17 IV 17 17 17.2 17.2 17.2 17.2 17.2 17.2 17.2 17.2 17.2 17.3 17.3 17.3 17.3 17.3 17.3 17.3 17.3 17.3 17.3 17.4 17.4 17.4 17.4 17.4 17.4 17.4 17.4 17.4 17.4 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.6 17.6 17.6 17.6 17.6 17.6 17.6 17.6 17.6 17.6 Tans. 9.903369 903828 903888 909144 909402 909880 909918 910177 910435 910593 910951 9.911209 911487 911724 911982 912240 912498 912756 913014 913271 913529 9.913787 914044 914302 914560 914817 915075 915332 915590 915847 916104 916362 916619 916877 917134 917391 917648 917905 918163 918420 918677 9.918934 919191 919448 919705 919962 920219 920476 920733 920990 921247 9.921503 921760 922017 922274 922530 922787 923044 923300 923557 923813 Co tang. D. W 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 43.0 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.9 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.8 42.7 Co cam 10.091631 091372 091114 090856 090598 090340 090082 089823 089565 089307 089049 10.088791 088533 088276 088018 ' 087760 087502 087244 086986 086729 086471 10-086213 085956 085698 085440 085183 034925 084668 084410 084153 083896 10- 083638 083381 083123 082866 082609 082352 082095 081837 081580 081323 10-081066 080809 080552 030295 080038 079781 079524 079267 079010!! 0787531; 10.078497|| 07824011 077983! 077726 j! 077470J' 077213 076956 076700 076443 076187 62932 (.2955 6297? 63000J 63022 63045 63058 63090 63113 63135 63158 93180 63203 63225 63248 63271 63293 63316 63338 63361 63383 63405 63428 63451 63473 63496 63518 63540 63563 63535 63608 63630 63653 63675 63698 63720 63742 63765 63787 63810 63832 63854 6387 i 63899 63922 63944 63966 77715 77696 77678 77660 77641 77623 77605 77586 77568 77550 77531 77513 77494 77476 77458 77439 77421 77402 77384 77366 77347 77329 77310 77292 77273 255 77236 77218 77199 77181 77162 77144 77125 77107 77088 77070 77051 77033 77014 76996 76977 76959 76940 76921 76903 76884 7o868 63985176847 64011 76828 Tan!?. 64033 64056 64078 64100 64123 64145 64167 64190 64212 64234 64256 64279 N. coft 76810 76791 76772 76754 76735 76717 76698 76679 76661 76642 76623 76604 .N.Hinc. 50 Degrees. TABLE II. Log. Sines and Tangents. (40°) Natural Sines. Gl 1 2 3 4 5 6 7 8 9 10 11 is 13 14 15 16 17 18 19 £20 •21 22 ■23 •24 •25 86 2? 28 29 30 31 32 33 34 86 30 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 00 Sine. 9.803087 808218 808308 808519 808669 808819 808969 809119 809269 809419 809569 9.809718 809868 810017 810167 810316 810465 810614 810763 810912 811081 9.811210 811358 811507 811655 811804 811952 812100 812248 812396 812544 .812692 812840 812988 813135 813283 813430 813578 813725 813872 814019 .814166 814313 814460 814607 814753 814900 815046 815193 815339 815485 .815631 815778 815924 816069 816215 816361 816507 816652 816798 816943 Cosine. • D. 10" 25.1 25.1 25.1 25.0 25.0 25.0 25.0 25.0 25.0 24.9 24.9 24.9 24.9 24.9 24.9 24.8 24.8 24.8 24.8 24.8 24.8 24.8 24.7 24.7 24.7 24.7 24.7 24.7 24.7 24.6 24.6 24.6 24.6 24.6 24.6 24.6 24.5 24.5 24 24 24 24 24 24 24 24.4 24.4 24 24 24 24 24 24 24.3 24.3 24.3 24.3 24.2 24.2 24.2 Cosine. |D. 10" .884254 884148 884042 883936 883829 883723 883617 883510 883404 883297 883191 ,883084 882977 882871 882764 882657 882550 882443 882336 882229 882121 .882014 881907 881799 881692 881584 881477 881369 881261 881153 881046 .880938 880830 880722 880613 880505 860397 880289 880180 880072 879963 .879855 879746 879637 879529 879420 879311 879202 879093 878984 878875 .878766 878656 878547 878438 878328 878219 878109 877999 877890 877780 "Stile. 7.7 7.7 7.7 7.7 7.7 7.7 7.7 7.9 7.9 7.9 7.9 7.9 7.9 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8.0 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8.2 9. 8 8 a 8.2 8.3 8.3 8.3 8.3 Tarn .923813 924070 924327 924583 924840 925096 925352 925609 925865 926122 926378 .926634 926890 927147 927403 927659 927915 928171 928427 928683 928940 .929196 929452 929708 929964 930220 930475 930731 930987 931243 931499 '•931755 932010 932266 932522 932778 933033 933289 933545 933800 934056 .934311 934567 934823 935078 935333 935589 935844 936100 936355 936610 1.936866 937121 937376 937632 937887 938142 938398 938653 938908 939163 Cotang. D. 10' 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.7 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.6 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 Cotang. hN.sine 10.076187 075930 075673 075417 076160 074904 074648 074391 074135 073878 073622 10.073366 073110 072853 072597 072341 072085 071829 071573 071317 071060 10.070804 070548 070292 070036 069780 069525 069269 069013 068767 068501 10.068245 067990 067734 067478 067222 066967 066711 066455 066200 065944 10.065689 065433 065177 064922 064667 064411 064156 063900 063645 063390 10.063134 062879 062624 062368 062113 061858 061602 061347 061092 060837 64279 64301 64323 64346 64368 64390 64412. 64435 64457 64479 64501 64524 64546 64568 64590 64612 64635 64657 64679 64701 64723 64746 64768 64790 64812 64834 64856 64878 64901 64923 64945 64967 . 64989 1165011 65033 65055 65077 65100 65122 65144 65166 65188 65210 65232 65254 65276 65298 65320 65342 65364 65386 65408 65430 i 1 65452 '65474 v 65496 i! 65518 65540 ! ! 65562 |65§84 I 656 06 N, cos. 76604 76586 6567 76548 76530 76511 76492 76473 76455 76436 76417 76398 76380 76361 76342 76323 6304 76286 6267 76248 76229 76210 76192 76173 76154 76135 76116 76097 76078 76059 76041 76022 76003 75984 75965 75946 75927 75908 75889 75870 75851 75832 75813 75794 75775 75/56 75738 76719 75700 75680 75661 75642 75623 75604 75585 75566 75547 75528 75509 75490 75471 Tang. N. eos. N.sine, 49 Degrees. 6:2 Log. Sines and Tangents. (41°) Natural Sines. TABLE IT. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 •2-2 23 24 25 2o 27 28 29 30 31 32 33 31- 35 36 3? 38 39 40 41 42 43 44 45 46 47 48 4!) 50 51 52 53 54 55 50 57 58 59 60 9.816943 817088 817233 817379 817524 817668 817813 817958 818103 818247 818392 9.818536 818681 818825 818969 819113 819257 819401 819545 819689 819832 9.819976 820120 820263 820403 820550 820893 820836 820979 821122 821265 9.821407 821550 821693 821835 821977 822120 822262 822404 822546 822688 9.822830 822972 823114 823255 823397 823539 823680 823821 823963 824104 9.824245 824386 824527 824668 824808 824949 825090 825230 825371 825511 Cosine. U. 10 24.2 24.2 24.2 24.2 24.1 24.1 24.1 24.1 24.1 24.1 24.1 24.0 24.0 24.0 24.0 24.0 24.0 24.0 23.9 23.9 23.9 23.9 23.9 23.9 23.9 23.8 23.8 23.8 23.8 23.8 23.8 23.8 23.8 23.7 23.7 23.7 23.7 23.7 23.7 23.7 23.6 23.6 23.6 23.6 23.6 23.6 23.6 23.5 23.5 23.5 23.5 23.5 23.5 23.5 23.4 23.4 23.4 23.4 23.4 23.4 Cosine. 9.877780 877670 877560 877450 877340 877230 877120 877010 876899 876789 876678 9.876568 876457 876347 876236 876125 876014 875904 875793 875682 875571 9.875459 875348 875237 875126 875014 874903 874791 874680 874568 874456 9.874344 874232 874121 874009 873896 873784 873672 873560 873448 873335 9.873223 873110 872998 872885 872772 872659 872547 872434 872321 872208 9.872095 871981 871868 871755 871641 871528 871414 871301 871187 871073 D. 10" Sine. 18.3 18.3 18.3 18.3 18 18 18 18 18 18 18 18.4 18.4 18.4 18.5 18.5 18.5 18.5 18.5 18.5 18.5 18.5 18.5 18.5 18.6 18.6 18.6 18.6 18-. 6 18.6 18.6 18.6 18.7 18.7 18.7 18.7 18.7 18.7 18.7 18.7 18.7 18.7 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.9 18.9 18.9 18.9 18.9 18.9 18.9 18.9 18.9 Tan?:. 939163 939418 939673 939928 940183 940438 940394 940949 941204 941458 941714 9.941968 942223 942478 942733 942988 943243 943498 943752 944007 944252 944517 944771 945026 945281 945535 945790 948045 946299 948554 946803 9.947053 947318 947572 947826 948081 948336 948590 948844 949099 949353 9.949607 949862 950116 950370 950625 950879 951133 951388 951642 951898 1.952150 952405 952659 952913 953167 953421 953675 953929 954183 951437 Cotang. D. 10" 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.5 42.4 42.4 42.4 42.4 42.4 42.4 42 42 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 42.4 Cotang. ' N. sine. 10.0308371! 65606 060582 1 165628 080327 165650 060072 i 65672 059817 1 165694 0595621,65716 059306 |! 65738 059051 ! 65759 75337 X. cos. 754 75452 75433 75414 75395 75375 75356 058796'! 65781^ 058542! 65803 058286 l| 65825 10.0580321 65847 057777: 65869 057522 i 65891 057267 ,65913 057012 j! 65935 056757 !' 65956 42 42 42 42 42 42 42 42 42.3 42.3 42.3 42.3 056502 165978 056248 66000 055993 66022 055738 ','66044 10.055483 66066 055229J 66088 054974 66109 054719 63131 054465 '66153 054210 66175 053955 6819 053701 | 66218 053446 '66240 053192 : 66262 10.052937! 66284 052682 ! 68308 052428! 1 6632 052174 166349 051919i 166371 051664 |! 66393 051410! 66414 051156 050901 050647 10.050393 050138 049884 049630 049375 049121 048867 048612 048358 048104 ! 10.047850' 047595 I 047341 i 047087 : 046833 : 046579 046325 046071 045817 045563 ! Tana. ! 66436 66458 ! 66480 66501 ; 66523 66545 66566 ! 66588 66610 66832 6665 66675 66697 66718 66740 66762 66783 66805 66827 66848 66870 66891 66913 N. cos 75318 75299 75280 75261 75241 75222 75203 75184 75165 75146 75126 75107 75083 75069 75050 75030 75011 74992 74973 74353 74934 74915 74896 74876 74857 74838 74818 74799 74780 74760 74741 74722 74703 74683 74663 4644 4625 74605 74586 74567 74548 74522 74509 74489 74470 74451 74431 74412 74392 74373 74353 74334 74314 X.sine 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 36 84 38 82 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 48 Degrees. TABLE II. Log. Sines and Tangents. (42°) Natural Sines. 63 Sine. D. 10' Cosine. D. 10"| Tang. 23.4 23.3 23.3 825511 825651 825791 835931 \f%% 826971 -°- J 826211 826351 826491 826631 826/70 826910 .827049 827189 827328 827467 827606 827745 827834 828023 828162 828301 .828439 828578 828716 828855 828993 829131 829269 829407 829545 829683 .829821 829959 830097 830234 830372 830509 830846 I 830784 i 830921 | 831058 ! .831195, 831332 8314691 831606 J 831742 831879 832015 832152 832288 832425 832561 ; 832697 ; 832833 832969 833105 833241 li-l 833377 f- b 833512 H -° 833648 833783 23.3 2;. 3 23.3 .3.3 23.2 23.2 23.2 23.2 23.2 23.2 23.2 23.2 23.1 23.1 23.1 23.1 23.1 23.1 23.1 23.0 23.0 23.0 23.0 23.0 23.0 23.0 22.9 22.9 22.9 J22.9 |22.9 '22.9 22.9 22.9 22.8 22.8 22.8 22.8 22.8 22.8 22.8 22.8 22.7 22.7 22.7 22.7 22.7 22.7 22.7 22.6 22.6 22.6 Cosine .871073 870960 870846 870732 870318 870504 '870390 870276 870161 870047 839933 .869818 859704 869589 869474 889360 869245 869130 869015 8J8900 868785 .868670 868555 868440 868324 868209 868093 857978 867862 867747 867631 .867515 867399 867283 867167 867051 866935 866819 866703 866586 866470 .866353 866237 866120 866004 865887 865770 865653 865536 865419 865302 1.865185 865068 864950 864833 864716 864598 864481 864363 864245 864127 Sine. 19.0 19.0 19.0 19.0 19.0 19.0 19.0 19.0 19.0 19.1 19.1 19.1 19.1 19.1 19.1 19.1 19.1 19.1 19.2 19.2 19.2 19.2 19.2 19.2 19.2 19.2 19.2 19.3 19.3 19.3 19.3 19.3 19.3 19.3 19.3 19.3 19.4 19.4 19.4 19.4 19.4 19.4 19.4 19.4 19.5 19.5 19.5 19.5 19.5 19.5 19.5 19.5 19.5 19.5 19.6 19.6 19.6 19.6 19.6 19.6 .954437 954691 954945 955200 955454 955707 955961 956215 956469 956723 956977 .957231 957485 957739 957993 958246 -958500 958754 959262 959516 .959769 960023 960277 960531 960784 961038 961291 961545 961799 962052 .962306 962560 962813 963067 963320 963574 963827 964081 964335 964588 .964842 965095 965349 965602 965855 966109 966362 966616 966869 967123 .967376 967629 967883 968136 968389 968643 968896 969149 969403 969656 Cotan" D. 10" 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42 42 42 4-2 42 42 42 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.3 42.2 42.2 42.2 42 42 42 42 42 42.2 42.2 42.2 42.2 42.2 42.2 42.2 42.2 42.2 42.2 42.2 42.2 Cotang. j N. sine. IN. cos. 10.045563 045309 045055 044800 044546 044293 044039 043785 043531 043277 043023 10.042769 042515 042261 042007 041754 041500 041246 040992 040738 040484 10.040231 039977 039723 039469 039216 038962 038709 038455 038201 037948 10.037694 037440 037187 036933 036680 036426 036173 035919 035665 035412 10.035158 034905 034651 034398 034145 033891 033638 033384 033131 032877 10.032624 032371 032117 031864 031611 031357 031104 030851 030597 030344 j] 66913174314 i! 66935174295 j! 66956 174276 166978174256 i! 66999 74237 67021 ! ! 67043 ! 67064 1 67086 67107 67129 67151 67172 67194 74217 74198 74178 74159 74139 74120 74100 74080 .74061 67215 !74041 67237 174022 67258 j 74002 67280173983 67301 73963 67323 ! 67344 I 67366 ! 67387 ! 67409 I ! 67430 j ! 67452 | '67473 I 67495 167516 ! 67538 j 67559 ,167580 ! 167602 ; ! 67623 167645 1 1 67666 1 1 67688 I S67709 | i 67730 | | 67752 67773 67795 67816 67837 67859 67880 67901 67923 67944 67965 67987 68008 68029 68051 68072 68093 68115 68136 68157 68179 68200 Tang. || N. cos. N.sine 73944 73924 73904 73885 73865 73846 73826 73806 73787 73767 73747 73728 73708 73688 73669 73649 73629 73610 73590 73570 73551 73531 73511 73491 73472 73452 73432 73413 73393 73373 73E53 73333 73314 73294 73274 73254 73234 73215 73195 73175 73155 73135 47 Degrees. 64 Log. Sines and Tangents. (43°) Natural Sines. TABLE II. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 30 •21 22 23 24 25 36 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 40 47 48 49 50 51 52 53 54 55 5G 57 58 59 00 Sine. 9.833783 833919 834054 834189 834325 834460 834595 834730 834865 834999 835134 835269 835403 835538 835672 835807 835941 836075 836209 836343 836477 836611 836745 836878 837012 837146 837279 837412 837546 837679 837812 9.837945 838078 838211 838344 838477 838610 838742 838875 839007 839140 .839272 839404 839536 839668 839800 839932 840084 840190 840328 840459 .840591 840722 840854 840985 841116 841247 841378 841509 841640 841771 O sine. D. 10" 22.6 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.5 22.4 22.4 22.4 22.4 22.4 22.4 22.4 22.4 Cosine. 22 22 22 22 22 22 22.3 22.2 22.2 22.2 22.2 22.2 22.2 22.2 22.2 22.1 22.1 22.1 22.1 22.1 22.1 22.1 22.1 22.0 22.0 22.0 22.0 22.0 22.0 22.0 21.9 21.9 21.9 21.9 21.9 21.9 21.9 21.9 21.8 21.8 21.8 21.8 21.8 >. 864127 864010 863892 863774 863656 863538 863419 863301 863183 863064 862946 .882827 862709 862590 862471 862353 862234 •862115 861998 861877 861758 .861638 851519 861400 861280 861161 861041 860922 860802 860682 860562 .860442 860322 860202 860082 859962 859842 859721 859601 859480 859360 859239 859119 858998 858877 858756 858635 858514 858393 858272 858151 ,858029 857903 857786 857665 857543 857422 857300 857178 357056 855934 "line. D. 10" 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19. 19, 19, 19, 19, 19, 19, 19. 19, 19. 19. 19, 19. 19. 19. 20. 20. 20. 20. 20. ■JO. 20 20. 20. 20. 20. 20. 20. 20. 20. 20. 20. 20. 20. 2D. 20. 20. 20. 20. 20. 20. 20. 20. 20. 20. 20. Tang. |D. 10" 9.969656 969909 970162 970416 970669 970922 971175 971429 971682 971935 972188 5.972441 972694 972948 973201 973454 973707 973960 974213 974466 974719 ). 974973 975226 975479 975732 975985 976238 976491 976744 976997 i 977250 ). 977503 977756 978009 978262 978515 978768 979021 979274 979527 979780 1.980033 980286 980538 980791 981044 981297 981550 981803 982056 982309 '.982562 982814 983067 933320 983573 983826 934079 984331 984584 984837 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42 42, 42, 42, 42, 42. 12. 42. 42. 42. 42. 42 42 42 42 42 42 42 42 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. 42. Cotang. | |N .sine. N 68200 68221 68242 68264 68285 68306 68327 68349 68370 68391 68412 68434 68455 68476 68497 68518 10.030344 030091 029838 029584 029331 029078 028825 028571 028318 028065 027812 10.027559 027306 027052 026799 026546 026293 026040 025787 025534 025281 10.025027 024774 024521 02426a 024015 023762 023509 023256 023003 022750 | 10.022497 022244 021991 021738 021485 021232 020979 020726 73135 73116 73096 73076 73056 73036 73016 72996 72976 72957 72937 2917 72897 72877 72857 72837 68539172817 68561 J72797 68582 68603 68624 68645 68666 68688 72777 72757 72737 72717 72697 7 68709J72657 6873072637 Cotany 68772 72597 68793172577 6881472557 68835 72537 68857 72517 6887872497 6889972477 68920 72457 6894172437 68962 72417 68983,72397 6900472377 020473 69025 72357 020220! 6904672337 10.019967' 69087 72317 019714 69088 72297 019462 J 69109 72277 019209 ! 1 69130 72257 018956^6915172236 018703 I 6917272216 018450 |69193 72196 018197: 69214 72176 017944! 69235 72156 017691 69256 72136 10.017438 i 69277 72116 017186 16929872095 016933 : 69319 72075 016680,1 69340 72055 016427 6936172035 016174 69382 72015 015921 169403 71995 015669 69424 71974 015416 69445 71954 015163 6946671934 Tan":. N. cos. |N. sine. 60 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 39 28 27 26 26 24 23 33 21 20 19 IS 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 46 Degrees. TABLE II. Log. Sines and Tangents. (44°) Natural Sines. 05 Sim;. ). 841771 841902 842033 842163 842294 842424 842555 842685 842815 842946 843076 • .843206 843336 843466 843595 843725 843855 843984 844114 844243 844372 1.844502 844631 844760 844889 845018 845147 845276 845405 845533 845662 .845790 845919 846047 846175 846304 846432 846560 846688 846816 846944 .847071 847199 847327 847454 847582 847709 847836 847964 848091 848218 .848345 848472 848599 848726 848852 848979 849106 849232 849359 849485 Cosine. D. 10" Cosine. ,856934 856812 856690 856568 856446 856323 856201 856078 855956 855833 855711 ,855588 855465 855342, 855219 855096 854973 854850 854727 854603 854480 854356 854233 854109 853986 853862 853738 853614 853490 853366 853242 853118 852994 852869 852745 852620 852496 852371 852247 852122 851997 851872 851747 851622 851497 851372 851246 851121 850996 850870 850745 9.850619 850493 850368 850242 850116 849990 849864 849738 849611 849485 Sine. • D. 10" Till!'. 984837 985090 985343 985596 985848 986101 986354 986607 986860 987112 987365 9.987618 987871 988123 988376 988629 988882 989134 989387 989640 989893 990145 990398 990651 990903 991156 991409 991662 991914 992167 992420 992672 992925 993178 993430 993683 993936 994189 994441 994694 994947 995199 995452 995705 995957 996210 996463 996715 996968 997221 997473 9.997726 997979 998231 998484 998737 999242 999495 999748 10.000000 Cotan» D>10" 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42.1 42 42 42 42 42 42.1 42.1 42.1 Cotang. N. sine. N. 10.015163 014910 014657 014404 014152 013899 013646 013393 013140 012888 012635 I 10.012382 012129 011877 011624 011371 011118 010866 010613 010360 010107 10.009855 009602 009349 009097 008844 008591 008338 008086 007833 007580 10-007328 007075 006822 006570 006317 006064 005811 005559 005306 005053 1 70298 10-004801 | 70319 004548 [ I 70339 004295 I 70360 004043 I 70381 003790! 70401 003537 I 70422 69466 6948 69503 69529 69549 69570 69591 69612 69633 69654 69675 69696 697 69737 69758 69779 69800 69821 69842 69862 69883 69904 69925 69946 69966 69987 70008 70029 70049 70070 70091 70112 70132 70153 70174 70195 70215 70236 70257 70277 003285 003032 002779 002527 10-002274 002021 001769 001516 001263 001011 000758 000505 000253 000000 | 70443 70463 70484 70505 70525 70546 70567 70587 70608 70628 70649 70670 70690 70711 Tarn 71934 71914 71894 71873 71853 71833 71813 71792 71772 71752 71732 71711 71691 71671 71650 71630 71610 71590 71569 71549 71529 71508 71488 71468 71447 71427 71407 71386 71366 71345 71325 71305 71284 71264 71243 71223 71203 71182 71162 71141 71121 71100 71080 7.1059 71039 71019 70998 0978 70957 0937 70916 70896 70875 70855 0834 70813 0793 0772 0752 0731 70711 N. cos. N.sine. 45 Decrees. 66 LOGARITHMS TABLE III. LOGARITHMS OF NUMBERS. From 1 to 200, INCLUDING TWELVE DECIMAL PLACES. N. Log. |K. Log. 1 N. Log. 1 oooooo oooooo 41 612783 856720 . 623249 290398 633468 455580 \ 81 908485 018879 2 301029 995864 42 82 913813 852384 3 477121 254720 43 83 919078 092376 4 602059 991328 44 643452 676486 84 924279 286062 5 698970 004336 46 653212 513775 85 929418 925714 6 778151 250384 46 662757 831682 86 934498 451244 7 845098 010014 47 672097 857926 87 939519 252619 8 *903089 986992 48 681241 237376 88 944482 672150 9 954242 509439 49 690193 080028 89 949390 006645 10 Same as to 1. 60 Same as to 5. 90 Same as to 9. 11 041392 685158 51 707570 176098 91 959041 392321 12 079181 246018 52 716003 343635 92 963787 827346 13 113943 352307 63 724275 869601 93 968482 948554 14 146128 035678 64 732393 759823 94 973127 853600 15 176091 259056 55 740382 689494 95 977723 605889 16 204119 982656 56 748188 02700S 95 982271 233040 17 230448 921378 57 755874 855672 97 986771 734266 18 255272 505103 68 763427 993563 98 991226 075692 19 278753 600953 69 770852 011642 99 995635 194598 20 Same as to 2. 60 Same as to 6 100 Same as to 10. 21 322219 2947 61 785329 835011 101 004321 373783 22 342422 680822 62 792391 699498 102 008600 171762 23 361727 836018 63 799340 549453 103 012837 224705 24 380211 241712 64 808179 973984 104 017033 339299 25 397940 008672 65 812913 356643 105 021189 299070 26 414973 347971 66 819543 935542 103 025305 865265 27 431363 764159 67 826074 802701 107 029383 777685 28 447158 031342 68 832508 912706 108 033423 755487 29 462397 997899 69 838849 090737 109 037426 497941 30 St- me as to 3. 70 Same as to 7. 110 Same as to 11. 31 491361 693834 71 851258 348719 111 045322 978787 32 505149 978320 72 857332 496431 112 049218 022670 33 518513 939878 73 863322 860120 113 053078 443483 34 531478 917042 74 869231 719731 114 056904 851336 35 544068 044350 75 875061 263392 115 060397 840354 36 556302 500767 76 880813 592281 116 054457 989227 37 568201 724067 77 886490 725172 117 058185 861746 38 579783 596617 78 892094 602690 118 071882 007306 39 591054 607026 79 897627 091290 119 075546 961393 40 Same as to 4. 80 Same as to 8. 120 Same as to 12. OF NUMBERS. 67 Is. Log._ 082785 370316 N. Log. N. , Log 121 148 170261 715395 175 243038 048686 122 086359 830875 149 173186 268412 176 245512 667814 123 08990o 111439 150 176091 259056 177 247973 266362 124 093421 685162 151 178976 947293 178 250420 002309 125 096910 013008 ; 152 181843 587945 179 25:2853 030980 126 100370 545118 153 184691 430818 180 255272 505103 127 103803 720956 1 154 187520 720836 181 257678 574869 128 107209 969648 ! 155 190331 698170 182 260071 387985 129 110589 710-299 156 193124 588354 183 262451 089730 130 Same as to 13. 157 195899 652409 184 264817 823010 131 117271 295656 158 198857 086954 i 185 267171 728403 132 120573 931206 159 201397 124320 186 269512 944218 133 123851 640967 160 204119 982656 187 271841 606536 134 127104 798365 161 208825 876032 188 274157 849284 135 130333 768495 162 209515 014543 139 276461 804173 136 133538 908370 163 212187 604404 190 278753 600953 137 136720 567156 164 214843 848048 191 281033 367248 138 139879 086401 165 217483 944214 192 283301 228704 139 143014 800254 166 220108 088040 193 285557 309008 140 146128 035678 167 222716 471148 194 287801 729930 141 149219 112655 163 225309 281726 195 290034 611362 142 152288 3443S3 169 227886 704614 196 292256 071356 143 155336 037465 170 230448 921378 197 294466 226162 144 158362 492095 171 232996 110392 198 296665 190262 145 161368 002235 172 235528 446908 199 298853 076410 146 164352 855784 173 238046 103129 147 167317 334748 174 240549 248283 LOGARITHMS OF THE PRIME NUMBERS From 200 to 1543, INCLUDING TWELVE DECIMAL PLACES. N. 201 203 207 209 211 223 227 229 233 239 241 251 257 263 269 271 Loi 303196 057420 307496 037913 315970 345457 320146 286111 324282 455298 348304 863048 356025 857193 359835 482340 367355 921020 378397 900948 3820 T 7 042575 399873 721481 409933 123331 419955 748490 429752 280002 432989 290874 N, 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 442479 769064 448708 319905 451786 435524 466867 620D54 487138 375477 492760 389027 495544 337546 501059 262218 519827 993776 527629 900871 540329 474791 542825 426959 r 47774 705388 555094 448578 664666 004252 571703 831809 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 Lo^. 578839 209968 583198 773968 589949 601326 598790 508763 603144 372020 611723 308007 622214 U22966 624282 095836 634477 270161 636487 896353 642424 520242 646403 726223 652246 341003 659916 200070 663700 925390 665580 991018 - - 68 LOGARITHMS N. Log. N. Log. 1171 Log. uo»566 fc950:2 467 6J9blo briU5b6 "821 914343 157119 479 680335 513414 823 915399 835212 1181 0/2249 807613 487 68/528 961215 827 917505 509553 1137 074450 718955 491 691081 492123 829 918554 530550 1193 076640 443670 499 69810 545623 8S9 923761 960829 1201 0.9543 007385 503 701567 985056 853 930949 031168 1213 083830 800345 509 706717 782337 857 932980 821923 1217 085290 678210 521 716S37 723300 859 933993 163331 1223 087426 458017 623 718501 688867 863 936010 795715 1229 089551 882866 641 733197 255107 877 942999 593356 1231 090258 052912 547 737987 326333 881 944975 908412 1237 092369 699609 657 745855 195174 883 945960 703578 1249 096562 438356 563 750508 394851 887 947923 619832 1259 100025 729204 669 755112 26639 > 907 957607 287060 1277 103190 896808 571 756636 1Q8213 911 959518 376973 1279 106870 542480 577 761175 813156 919 963316 511386 1283 108226 656362 687 768638 101248 929 968015 713994 1289 110252 917337 593 773U54 693364 937 971739 590888 1291 110926 242517 599 777426 822389 941 973589 623427 1297 112939 986066 601 778874 472002 947 976349 979003 1301 114277 296540 607 783138 691075 953 979092 900838 1303 114944 415712 613 787460 474618 967 985426 474083 1307 116275 587564 617 790285 164033 9/1 987219 229908 1319 120244 795568 619 791690 649020 977 989894 563719 1321 120902 817604 631 800029 359244 9£3 992553 617832 1327 122870 922849 641 806858 029519 991 996073 654485 1361 133858 125188 643 808210 972924 997 998695 158312 1367 135768 514554 647 810904 280669 1009 003891 166237 1373 137670 537223 653 814913 181275 1013 005609 445360 1381 140193 678544 659 818885 414594P 1019 008174 1840J6 1399 145817 714122 661 810201 459486 1021 009025 742087 1409 148910 994096 673 828015 064224 1031 013258 665284 1423 153204 896557 677 830588 668685 1033 014100 321520 1427 154424 012366 683 834420 703682 1039 016615 547557 1429 155032 228774 691 839478 047374 1049 020775 488194 1433 156246 402184 701 845718 017967 1051 021602 716028 1439 158060 793919 709 850646 235183 1061 025715 383901 1447 160468 531109 719 856728 890383 1063 026533 264523 1451 161667 412427 727 861534 410859 1069 028977 705209 1453 162265 614286 733 865103 974742 1087 036229 544036 1459 164055 291883 739 868644 488395 1091 037824 750588 1471 167612 672629 743 870988 813761 1093 038620 161950 1481 170555 058512 751 855639 937004 1097 040206 627575 1483 171141 151014 757 879095 879500 1103 042595 512440 1487 172310 968489 761 881384 656771 1109 044931 546119 1489 172894 731332 ! 769 885926 339801 1117 048053 173116 1493 174059 807708 773 888179 493918 1123 050379 756261 1499 175801 632866 787 895974 732359 1129 052693 941926 1511 179264 464329 797 901458 321396 1151 031075 323630 1523 182699 903324 809 907948 521612 1153 061829 307295 1531 184975 190807 811 909020 854211 1163 065579 714728 1543 188365 926053 OF NUMBERS AUXILIARY LOGARITHMS, 1ST. 1.009 1.008 1.007 1.006 1 . 005 1.004 1.003 1.002 1.001 Log. 1 ff. 003891166237 >, 1.0009 003460532110 1.0008 003029470554 1.0007 002598080685 1.0003 0021660ol756 \a 1.0005 001733712775 1.0004 001300933020 1.0003 000867721529 1.0002 000434077479 J 1.0001 Lost. 000390689248 000347296C84 000303899784 000260-198547 000217092970 000173683057 000130268804 000086850211 000043427277 c N. 1 .00009 1 .00008 1 .00007 1 .00006 l .00005 1 .00004 1 .00003 1 .00002 1 00001 Log- 000039083266 000034740691 000030398072 000026055410 000021712704 000017371430 000013028638 000008685802 000004342923 L i 1ST. f .000009 i . 000008 i .000007 i .000006 i .000005 i .000004 i 000003 i 000002 i 000001 Log. 000003908628 000003474338 000003040047 000002605756 000002171464 000001737173 000001302880 000000868587 000000434294 1.00000001 1.000000001 1.0000000001 Log. 000000043429 000000004343 000000000434 000000000043 (n) (o) (P) (q) »i=0.4342944819 log. —1.637784298. By the preceding tables — and the auxiliaries A, B, and C, we can find the logarithm of any number, true to at leap.t ten decimal places. But some may prefer to use the following direct formula, which may be found in any of the standard works on algebra: Log. (2-j-l)=log.2+0.8685889638^_J- \ The result will be true to twelve decimal places, if z be over 2000. The log. of composite numbers can be determined by the combination of logarithms, already in the table, and the prime numbers from the formula. Thus, the number 3083 is a prime number, find its loga- rithm. We first find the log. of the number 3082. By factoring, we discover that this is the product of 46 into 67. 70 NUMBERS Log. 46, Log. 67, Log. 3082 Log. 3083=3.4888326343- 1.6627578316 1.8260748027 3.4888326343 0.8685889638 6165 NUMBERS AND THEIR LOGARITHMS, OFTEN USED IN COMPUTATIONS. Log. 3.14159265 0.4971499 Circumference of a circle to dia. 1 ) Surface of a sphere to diameter 1 Area of a circle to radius 1 Area of a circle to diameter 1 =; .7853982 —1.8950899 Capacity of a sphere to diameter 1 = .5235988—1.7189986 Capacity of a sphere to radius 1 =4.1887902 0.6220886 I- Arc of any circle equal to the radius =57°29578 1.7581226 Arc equal to radius expressed in sec. =206264"8 5.3144251 Length of a degree, (radius unity) = .01745329 —2.2418773 12 hours expressed in seconds, = 43200 4.6354837 Complement of the same, =0.00002315 —5.3645163 360 degrees expressed in seconds, = 1296000 6.1 126050 • A gallon of distilled water, when the temperature is 62 Fahrenheit, and Barometer 30 inches, is 277. r 2 -VV cubi< inches. 10 7277.274=16.651542 nearly. 4 ■I 277.274 .775398 = 18.78925284 7231=15.198684. 7282=16.792855. = 18.948708. .785398 The French Metre=3.2808992, English feet linear mea- sure, =39.3707904 inches, the length of a pendulum vi- brating seconds. ^ OF THE ^j\ UNIVERSITY ) / J ryiM4tf^ SMS*: ■ i* 'HiiBi