UC-NRLF $B 530 730 \ /.v ■4:U4^. L '7 r^ i^***^> '^ ^ /^xf ^^ ^V^ ELEMENTS OF GEOMETRY AND TRIGONOMETRY. TRANSLATED FROM THE FRENCH OF A. M. LEGENDRE, MEMBER OF THE INSTITUTE AND OF THE LEGION OF HONOUR, AND OF THE ROYAL SOCIETIES OF LONDON AND EDINBURGH, &C. BY DAVID BREWSTER, LL. D. FKLLOW OF THE ROYAL SOCIETY OF LONDON, AND SECRETARY TO THE ROYAL SOCIETY Of EDINBURGH, &C. &C. REVISED AND ABRIDGED BY CHARLES DAVIES, :| PROFESSOR OF MATHEMATICS * IN THE MILITARY .ACADEMY, AND AUTHOR OF tHE COMMON SCHOOL ARITHMETIC, DESCRIPTIVE GEOMETRY, ELEMENTS OF SURVEYING, AND A TREATISE ON SHADOWS AND PERSPECTIVE. FOURTH EDITION. NEW-YORK: PUBLISHED BY HARPER AND BROTHERS, NO. 82 CLIFF-STREET. Stereotyped by A. Chandler. 1834. .v>....^v,^,.0 ^■^"'^^^ Entered according to the Act of Congress, in the year one thousand eight hundred and thirty-four, by Charles Da vies, in the Clerk's Office of the Dis- trict Court of the United States, for the Southern District of New- York. PREFACE TO THE FOURTH AMERICAN EDITION. The Editor, in offering to the public Dr. Brewster's translation of Legendre's Geometry under its present form, is fully impressed with the responsibility he assumes in making alterations in a work of such de- served celebrity. In the original work, as well as in the translations of Dr. Brewster and Professor Farrar, the proposi- tions are not enunciated in general terms, but with reference to, and by the aid of, the particular diagrams used for the demonstrations. It is believed that this departure from the method of Euclid has been gene- rally regretted. The propositions of Geometry are general truths, and as such, should be stated in gene- ral terms, and without reference to particular figures. The method of enunciating them by the aid of particu- lar diagrams seems to have been adopted to avoid the difiiculty which beginners experience in comprehend- ing abstract propositions. But in avoiding this diffi- culty, and thus lessening, at first, the intellectual labour, the faculty of abstraction, which it is one of the primary objects of the study of Geometry to strengthen, remains, to a certain extent, unimproved. 3037 lis iv :* PREFACE. -Vu^ Besides the alterations in the enunciation of the propositions, others of considerable importance have also been made in the present edition. The propo- sition in Book V., which proves that a polygon and circle may be made to coincide so nearly, as to differ from each other by less than any assignable quantity, has been taken from the Edinburgh Encyclopedia. It is proved in the corollaries that a polygon of an infinite number of sides becomes a circle, and this principle is made the basis of several important de- monstrations in Book VIII. Book II.,on Ratios and Proportions, has been partly adopted from the Encyclopedia Metropolitana, and will, it is believed, supply a deficiency in the original work. Very considerable alterations have also been made in the manner of treating the subjects of Plane and Spherical Trigonometry. It has also been thought best to pubhsh with the present edition a table of logarithms and logarithmic sines. Military Academy, West Point, March, 1834. CONTENTS. BOOK I. The principles, 9 * ^^ ^ BOOK II. Ratios and Proportions, 34 BOOK III. The Circle and the Measurement of Angles, - - - 41 Problems relating to the First and Third Books, - - 57 BOOK IV. The Proportions of Figures and the Measurement of Areas, - 68 Problems relating to the Fourth Book, - - . - 98 BOOK V. Regular Polygons eind the Measurement of the Circle, - 109 BOOK VI. Planes and SoHd Angles, 126 BOOK VII. Polyedrons, 142 BOOK VIII. The three round bodies, 166 BOOK IX. Of Spherical Triangles and Spherical Polygons, - - - 186 APPENDIX. The regular Polyedrons, - - ... - 205 VI CONTENTS. Tan PLANE TRIGONOMETRY. Division of the Circumference, General Ideas relating to the Trigonometrical Lines, Theorems and Formulas relating to the Sines, Cosines, gents, (fee. ...... Construction and Description of the Tables, Description of Table of Logarithms, - Description of Table of Logarithmic Sines, Principles for the Solution of Rectilineal Triangles, Solution of Rectilineal Triangles by Logarithms, Solution of Right angled Triangles, Solution of Triangles in general, . - - SPHERICAL TRIGONOMETRY. First principles, _-.----- Napier's Circular Parts, Solution of Right angled Spherical Triangles by Logarithms; Quadrantal Triangles, i - - - . - . Solution of Oblique angled Triangles by Logarithms, - 207 208 215 223 224 228 231 235 237 238 246 252 255 257 259 AN INDEX SHOWING THE PROPOSITIONS OF LEGENDRE WHICH CORRESPOND TO THE PRINCIPAL PROPOSITIONS OF THE FIRST SIX BOOKS OF EUCLID. Euclid. Legendre. Euclid. Legendre. Euclid. Legendre. Book I. Book I. Cor.2. of 32 33 34 35 Prop. 27 30 28 Prop. 26 28 29 31 Prop. 15 5 5 5 Cor.2. ?To i &3. r^ Prop. 4 5 Cor. of 5 6 Prop. 5 11 Cor. of 11 12 Book IV. 1 8 13 10 1 36 37 1 Cor 2 of 2 Book IV. 14 3 38 Cor. 2. of 2 35 28 15 Cor. l.< ,. &2. \^^ 16 17 4 4 47 2 11 8 36 30 Sch. of 4 5 Cor. of 25 I 25 Book VI. 5 Cor. 1. of 4 I Cor. of 6 Book II. 1 4 18 13 12 13 19 20 21 24 13 13 12 2 S 15 i 16 17 18 7 8 9 Book III. Book III. 3 4 Prop. 3 Prop. 6 25 9 10 Cor. of 7 5 19 26 6 11 Cor. of 14 6 20 27 Cor'. 1. of 19 12 Cor. of 14 8 22 28 Cor. 2. of 19 14 8 14 ) 25 I Cor. of 15 29 \ Cor.2. & I 4. of 20 15 2 15 18 9 19 25 30 22 20 18 20 5 26 I 27' ICor.l.of32 26 21 Cor. of 18 ELEMENTS OF GEOMETRY. BOOK I. THE PRINCIPLES. Definitions, 1. Geometry is the science which has for its object the measurement of extension. Extension has three dimensions, length, breadth, and height, or thickness. 2. A line is length without breadth, or thickness. The extremities of a line are called points : a point, there- fore, has neither length, breadth, nor thickness, but position only. 3. A straight line is the shortest distance from one point to another. 4. Every line which is not straight, or composed of straight lines, is a curved line. Thus, AB is a straight line ; ACDB is a broken line, or one composed of straight A.^ lines ; and AEB is a curved line. The word line, when used alone, will designate a straight line ; and the word curve, a curved line. 5. A surface is that which has length and breadth, without height or thickness. 6. A plane is a surface, in which, if two points be assumed at pleasure, and connected by a straight line, that line will lie wholly in the surface. 7. Every surface, which is not a plane surface, or composed of plane surfaces, is a curved surface. 8. A solid or body is that which has length, breadth, and thickness ; and therefore combines the three dimensions of extension. 2 10 GEOMETRY. 9. When two straight lines, AB, AC, meet each other, their inchnation or opening is call- ed an angle, which is greater or less as the lines are moreor less inclined or opened. The point of intersection A is the vertex of the j^ angle, and the lines AB, AC, are its sides. The angle is sometimes designated simply by the letter at the vertex A ; sometimes by the three letters BAC, or CAB, the letter at the vertex being always placed in the middle. Angles, like all other quantities, are susceptible of addition, subtraction, multiplication, and division. Thus the angle DCE is the sum of the two angles DCB, BCE ; and the an- gle DCB is the difference of the two angles DCE, BCE. 10. When a straight line AB meets another straight line CD, so as to make the adjacent angles BAC, BAD, equal to each other, each of those angles is called a right angle ; and the line AB is said to be perpendicular to CD. : 11. Every angle BAC, less than dP^ right angle, is an acute angle ; and every angle DEF, greater than a right angle, is an obtuse angle. -F 12. Two lines are said to be parallel, when being situated in the same plane, they cannot meet, how far soever, either way, both of them be produced. 13. A plane figure is a plane terminated on all sides by lines. If the lines are straight, the space they enclose is called a rectilineal figure, or polygon, and the lines themselves, taken together, form the contour, or perimeter of the polygon. 14. The polygon of three sides, the simplest of all, is called a triangle ; that of four sides, a quadrilateral ; that of five, a pentagon ; that of six, a hexagon ; that of seven, a heptagon ; that of eight, an octagon ; that of nine^ a nonagon ; that of ten, a decagon ; and that of twelve, a dodecagon. BOOK I. 11 15. An equilateral triangle is one which has its three sides equal ; an isosceles triangle, one which has two of its sides equal ; a scalene triangle, one which has its three sides unequal. 16. A right-angled triangle is one which has a right angle. The side opposite the right angle is called the hypothenuse. Thus, ^in the triangle ABC, right-angled at A, the side BC is the hypothenuse. 1 7. Among the quadrilaterals, we distinguish : The square, which has its sides equal, and its an- gles right-angles. The rectangle, which has its angles right an- gles, without having its sides equal. The parallelogram, or rhomboid, which has its opposite sides parallel. The rhombus, or lozenge, which has its sides equal, without having its angles right angles. And lastly, the trapezoid, only two of whose sides are parallel. 18. A diagonal is a line which joins the ver- tices of two angles not adjacent to each other. Thus, AF, AE, AD, AC, are diagonals. 19. Axi equilateral polygon is one which has all its sides equal ; an equiangular polygon, one which has all its angles equal. 20. Two polygons are mutually equilateral, when they have their sides equal each to each, and placed in the same order ; 12 R f r GEOMETRY. that is to say, when following their perimeters in the same di- rection, the first side of the one is equal to the first side of the other, the second of the one to the second of the other, the third to the third, and so on. The phrase, mutually equian- gular, has a corresponding signification, with respect to the angles. In both cases, the equal sides, or the equal angles, are named homologous sides or angles. Definitions of terms employed in Geometry, An axiom is a self-evident proposition. A theorem is a truth, which becomes evident by means of a train of reasoning called a demonstration, A problem is a question proposed, which requires a solu- tion, A lemma is a subsidiary truth, employed for the demonstra- tion of a theorem, or the solution of a problem. The common name, proposition, is applied indifferently, to theorems, problems, and lemmas. A corollary is an obvious consequence, deduced from one or several propositions. A scholium is a remark on one or several preceding propo- sitions, which tends to point out their connexion, their use, their restriction, or their extension. A hypothesis is a supposition, made either in the enunciation of a proposition, or in the course of a demonstration. Explanation of the symbols to be employed. The sign = is the sign of equality; thus, the expression A=B, signifies that A is equal to B. To signify that A is smaller than B, the expression A<-B is used. To signify that A is greater than B, the expression A>B is used ; the smaller quantity being always at the vertex of the angle. The sign + is called plus : it indicates addition. The sign — is called minus : it indicates subtraction. Thus, A + B, represents the sum of the quantities A and B ; A — B represents their difference, or what remains after B is taken from A ; and A — B + C, or A + C — B, signifies that A and C are to be added together, and that B is to be subtracted from their sum. BOOK I. 13 The sign x indicates multiplication : thus, A x B represents the product of A and B. Instead of the sign x , a point is sometimes employed ; thus, A.B is the same thing as A x B. The same product is also designated without any intermediate sign, by AB ; but this expression should not be employed, when there is any danger of confounding it with that of the line AB, which expresses the distance between the points A and B. The expression A x (B + C — D) represents the product of A by the quantity B + C — D. If A + B were to be multiplied by A — B + C, the product would be indicated thus, (A + B)x (A — B + C), whatever is enclosed within the curved lines, being considered as a single quantity. A number placed before a line, or a quantity, serves as a multiplier to that line or quantity ; thus, 3AB signifies that the line AB is taken three times ; | A signifies the half of the angle A. The square of the line AB is designated by AB^ ; its cube by ABl What is meant by the square and cube of a line, will be explained in its proper place. The sign V indicates a root to be extracted ; thus \/2 means the square-root of 2 ; V A x B means the square-root of the product of A and B. Axioms. 1. Things which are equal to the same thing, are equal to each other. 2. If equals be added to equals, the wholes will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals, the wholes will be un- equal. 5. If equals be taken from unequals, the remainders will be unequal. 6. Things which are double of the same thing, are equal to each other. 7. Things which are halves of the same thing, are equal to each other. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. All right angles are equal to each other. 11. From one point to another only one straight line can be drawn. 12. Through the same point, only one straight line can be drawn which shall be parallel to a given line. 13. Magnitudes, which being applied to each other, coincide throughout their whole extent, are equal. B 14 GEOMETRY. PROPOSITION I. THEOREM. If one straight line meet another straight line, the sum of the two adjacent angles will he equal to two right angles. Let the straight line DC meet the straight line AB at C, then will the angle ACD + the angle DCB, be equal to two right angles. At the point C, erect CE perpendicular to AB. The angle ACD is the sum of the an-•^^ gles ACE, ECD: therefore ACD -f DCB is the sum of the three angles ACE, ECD, DCB : but the first of these three angles is a right angle, and the other two make up the right angle ECB ; hence, the sum of the two an- gles ACD and DCB, is equal to two right angles. Cor. 1. If one of the angles ACD, DCB, is a right angle, the other must be a right angle also. Cor, 2. If the line DE is perpendicular to AB, reciprocally, AB will be perpendicu- lar to DE. For, since DE is perpendicular to AB, the angle ACD must be equal to its adjacent an- gle DCB, and both of them must be right angles (Def 10.). But since ACD is a right angle, its adjacent angle ACE must also be a right angle (Cor. 1.). Hence the angle ACD is equal to the angle ACE, (Ax. 10.) : therefore AB is perpendicular to DE. Cor, 3. The sum of all the successive angles, BAC, CAD, DAE, EAF, formed on the same side of the straight line BF, is equal to two right angles ; for their sum is equal to that of the two adjacent an- gles, BAC, CAR ^ PROPOSITION II. THEOREM. Tv)o straight lines, which have two points common, coincide with # •." eac^ other throughout their whole extent, and form one and .*r-. the same straight line. Let A and B be the two common points. In the first place it is evident that the two lines must coincide entirely between A and B, for otherwise there would be two straight lines between A and B, which is impossible (Ax. 11.). Sup- JL B BOOK I. 15 pose, however, that on being produced, these lines begin to separate at C, the one becoming CD, the other CE. From the point C draw the Hne CF, making with AC the right angle ACF. Now, since ACD is a straight line, the angle FCD will be a right angle (Prop. I. Cor. 1.) ; and since ACE is a straight line, the angle FCE will likewise be a right angle. Hence, the angle FCD is equal to the angle FCE (Ax. 10.); which can only be the case when the lines CD and CE coincide : there- fore, the straight lines which have two points A and B com- mon, cannot separate at any point, when produced ; hence they . form one and the same straight line. PROPOSITION III. THEOREM. If a straight line meet two other straight lines at a common point, making the sum of the two adjacent angles equal to two right angles, the two straight lines which are met, will form one and the same straight line. Let the straight line CD meet the two lines AC, CB, at their common point C, making the sum of the two adjacent angles DCA, DCB, equal to 57 two right angles ; then will CB be the prolongation of AC, or AC and CB will form one and the same straight line. For, if CB is not the prolongation of AC, let CE be that pro- longation: then the line ACE being straight, the sum of the angles ACD, DCE, will be equal to two right angles (Prop. I.). But by hypothesis, the sum of the angles ACD, DCB, is also equal to two right angles: therefore, ACD + DCE must be equal to ACD + DCB ; and taking away the angle ACD from each, there remains the angle DCE equal to the angle DCB, which can only be the case when the lines CE and CB coincide ; hence, AC, CB, form one and the same straight line. PROPOSITION IV. THEOREM. When two straight lines intersect each other, the opposite or ver- tical angles, which they form, are equal. 16 GEOMETRY. Let AB and DE be two straight^ lines, intersecting each other at C ; then will the angle ECB be equal to the angle ACD, and the angle ACE to the angle DCB. TO B" For, since the straight line DE is met by the straight line AC, the sum of the angles ACE, ACD, is equal to two right angles (Prop. L) ; and since the straight line AB, is met by the straight line EC, the sum of the angles ACE and ECB, is equal to two right angles: hence the sum ACE + ACD is equal to the sum ACE + ECB (Ax. 1.). Take away from both, the com- mon angle ACE, there remains the angle ACD, equal to its opposite or vertical angle ECB (Ax. 3.). Scholium, The four angles formed about a point by two straight lines, which intersect each other, are together equal to four right angles : for the sum of the two angles ACE, ECB, is equal to two right angles ; and the sum of the other two, ACD, DCB, is also equal to two right angles : therefore, the sum of the four is equal to four right angles. In general, if any number of straight lines CA, CB, CD, &c. meet in a point C, the ^ sum of all the successive angles ACB,BCD, DCE, ECF, FCA, will be equal to four right angles : for, if four right angles were formed about the point C,by two lines per- pendicular to each other, the same space would be occupied by the four right angles, as by the succes- sive angles ACB, BCD, DCE, ECF, FCA. PROPOSITION V. THEOREM. If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will he equal. Let the side ED be equal lo the side BA, the side DF to the side AC, and the an- gle D to the angle A ; then will the triangle EDF be equal to the triangle BAC. ^. £ !F S C ^ For, these triangles may be so applied to each other, that they shall exactly coincide. Let the triangle EDF, be placed upon the triangle BAC, so that the point E shall fall upon B, and the side ED on the equal side BA ; then, since the angle D is equal to the angle A, the side DF will take the direction AC. But BOOK I. rt DF is equal to AC ; therefore, the point F will fall on C, and the third side EF, will coincide with the third side BC (Ax. 11.): therefore, the triangle EDF is equal to the triangle BAG (Ax. 13.). Cor, When two triangles have these three things equal, namely, the side ED=BA, the side DF=AC, and the angle D=A, the remaining three are also respectively equal, namely, the side EF=BC, the angle E=B, and the angle F=C PROPOSITION VI. THEOREM. If two triangles have two angles and the included side of the one, equal to two angles and the included side of the other, each to each, the two triangles will he equal. Let the angle E be equal to the angle B, the angle F to the angle C, and the in- cluded side EF to the in- cluded side BC ; then will the triangle EDF be equal to the triangle BAC. ^ ^ ^ ^ For to apply the one to the other, let the side EF be placed on its equal BC, the point E falling on B, and the point F on C; then, since the angle Eis equal to the angle B, the side ED will take the direction BA ; and hence the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to Jhe angle C, the line FD will take the direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, falling at the same time in the two straight lines B A and C A, must fall at their intersection A : hence, the two triangles EDF, BAC, coincide with each other, and are therefore equal (Ax. 13.). Cor, Whenever,in two triangles,these three things are equal, namely, the angle E=B, the angle F=C, and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, namely, the angle D=A, the side ED=BA, and the side DF=AC. Scholium. Two triangles are said to be equal, when being applied to each other, they will exactly coincide (Ax. 13.). Hence, equal triangles have their like parts equal, each to each, since those parts must coincide with each other. The converse of this proposition is also true, namely, that two triangles which have all the parts of the one equal to the parts of the other, each B*3 18 GEOMETRY. to eachf are equal ; for they may be applied to each other, and the equal parts will mutually coincide. PROPOSITION VII. THEOREM. The sum of any two sides of a triangle, is greater than the third side. Let ABC be a triangle : then will the sum of two of its sides, as AC, CB, be greater than the third side AB. For, the line AB is the shortest dis- tance between the points A and B D ; then will BC> EF. Make the angle CAGB =B; take AG=DE, and draw CG, The BOOK I. m triangle GAC is equal to DEF, since, by construction, they have an equal angle in each, contained by equal sides, (Prop. V.) ; therefore CG is equal to EF. Now, there may be three cases in the proposition, according as the point G falls without the triangle ABC, or upon its base BC, or within it. First Case. The straight hne GCEF, B Scholium. Conversely, if two sides BA, AC, of the triangle BAC, are equal to the two ED, DF,of the triangle EDF, each to each, while the third side BC of the first triangle is greater than the third side EF of the second ; then will the an- gle BAC of the first triangle, be greater than the angle EDF of the second. For, if not, the angle BAC must be equal to EDF, or less than it. In the first case, the side BC would be equal to EF, (Prop. V. Cor.) ; in the second, CB would be less than EF ; but eitherof these results contradicts the hypothesis : therefore, BAC is greater than EDF, PROPOSITION X. THEOREM. /f two triangles have the three sides of the one equal to the three sides of the other^ each to each, the three angles will also he equal, each to each, and the triangles themselves will he equal. JIO GEOMETRY. Let the side ED=BA, the side EF=BC, and the side DF=AC ; then will the angle D=A, the angle E=B, and the angle F = C. E TB C For, if the angle D were greater than A, while the sides ED, DF, were equal to BA, AC, each to each, it would fol- low, by the last proposition, that the side EF must be greater than BC ; and if the angle t> were less than A, it would follow, that the side EF must be less than BC : but EF is equal to BC, by hypothesis ; therefore, the angle D can neither be greater nor less than A ; therefore it must be equal to it. In the same manner it may be shown that the angle E is equal to B, and the angle F to C : hence the two triangles are equal (Prop. VI. Sch.). Scholium. It may be observed that the equal angles lie op- posite the equal sides : thus, the equal angles D and A, He op- posite the equal sides EF and BC. PROPOSITION XI. THEOREM. In an isosceles triangle, the angles opposite the equal sides are equal. I>et the side BA be equal to the side AC ; then will the angle C be equal to the angle B. For, join the vertex A, and D the middle point of the base BC. Then, the triangles BAD, DAC, will have all the sides of the one equal to those of the other, each to each ; for BA is equal to AC,^" by hypothesis ; AD is common, and BD is equal to DC by construction : therefore, by the last proposition, the angle B is equal to the angle C. Cor. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal. * Scholium. The equality of the triangles BAD, DAC, proves also that the angle BAD, is equal to DAC, and BDA to ADC, hence the latter two are right angles ; therefore, the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to the base, and divides the angle at the vertex into two equal parts. In a triangle which is not isosceles, any side may be assumed indifferently as the base ; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however, BOOK I. Si that side is generally assumed as the base, which is not equal to either of the other two. PROPOSITION XII. THEOREM. Conversely, if two angles of a triangle are equal, the sides oppo' site them are also equal, and the triangle is isosceles. Let the angle ABC be equal to the angle ACB ; then will the side AC be equal to the side AB. For, if these sides are not equal, suppose AB to be the greater. Then, take BD equal to AC, and draw CD. Now, in the two triangles BDC, BAC, we have BD=AC, by construction ; the angle B equal to the angle ACB, by hypothesis ',]^J. and the side BC common : therefore, the two triangles, BDC, BAC, have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each : hence they are equal (Prop. V.). But the part cannot be equal to the whole (Ax. 8.) ; hence, there is no inequality between the sides BA, AC ; therefore, the triangle BAC is isosceles. PROPOSITION XIII. THEOREM. The greater side of everi^ triangle is opposite to the greater an- gle ; and conversely, the greater angle is opposite to the greater side. First, hei the angle C be greater than the angle B ; then will the side AB, opposite C, be greater than AC, opposite B. For, make the angle BCD=B. Then, in the triangle CDB, we shall haveCD=BD (Prop. XII.). Now, the side AC < AD + CD; butAD+CD=C' AD + DB=AB: therefore ACAC; then will the angle C, opposite to AB, be greater than the angle B, opposite to AC. For, if the angle CAC, the angle C must be greater than B. 22 GEOMETRY. PROPOSITION XIV. THEOREM. From a given pointy without a straight line, only one perpendicu- lar can he drawn to that line. Let A be the point, and DE the given line. Let us suppose that we can draw two perpendiculars, AB, AC. Produce either of them, as AB, till BF is equal to AB, and D^ draw FC. Then, the two triangles CAB, CBF, will be equal: for, the angles CBA, and CBF are right angles, the side CB is ^I* common, and the side AB equal to BF, by construction ; there- fore, the triangles are equal, and the angle ACB=BCF (Prop. V. Cor.). But the angle ACB is a right angle, by hypothesis ; therefore, BCF must likewise be a right angle. But if the adja- cent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (Prop. IIL) : from whence it fol- lows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Ax. 11.) J hence, two perpendiculars cannot be drawn from the same point to the same straight line. Scholium. At a given point C, in the line ^1 AB, it is equally impossible to erect two per- pendiculars to that line. For, if CD, CE, were those two perpendiculars, the angles BCD, BCE, would both be right angles:— hence they would be equal (Ax. 10.); and the line CD would coincide with CE ; otherwise, a part would be equal to the whole, which is impossible (Ax. 8.). PROPOSITION XV. THEOREM. If from a point without a straight line, a perpendicular he let fall on the line, and ohlique lines he drawn to different points : 1st, The perpendicular ivill be shorter than any ohlique line. 2d, Any tioo ohlique lines, drawn on different sides of the perpen- dicular, cutting off equal distances on the other line, will he equal. 2d, Of two ohlique lines, drawn at pleasure, that which is farther from the peipendicular will he the longer. BOOK I. 23 Let A be the given point, DE the given line, AB the perpendicular, and AD, AC, AE, the oblique lines. Produce the perpendicular AB till BF is equal to AB, and draw FC, FD. D'" First. The triangle BCF, is equal to the triangle BCA, for they have the right angle CBF=CBA, the side CB common, and the "^-^ side BF=BA ; hence the third sides, CF and CA are equal (Prop. V. Cor.). But ABF, being a straight line, is shorter than ACF, v^^hich is a broken line (Def. 3.) ; therefore, AB, the half of ABF, is shorter than AC, the half of ACF ; hence, the per- pendicular is shorter than any obhque line. Secondly. Let us suppose BC=BE; then will the triangle CAB be equal to the the triangle BAE ; for BC=BE,the side AB is common, and the angle CBA=ABE ; hence the sides AC and AE are equal (Prop. V. Cor.) : therefore, two oblique, lines, equally distant from the perpendicular, are equal. Thirdly. In the triangle DFA, the sum of the lines AC, CF, is less than the sum of the sides AD, DF (Prop. VIII.) ; there- fore, AC, the half of the line ACF, is shorter than AD, the half of the line ADF : therefore, the obhque line, which is farther from the perpendicular, is longer than the one which is nearer. Cor. \. The perpendicular measures the shortest distance of a point from a line. Cor. 2. From the same point to the same straight line, only two equal straight lines can be drawn \ for, if there could be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impossible. PROPOSITION XVI. THEOREM. If from the middle point of a straight line, a perpendicular he drawn to this line ; \st, Every point of the perpendicular will he equally distant from the extremities of the line, 2d, Every point, without the perpendicular, will he unequally dis- tant from those extremities. 124 GEOMETRY. Let AB be the given straight line, C the middle point, and ECF the perpendicular. First, Since AC = CB, the two oblique lines AD, DB, are equally distant from the perpen- dicular, and therefore equal (Prop. XV.). So, likewise, are the two oblique lines AE, EB, the^^ two AF, FB, and so on. Therefore every point in the perpendicular is equally distant from the extremities A and B. Secondly, Let I be a point out of the perpen- dicular. If lA and IB be drawn, one of these lines will cut the perpendicular in D ; from which, drawing DB, we shall have DB=DA. But the straight line IB is less than ID + DB, and ID + DB=ID + DA-IA; therefore, IBCI; hence of two unequal chords, the less is the farther from the centre. PROPOSITION IX. THEOREM. A straight line perpendicular to a radius, at its extremity, is a tangent to the circumference. Let BD be perpendicular to the B. radius C A, at its extremity A ; then will it be tangent to the circumfe- rence. For, every oblique hne CE, is longer than the perpendicular CA (Book I. Prop. XV.); hence the point E is without the circle ; therefore, BD has no point but A common to it and the circumference ; consequently BD is a tangent (Def. 8.). :i8 GEOMETRY. Scholium. At a given point A, only one tangent AD can be drawn to the circumference ; for, if another could be drawn, it would not be perpendicular to the radius CA (Book I. Prop, XIV. Sch.) ; hence in reference to this new tangent, the radius AC would be an oblique line, and the perpendicular let fall from the centre upon this tangent would be shorter than CA ; hence this supposed tangent would enter the circle, and be a secant. PROPOSITION X. THEOREM. HQ ; in other words, Two parallels intercept equal arcs on the circumfirence. There may be three cases. First. If the two parallels are se- cants, draw the radius CH perpendicu- lar to the chord MP. It will, at the same time be perpendicular to NQ (Book I.Prop.XX.Cor. 1 .) ; therefore, the point H will be at once the middle of the arc MHP, and of the arc NHQ (Prop. VI.) ; therefore, we shall have the arc MH=HP, and the arc NH = HQ ; and therefore MH— NH=HP- MN=PQ. Second. When, of the two paral- lels AB, DE, one is a secant, the other a tangent, draw the radius CH to the point of contact H ; it will be perpendicular to the tangent DE (Prop. IX.), and also to its parallel MP. But, since CH is perpendicular to the chord MP, the point H must be the middle of the arc MHP (Prop. VI.) ; therefore the arcs MH, HP, in- cluded between the parallels AB, DE, are equal. Third. If the two parallels DE, IL, are tangents, the one at H, the other at K, draw the parallel secant AB ; and, from what has just been shown, we shall have MH=HP, MK=KP; and hence the whole arc HMK=HPK. It is farther evident that each of these arcs is a semicircumference. BOOK IIL 49 PROPOSITION XI. THEOREM. If two circles cut each other in two points, the line which passes through their centres, will he perpendicular to the chord which joins the points of intersection, and will divide it into two equal parts. For, let the line AB join the points of intersection. It will be a common chord to the two circles. Now if a perpendicular be erected from the middle of this chord, it will pass through each of the two centres C and D (Prop. VI. Sch.). But no more than one straight line can be drawn through two points ; hence the straight hne, which passes through the centres, will bisect the chord at right angles. PROPOSITION Xn. THEOREM. If the distance between the centres of two circles is less than the sum of the radii, the greater radius being at the same time less than the sum of the smaller and the distance between the centres, the two circumferences will cut each other. For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD < AC + AD, but also the greater radius AD< AC + CD (Book I. Prop. VII.). And, whenever the triangle CAD can be constructed, it is plain that the circles described from the centres C and D, will cut each other in A and B. 5a GEOMETRY. PROPOSITION XIII. THEOREM. Ij the distance between the centres of two circles is equal to the sum of their radiif the two circles will touch each other exter- nally. Let C and D be the centres at a distance from each other equal to CA + AD. The circles will evidently have the point A common, and they w^ill have no other; because, if they had two points common, the distance between their centres must be less than the sum of their radii. PROPOSITION XIV. THEOREM. If the distance between the centres of two circles is equal to the difference of their radiif the two circles will touch each other internally. Let C and D be the centres at a dis- tance from each other equal to AD — CA. It is evident, as before, that they will have the point A common ; they can have no other; because, if they had, the greater radius AD must be less than the sum of the radius AC and the distanceCD between the centres (Prop. XIL) ; which is contraiy to the supposition. Cor. Hence, if two circles touch each other, either exter- nally or internally, their centres and the point of contact will be in the same right line. Scholium. All circles which have their centres on the right line AD, and which pass through the point A, are tangent to each other. For, they have only the point A common, and it through the point A, AE be drawn perpendicular to AD, the straight line AE will be a common tangent to all the circles. BOOK III. 61 PROPOSITION XV. THEOREM. In the same circle, or in equal circles, equal angles having their vertices at the centre, intercept equal arcs on the circumference : and conversely, -if the arcs intercepted are equal, the angles contained by the radii will also he equal. Let C and C be the centres of equal circles, and the angle ACB=DCE. First. Since the angles ACB, DCE, are equal, they may be placed upon each other ; and since their sides are equal, the point A will evidently fall on D, and the point B on E. But, in that case, the arc AB must also fall on the arc DE ; for if the arcs did not exactly coincide, there would, in the one or the other, be points unequally distant from the centre ; which is impossible : hence the arc AB is equal to DE. Secondly. If we suppose AB=DE, the angle ACB will be equal to DCE» For, if these angles are not equal, suppose ACB to be the greater, and let ACI be taken equal to DCE. From what has just been shown, we shall have AI=DE : but, by hypothesis, AB is equal to DE ; hence AI must be equal to AB, or a part to the whole, which is absurd (Ax. 8.) : hence, the angle ACB is equal to DCE. PROPOSITION XVI. THEOREM. In the same circle, or in equal circles, if two angles at the centre are to each other in the proportion of two whole numbers, the intercepted arcs will be to each other in the proportion of the same numbers, and we shall have the angle to the angle, as the corresponding arc to the corresponding arc. 52 GEOMETRY. Suppose, for example, that the angles ACB, DCE, are to each other as 7 is to 4 ; or, which is the same thing, suppose that the angle M, which may serve as a common measure, is contained 7 times in the angle ACB, and 4 times in DCE. C The seven partial angles ACw, twCti, nCp, &c., into which ACB is divided, being each equal to any of the four partial angles into which DCE is divided ; each of the partial arcs Am, mn, np, &c., will be equal to each of the partial arcs Do', xy, &c. (Prop. XV.). Therefore the whole arc AB will be to the whole arc DE, as 7 is to 4. But the same reasoning would evidently apply, if in place of 7 and 4 any numbers whatever were employed ; hence, if the ratio of the angles ACB, DCE, can be expressed in whole numbers, the arcs AB, DE, will be to each other as the angles ACB, DCE. Scholium, Conversely, if the arcs, AB, DE, are to each other as two whole numbers, the angles ACB, DCE will be to each other as the same whole numbers, and we shall have ACB : DCE : : AB : DE. For the partial arcs. Aw, mw, &c. and Dx, xy, &c., being equal, the partial angles ACm, mCn, &c. and DCo;, xCy, &c. will also be equal. PROPOSITION XVII. THEOREM. Whatever be the ratio of two angles, they will always he to each other as the arcs intercepted between their sides ; the arcs being described from the vertices of the angles as centres, with equal radii. Let ACB be the greater and ACD the less angle. Let the less angle be placed on the greater. If the propo- sition is not true, the angle ACB will be to the angle ACD as the arc AB is to an arc greater or less than AD. Suppose this arc to be greater, and let it be represented by AO ; we shall thus have, the angle ACB : angle ACD : : arc AB : arc AO. Next conceive the ai*c BOOK III. 63 AB to be divided into equal parts, each of which is less than DO ; there will be at least one point of division between D and O ; let I be that point ; and draw CI. The arcs AB, AI, will be to each other as two whole numbers, and by the preceding theorem, we shall have, the angle ACB : angle ACI : : arc AB : arc AI. Comparing these two proportions with each other, we see that the antecedents are the same : hence, the conse- quents are proportional (Book II. Prop. IV.) ; and thus we find the angle ACD : angle ACI : : arc AO : arc AI. But the arc AO is greater than the arc AI ; hence, if this proportion is true, the angle ACD must be greater than the angle ACI : on the contrary, however, it is less ; hence the angle ACB cannot be to the angle ACD as the arc AB is to an arc greater than AD. By a process of reasoning entirely similar, it may be shown that the fourth term of the proportion cannot be less than AD ; hence it is AD itself ; therefore we have Angle ACB : angle ACD : : arc AB : arc AD. Cor. Since the angle at the centre of a circle, and the arc intercepted by its sides, have such a connexion, that if the one be augmented or diminished in any ratio, the other will be augmented or diminished in the same ratio, we are authorized to establish the one of those magnitudes as the measure of the other ; and we shall henceforth assume the arc AB as the mea- sure of the angle ACB. It is only necessary that, in the com- parison of angles with each other, the arcs which serve to measure them, be described with equal radii, as is imphed in all the foregoing propositions. Scholium 1. It appears most natural to measure a quantity by a quantity of the same species ; and upon this principle it would be convenient to refer all angles to the right angle ; which, being made the unit of measure, an acute angle would be expressed by some number between and 1 ; an obtuse an- gle by some number between 1 and 2, This mode of express- ing angles would not, however, be the most convenient in practice. It has been found more simple to measure them by arcs of a circle, on account of the facility with which arcs can be made equal to given arcs, and for various other reasons. At all events, if the measurement of angles by arcs of a circle is in any degree indirect, it is still equally easy to obtain the direct and absolute measure by this method ; since, on comparing the arc which serves as a measure to any an- gle, with the fourth part of the circumference, we find the ratio of the given angle to a right angle, which is the absolute measure. 54 GEOMETRY. Scholium 2. All that has been demonstrated in the last three propositions, concerning the comparison of angles with arcs, holds true equally, if applied to the comparison of sectors with arcs ; for sectors are not only equal when their angles are so, but are in all respects proportional to their angles ; hence, two sectors ACB, ACD, taken in the same circle, or in equal circles^ are to each other as the arcs AB, AD, the bases of those sectors. It is hence evident that the arcs of the circle, which serve as a measure of the different angles, are proportional to the different sectors, in the same circle, or in equal circles. PROPOSITION XVIII. THEOREM. An inscribed angle is measured by half the arc included between its sides. Let BAD be an inscribed angle, and let us first suppose that the centre of the cir- cle lies within the angle BAD. Draw the diameter AE, and the radii CB, CD. The angle BCE, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC (Book I. Prop. XXV. Cor. 6.) : but the triangle BAC being isosceles, the angle CAB is equal to ABC ; hence the angle BCE is double of BAC. Since BCE lies at the centre, it is measured by the arc BE ; hence BAC will be measured by the half of BE. For a like reason, the angle CAD will be measured by the half of ED ; hence BAC + CAD, or BAD will be measured l?y half of BE + ED, or of BED. Suppose, in the second place, that the centre C lies without the angle BAD. Then drawing the diameter AE, the angle BAE will be measured by the half of BE ; the angle DAE by the half of DE : hence their difi'erence BAD will be measured by the half of BE minus the half of ED, or by the half of BD. Hence every inscribed angle is measured by half of the arc included between its sides* BOOK III. 55 Cor. 1. All the angles BAG, BDC, BEC, inscribed in the same segment are equal ; because they are all measured by the half of the same arc BOC. Cor. 2. Every angle BAD, inscribed in a semicircle is a right angle ; because it is mea- sured by half the semicircumference BOD, that is, by the fourth part of the whole cir- cumference. Cor. 3. Every angle BAG, inscribed in a segment greater than a semicircle, is an acute angle ; for it is measured by half of the arc BOG, less than a semicircumference. And every angle BOG, inscribed in a segment less than a semicircle, is an obtuse angle ; for it is measured by half of the arc B BAG, greater than a semicircumference. Cor. 4. The opposite angles A and G, of an inscribed quadrilateral ABGD, are to- gether equal to two right angles : for the an- gle BAD is measured by half the arc BGD, the angle BGD is measured by half the arc BAD ; hence the two angles BAD, BGD, ta- ken together, are measured by the half of the circumference ; hence their sum is equal to two right angles. PROPOSITION XIX. THEOREM. The angle formed hy two chords, which intersect each other, is measured by half the sum of the arcs included between its sides. 56 GEOMETRY. Let AB, CD, be two chords intersecting each other at E : then will the angle AEC, or DEB, be measured by half of AC + DB. Draw AF parallel to DC : then will the arc DF be equal to A(^ (Prop. X.) ; and the angle FAB equal to the angle DEB (Book I. Prop. XX. Cor. 3.). But the angle FAB is measured by half the arc FDB (Prop. XVIII.); therefore, DEB is measured by half of FDB ; that is, by half of DB + DF, or half of DB + AC. In the same manner it might be proved that the angle AED is measured by half of AFD+BC. PROPOSITION XX. THEOREM. The angle formed hy two secants, is measured by half the diffe- rence of the arcs included between its sides. Let AB, AC, be two secants : then will the angle BAC be measured by half the difference of the arcs BEC and DF. Draw DE parallel to AC : then will the arc EC be equal to DF, and the angle BDE equal to the angle BAC. But BDE is measured by half the arc BE ; hence, BAC is also measured by half the arc BE ; that is, by half the difference of BEC and EC, or half the difference of BEC and DF. PROPOSITION XXI. THEOREM. The angle formed by a tangent and a chord, is measured by half of the arc included between its sides. BOOK III. 57 Let BE be the tangent, and AC the chord. From A, the point of contact, draw the diameter AD. The angle BAD is a right angle (Prop. IX.), and is measured by half the semi circumference AMD ; the angle DAC is measured by the half of DC: hence, BAD + DAC, or BAC, is measured by the half of AMD plus the half of DC, or by half the whole arc AMDC. It might be shown, by taking the difference between the an- gles DAE, DAC, that the angle CAE is measured by half the arc AC, included between its sides. '.m9^9* PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS. PROBLEM I. To divide a given straight line into two equal parts. X> 4^ Let AB be the given straight line. " From the points A and B as centres, with a radius greater than the half of AB, describe two arcs cutting each other in D ; the point D will be equally distant from A and B. Find, in like manner, above or beneath the line AB, -j a second point E, equally distant from the " points A and B ; through the two points D and E, draw the line DE : it will bisect the hne AB in C. For, the two points D and E, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle of AB (Book I. Prop. XVI. Cor.). But only one straight line can pass through two given points ; hence the line DE must itself be that perpendicular, which divides AB into two equal parts at the point C. ><^ 8 58 GEOMETRY. ^ * PROBLEM II. At a given point, in a given straight line, to erect a perpendicu- lar to this line. >^ Let A be the given point, and BC the given Hne. Take the points B and C at equal dis- tances from A ; then from the points B and C as centres, with a radius greater than — t— BA, describe two arcs intersecting each other in D ; draw AD : it will be the perpendicular required. For, the point D, being equally distant from B and C, must be in the perpendicular raised from the middle of BC (Book L Prop. XVI.) ; and since two points determine a line, AD is that perpendicular. Scholium. The same construction serves for making a right angle BAD, at a given point A, on a given straight line BC. PROBLEM III. From a given point, without a straight line, to let fall a perpen- dicular on this line. Let A be the point, and BD the straight line. From the point A as a centre, and with a radius sufficiently great, describe an arc cutting the line BD in the two points B and D ; then mark a point E, equally distant from the points B and D, and draw AE : it will be the perpendicular required. For, the two points A and E are each equally distant from the points B and D ; hence the line AE is a perpendicular passing through the middle of BD (Book L Prop. XVL Cor.). PROBLEM IV. At a point in a given line, to make an angle equal to a given angle. BOOK III. 69 Let A be the given point, AB the given line, and IKL the given angle. From the vertex K, as a cen- t o<^ tre, with any radius, describe the ^^^ ^^^\ arc IL, terminating in the two ^y^ \ ^y^ \ sides of the angle. From the E-^^- XA B point A as a centre, with a dis- tance AB, equal to KI, describe the indefinite arc BO ; then take a radius equal to the chord LI, with which, from the point B as a centre, describe an arc cutting the indefinite arc BO, in D ; draw AD ; and the angle DAB will be equal to the given angle K. For, the two arcs BD, LI, have equal radii, and equal chords ; hence they are equal (Prop. IV.) ; therefore the angles BAD, IKL, measured by them, are equal. PROBLEM V. To divide a given arc, or a given angle, into two equal parts. First. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with the same radius, describe two arcs cutting each other in D ; through the point D and the centre C, draw CD : it will bisect the arc AB in the point E. For, the two points C and D are each equally distant from the extremities A and B of the chord AB ; hence the line CD bi- sects the chord at right angles (Book I. Prop. XVI. Cor.) ; hence, it bisects the arc AB in the point E (Prop. VI.). Secondly. Let it be required to divide the angle ACB into two equal parts. We begin by describing, from the vertex C as a centre, the arc AEB ; which is then bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts. Scholium. By the same construction, each of the halves AE, EB, may be divided into two equal parts ; and thus, by successive subdivisions, a given angle, or a given arc may be divided into four equal parts, into eight, into sixteen, and so on. 60 GEOMETRY. PROBLEM VI. Through a given point, to draw a parallel to a given straight line. Let A be the given point, and BC the given line. B From the point A as a centre, with a radius greater than the shortest dis- tance from A to BC, describe the in- definite arc EO ; from the point E as a centre, with the same radius, describe the arc AF ; make ED=AF, and draw AD : this will be the parallel required. For, drawing AE, the alternate angles AEF, EAD, are evi- dently equal ; therefore, the lines AD, EF, are parallel (Book I. Prop. XIX. Cor. 1.). E Dl o PROBLEM VII. Two angles of a triangle being given, to find the third. Draw the indefinite line DEF; at any point as E, make the an- gle DEC equal to one of the given angles, and the angle CEH equal to the other : the remaining angle HEF will be the third angle required ; be- cause those three angles are together equal to two right angles (Book I. Prop. I. and XXV). problem' VIII. Two sides of a triangle, and the angle which they contain, being given, to describe the triangle. Let the lines B and C be equal to the given sides, and A the given an- gle. Having drawn the indefinite line DE, at the point D, make the angle EDF equal to the given angle A ; then take DG=B, DH=C, and draw GH ; DGH will be the triangle required (Book L Prop. V.). BOOK III. 61 PROBLEM IX. A side and two angles of a triangle being given, to describe the triangle. The two angles will either be both ad- jacent to the given side, or the one adja- cent, and the other opposite : in the lat- ter case, find the third angle (Prob. VII.) ; and the two adjacent angles will thus be known : draw the straight line DE equal to the given side : at the point D, make an angle EDF equal to one of the adjacent angles, and at E, an angle DEG equal to the other ; the two lines DF, EG, will cut each other in H ; and DEH will be the triangle required (Book I. Prop. VI.). PROBLEM X. The three sides of a triangle being given, to describe the triangle^ Let A, B, and C, be the sides. Draw DE equal to the side A ; from the point E as a centre, with a radius equal to the second side B, describe an arc ; from D as a cen- tre, with a radius equal to the third side C, describe another arc inter- secting the former in F ; draw DF, EF ; and DEF will be the triangle required (Book I. Prop. X.). Scholium. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other : but the solution will always be possible, when the sum of two sides, any how taken, is greater than the third. 62 GEOMETRY. PROBLEM XI. Two sides of a triangle, and the angle opposite one of them, being given, to describe the triangle. Let A and B be the given sides, and C the given angle. There are two cases. First. When the angle C is a right angle, or when it is obtuse, make the angle EDF=C; take DE=A ; from the point E as a centre, with a radius equal to the given side B, describe an arc cutting DF in F; draw EF : then DEF will be the triangle required. In this first case, the side B must be greater than A ; for the angle C, being a right angle, or an obtuse an- gle, is the greatest angle of the tri- angle, and the side opposite to it must, therefore, also be the greatest (Book I. Prop. XIII.). Ai \ Secondly. If the angle C is acute, and B greater than A, the same construction will again ap- ply, and DEF will be the triangle required. But if the angle C is acute, and the side B less than A, then the arc described from the centre E, with the radius EF=B, will cut the side DF in two points F and G, lying on the same side of D : hence there will be two triangles DEF, DEG, either of which wiU satisfy the conditions of the pro- blem. Scholium. If the arc described with E as a centre, should be tangent to the line DG, the triangle would be right angled, and there would be but one solution. The problem would be impossible in all cases, if the side B were less than the perpen- dicular let fall from E on the line DF. 4- BOOK III. PROBLEM XII. The adjacent sides of a parallelogram^ with the angle which they contain^ being given^ to describe the parqllelogram. Let A and B be the given sides, and C the given angle. Draw the line DE=A; at the point D, make the angle EDF= C ; take DF=B ; describe two arcs, the one from F as a cen- tre, with a radius FG=DE, the other from E as a centre, with a radius EG=DF; to the point G, where these arcs intersect each other, draw FG, EG ; DEGF will be the parallelogram required. For, the opposite sides are equal, by construction ; hence the figure is a parallelogram (Book I. Prop. XXIX.) : and it is formed with the given sides and the given angle. Cor. If the given angle is a right angle, the figure will be a rectangle ; if, in addition to this, the sides are equal, it will be a square. < PROBLEM Xin. To find the centre of a given circle or arc. Take three points, A, B, C, any where in the circumference, or the arc; draw AB,BC, or suppose them to be drawn ; bisect those two lines by the perpendiculars DE, FG : the point O, where these perpen- diculars meet, will be the centre sought (Prop. VI. Sch.). Scholium. The same construc- tion serves for making a circum- ference pass through three given points A, B, C ; and also for describing a circumference, in which, a given triangle ABC shall be inscribed. ^ir- ^^■. 64 GEOMETRY. PROBLEM XIV. Through a given pointy to draw a tangent to a given circle. If the given point A lies in the circum- ^ ference, draw the radius CA, and erect AD perpendicular to it : AD will be the tangent required (Prop. IX.). ^ . If the point A lies without the circle, join A and the centre, by the straight line CA : bisect CA in O ; from O as a centre, with the radius OC, describe a circumference intersecting the given cir- cumference in B ; draw AB : this will be the tangent required. For, drawing CB, the angle CBA be- ing inscribed in a semicircle is a right angle (Prop. XVIII. Cor. 2.) ; therefore AB is a perpendicular at the extremity of the radius CB ; therefore it is a tan- gent. Scholium. When the point A lies without the circle, there will evidently be always two equal tangents AB, AD, passing through the point A : they are equal, because the right angled triangles CBA, CDA, have the hypothenuse CA common, and the side CB = CD; hence they are equal (Book I. Prop. XVII.); hence AD is equal to AB, and also the angle CAD to CAB. And as there can be but one line bisecting the angle BAC, it follows, that the line which bisects the angle formed by two tangents^ must pass through the centre of the circle. PROBLEM XV. To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect the angles A and B, by the lines AO and BO, meeting in the point O ; from the point O, let fall the perpendiculars OD, OE, OF, on the three sides of the triangle: these perpendiculars will all be equal. For, by construe- BOOK III. ^ 66 tion, we have the angle DAO=OAF, the right angle ADO = AFO ; hence the third angle AOD is equal to the third AOF (Book I. Prop. XXV. Cor. 2.). Moreover, the side AO is com- mon to the two triangles AOD, AOF ; and the angles adjacent to the equal side are equal : hence the triangles themselves are equal (Book I. Prop. VI.) ; and DO is equal to OF. In the same manner it may be shown that the two triangles BOD, BOE, are equal ; therefore OD is equal to OE ; therefore the three perpendiculars OD, OE, OF, are all equal. Now, if from the point O as a centre, with the radius OD, a circle be described, this circle will evidently be inscribed in the triangle ABC ; for the side AB, being perpendicular to the radius at its extremity, is a tangent ; and the same thing is true of the sides BC, AC. ' - Scholium. The three lines which bisect the angles of a tri- angle meet in the same point. PROBLEM XVI. On a given straight line to describe a segment that shall contain a given cCngle ; that is to say, a segment such, that all the an- gles inscribed in it, shall be equal to the given angle. Let AB be the given straight Hne, and C the given angle. Produce AB towards D ; at the point B, make the angle DBE = C ; draw BO perpendicular to BE, and GO perpen- dicular to AB, through the middle point G ; and from the point O, where these perpendiculars meet, as a centre, with a dis- tance OB, describe a circle : the required segment will be AMB. For, since BF is a perpendicular at the extremity of the radius OB, it is a tangent, and the angle ABF is measured by half the arc AKB (Prop. XXL). Also, the angle AMB, being an inscribed angle, is measured by half the arc AKB : hence we have AMB=ABF=EBD = C ; hence all the angles in- scribed in the segment AMB are equal to the given angle C. F*9 66 GEOMETRY. Scholium. If the given angle were a right angle, the required segment would be a semicircle^ described on AB as a diameter. PROBLEM XVII. To find the numerical ratio of two given straight lines, these lines being supposed to have a common measure. Let Afi and CD be the given lines. A , C From the greater AB cut off a part equal to the less CD, as many times as possible ; for example, twice, with the remainder BE. From the line CD, cut off a part equal to the re- mainder BE, as many times as possible ; once, for ex- ample, with the remainder DP. From the first remainder BE, cut off a part equal to the second DF, as many times as possible ; once, for example, with the remainder BG. From the second remainder DF, cut off a part equal J.0. to BG the third, as many times as possible. Continue this process, till a remainder occurs, which "° is contained exactly a certain number of times in the preced- ing one. Then this last remainder will be the common measure of the proposed lines ; and regarding it as unity, we shall easily find the values of the preceding remainders ; and at last, those of the two proposed lines, and hence their ratio in numbers. Suppose, for instance, we find GB to be contained exactly twice in FD ; BG will be the common measure of the two pro- posed hues. Put BG=1 ; we shall have FD=2 : but EB con- tains FD once, plus GB ; therefore we have EB=3 ; CD con- tains EB once, plus FI) ; therefore we have CD=5 : and, lastly, AB contains CD twice, plus EB ; therefore we have AB = 13 ; hence the ratio of the lines is that of 13 to 5. If the line CD were taken for unity, the line AB would be '/ ; if AB were taken for unity, CD would be /g . --B Scholium. The method just explained is the same as that employed in arithmetic to find the common divisor of two num- bers : it has no need, therefore, of any other demonstration. How far soever the operation be continued, it is possible that no remainder may ever be found, which shall be contained an exact number of times in the preceding one. When this happens, the two lines have no common measure, and are said to be incommensurable. An instance of this will be seen after- BOOK III. 67 wards, in the ratio of the diagonal to the side of the square. In those cases, therefore, the exact ratio in numbers cannot be found ; but, by neglecting the last remainder, an approximate ratio will be obtained, more or less correct, according as the operation has been continued a greater or less number of times. PROBLEM XVIII. Two angles being given, to find their common measure, if they have one, and by means of it, their ratio in numbers. Let A and B be the given an- With equal radii describe the arcs CD, EF, to serve as mea- sures for the angles : proceed afterwards in the comparison of the arcs CD, EF, as in the last problem, since an arc may be cut off from an arc of the same radius, as a straight line from a straight line. We shall thus arrive at the common measure of the arcs CD, EF, if they have one, and thereby at their ratio in numbers. This ratio will be the same as that of the given angles (Prop. XVII.) ; and if DO is the common measure of the arcs, DAO will be that of the angles. Scholium. According to this method, the absolute value of an angle may be found by comparing the arc which measures it to the whole circumference. If the arc CD, for example, is to the circumference, as 3 is to 25, the angle A will be 2^5 of four right angles, or ^| of one right angle. It may also happen, that the arcs compared have no com- mon measure ; in which case, the numerical ratios of the angles will only be found approximatively with more or less correct- ness, according as the operation has been continued a greater or less number of times. 68 GEOMETRY. BOOK IV. OE THE PROPORTIONS OF FIGURES, AND THE MEASUREMENT OF AREAS. Definitions. 1. Similar figures are those which have the angles of the one equal to the angles of the other, each to each, and the sides about the equal angles proportional. 2. Any two sid(ite, or any two angles, which have like po- sitions in two similar figures, are called homologous sides or angles. 3. In two different circles, similar arcSf sectors^ or segments f are those which correspond to equal angles at the centre. Thus, if the angles A and O are equal, the arc BC will be similar to DE, the sector BAG to the sector DOE, and the segment whose chord is BC, to the seg- ment whose chord is DE. 4. The base of any rectilineal figure, is the side on which the figure is supposed to stand. 5. The altitude of a triangle is the per- pendicular let fall from the vertex of an angle on the opposite side, taken as a base. Thus, AD is the altitude of the triangle BAG D E P B DEC 6. The altitude of a parallelogram is the perpendicular which measures the distance between two opposite sides taken as bases. Thus, EF is the altitude of the parallelo- j gram DB. 7. The altitude of a trapezoid is the per- pendicular drawn between its two parallel sides. Thus, EF is the altitude of the trape- zoid DB. jtV. JtJ' u 8. The area and surface of a figure, are terms very nearly synonymous. The area designates more particularly the super- ficial content of the figure. The area is expressed numeri- L P B BOOK IV. 69 cally by the number of times which the figure contains some other area, that is assumed for its measuring unit. 9. Figures have equal areas, when they contain the same measuring unit an equal number of times. 10. Figures which have equal areas are called equivalent. The term equal, when applied to figures, designates those which are equal in every respect, and which being applied to each other will coincide in all their parts (Ax. 13.) : the term equi- valent implies an equality in one respect only ; namely, an equality between the measures of figures. We may here premise, that several of the demonstrations are grounded on some of the simpler operations of algebra, which are themselves dependent on admitted axioms. Thus, if we have A=B + C, and if each member is multiplied by the same quantity M, we may infer that AxM=BxM + CxM; in like manner, if we have, A=B-|-C, and D=E — C, and if the equal quantities are added together, then expunging the + C and — C, which destroy each other, we infer that A + D=B-{- E, and so of others. AH this is evident enough of itself; but in cases of difficulty, it will be useful to consult some agebrai- cal treatise, and thus to combine the study of the two sciences. PROPOSITION I. THEOREM. Parallelograms which have equal bases and equal altitudes, are equivalent. Let AB be the common base of-j) CT EDPCE the two parallelograms ABCD, V y / Vl T? ABEF: and since they are sup- \ /\ / \ / \ / posed to have the same altitude, v_V V V their upper bases DC, FE, will be ^ ^ A B both situated in one straight line parallel to AB. Now, from the nature of parallelograms, we have AD=BC, andAF=BE; for the same reason, we have DC=AB, and FE=AB; hence DC=FE: hence, if DC and FE be taken away from the same line DE, the remainders CE and DF will be equal : hence it follows that the triangles DAF, CBE, are mutually eqilateral, and consequently equal (Book I. Prop. X.). But if from the quadrilateral ABED, we take away the tri- angle ADF, there will remain the parallelogram ABEF ; and if from the same quadrilateral ABED, we take away the equal triangle CBE, there will remain the parallelogram ABCD. 70 GEOMETRY. Hence these two parallelograms ABCD, ABEF, which have the same base and altitude, are equivalent. Cor. Every parallelogram is equivalent to the rectangle I which has the same base and the same altitude. PROPOSITION II. THEOREM. Every triangle is half the parallelogram which has the same hose and the same altitude. Let ABCD be a parallelo- gram, and ABE a triangle, having the same base AB, and the same altitude : then will the triangle be half the parallelogram. •^ -^ For, since the triangle and the parallelogram have the same altitude, the vertex E of the triangle, will be in the line EC, par- allel to the base AB, Produce BA, and from E draw EF parallel to AD. The triangle FBE is half the parallelogram FC, and the triangle FAE half the parallelogram FD (Book I. Prop. XXVIII. Cor.). Now, if from the parallelogram FC, there be taken the par- allelogram FD, there will remain the parallelogram AC : and if from the triangle FBE, which is half the first parallelogram, there be taken the triangle FAE, half the second, there will re- main the triangle ABE, equal to half the parallelogram AC. Cor 1. Hence a triangle ABE is half of the rectangle ABGH, which has the same base AB, and the same altitude AH : for the rectangle ABGH is equivalent to the parallelogram ABCD (Prop. I. Cor.). Cor. 2. All triangles, which have equal bases and altitudes, are equivalent, being halves of equivalent parallelograms. PROPOSITION III. THEOREM. Two rectangles having the same altitude^ are to each other as their bases. m. BOOK IV. 71 D !& C 1 A E B Let ABCD, AEFD, be two rectan- gles having the common altitude AD : they are to each other as their bases AB, AE. Suppose, first, that the bases are commensurable, and are to each other, for example, as the numbers 7 and 4. If AB be divided into 7 equal parts, AE will contain 4 of those parts : at each point of division erect a perpendicular to the base ; seven partial rect- angles will thus be formed, all equal to each other, because all have the same base and altitude. The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four; hence the rectangle ABCD is to AEFD as 7 is to 4, or as AB is to AE. The same reasoning may be applied to any other ratio equally with that of 7 to 4 : hence, whatever be that ratio, if its terms be commensurable, we shall have ABCD : AEFD : : AB : AE. Suppose, in the second place, that the bases AB, AE, are incommensurable : it is to be shown that we shall still have ABCD : AEFD : : AB : AE. For if not, the first three terms continuing the same, the fourth must be greater or less than AE. Suppose it to be greater, and that we have ABCD : AEFD : : AB : AO. Divide the line AB into equal parts, each less than EO. There will be at least one point I of division between E and O : from this point draw IK perpendicular to AI : the bases AB, AI, will be commensurable, and thus, from what is proved above, we shall have ABCD : AIKD : : AB : AI. But by hypothesis we have ABCD : AEFD : : AB : AO. In these two proportions the antecedents are equal ; hence the consequents are proportional (Book II. Prop. IV.) ; and we find AIKD : AEFD : : AI : AO. But AO is greater than AI ; hence, if this proportion is cor- rect, the rectangle AEFD must be greater than AIKD : on the contrary, however, it is less ; hence the proportion is im- possible ; therefore ABCD cannot be to AEFD, as AB is to a line greater than AE. EIOB n GEOMETRY. Exactly in the same manner, it may be shown that the fourth term of the proportion cannot be less than AE ; therefore it is equal to AE. Hence, whatever be the ratio of the bases, two rectangles ABCD, AEFD, of the same altitude, are to each other as their bases AB, AE. PROPOSITION IV. THEOREM. Any two rectangles are to each other as the products of their bases multiplied by their altitudes. Let ABCD, AEGF, be two rectangles ; then will the rect- angle, ABCD : AEGF : : AB.AD : AF.AE. Having placed the two rectangles, so that the angles at A are vertical, produce the sides GE, CD, till they meet in H. The two rectangles ABCD, AEHD, having the same al- titude AD, are to each other as their bases AB, AE : in like manner the two rectangles AEHD, AEGF, having the same altitude AE, are to each other as their bases AD, AF : thus we have the two proportions, ABCD : AEHD : : AB ; AE, AEHD : AEGF : : AD : AF. Multiplying the corresponding terms of these proportions together, and observing that the term AEHD may be omit- ted, since it is a multiplier of both the antecedent and the con- sequent, we shall have ABCD : AEGF : : ABxAD : AExAF. Scholium, Hence the product of the base by the altitude may be assumed as the measure of a rectangle, provided we under- stand by this product, the product of two numbers, one of which is the number of linear units contained in the base, the other the number of linear units contained in the altitude. This product will give the number of superficial units in the surface ; because, for one unit in height, there are as many superficial units as there are linear units in the base ; for two units in height twice as many ; for three units in height, three times as many, &c. Still this measure is not absolute, but relative : it supposes IT D c ■R B G -A. BOOK IV. 73 that the area of any other rectangle is computed m a similar manner, by measuring its sides with the same linear unit ; a second product is thus obtained, and the ratio of the two pro- ducts is the same as that of the rectangles, agreeably to the proposition just demonstrated. For example, if the base of the rectangle A contains three units, and its altitude ten, that rectangle will be represented by the number 3 x 10, or 30, a number which signifies nothing while thus isolated ; but if there is a second rectangle B, the base of which contains twelve units, and the altitude seven, this second rectangle will be represented by the number 12x7 = 84 ; and we shall hence be entitled to conclude that the two rectangles are to each other as 30 is to 84 ; and therefore, if the rectangle A were to be assumed as the unit of measurement in surfaces, the rectangle B would then have ^^ for its absolute measure, in other words, it would be equal to || of a super- ficial unit. . . It is more common and more simple, to assume the square as the unit of surface ; and to se- lect that square, whose side is the unit of length. In this case the measurement which we have regarded merely as relative, becomes absolute : the number 30, for instance, by which the rectangle A was measured, now represents 30 superficial units, or 30 of those squares, which have each of their sides equal to unity, as the diagram exhibits. In geometry the product of two lines frequently means the same thing as their rectangle, and this expression has passed into arithmetic, where it serves to designate the product of two unequal numbers, the expression square being employed to designate the product of a number multiplied by itself. The arithmetical squares of 1, 2, 3, &c. are 1, 4, 9, &c. So Hkewise, the geometrical square constructed on a double line is evidently four times greater than the square on a single One ; on a triple line it is nine times great- er, &c. 10 -A 74 / GEOMETRY. PROPOSITION V. THEOREM. The area of any parallelogram is equal to the product of its base by its altitude. For, the parallelogram ABCD is equivalent jF D EC to the rectangle ABEF, which has the same base AB, and the same altitude BE (Prop. I. Cor.) : but this rectangle is measured by AB X BE (Prop. IV. Sch.) ; therefore, AB x BE A. B is equal to the area of the parallelogram ABCD. Cor. Parallelograms of the same base are to each other as their altitudes ; and parallelograms of the same altitude are to each other as their bases : for, let B be the common base, and C and D the altitudes of two parallelograms : then, BxC : BxD : : C : D, (Book II. Prop. VII.) And if A and B be the bases, and C the common altitude, we shall have AxC : BxC : : A : B. And parallelograms, generally, are to each other as the pro- ducts of their bases and -altitudes. PROPOSITION VI. THEOREM The area of a triangle is equal to the product of its base by half its al^i'^'de. For, the triangle ABC is half of the par- allelogram ABCE, which has the same base BC, and the same altitude AD (Prop. II.) ; but the area of the paralfelogram is equal to BC X AD (Prop. V.) ; hence that of the trian- gle must be iBC x AD, or BC x \KD. Cor. Two triangles of the same altitude are to each other as their bases, and two triangles of the same base are to each other as their altitudes. And triangles generally, are to each other, as the products of their bases and altitudes. BOOK IV. 75 PROPOSITION VII. THEOREM. 2%e area of a trapezoid is equal to its altitude multiplied by the half sum of its parallel bases. Let ABCD be a trapezoid, EF its alti- tude, AB and CD its parallel bases; then will its area be equal to EF x i(AB + CD). Through I, the middle point of the side BC, draw KL parallel to the opposite side AD ; and produce DC tHl it meets KL. In the triangles IBL, ICK, we have the side IB=IC, by construction; the angle LIB = CIK; and since CK and BL are parallel, the angle IBL=ICK (BookL Prop. XX. Cor. 2.); hence the triangles are equal (Book L Prop. VL); therefore, the trapezoid ABCD is equivalent to the parallelogram ADKL, and is measured by EF x AL. But we have AL = DK ; and since the triangles IBL and KCI are equal, the side BL=:CK: hence, AB4- CD =^AL + DK=2AL ; hence AL is the half sum of, the bases AB, CD ; hence the area of the trapezoid ABCD, is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result which is expressed thus : ABCD=EF x :^5±J^. Scholium. If through I, the middle point of BC, the line IH be drawn parallel to the base AB, the point H will also be the middle of AD. For, since the figure AHIL is a parallelogram, as also DHIK, their opposite sides being parallel, we have AH=1L, and I)H=IK; but since the triangles BIL, CIK, are equal, we already have IL=IK; therefore, AH=DH. It may be observed, that the line HI=AL is equal to i ; hence the area of the trapezoid may also be ex- pressed by EF X HI : it is therefore equal to the altitude of the trapezoid multiplied by the line which connects the middle points of its inclined sides. 76 GEOMETRY. PROPOSITION VIII. THEOREM. If a line is divided into two parts, the square described on the whole line is equivalent to the sum of the squares described on the parts, together with twice the rectangle contained by the parts. Let AC be the line, and B the point of division ; then, is AC2 or (AB + BC)2=AB2+BC2+2ABxBC. Construct the square ACDE ; take AF= 1 2 H I ) AB ; draw FG parallel to AC, and ^H par- allel to AE. The square ACDE is made up of four parts ; the first ABIF is the square described on AB, since we made AF= AB : the second IDGH is K "Q C the square described on IG, or BC ; for since we have AC = AE and AB=AF, the difference, AC — AB must be equal to the difference AE — AF, which gives BC=EF ; butIG is equal to BC, and DG to EF, since the lines are parallel ; therefore IGDH is equal to a square described on BC. And those two squares being taken away from the whole square, there re- mains the two rectangles BCGI, EFIH, each of which is mea- sured by AB X BC : hence the large square is equivalent to the two small squares, together with the two rectangles. Cor. If the line AC were divided into two equal parts, the two rectangles EI, IC, would become squares, and the square described on the whole line would he equivalent to four times the square described on half the line. Scholium, This property is equivalent to the property de- monstrated in algebra, in obtaining the square of a binominal; which is expressed thus : (a+6)2=:aH2a6 + 6^ PROPOSITION IX. THEOREM. The square described on the difference of two lines, is equivalent to the sum of the squares described on the lines, minus twice the rectangle contained by the lines. BOOK IV. 77 Let AB and BC be two lines, AC their difference ; then is AC2, or (AB— BC)2=:AB2+BC2— 2ABxBC. Describe the square ABIF ; take AE JL IP G- I =AC ; draw CG parallel to to BI, HK parallel to AB, and complete the square EFLK. The two rectangles GBIG, GLKD, are each measured by AB x BC ; take them away from the whole figure ABILKEA, which is equivalent to AB^+BC^ and there will evidently remain the square ACDE; hence the theorem is true. Scholium. This proposition is equivalent to the algebraical formula, {a-^f=d'-^2ah-^h\ K E D A C B PROPOSITION X. THEOREM. The rectangle contained by the sum and the difference of two lines, is equivalent to the difference of the squares of those lines. O I Let AB, BC, be two lines ; then, will (AB+BC) X (AB— BC)=rAB2— BC2. On AB and AC, describe the squares -p, ABIF, ACDE ; produce ABtill the pro- duced part BK is equal to BC ; and complete the rectangle AKLE. The base AK of the rectangle EK, is the sum of the two lines AB, BC ; its altitude AE is the difference of the same lines ; therefore the rectangle AKLE is equal to (AB + BC) x (AB— BC). But this rectangle is composed of the two parts ABHE 4-BHLK ; and the part BHLK is equal to the rectangle EDGF, because BH is equal to DE, and BK to EF ; hence AKLE is equal to ABHE + EDGF. These two parts make up the square ABIF minus the square DHIG, which latter is equal to a square described on BC : hence we have (AB+BC) X (AB— BC)-AB2-^BC2 Scholium. This proposition is equivalent to the algebraical formula, {a + b)x (a — b) =a^ — b^. E h: X D L J ^ ( J I 1 I L G* 78 OEOMETRY. PROPOSITION XI. THEOREM. The square described on the hypothenuse of a right angled tri- angle is equivalent to the sum of the squares described on the other two sides. Let the triangle ABC be right angled at A. Having described squares on the three sides, let fall from A, on the hypothenuse, the perpendicular AD, which produce to E; and draw the diagonals AF, CH. The angle ABF is made up of the angle ABC, together with the right angle CBF ; the angle CBH is made up of the same angle ABC, together with the right angle ABH ; hence the -^ e angle ABF is equal to HBC. But we have AB=BH, being sides of the same square ; and BF=:BC, for the same reason : therefore the triangles ABF, HBC, have two sides and the in- cluded angle in each equal ; therefore they are themselves equal (Book I. Prop. V.). The triangle ABF is half of the rectangle BE, because they have the same base BF, and the same altitude BD (Prop. II. Cor. 1.). The triangle HBC is in like manner half of the square AH : for the angles BAC, BAL, being both right angles, AC and AL form one and the same straight line parallel to HB (Book I. Prop. III.) ; and consequently the triangle HBC, and the square AH, which have the common base BH, have also the common altitude AB ; hence the triangle is half of the square. The triangle ABF has already been proved equal to the tri- angle HBC ; hence the rectangle BDEF, which is double of the triangle ABF, must be equivalent to the square AH, which is double of the triangle HBC. In the same manner it may be proved, that the rectangle CDEG is equivalent to the square AI. But the two rectangles BDEF, CDEG, taken together, make up the square BCQF : therefore the square BCGF, de- scribed on the hypothenuse, is equivalent to the sum of the squares ABHL, ACIK, described on the two other sides ; in other words, BC-r-ABHAC^. BOOK IV. 79 Cor. 1. Hence the square of one of the sides of a right an- gled triangle is equivalent to the square of the hypothenuse diminished by the square of the other side ; which is thus ex- pressed : AB2=BC2—AC2. • Cor. 2. It has just been shown that the square AH is equi- valent to the rectangle BDEF ; but by reason of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is. to the base BD ; therefore we have BC2 : AB2 : : BC : BD. Hence the square of the hypothenuse is to the square of one of the sides about the right angle, as the hypothenuse is to the seg- ment adjacent to that side: The word segment here denotes that part of tlie hypothenuse, which is cut off by the perpen- dicular let fall from the right angle : thus BD is the segment adjacent to the side AB ; and DC is the segment adjacent to the side AG. We might have, in like manner, BC2 : AC2 : : BC : CD. Cor. 3. The rectangles BDEF, DCGE, having Hkewise the same altitude, are to each other as their bases BD, CD. But these rectangles are equivalent to the squares AH, AI ; there- fore we have AB^ : AC^ : : BD : DO. Hence the squares of the two sides containing the right angle, are to each other as the segments of the hypothenuse which lie ^adjacent to those sides. Cor. 4. Let ABCD be a square, and AC its ff diagonal : the triangle ABC being right an- gled and isosceles, we shall have AC^— AB^+ BC^=2AB^: hence the square described on the diagonal AC, is double of the square described on the side AB. This property may be exhibited more plainly, by drawing parallels to BD, through the points A and C, and parallels to AC, through the points B and D. A new square EFGH will thus be formed, equal to the square of AC. Now EFGH evidently contains eight triangles each equal to ABE ; and ABCD contains four such triangles : hence EFGH is double of ABCD. Since we have AC" ; AB^ : : 2 : 1 ; by extracting the square roots, we shall have AC : AB : : V2 : 1 ; hence, the diagonal of a square is incommensurable with its side ; a pro- perty which will be explained more fully in another place. 80 GEOMETRY. PROPOSITION XII. THEOREM. In every tinangle, the square of a side opposite an acute angle is less than the sum, of the squares of the other two sides ^ hy twice the rectangle contained hy the base and the distance from the acute angle to the foot of the perpendicular let fall from thb opposite angle on the base, or on the base produced. Let ABC be a triangle, and AD perpendicular to the base CB ; then will AB2= ACH BC^— 2BC x CD, There are two cases. First. When the perpendicular falls within the triangle ABC, we have BD==:BC— CD, and consequently BD^-BCHCD^— 2BC xCD (Prop. IX.). Adding AD^ to each, and observing that the right angled trian- gles ABD, ADC, give AD^+BW=AB\ and AD2+CD2=AC2, we have AB^^BC^-f ^ AC2— 2BC X CD. ® Secondly. When the perpendicular AD falls without the triangle ABC, we have'BD = CD— BC ; and consequently BD^^CD^^- BC^— 2CD X BC (Prop. IX.). Adding AD^ to both, we find, as before, AB2=BC2+AC2 — 2BCxCD. PROPOSITION XIII. THEOREM. In every obtuse angled triangle, the square of the side opposite the obtuse angle is greater than the sum of the squares of the other two sides by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicular let fall from the opposite angle on the base produced. Let ACB be a triangle, C the obtuse angle, and AD perpen- dicular to BC produced ; then will AB2^AC2+BCH2BCx CD. The perpendicular cannot fall within the triangle ; for, if it fell at any point such as E, there would be in the triangle ACE, the right angle E, and the obtuse angle C, which is impossible (Book L Prop, XXV. Cor. 3.) ; BOOK IV. 81 hence the perpendicular falls without ; and we have BD=BC 4- CD. From this there results BD2=BC2 + CW+ 2BC x CD (Prop. VIII.). Adding AD^ to both, and reducing the sums as in the last theorem, we find AB^^BCH ACHsBC x CD. Scholium. The right angled triangle is the only one in which the squares described on the two sides are together equivalent to the square described on the tUird ; for if the angle contained by the two sides is acute, the sum of their squares will be greater than the square of the opposite side ; if obtuse, it will be less. PROPOSITION XIV. THEOREM. In any triangle, if a straight line be dravm from the vertex to the middle of the base, twice the square of this line, together with twice the square of half the base, is equivalent to the sum of the squares of the other two sides of the triangle. Let ABC be any triangle, and AE a line drawn to the mid- dle of the base BC ; then will 2AE2-f2BE2=AB2+AC2. On BC, let fail the perpendicular AD. Then, by Prop. XII. AC2=AEHEC2_2EC X ED. And by Prop. XIII. AB2-AEHEB2+2EB x ED. b"" ED" Hence, by adding, and observing that EB and EC are equal, we*^ have. AB2 + AC2-2 AE2 + 2EB2. Cor. Hence, in every parallelogram the squares of the sides are together equivalent to the squares of the diagonals. For the diagonals AC, BD, bisect each jg q other (Book I. Prop. XXXI.) ; consequently the triangle ABC gives AB24-BC2=:2AE2+2BE2. The triangle ADC gives, in like manner. AD^ + DC2r=2 AEH 2DE2. Adding the corresponding members together, and observing that BE and DE are equal, we shall have AB2+ADHDCHBC2r=4AE2+4DE2 But 4AE2 is the square of 2AE, or of AC ; 4DE^ is the squares of BD (Prop. VIII. Cor.) : hence the squares of the sides are together equivalent to the squares of the diagonals. 82 GEOMETRY. PROPOSITION XV. THEOREM. If a line he drawn parallel to the base of a triangle^ it will divide the other sides proportionally. Let ABC be a triangle, ami DE a straight line drawn par- allel to the base BC ; then will AD : DB : : AE : EC. Draw BE and DC. The two triangles BDE, DEC having the same base DE, and the same altitude, since both Iheir vertices lie in a line parallel to the base, are equivalent (Prop. II. Cor. 2.). . ' The triangles ADE, BDE, whose common vertex is E, have the same altitude, and are to each other as their bases (Prop. VI. Cor.) ; hence we have ADE : BDE : : AD : DB. The triangles ADE, DEC, whose common vertex is D, have also the same altitude, and are to each other as their bases ; hence ADE : DEC : : AE : EC. But the triangles BDE, DEC, are equivalent ; and therefore, we have (Book II. Prop. IV. Cor.) AD : DB : : AE : EC. Cor. 1. Hence, by composition, we have AD + DB : AD : : AE + EC : AE, or AB : AD : : AC : AE ; and also AB : BD : : AC : CE. Cor. 2. If between two straight lines AB, CD, any number of parallels AC, EF, GH, BD, &c. be drawn, those straight lines will be cut proportionally, and we shall have AE : CF : : EG : FH : GB : HD. For, let O be the point where AB and CD meet. In the triangle OEF, the line AC being drawn parallel to the base EF, we shall have OE : AE : : OF : CF, or OE ; OF : : AE ; CF. In the triangle OGH, we shall likewise have OE : EG : : OF : FH,brOE : OF : : EG ; FH. And by reason of the common ratio OE : OF, those two proportions give AE : CF : : EG : FH. It may.be proved in the same manner, that EG : FH : : GB : HD, and so on ; hence the lines AB, CD, are cut proportionally by the parallels AC, EF, GH, &c. BOOK IV. 83 PROPOSITION XVI. THEOREM. Conversely, if two sides of a triangle are cut proportionally by a straight line, this straight line will be parallel to the third side. In the triangle ABC, let the line DE be drawn, making AD : DB : t AE : EC : then will DE be parallel to BC. For, if DE is not parallel to BC, draw DO paral- lel to it. Then, by the preceding theorem, we shall have AD : PB : : AO : OC. But by hypothe- sis, we have AD : DB : : AE : EC ; hence we must have AO : OC : : AE ; EC,or AO i AE : : OC : EC ; an impossible result, since AO, the one antecedent, is less than its consequent AE, and OC, the other antecedent, is greater than its consequent EC. Hence the parallel to BC, drawn from the point D, cahnot differ from DE ; hence DE is that parallel. Scholium, The same conclusion would be true, if the pro- portion AB ; AD : : AC : AE were the proposed one. For this proportion would give AB — AD : AD : : AC — AE : AE, or BD : AD : : CE : AE. PROPOSITION XVII. THEOREM. The line which bisects the vertical angle of a triangle, divides the base into two segments, which are proportional to the adjacent sides. In the triangle ACB, let AD be drawn, bisecting the angle CAB; then will BD : CD : : AB : AC. Through the point C, draw CjE e parallel to AD till it meets BA K.^. produced. In the triangle BCE, the line AD is parallel to the base CE ; hence we have the proportion (Prop. XV.), * BD : DC : : AB : AE. But the triangle ACE is isos- celes : for, since AD, CE are parallel, we have the angle ACE = DAC, and the angle AEC—BAD (Book I. Prop. XX. Cor. 2 & 3.) ; but, by hypothesis, DAC=zBAD ; hence the an- gle ACErr:AEC, and consequently AEmAC (Book I. Prop. XII.). In place of AE in the above proportion, substitute AC, and we shall have BD : DC : : AB : AC. 84 - GEOMETRY. PROPOSITION XVIII. THEOREM. Tmo.equiangular triangles have their homologous sides propor- tional, and are similar. Let ABC, CDE be two triangles which have their angles equal each to each, namely, BAC=r:CDE, ABC=:DCE and ACB=: DEC ; then the homologous sides, or the sides adjacent to the equal angles, will be proportional, so that we shall have BC : CE : : AB : CD : : AC : DE. Place .the homologous sides BC, CE in the same straight line ; and produce the sides BA, ED, till they meet in F. Since BCE is a straight hne, and the angle BCA is equal to CED, it follows that AC is parallel to DE (Book I. Prop. XIX. Cor. 2.). In like maimer, since the ai^le ABC is equal to DCE, the line AB is parallel to DC. Hence the figure ACDF is a parallelogram. In the triangle BFE, the line AC is parallel to the base FE ; hence we have BC : CE : : BA : AF (Prop. XV.) ; or put- ting CD in the place of its equal AF, BC : CE : : BA : CD. In the same triangle BEF, CD is parallel to BF which may be considered as the base ; and we have the proportion BC : CE : : FD : DE ; or putting AC in the place of its equal FD, BC : CE : : AC : DE. And finally, since both these proportions contain the same ratio BC : CE, we have AC : DE : : BA : CD. Thus the equiangular triangles BAC, CED, have their ho- mologous sides proportional. But two figures are similar when they have their angles equal, each to each, and their homolo- gous sides proportional (Def. 1.) ; consequently the equiangu- lar triangles BAC, CED, are two similar figures. Cor. For the similarity of two triangles, it is enough that they have two angles equal, each to each ; since then, the third will also be equal, in both, and the two triangles will be equiangular. (? BOOK IV. 86 Scholium, Observe, that in similar triangles, the homolo- gous sides are opposite to the equal angles ; thus the angle ACB being equal to DEC, the side AB is homologous to DC ; in like manner, AC and DE are homologous, because opposite to the equal angles ABC, DCE. When the homologous sides are de- termined, it is easy to form the proportions : AB : DC : : AC : DE : : BC : CE. PROPOSITION XIX. THEOREM. Two triangles, which have their homologous sides proportional, are equiangular and similar. In the two triangles BAC, DEF, suppose we have BC : EF : : AB : DE : : AC : DF; then will the triangles ABC, DEF have their an- gles equal, namely, A=D, B=E, At the point E, make the angle FEG=B, and at F, the angle EFG=C ; the third G will be equal to the third A, and the two triangles ABC, EFG will be equiangular (Book I. Prop. XXV. Cor. 2.). Therefore, by the last theorem, we shall have BC : EF : : AB : EG ; but, by hypothesis, we have BC : EF : : AB : DE; hence EG = DE. By the same theorem, we shall also have BC : EF : : AC ; FG; and by hypothesis, we have BC : EF •: : AC : DF ; hence FG=DF. Hence the triangles EGF, DEF, having their three sides equal, each to each, are themselves equal (Book I. Prop. X.). But by construction, the triangles EGF " and ABC are equiangular ; hence DEF and ABC are also equian- gular and similar. Scholium 1. By the last two propositions, it appears that in triangles, equahty among the angles is a consequence of pro- portionahty among the sides, and conversely ; so that either of those conditions sufficiently determines the similarity of two triangles. The case is dilferent with regard to figures of more than three sides : even in quadrilaterals, the proportion between the sides may be altered without altering the angles, or the angles may be altered without altering the proportion between the sides ; and thus proportionality among the sides cannot be a consequence of equality among the angles of two quadrilaterals, or vice versa. It is evident, for example, that H 86 GEOMETRY. by drawing EF parallel to BC, the angles of the quadrilateral AEFD, are made equal to those of ABCD, though the proportion be- tween the sides is different ; and, in like nian- ner, without changing the four sides AB, BC, CD, AD, we can make the point B approach D or recede from it, which will change the angles. Scholium 2. The two preceding propositions, which in strict- ness form but one, together with that relating to the square of the hypothenuse, are the most important aqd fertile in results of any in geometry : they are almost sufficient of themselves for every application to subsequent reasoning, and for solving every problem. The reason is, that all figures may be divided into triangles, and any triangle into two right angled triangles. Thus the general properties of triangles include, by implica- tion, those of all figures. PROPOSITION XX. THEOREM. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing those angles proportional, are similar. In the two triangles ABC, DEF, let the angles A and D be equal ; then, if AB : DE : : AC : DF, the two trian- gles will be similar. Take AG=DE, and draw GH paral- lel to BC. The angle AGH will be equal to the angle ABC (Book I. Prop. XX. Cor 3.) ; and the triangles AGH, ABC, will be equiangular : hence we shall have AB : AG : : AC : AH. But by hypo- thesis, we have AB : DE : : AC : DF ; and by construction, AG=DE: hence AH=DF. The two triangles AGH, DEF, have an equal angle included between equal sides ; therefore they are equal j but the triangle AGH is similar to ABC ; there- fore DEF is also similar to ABC. BOOK IV. 87 PROPOSITION XXI. THEOREM. Two triangles, which have their homologous sides parallel^ or perpendicular to each other, are similar. Let BAG, EDF, be two triangles. First. If the side AB is parallel to DE, and BC to EF, the angle ABC will be equal to DEF (Book I. Prop. XXIV.) ; if AC is parallel to DF, the angle ACB will be equal to DFE, and- also BACtoEDF; hence the triangles ABC, DEF, are equiangular; consequently they are similar (Prop. XVIII.). Secondly. If the side DE is perpen- dicular to AB, and the side DF to AC, the two angles I and H of the quadri- lateral AIDH will be right angles ; and since all the four angles are together equal to four right angles (Book I. Prop. XXVI. Cor. l.),the remaining two I AH, IDH, will be together equal to two right ^ angles. But the two angles EDF, IDH, are also equal to two right angles : hence the angle EDF is equal to lAH or BAC. In like manner, if the third side EF is perpendicular to the third side BC, it may be shown that the angle DFE is equal to C, and DEF to B : hence the triangles ABC, DEF, which have the sides of the one perpendicular to the corresponding sides of the other, are equiangular and similar. Scholium. In the case of the sides being parallel, the homolo- gous sides are the parallel ones: in the case of their being per- pendicular, the homologous sides are the perpendicular ones. Thus in the latter case DE is homologous with AB, DF with AC, and EF with BC. The case of the perpendicular sides might present a rela- tive position of the two triangles different from that exhibited in the diagram. But we might always conceive a triangle DEF to be constructed within the triangle ABC, and such that its sides should be parallel to those of the triangle compared with ABC ; and then the demonstration given in the text would apply. 88 GEOMETRY. PROPOSITION XXII. THEOREM. In any triangle, if a line he drawn parallel to the base, then, all lines drawn from the vertex will divide the base and the par- allel into proportional parts. Let DE be parallel to the base BC, and the other lines drawn as in the figure ; then will DI : BF : : IK : FG : : KL : GH. For, since DI is parallel to BF, the triangles ADI and ABF are equiangu- lar ; and we have DI : BF : : AT : AF ; and since IK is parallel to FG, we have in like manner AI : AF : : IK : FG; hence, the ratio AI i AF being common, we shall have DI : BF : : IK ; FG. In the same manner we shall find IK : FG : : KL : GH ; and so with the other segments : hence the line DE is divided at the points I, K, L, in the same proportion, as the base BC, at the points F, G, H. Cor. Therefore if BC were divided into equal parts at the points F, G, H, the parallel DE would also be divided into equal parts at the points I, K, L. PROPOSITION XXIII. THEOREM. If from the right angle of a right angled triangle, a perpendicu- lar be let fall on the hypothenuse ; then, 1st. The two partial triangles thus foimed, will be similar to each other, and to the whole triangle. 2d. Either side including the right angle will be a mean propor- tional betiveen the hypothenuse and the adjacent segment. 2d. The perpendicular will be a mean proportional between the two segments of the hypothenuse. Let BAC be a right angled triangle, and AD perpendicular to the hypothenuse BC. First. The triangles BAD and BAC have the common angle B, the right angle BDA=BAC, and therefore the third angle BAD of the one, equal to the third angle C, of the other (Book I. Prop. XXV. Cor 2.) : hence those two triangles are equiangular and BOOK IV. 89 similar. In the same manner it may be shown that the trian- gles DAC and BAG are similar ; hence all the triangles are equiangular and similar. Secondly, The triangles BAD, BAG, being similar, their homologous sides are proportional. But BD in the small tri- angle, and BA in the large one, are homologous sides, because they lie opposite the equal angles BAD, BGA ; the hypothe- nuse BA of the small triangle is homologous with the hypo- thenuse BG of the large triangle : hence the proportion BD : BA : : BA : BG. By the same reasoning, we should find DG : AG : ; AG : BG ; lience, each of the sides AB, AG, is a mean proportional between the hypothenuse and the segment adjacent to that side. Thirdly. Since the triangles ABD, ADG, are similar, by comparing their homologous sides, we have BD : AD : : AD : DC ; hence, the perpendicular AD is a mean proportional between the segments BD, DG, of the hypothenuse. Scholium. Since BD : AB : : AB : BG, the product of the extremes will be equal to that of the means, or AB^==BD.BG. For the same reason we have AG^=DG.BG ; therefore AB^+ AG2=BD.BG + DG.BG= (BD + DG).BG=BG.BG=BG2; or the square described on the hypothenuse BG is equivalent to the squares described on the two sides AB, AG. Thus we again arrive at the property of the square of the hypothenuse, by a path very different from that which formerly conducted us* to it : and thus it appears that, strictly speaking, the property of the square of the hypothenuse, is a consequence of the more general property, that the sides of equiangular triangles are proportional. Thus the fundamental propositions of geometry are reduced, as it were, to this single one, that equiangular tri- angles have their homologous sides proportional. It happens frequently, as in this instance, that by deducing consequences from one or more propositions, we are led back to some proposition already proved. In fact, the chief charac- teristic of geometrical theorems, and one indubitable proof of their certainty is, that, however we combine them together, provided only our reasoning be correct, the results we obtain are always perfectly accurate. The case would be different, if any proposition were false or only approximately true : it would frequently happen that on combining the propositions together, the error would increase and become perceptible. Examples of this are to be seen in all the demonstrations, in which the reductio ad absiirdum is employed. In such demon- strations, where the object is to show that two quantities are equal, we proceed by showing that if there existed the smallest HM2 90 GEOMETRY. inequality between the quantities, a train of accurate reason- ing would lead us to a manifest and palpable absurdity; from which we are forced to conclude that the two quantities are equal. Cor. If from a point A, in the circumference of a circle, two chords AB, AC, be drawn to the extremities of a diameter BC, the triangle BAG will be right angled at A (Book III. Prop. ^ ^ ^ XVIII. Gor. 2.) ; hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, or what is the same, AD^fuBD.DC. Hence also, in the second place, the chord AB is a mean pro- portional between the diameter BC and the adjacent segment BD, or, what is the same, AB^=BD.BG. In like manner, we have AC2~CD.BC ; hence AB^ : AC^ : : BD : DC ; and com- paring AB2 and AC^, to BC^, we have AB^ : BC^: : BD : BC, and AC^ : BC^ : : DC : BC. Those proportions between the squares of the sides compared with each other, or with the square of the hypothenuse, have already been given in the third and fourth corollaries of Prop. XI. PROPOSITION XXIV. THEOREM. Two triangles having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles. In the two triangles ABC, ADE, let the angle A be equal to the angle A ; then will the triangle ABC : ADE : : AB.AC ; AD.AE. Draw BE. The triangles ABE, ADE, having the com- mon vertex E, have the same altitude, and consequently are to each other as their bases (Prop. VI. Cor.) : that is, ABE : ADE : : AB • AD. In like manner, ABC : ABE : : AC : AE, Multiply together the corresponding terms of these proportions, omitting the common term ABE ; we have ABC ; ADE : AB.AC : AD.AE. BOOK IV 91 Cor. Hence the two triangles would be equivalent, if the rectangle AB.AC were equal to the rectangle AD.AE, or if we had AB : AD : : AE : AC ; which would happen if DC were parallel to BE. PROPOSITION XXV. THEOREM. Two similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEF, be two similar trian- gles, having the angle A equal to D, and the angle B=E. ' Then, first, by reason of the eqqal an- gles A and D, according to the last pro- position, we shall have ABC : DEF : ; AB.AC : DE.DF. Also, because the triangles are similar, AB : DE : : AC : DF, And multiplying the terms of this proportion by the corres- ponding terms of the identical proportion, AC : DF : : AC : DF, there will result AB.AC : DE.DF : : AC^ : DF. Consequently, ABC : DEF ; : AC^ : DP. Therefore, two similar triangles ABC, DEF, are to each other as the squares described on their homologous sides AC, DF, or as the squiares of any other two homologous sides. PROPOSITION XXVI. THEOREM. Two similar polygons are composed of the same number of tri^ angleSf similar each to each, and similarly situated. 92 GEGlVfETRY. Let ABCDE, FGHIK, be two similar polygons. From any angle A, in c the polygon ABCDE, draw diagonals AC, AD to the other angles. From the homologous angle F, in the other polygon -FGHIK, draw diagonals FH, FI to the other an- gles. These polygons being similar, the angles ABC, FGH, which are homologous, must be equal, and the sides AB, BC, must also be proportional to FG, GH, that is, AB : FG : : BC : GH (Def. 1.). Wherefore the triangles ABC, FGH, have each an equal angle, contained between proportional sides ; hence they are similar (Prop. XX.) ; therefore the angle BC A is equal to GHF. Take away these equal angles from the equal angles BCD, GHI, and there remains ACD=FHL But since the triangles ABC, FGH, are similar, we have AC : FH : : BC : GH ; and, since the polygons are similar, BC : GH : : CD : HI ; hence AC : FH : ; CD : HI. But the angle ACD, we already know, is equal to FHI ; hence the triangles ACD, FHI, have an equal angle in each, included between proportional sides, and are consequently similar (Prop. XX.). In the same manner it might be shown that all. the remaining triangles are similar, whatever be the number of sides in the polygons pro- posed : therefore two similar polygons are composed of the same number of triangles, similar, and similarly situated. Scholium. The converse of the proposition is equally true : If two polygons are composed of the same number of triangles similar and similarly situated^ those two polygons will he similar. For, the similarity of the respective triangles will give the angles, ABC = FGH, BCA=GHF, ACD = FHI : hence BCD = GHI, likewise CDE=HIK, &c. Moreover we shall have AB : FG : : BC : GH : : AC : FH : : CD : HI, ifec.; hence the two polygons have their angles equal and their sides pro- portional ; consequently they are similar. - PROPOSITION XXVII. THEOREM. The contours or perimeters of similar polygons are to each other as the homologous sides : and the areas are to each other as the squares described on those sides. BOOK IV. 98 First. Since, by the nature of similar figures, we have AB : FG : : BC : GH ::CD : HI, &c. we conclude from this series of equal ratios that the sum of the ante- cedents AB + BC + CD, &c., which makes up the perimeter of the first polygon, is to the sum of the consequents FG+GH + HI, &c., which makes up the perimeter of the second polygon, as any one antecedent is to its consequent ; and therefore, as the side AB is to its cor- responding side FG (Book II. Prop. X.). Secondly. Since the triangles ABC, FGH are similar, we shall have the triangle ABC : FGH : : AC^ : FH^ (Prop. XXV.) ; and in like manner, from the similar triangles ACD, FHI, we shall have ACD : FHI : : AC^ : FH^; therefore, by reason of the common ratio, AC^ : FH^ we have ABC : FGH : : ACD : FHI. By the same mode of reasoning, we should find ACD : FHI : : ADE : FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents ABC + ACD + ADE, or the polygon ABCDE, is to the sum of the consequents FGH + FHI + FIK, or to the polygon FGHIK, as one antecedent ABC, is to its consequent FGH, or as AB^ is to FG^ (Prop. XXV;) ; hence the areas of similar poly- gons are to each other as the squares described on the homolo- gous sides. Cor^ If three similar figures were constructed, on the three sides of a right angled triangle, the figure on the hypo- thenuse would be equivalent to the sum of the other two : for the three figures are proportional to the squares of their homologous sides ; but the square-pf the hypothenuse is equivalent to the sum of the squares of the two other sides ; hence, &c. PROPOSITION XXVIII. THEOREM. The segments of two chords, which intersect each other in a circle, are reciprocally proportional. 94 GEOMETRY. Let the chords AB and CD intersect at O : then will AO : DO : : OC : OB. Draw AC and BD. In the triangles ACO, BOD, the angles at O are equal, being verti- cal ; the angle A is equal to the angle D, be- cause both are inscribed in the same segment (Book III. Prop. XVIII. Cor. 1.) ; for the same reason the angle CrrB ; the triangles are there- fore similar, and the homologous sides give the proportion AO : DO : : CO : OB. Cor. Therefore AO.OB=DO.CO: hence th^ rectangle under the two segments of the one chord is equal to the rect- angle under the two segments of the other. PROPOSITION XXIX. THEOREM. If from the same point without a circle, two secants he drawn terminating in the concave arc, the whole secants will he recip- rocally proportional to their external segments. Let the secants OB, OC, be drawn from the point O : then will OB : OC : : OD : OA. For, drawing AC, BD, the triangles OAC, OBD have the angle O common ; likewise the angle B=C (Book III. Prop. XVIII. Cor. 1.); these triangles are therefore similar; and their homologous sides give thje proportion, OB : OC : : OD : OA. Cor. Hence the rectangle OA.OB is equal to the rectangle OC.OD. . ^ Scholium. This proposition, it may be observed, bears a great analogy to the preceding, and differs from it only as the two chords AB, CD, instead of intersecting each other within, cut each other without the circle. The following proposition may also be regarded as a particular case of the proposition just demonstrated. BOOK IV. 95 PROPOSITION XXX. THEOREM. If from the same point without a circle, a tangent and a secant he drawn, the tangent will he a mean proportional hetween the secant and its external segment. From the point O, let the tangent OA, and the secant OC be be drawn ; then will OC : OA : : OA : OD, or 0A2==0G.0D. For, drawing AD and AC, the triangles O OAD, OAC, have the angle O common ; also the angle OAD, formed by a tangent and a chord, has for its measure half of the arc AD (Book III. Prop. XXI.) ; and the angle C has the same measure : hence the angle OAD= C ; therefore the two triangles are similar, and we have the proportion OC : OA ; : AO : OD, which gives OA^=OC.OD. PROPOSITION XXXI. • THEOREM. If either angle of a triangle he bisected hy a line terminating in the opposite side, the rectangle of the sides including the bi- sected angle, is equivalent to the square of the bisecting line together with the rectangle contained by the segments of the third side. In the triangle BAC, let AD bisect the angle A ; then will AB.AC=AD2-fBD.DC. Describe a circle through the three points A, B, C ; produce AD till it meets the cir- cumference, and draw CE. The triangle BAD is similar iq the trian- gle EAC ; for, by hypothesis, tlie angle BAD=EAC; also the angle B=E, since they are both measured by half of the arc AC ; hence these triangles are similar, and the homologous sides give the proportion BA : AE : : AD : AC ; hence BA.AC=AE.AD ; but AE=AD + DE, and multi- plying each of these equals by AD, we have AE.AD=AD^+ AD.DE; now AD.DE=BD.DC (Prop. XXVIII.); hence, finally, BA.AC=AD2+BD.DC. GEOMETRY. PROPOSITION XXXIL THEOREM. In every triangle, the rectangle contained by two sides is equiva- lent to the rectangle contained hy the diameter of the circum- scribed circle f and the perpendicular let fall upon the third- side. In the triangle ABC, let AD be drawn perpendicular to BC ; and let EC be- the diameter of the circumscribed circle ; then will AB.AC=:AD.CE. For, drawing AE, the triangles ABD, AEC, are right angled, the one at D, the other at A: also the angle B=E ; these tri- angles are therefore similar, and they give the proportion AB : CE : : AD : AC ; and hence AB.AC=CE.AD. Cor, If these equal quantities be multiplied by the same quantity BC, there will result AB.AC.BC=CE.AD.BC ; now AD.BC is double of the area of the triangle (Prop. VL) ; there- fore the product of three sid^s of a triangle is equal to its area multiplied by twice the diameter of the circumscribed- circle. " The product of three lines is sometimes called a solid, for a reason that shall be seen afterwards. Its value is easily con- ceived, by imagining that the lines are reduced into numbers, and multiplying these numbers together. Scholium. It may also be demonstrated, that the area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle. . For, the triangles AOB, BOC, AOC, which have a common vertex at O, have for their com- mon altitude the radius of the inscribed circle ; hence the sum of these triangles will be equal to the sum of the bases AB, BC, . AC, multiplied by half the radius OD ; hence the area of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle. BOOK IV. 97 PROPOSITION XXXIII. THEOREM. In every quadrilateral inscribed in a circle^ the rectangle of the two diagonals is equivalent to the sum of the rectangles of the opposite sides. In the quadrilateral ABCD, we shall have AC.BD = AB.CD + AD.BC. Take the arc CO— AD, and draw BO meeting the diagonal AC in I. The angle ABD=CBI, since the one has for its measure half of the arc AD, and the other, half of CO, equal to AD ; the angle ADB=:BCI, because they are both inscribed in the same segment AOB ; hence the triangle ABD is similar to the triangle IBC, and we have the proportion AD : CI : : BD : BC ; hence AD.BC =CI.BD. Again, the triangle ABl is similar to the triangle BDC ; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc AO==DC ; hence the angle ABI is equal to DBC ; also the angle BAI to BDC, because they are in- scribed in the same segment ; hence the triangles ABI, DBC, are similar, and the homologous sides give the proportion AB : BD : : AI : CD ; hence AB.CD=AI.BD. Adding the two results obtained, and observing that AI.BD + CI.BD = (AI + CI).BDr=AC.BD, we shall have AD.BC +AB.CD=AC.BD. 113 GEOMETRY. PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM I. To divide a given straight line into any number of equal parts, or into parts proportional to given lines. First, Let it be proposed to divide the line AB into five equal parts. Through the ex- tremity A, draw the indefinite straight line AG ; and taking AC of any magnitude, apply it five times upon AG ; join the last point of division G, and the extremity B, by the straight line GB ; then draw CI parallel to GB : AI will be the fifth part of the line AB ; and thus, by applying AI five times upon AB, the line AB will be divided into five equal parts. For, since CI is parallel to GB, the sides AG, AB, are cut proportionally in C and I (Prop. XV.). But AC is the fifth part of AG, hence AI is the fifth part of AB, Secondly. Let it be pro- posed to divide the line AB into parts proportional to the given lines P, Q, R. Through A, draw the indefi- nite line AG ; make AC = P, CD=Q, DE=R; join the extremities E and B ; and through the points C, D, draw CI, DF, parallel to EB ; the line AB will be divided into parts AI, IF, FB, proportional to the given lines P, Q,R. For, by reason of the para..cis CI, DF, EB, the parts AI, IF, FB, are proportional to the parts AC, CD, DE ; and by construction, these are equal to the given lines P, Q, R. BOOK IV. 99 PROBLEM II. To find a fourth proportional to three given lines. A, B, C. Draw the two indefi- nite lines DE, DF, form- ing any angle with each other. Upon DE take DA=A, and DB=:B; upon DF take DC=C ; draw AC ; and through the point B, draw BX parallel to AC ; DX will be the fourth proportional required ; for, since BX is parallel to AC, we have the proportion DA : DB : : DC : DX ; now the first three terms of this pro- portion are equal to the three given lines : consequently DX is the fourth proportional required. Cor. A third proportional to two given lines A, B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines A, B, B. PROBLEM III. To find a mean proportional between two given lines A and B. Upon the indefinite line DF, take DE=A, and EF=:B ; upon the whole line DF, as a diameter, describe the semicircle DGF ; at the point E, erect upon the diameter the perpen- dicular EG meeting the circumfe- rence in G ; EG will be the mean proportional required. For, the perpendicular EG, let fall from a point in the cir- cumference upon the diameter, is a mean proportional between DE, EF, the two segments of the diameter (Prop. XXIII. Cor.) ; and these segments are equal to the given lines A andB. PROBLEM IV. To divide a given line into two partSy such that the greater part shall be a mean proportional between the whole line and the other part. lt)0 GEOMETRY. Let AB be the given line. At the extremity B of the line AB, erect the perpendicular BC equal to the half of AB ; from the point C, as a centre, with the ra- dius CB, describe a semicircle ; draw AC cutting the circumfe- rence in D ; and take AF=AD : ^ E B the line AB will be divided at the point F in the manner re- quired ; that is, we shall have AB : AF : : AF : FB. For, AB being perpendicular to the radius at its extremity, is a tangent ; and if AC be produced till it again meets the circumference in E, we shall have AE ; AB : : AB : AD (Prop. XXX.) ; hence, by division, AE— AB : AB : : AB— AD : AD. But since the radius is the half of AB, the diame- ter DE is equal to AB, and consequently AE — AB=AD=AF ; also, because AF=AD, we have AB — AD=FB ; hence AF : AB : : FB : AD or AF ; whence, by exchanging the extremes for the means, AB : AF : : AF : FB. SchAium. This sort of division of the line AB is called di- vision in extreme and mean ratio : the use of it will be per- ceived in a future part of the work. It may further be observed, that the secant AE is divided in extreme and mean ratio at t(he point D ; for, since AB=DE, we have AE : DE : : DE : AD. PROBLEM V. Through a given pointy in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal. Let BCD be the given angle, and A the given point. Through the point A, draw AE paral- lel to CD, make BE=CE, and through the points B and A draw BAD ; this will be the line required. For, AE being parallel to CD, we have BE : EC : : BA : AD ; but BE=EC ; therefore BA=AD. BOOK IV. PROBLEM VI. 101 To describe a square that shall he equivalent to a given parallelo- gram, or to a given triangle. First. Let ABCD be X Y the given parallelogram, AB its base, DE its alti- tude : between AB and DE find a mean propor- tional XY ; then will the square described upon XY be equivalent to the parallelogram ABCD. For, by construction, AB : XY : : XY : DE ; therefore, XY^= AB.DE ; but AB.DE is the measure of the parallelogram, and XY^ that of the square ; consequently, they are equiva- lent Secondly. Let ABC be the given triangle, BC its base, AD its altitude : find a mean proportional between BC and the half of AD, and let XY be that mean ; the square de- scribed upon XY will be equi- valent to the triangle ABC. For, since BC : XY : : XY : ^AD, it follows that XY2= BC.^AD ; hence the square described upon XY is equivalent to the triangle ABC. PROBLEM VII. Upon a given line, to describe a rectangle that shall be equiva- lent to a given rectangle. Let AD be the line, and ABFC the given rectangle. Find a fourth propor- tional to the three lines AD, AB, AC, and let AX be that fourth propor- tional ; a rectangle con- structed with the lines AD and AX will be equi- valent to the rectangle ABFC. For, since AD : AB : : AC : AX, it follows that AD. AX = AB.AC ; hence the rectangle ADEX is equivalent to the rect- angle ABFC. im^ / " * * ' ' GEOMETRY. PROBLEM VIII. To find two lines whose ratio shall be the same as the ratio of two rectangles contained by given lines. Let A.B, CD, be the rectangles contained by the given lines A, B, C, andD. Find X, a fourth proportional to the three j^ lines B, C, D ; then will the two lines A and X have the same ratio to each other as the _ ^ rectangles A.B and CD. ^ • ' For, since B : C : : D : X, it follows that ^ i -* CD=B.X ; hence A.B : CD : : A.B : B.X Xi » : : A : X. Cor, Hence to obtain the ratio of the squares described upon the given lines A and C, find a third proportional X to the lines A and C, so that A : C : : C : X ; you will then have A.X=:C2, or A2.X=zA.C2 ; hen6e A2 : C2 : : A : X. PROBLEM IX. To find a triangle that shall be equivalent to a given polygon. Let ABCDE be the given polygon. Draw first the diagonal CE cutting off the triangle CDE ; through the point D, draw DF parallel to CE, and meet- ing AE produced ; draw CF : the poly- gon ABCDE will be equivalent to the polygon ABCF, which has one side less than the original polygon. For, the triangles CDE, CFE, have the base CE common^ they have also the same altitude, since their vertices D and F,. are situated in a line DF parallel to the base : these triangles are therefore equivalent (Prop. II. Cor. 2.). Add to each of them the figure ABCE, and there will result the polygon ABCDE, equivalent to the polygon ABCF. The angle B may in like manner be cut off, by substituting for the triangle ABC the equivalent triangle AGC, and thus the pentagon ABCDE will be changed into an equivalent trir angle GCF. The same process may be applied to every other figure ; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at last be found. BOOK IV. 103 Scholium. We have already seen that every triangle may be changed into an equivalent square (Prob. VI.) ; and thus a square may always be found equivalent to a given rectilineal figure, which operation is called squaring the rectilineal figure, or finding the quadrature of it. The problem of the quadrature of the circle, consists in find- ing a square equivalent to a circle whose diameter is given. PROBLEM X. To find the side of a square which shall be equivalent to the sum or the difference of two given squares. Let A and B be the sides of the given squares. First. If it is required to find a square equivalent to the sum of these squares, draw the two indefi- nite lines ED, EF, at right angles to each other; take ED = A, and EG=B ; draw DG : this will be the side of the square re- quired. For the triangle DEG being right angled, the square de- scribed upon DG is equivalent to the sum of the squares upon ED and EG. Secondly. If it is required to find a square equivalent to the difference of the given squares, form in the same manner the right angle FEH ; take GE equal to the shorter of the sides A and B ; from the point G as a centre, with a radius GH, equal to the other side, describe an arc cutting EH in H : the square described upon EH will be equivalent to the difference of the squares described upon the lines A and B. For the triangle GEH is right angled, the hypothenuse GH=A, and the side GE=B; hence the square described upon EH, is equivalent to the difference of the squares A andB. Scholium. A square may thus be found, equivalent to the sum of any number of squares ; for a similar construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. It would be the same, if any of the squares were to be subtracted from the sura oi the others. lQ4t GEOMETRY. PROBLEM XI, To find a square which shall he to a given square as a given line to a given line. Let AC be the given D square, and M and N the given hnes. Upon the indefinite line EG, take EF=M, and FG=N ; upon EG as a diameter describe A. a semicircle, and at the point F erect the perpendicular FH. From the point H, draw the chords HG, HE, which produce indefinitely : upon the first, take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG ; HI will be the side of the square required. For, by reason of the parallels KI, GE, we have HI : HK : : HE : HG; hence, HP : HK^ ; : HE^ : HG^: but in the right angled triangle EHG, the square of HE is to the square of HG as the segment EF is to the segment FG (Prop. XI. Cor. 3.), or as M is to N ; hence HP : HK^ : ; M : N. But HK=AB ; therefore the square described upon HI is to the square described upon AB as M is to N. PROBLEM XIL Upon a given line, to describe a polygon similar to a given polygon. Let FG be the given line, and AEDCB the given polygon. In the given polygon, draw the diagonals AC, AD ; at the point F make the angle GFH=: BAC, and at the point G the angle FGH=ABC ; the lines FH, GH will cut each other in H, and FGH will be a triangle similar to ABC. In the same manner upon FH, homologous to AC, describe the triangle FIH similar to ADC ; and upon FI, homologous to AD, describe the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE, as required. For, these two polygons are composed of the same number of triangles, which are similai' and similarly situated (Prop. XXVL Sch.). BOOK IV. 105 PROBLEM XIII. T\uo similar figures being given, to describe a similar figure which shall be equivalent to their sum or their difference. Let A and B be two homologous sides of the given figures. Find a square equivalent to the sum or to the difference of the squares described upon A and B ; let X be the side of that square ; then will X in the figure required, be the side which is homologous to the sides A and B in the given figures. The figure itself may then be constructed on X, by the last problem. For, the similar figures are as the squares of their homolo- gous sides ; now the square of the side X is equivalent to the sum, or to the d.ierence of the squares described upon the homologous sides A and B ; therefore the figure described upon the side X is equivalent to the sum, or to the difference of the similar figures described upon the sides A and B. PROBLEM XIV. To describe a figure similar to a given figure, and bearing to it the given ratio of M to N. Let A be a side of the given figure, X the homologous side of the figure required. The square of X must be to the square of A, as M is to N : hence X will be found by (Prob. XL), and knowing X, the rest will be accompKshed by (Prob. XIL). 14 106 GEOMETRY. PROBLEM XV. To construct a figure similar to the figure P, and equivalent to the figure Q. Find M, the side of a square equivalent to the figure P, and N, the side of a square equiva- lent to the figure Q. Let X be a fourth proportional to the three given lines, M, N, AB ; upon the side X, homologous to AB, describe a figure similar to the figure P ; it will also be equiva- lent to the figure Q. For, calling Y the figure described upon the side X, we have P : Y : : AB2 : X^ ; but by construction, AB : X : : M : N, or AW : X^ : : M^ : N2 ; hence P : Y : : M^ : N'^. But by construction also, M2=P and N2=Q; therefore P : Y : : P : Q; consequently Y=Q; hence the figure Y is similar to the figure P, and equivalent to the figure Q. PROBLEM XVI. To construct a rectangle equivalent to a given square^ and having the sum of its adjacent sides equal to a given line. Let C be the square, and AB equal to the sum of the sides of the required rectangle. Upon AB as a diame- ter, describe a semicir- cle ; draw the line DE parallel to the diameter, at a distance AD from it, equal to the side of the i\. T?B given square C ; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter ; AF and FB will be the sides of the rectangle required. For their sum is equal to AB ; and their rectangle AF.FB is equivalent to the square of EF, or to the square of AD ; hence that rectangle is equivalent to the given square C. Scholium. To render the problem possible, the distance AD must not exceed the radius ; that is, the side of the square C must not exceed the half of the line AB. BOOK IV. 107 PROBLEM XVII. To construct a rectangle that shall he equivalent to a given square, and the difference of whose adjacent sides shall he equal to a given line. Suppose C equal to the given square, and AB the difference of the sides. Upon the given line AB as a diame- ter, describe a semicircle : at the ex- tremity of the diameter draw the tan- gent AD, equal to the side of the square C ; through the point D and the centre O draw the secant DF ; then will DE and DF be the adjacent sides of the rectangle required. For, first, the difference of these sides is equal to the diameter EF or AB ; secondly, the rectangle DE, DF, is equal to AD^ (Prop. XXX.) ; hence that rectangle is equivalent to the given square C. PROBLEM XVIII. To find the common measure, if there is one, between the diagonal and the side of a square. Let ABCG be any square what- ever, and AC its diagonal. We must first apply CB upon CA, as often as it may be contained there. For this purpose, let the semicircle DBE be described, from the centre C, with the radius CB. It is evident that CB is contained once in AC, with the remainder AD ; the result of the first operation is therefore the quotient 1, with the remainder AD, which lat- ter must now be compared with BC, or its equal AB. We might here take AF=AD, and actually apply it upon AB ; we should find it to be contained twice with a remain- der : but as that remainder, and those which succeed it, con- 108 GEOMETRY. linue diminishing, and would soon elude our comparisons by their mi- nuteness, this would be but an imper- fect mechanical method, from which no conclusion could be obtained to determine whether the lines AC, CB, have or have not a common measure. There is a very simple way, however, of avoiding these decreasing lines, and obtaining the result, by operating only upon lines which remain always of the same magnitude. The angle ABC being a right angle, AB is a tangent, and AE a secant drawn from the same point ; so that AD : AB : : AB : AE (Prop. XXX.). Hence in the second operation, when AD is compared with AB, the ratio of AB to AE may be taken instead of that of AD to AB ; now AB, or its equal CD, is con- tained twice in AE, with the remainder AD ; the result of the second operation is therefore the quotient 2 with the remain- der AD, which must be compared with AB. Thus the third operation again consists in comparing AD with AB, and may be reduced in the same manner to the com- parison of AB or its equal CD with AE ; from which there will again be obtained 2 for the quotient, and AD for the re- mainder. Hence, it is evident that the process will never terminate ; and therefore there is no common measure between the diago- nal and the side of a square : a truth which was already known by arithmetic, since these two lines are to each other : : \/2 : 1 (Prop. XI. Cor. 4.), but which acquires a greater degree of clearness by the geometrical investigation. BOOK V. 109 BOOK V. REGULAR POLYGONS, AND THE MEASUREMENT OF THE CIRCLE. Befinitioru A Polygon, which is at once equilateral and equiangular, is called a regular polygon. Regular polygons may have any number of sides : the equi- lateral triangle is one of three sides ; the square is one of Jour. PROPOSITION I. THEOREM. Two regular polygons of the same number of sides are similar figures. Suppose, for example, that ABCDEF, ahcdef are two regular; hexa- gons. The sum of all the angles is the same in both figures,being in each equal to eight right angles (Book I. Prop. XXVI. Cor. 3.). The angle A is the sixth part of that sum ; so is the angle a : hence the angles A and a are equal ; and for the same reason, the angles B and ft, the angles C and c, &c. are equal. Again, since the polygons are regular, the sides AB, BC, CD, &c. are equal, and likewise the sides ah, he, cd, &c. (Def.) ; it is plain that AB : ah : : BC : he : : CD : cd, &c. ; hence the two figures in question have their angles equal, and their ho- mologous sides proportional ; consequently they are similar (Book IV. Def. 1.). Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (Book IV. Prop. XXVII.). Scholium. The angle of a regular polygon, like the angle of an equiangular polygon, is determined by the number of its sides (Book I. Prop. XXVI.). K no GEOMETRY. % PROPOSITION II. THEOREM. Any regular polygon may he inscribed in a circle^ and circum- scribed about one. Let ABCDE &c. be a regular poly- gon : describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of EC : draw AO and OD. If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide ; for the side OP is common ; the angle OPC=OPB, each being a right angle ; hence the side PC will apply to its equal PB, and the point C will fall on B : besides, from the nature of the polygon, the angle PCD = PBA ; hence CD will take the direction BA ; and since CD = BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO ; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown, that the circle which passes through the three points B, C,.D, will also pass through the point E ; and so of all the rest : hence the circle which passes through the points A, B, C, passes also through the vertices of all the angles in the polygon, which is therefore inscribed in this circle. Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords ; they are therefore equally distant from the centre (Book III. Prop. VIII.) : hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its mid- dle point, and the circle will be inscribed in the polygon, or the polygon described about the circle. Scholium 1. The point O, the common centre of the in- scribed and circumscribed circles, may also be regarded as the centre of the polygon ; and upon this principle the angle AOB is called the angle at the centre^ being formed by two radii drawn to the extremities of the same side AB. Since all the chords AB, BC, CD, &c. are equal, all the an- gles at the centre must evidently be equal likewise ; and there- fore the value of each will be found by dividing four right an- gles by the number of sides of the polygon. BOOK V. Ill Scholium 2. To inscribe a regu- lar polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides : for the arcs being equal, the chords AB, BC, CD, &c. will also be equal ; hence likewise the triangles AOB, BOC, COD, must be equal, because the sides are equal each to each ; hence all the angles ABC, BCD, CDE, 6lc, will be equal ; hence the figure ABCDEH, will be a regular polygon. PROPOSITION III. PROBLEM. To inscribe a square in a given circle. Draw two diameters AC, BD, cut- ting each other at right angles ; join their extremities A, B, C, D : the figure ABCD will be a square. For the an- gles AOB, BOC, &c. being equal, the chords AB, BC, &c. are also equal : and the angles ABC, BCD, &c. being in semicircles, are right angles. Scholium. Since the triangle BCO is right angled and isos- celes, we have BC : BO : : V2 : 1 (Book IV. Prop. XL Cor. 4.) ; hence the side of the inscribed square is to the radius^ as the square root of 2, is to unity. PROPOSITION IV. PROBLEM. In a given circkf to inscribe a regular hexagon and an equilate- ral triangle. i 112 GEOMETRY. Suppose the problem solved, and that AB is a side of the in- scribed hexagon ; the radii AO, OB being drawn, the triangle AOB will be equilateral. For, the angle AOB is the sixth part of four right angles ; there- fore, taking the right angle for unity, we shall have AOB=|r: f : and the two other angles ABO, BAO, of the same trian- gle, are together equal to 2 — | =f ; and being mutually equal, each of them must be equal to | ; hence the triangle ABO is equilateral ; therefore the side of the inscribed hexagon is equal to the radius. Hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference ; which will bring us round to the point of beginning. And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles. Scholium, The figure ABCO is a parallelogram and even a rhombus, since AB=BC=CO=AO ; hence the sum of the SCJUCires of the diagonals AC^+BO^ is equivalent to the sum of the squares of the sides, that is, to 4AB^ or 4B0^ (Book IV. Prop XIV. Cor.) : and taking away BO^ from both, there will remain AC2=3B02; hence AC2 : BO^ : : 3 : l,or AC : BO : : V3 : 1 ; hence the side of the inscribed equilateral triangle is to the radius as the square root of three is to unity. PROPOSITION V. PROBLEM. In a given circle^ to inscribe a regular decagon^ then apentagon^ and also a regular polygon of fifteen sides. ♦»« BOOK V. 113 Divide the radius AO in extreme and mean ratio at the point M (Book IV. Prob. IV.) ; take the chord AB equal to OM the greater segment ; AB will be the side of the regular decagon, and will re- quire to be applied ten times to the circumference. For, drawing MB, we have by construction, AO : OM : : OM : AM ; or, since AB =0M, AO : AB : : AB : AM ; since the triangles ABO, AMB, have a common angle A, included between proportional sides, they are similar (Book IV. Prop. XX.). Now the fxiangle OAB being isosceles, AMB must be isosceles also, and AB==BM ; but AB=OM ; hence also MB r= OM ; hence the triangle BMO is isosceles. Again, the angle AMB being exterior to the isosceles trian- gle BMO, is double of the interior angle O (Book I. Prop. XXV. Cor. 6.) : but the angle AMB=MAB ; hence the trian- gle OAB is such, that each of the angles OAB or OBA, at its base, is double of O, the angle at its vertex ; hence the three angles of the triangle are together equal to five times the angle O, which consequently is the fifth part of the two right angles, or the tenth part of four ; hence the arc AB is the tenth part of the circumference, and the chord AB is the side of the reg- ular decagon. 2d. By joining the alternate corners of the regular decagon, the pentagon ACEGI will be formed, also regular. 3d. AB being still the side of the decagon, let AL be the side of a hexagon ; the arc BL will then, with reference to the whole circumference, be } — yV, or yV ; hence the chord BI^ will be the side of the regular polygon of fifteen sides, or pente- decagon. It is evident also, that the arc CL is the third of CB. Scholium. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the num- ber of sides : thus it is plain, that the square will enable us to in- scribe successively regular polygons of 8, 16, 32, &c. sides. And m like manner, by means of the hexagon, regular polygons of 12, 24, 48, i&c. sides may be inscribed ; by means of the deca- gon, polygons of 20, 40, 80, &lc. sides ; by means of the pente- decagon, polygons of 30, 60, 120, &c. sides. It is further evident, that any of the inscribed polygons will be less than the inscribed polygon of double the number of fiides, since a part is less than the whole. K* 15 114 GEOMETRY. ii^v;; PROPOSITION VI. PROBLEM. A regular inscHhed polygon being given, to circumscribe a sim- ilar polygon about the same circle. Let CBAFED be a regular polygon. At T, the middle point of the arc AB, apply the tangent GH, which will be parallel to AB (Book III. Prop. X.) ; do the same at the middle point of each of the arcs BC, CD, &c. ; these tangents, by their intersections, will form the regular circumscribed polygon GHIK &c. similar to the one inscribed. k: Q t Since T is the middle point of the arc BTA, and N the mid- dle point of the equal arc BNC, it follows, that BT=BN ; or that the vertex B of the inscribed polygon, is at the middle point of the arc NBT. Draw OH. The line OH will pass through the point B. For, the right angled triangles OTH, OHN, having the com- mon hypothenuse OH, and the side OT=:ON, must be equal (Book I. Prop. XVn.), and consequently the angle TOHrr: HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the pro- longation of OC ; and so with the rest. But, since GH is parallel to AB, and HI to BC, the angle GHI=ABC (Book I. Prop. XXIV.) ; in like manner HIK=: BCD ; and so with all the rest : hence the angles of the cir- cumscribed polygon are equal to those of the inscribed one. And further, by reason of these same parallels, we have GH : AB : : OH : OB, and HI : BC : : OH : OB ; therefore GH : AB : : HI : BC. But AB=BC, therefore GH=HI. For the same reason, HI = IK, &c.; hence the sides of the circum- scribed polygon are all equal ; hence this polygon is regular, and similar to the inscribed one. Cor. 1. Reciprocally, if the circumscribed polygon GHIK &c. were given, and the inscribed one ABC &c. were re- quired to be deduced from it, it would only be necessary to BOOK V. 115 draw from the angles G, H, I, &c lines OG, OH, &c. meeting the A, B, C, &c. ; then to join AB, BC, &c. ; this would form easier solution of this problem points of contact T, N, P, &c. which likewise would form an the circumscribed one. . of the given polygon, straight circumference in the points those points by the chords the inscribed polygon. An would be simply to join the by the chords TN, NP, &c. inscribed polygon similar to Cor. 2. Hence we may circumscribe about a circle any regular polygon, which can be inscribed within it, and con- versely. Cor, 3. It is plain that NH + HT~HT + TG=HG, one of the equal sides of the polygon. PROPOSITION VII. PROBLEM. A circle and regular circumscribed polygon being given, it is required to circumscribe the circle by another regular polygon having double the number of sides. i ^ /v E Let the circle whose centre is P, be circumscribed by the square CDEG : it is required to find a regular circumscribed octagon. Bisect the arcs AH, HB, BF, FA, and through the middle points c, dy a, b, draw tangents to the circle, and produce them till they meet the sides of the square : then will the figure ApHdB &c, be a regular octagon. For, having drawn P ( /\ P \ , P \ \} \ .^ J c 7t ^ k smce tne angle arJts is equal to the angle BP<2, each being half a right angle, the line Yd will fall on its equal P«, and the point d on the point a. But the angles Vdg, Faf, are right angles (Book HI. Prop. IX.) ; hence the line dg will take the direction of. The angles PB^, PB/, are also right angles ; hence B^ will take the direction B/*; therefore, the two quadrilaterals will coincide, and the point g will fall at/; hence, B^=B/, dg=af, and the angle dgB=Bfa. By applying in a similar manner, the quadrilate- rals PBfa, PF/i at the same point, yy I ^ Thus, the solid angle S, is formed by ^ y/ / \ the union of the planes ASB, BSC, CSD, V^.......U^..)c DSA. //"^ y Three planes at least, are requisite to ^- v form a sohd angle, -A B BOOK VI. 127 PROPOSITION I. THEOREM. A straight line cannot be partly in aplane^ and partly out of it. For, by the definition of a plane, when a straight line has two points common with a plane, it lies wholly in that plane. Scholium. To discover whether a surface is plane, it is ne- cessary to apply a straight line in different ways to that sur- face, and ascertain if it touches the surface throughout its whole extent. PROPOSITION II. THEOREM. Two straight lines, which intersect each other, lie in the same plane, and determine its position. Let AB, AC, be two straight lines' which intersect each other in A ; a plane may be conceived in which the straight line AB is found ; if this plane be turned round AB, until it pass through the point C, then the line AC, which has two of its points A and C, in this plane, lies wholly in it ; hence the position of the plane is determined by the single condition of containing the two straight lines AB, AC. Cor. 1. A triangle ABC, or three points A, B, C, not in a straight line, determine the position of a plane. Cor. 2. Hence also two parallels AB, CD, determine the position of a plane ; for, drawing the secant EF, the plane of the two straight lines AE, EF, is that of the parallels AB, CD. I PROPOSITION III. THEOREM. ff two planes cut each other, their common intersection will be a straight line. 128 GEOMETRY. Let the two planes AB, CD, cut each other. Draw the straight hne EF, joining any two points E and F in the common section of the two planes. This line will lie wholly in the plane AB, and also wholly in the plane CD (Book I. Def. 6.) : therefore it will be in both planes at once, and conse- quently is their common intersection. PROPOSITION IV. THEOREM. If a straight line he perpendicular io two straight lines at their point of intersection, it will be perpendicular to the plane of those lines. Let MN be the plane, of the two hnes BB, CC, and let AP be perpendicular to them at their point of intersection P ; then will AP be perpendicular to every line of the plane pass- ing through P, and consequently to the plane itself (Def. 1.). Through P, draw in the plane MN, any straight line as PQ, and through any point of this line, as Q, draw BQC, so that BQ shall be equal to QC (Book IV. Prob. V.) ; draw AB, AQ. AC. The base BC being divided into two equal parts at the point Q, the triangle BPC will give (Book IV. Prop. XIV.), PC2+PB2=2PQ2+2QC2. The triangle BAC will in like manner give, AC2+AB2=r2AQ2+2QC2. Taking the first equation from the second, and observing that the triangles APC, APB, which are both right angled at ' ^^^^ AC^— PC2= AP2, and AB^— PB2= AF ; we shall have AP2+AP2=2AQ2— 2PQ2. Therefore, by taking the halves of both, we have AP^=AQ2— PQ2, or AQ^^APHPQ^ ; hence the triangle APQ is right angled at P ; hence AP is per- pendicular to PQ. t BOOK VI. 129 Scholium. Thus it is evident, not only that a straight Hne may be perpendicular to all the straight lines which pass through its foot in a plane, but that it always must be so, when- ever it is perpendicular to two straight hues drawn in the plane ; which proves the first Definition to be accurate. Cor. 1. The perpendicular AP is shorter than any oblique line AQ ; therefore it measures the true distance from the point A to the plane MN. Cor. 2. At a given point P on a plane, it is impossible to erect more than one perpendicular to that plane ; for if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose intersection with the plane MN is PQ ; then these two perpendiculars would be per- pendicular to the line PQ, at the same point, and in the same plane, which is impossible (Book I. Prop. XIV. Sch.). It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane ; for let AP, AQ, be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible. PHOPOSITION V. THEOREM. r • If from a point without a plane, a perpendicular he drawn to the plane, and oblique lines he drawn to different points, \st. Any two ohlique lines equally distant from the perpendicular will he equal. 2d. Of any two ohlique lines unequally distant from the perpen- dicular, the more distant will he the longer. Let AP be perpendicular to the plane MN ; AB, AC, AD, oblique lines equally distant from the perpendicular, and AE a line more remote : then will AB=AC=AD; and AE will be greater than AD. For, the angles APB, APC, APD, being right angles, if we suppose the distances PB, PC, IfJ PD, to be equal to each other, the triangles APB, APC, APD, will have in each an equal angle contained by two equal sides; therefore they will be equal ; hence the hypothenuses, or the oblique lines AB, AC, AD, will be equal to each other. In like 17 130 GEOMETRY. manner, if the distance PE is greater than PD or its equal PB, the obHque line AE will evidently be greater than AB, or its equal AD. Cor. All the equal oblique lines, AB, AC, AD, &c. termi- nate in the circumference BCD, described from Pthe foot of the perpendicular as a centre ; therefore a point A being given out of a plane, the point P at which the perpendicular let fall from A would meet that plane, may be found by marking upon that plane three points B, C, D, equally dist^t from the pomt A, and then finding the centre of the circle which passes through these points ; this centre will be P, the point sought. Scholium, The angle ABP is called the inclination of the oblique line AB to the plane MN ; which inclination is evidently equal with respect to all such lines AB, AC, AD, as are equally distant from the perpendicular ; for all the triangles ABP, ACP, ADP, &c. are equal to each other. PROPOSITION VI. THEOREM. If from a point without a plane, a perpendicular he let fall on the plane, and from the foot of the perpendicular a perpendicular he drawn to any line of the plane, and from the point of inter- section a line he drawn to the first point, this latter line will he perpendicular to the line of the plane. Let AP be perpendicular to the plane NM, and PD perpendicular to BC ; then will AD be also perpen- dicular to BC. Take DB=DC, and draw PB, PC, AB, AC. Since DB=DC, the ob- lique line PB — PC : and with regard to the perpendicular AP, since PB = . PC, the oblique line AB = AC (Prop. V. Cor.) ; therefore the line AD has two of its points A and D equally distant from the extremities B and C ; therefore AD is a perpendicular to BC, at its middle point D (Book I. Prop. XVI. Cor.). EP' BOOK VI. 181 Cor. It is evident likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD. Scholium. The two lines AE, BC, afford an instance of two lines which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is at once perpendicular to the line AP and to the line BC, The distance PD is the shortest distance between them, because if we join any other two points, such as A and B, we shall have AB>AD, AD>PD; therefore AB>PD. The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other, because AE and the line drawn through one of its points parallel to BC would make with each other a right angle. In the same manner, the line AB and the line PD, which represent any two straight Knes not situated in the same plane, are supposed to form with each other the same angle, which would be formed by AB and a straight line parallel to PD drawn through one of the points of AB. PROPOSITION VII. THEOREM. If one of two parallel lines he perpendicular to a plane, the other will also be perpendicular to the same plane. Let the lines ED, AP, be parallel ; if AP is perpen- dicular to the plane NM, then will ED be also per- pendicular to it. Through the parallels AP, DE, pass a plane ; its inter- section with the plane MN will be PD^ in the plane MN "^ ^^ draw BC perpendicular to PD, and draw AD. By the Corollary of the preceding Theorem, BC is perpen- dicular to the pltuie APDE ; therefore the angle BDE is a right angle ; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE parallel to AP (Book I. Prop. XX. Cor. 1.) ; therefore the line DE is perpendicular to the two straight lines DP, DB ; consequently it is perpendicular to their plane MN (Prop. IV.j. i K^ :p ^ r l^S GEOMETRY. .. T 7 ^ /c A N- Cor. I. Conversely, if the straight lines AP, DE, are perpendicular to the same plane MN, they will be par- allel ; for if they be not so, draw through the point D, a line parallel to AP, this par- allel will be perpendicular to the plane MN ; therefore through the same point D more than one perpendicular might be erected to the same plane, which is impossible (Prop. IV. Cor. 2.). Cor. 2. Two lines A and B, parallel to a third C, are par- allel to each other ; for, conceive a plane perpendicular to the line C ; the lines A and B, being parallel to C, will be perpen- dicular to the same plane ; therefore, by the preceding Corol- lary, they will be parallel to each other. The three lines are supposed not to be in the same plane ; otherwise the proposition would be already known (Book I. Prop. XXII.). PROPOSITION VIII. THEOREM. If a straight line is parallel to a straight line drawn in a plane, it will hp parallel to thai plane. Let AB be parallel to CD of the plane NM ; then will it be parallel to the plane NM. For, if the line AB, which lies in the plane ABDC, could meet the plane MN, this could only be in some M :n point of the line CD, the common intersection of the two planes : but AB cannot meet CD, since they are parallel ; hence it will not nieet the plane MN ; hence it is parallel to that plane (Def 2.). PROPOSITION IX. THEOREM. Two planes which are perpendicular to the same straight line, are parallel to each other. BOOK VI. 133 Let the planes NM, QP, be per- - pendicular to the Hne AB, then will P they be parallel. For, if they can meet any where, let O be one of their common points, and draw OA, OB ; the line AB which is perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in that plane ; for the same reason AB is perpendicular to BO ; therefore OA and OB are two perpendiculars let fall from the same point O, upon the same straight line ; which is impossible (Book I. Prop. XIV.); therefore the planes MN, PQ, cannot meet each other ; conse- quently they are parallel. : ^ -A P V M- \"J \ "\ PROPOSITION X. THEOREM. If a "plane cut two parallel planes , the lines of intersection will he parallel. Let the parallel planes NM, QP, be intersected by the plane EH ; then will the lines of inter- section EF, GH, be parallel. For, if the Hues EF, GH, lying in the same plane, were not par- allel, they would meet each other when produced ; therefore, the planes MN, PQ, in which those lines lie, would also meet ; and hence the planes would not be psirallel. M E PROPOSITION XI. THEOREM. If two planes are parallel, a straight line which is perpendicular to one, is also perpendicular to the other. M 134 iM OMETRY. Let MN, PQ, be two parallel •--. M planes, and let AB be perpendicu- P ' ^ t X" ^ I IT lar to NM ; then will it also be per- pendicular to QP. Having drawn any line BC in the plane PQ, through the lines AB and BC, draw a plane ABC, inter- secting the plane MN in AD ; the intersection AD will be parallel to BC (Prop. X.) ; but the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD ; therefore also, to its parallel BC (Book I. Prop. XX. Cor. 1.): hence the line AB being perpendicular to any line BC, drawn through its foot in the plane PQ, is con- sequently perpendicular to that plane (Def. 1.). PROPOSITION XII. THEOREM. The parallels comprehended between two parallel planes are equal. Let MN, PQ, be two parallel planes, and FH, GE, two paral- lel lines : then will EG=FH. For, through the parallels EG, FH, draw the plane EGHF, in- tersecting the parallel planes in EF and GH. The intersections EF, GH, are parallel to each other (Prop. X.) ; so likewise are EG, FH ; therefore the figure EGHF is a parallelogram ; con- sequently, EG=FH. Cor. Hence it follows, that two parallel planes are every where equidistant : for, suppose EG were perpendicular to the plane PQ ; the parallel FH would also be perpendicular to it (Prop. VH.), and the two parallels would likewise be perpen- dicular to the plane MN (Prop. XI.) ; and being parallel, they will be equal, as shown by the Proposition. BOOK VJ.i 135 PROPOSITION XIII. THEOREM. If two angles, not situated in the same plane^ have their sides parallel and lying in the same direction, those angles will he equal and their planes will he parallel. Let the angles be CAE and DBF. Make AC=BD, AEzr BF ; and draw CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram ; therefore CD is equal and parallel to AB. For a similar rea- son, EF is equal and par- allel to AB ; hence also CD is equal and parallel to EF ; hence the figure CEFD is a parallelogram, and the side CE is equal and parallel to DF ; therefore the triangles CAE, DBF, have their corresponding sides equal ; therefore the angle CAE= DBF. Again, the plane ACE is parallel to the plane BDF. For suppose the plane drawn through the point A, parallel to BDF, were to meet the lines CD, EF, in points different from C and E, for instance in G and H ; then, the three lines AB, GD, FH, would be equal (Prop. XII.) : but the lines AB, CD, EF, are already known to be equal; hence. CD =GD, and FH=EF, which is absurd ; hence the plane ACE is parallel to BDF. Cor, If two parallel planes MN, PQ are met by two other planes CABD, EABF, the angles CAE, DBF, formed by the mtersections of the parallel planes will be equal ; for, the inter- section AC is parallel to BD, and AE to BF (Prop. X.) ; there- fore the angle CAE = DBF. PROPOSITION XIV. THEOREM. If three straight lines, not situated in the same plane, are equal and parallel, the opposite triangles formed hy joining the eX' tremities of these lines will he equal, and their planes will he parallel. 136 GEOMETRY. Let AB, CD, EF, be the lines. * Since AB is equal and parallel to CD, the figure ABDC is a parallelogram ; hence the side AC is equal and parallel to BD. For a like reason the sides AE, BF, are equal and parallel, as also CE, DF ; therefore the two triangles ACE, BDF, are equal ; hence, by the last Proposition, their planes are parallel. PROPOSITION XV. THEOREM. If two straight lines he cut by three parallel planes, they will he divided proportionally , Suppose the line AB to meet the parallel planes MN, PQ, RS, at the points A, E, B ; and the line CD to meet the same planes at the points C, F, D : we are now to show that AE : EB : : CF : FD. Draw AD meeting the plane PQ in G, and draw AC, EG, GF, BD ; the intersections EG, BD, of the parallel planes PQ, RS, by the plane ABD, are parallel (Prop. X.) ; therefore AE : EB : : AG : GD ; in like manner, the intersections AC, GF, being parallel, AG : GD : : CF : FD ; the ratio AG : GD is the same in both ; hence AE : EB : : CF : FD. H ^ - f \\ jsr + M ' V . / W " jl- — \ PROPOSITION XVI. THEOREM, If a line is perpendicular to a plane, every plane parsed through .% the perpendicular, will also he perpendicular to the plane. BOOK VI. 137 Let AP be perpendicular to the plane NM ; then will every plane passing through AP be perpendicu- lar to NM. Let BC be the intersection of the planes AB, MN ; in the plane MN, draw DE perpendicular to BP : then the line AP, being perpendicular to the plane MN, will be perpendicu- lar to each of the two straight lines BC, DE ; but the angle APD, formed by the two perpendicu- lars PA, PD, to the common intersection BP, measures the angle of the two planes AB, MN (Def. 4.) ; therefore, since that angle is a right angle, the two planes are perpendicular to each other. Scholium. When three straight lines, such as AP, BP, DP, are perpendicular to each other, each of those lines is perpen- dicular to the plane of the other two, and the three planes are perpendicular to each other. PROPOSITION XVII. THEOREM. If two planes are perpendicular to each other ^ a line draxon in one of them perpendicular to their common intersection^ will he perpendicular to the other plane. Let the plane AB be perpen- dicular, to NM ; then if the line AP be perpendicular to the inter- section BC, it will also be perpen- dicular to the plane NM. For, in the plane MN draw PD perpendicular to PB ; then, be- cause the planes are perpendicu- lar, the angle APD is a right an- gle ; therefore, the line AP is perpendicular to the two straight lines PB, PD ; therefore it is perpendicular to their plane MN (Prop. IV.). Cor. If the plane AB is perpendicular to the plane MN, and if at a point P of the common intersection we erect a perpen- dicular to the plane MN, that perpendicular will be in the plane AB ; for, if not, then, in the plane AB we might draw AP per- M* 18 138 GEOMETRY. pendicular to PB the common intersection, and this AP, at the same time, would be perpendicular to the plane MN; therefore at the same point P there would be two perpendiculars to the plane MN, which is impossible (Prop. IV. Cor. 2.). PROPOSITION XVIII. THEOREM. If two planes are perpendicular to a third plane, their common intersection will also he perpendicular to the third plane. ' Let the planes AB, AD, be per- pendicular to NM ; then will their intersection AP be perpendicular toNM. For, at the point P, erect a per- pendicular to the plane MN ; that perpendicular must be at once in the plane AB and in the plane AD (Prop. XVII. Cor.) ; therefore it is their common intersection AP. M M D ^ y E iif PROPOSITION XIX. THEOREM. If a solid angle is formed by three plane angles, the sum of any two of these angles will he greater than the third. The proposition requires demonstra- tion only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Therefore suppose the solid angle S to be formed by three plane angles ASB, ASC, BSC, whereof the angle ASB is the greatest; we are to show that ASB: Two parallelopipedons, having the same base and the same alti' tude, are equivalent. Let ABCD be the com- mon base of the two par- allelopipedons AG, AL ; since they have the same altitude, their upper bases EFGH,IKLM,willbein the same plane. Also the sides EF and AB will be equal and parallel, as well as IK and AB ; hence EF is equal and parallel to IK; for a like reason, GF is equal and parallel to LK. Let the sides EF, GH, be produced, and likewise KL, IM, till by their intersections they form the parallelogram NOPQ ; this parallelogram will evidently be equal to either of the bases EFGH, IKLM. Now if a third parallelopipedon be conceived, having for its lower base the parallelogram ABCD, and NOPQ for its upper, the third parallelopipedon will be equivalent to the parallelopipedon AG, since with the same lower base, their upper bases lie in the same plane and between the same parallels, GQ, FN (Prop. VIII.). For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon AL ; hence the two paral- lelopipedons AG, AL, which have the same base and the same altitude, are equivalent. PROPOSITION X. THEOREM. Any parallelopipedon may he changed into an equivalent rectan- gular parallelopipedon having the same altitudt and an equivalent base. BOOK VII. 153 Let AG be the par- allelopipedon proposed. From the points A, B, C, D,drawAI,BK,CL,DM, perpendicular tothe plane of the base ; you will thus form the parallelopipe- don AL equivalent to AG, and having its late- ral faces AK, BL, &c, rectangles. Hence if the base ABCD is a rectan- gle, AL will be a rectan- gular parallelopipedon equivalent to AG, and consequently, the parallelopipedon required. But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, and mQ IJ? OQ and NP perpendicular to the base ; you will then have the solid ABNO-IKPQ, which will be a rectangular parallelopipedon : for by construction, the bases ABNO, and IKPQ are rectangles ; so also are the lateral faces, the edges AT, OQ, &c. being perpendicular to the plane of the base ; hence the solid AV is a rectangular parallelopipedon. But the two parallelopipedons AP, AL may be con- ceived as having the same base ABKI and the same altitude AO : hence the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL,^ is again changed into an equivalent rectangular parallelopipe- don AP, having the same altitude AI, and a base ABNO equi- valent to the base ABCD. PROPOSITION XI. THEOREM. Two rectangular parallelopipedons^ which have the same hase^ are to each other as their altitudes. 154 GEOMETRY. Let the parallelopipedons AG, AL, have the same base BD ; then will they be to each other as their altitudes AE, AI. First, suppose the altitudes AE, AI, to be e H to each other as two whole numbers, as 15 is to 8, for example. Divide AE into 15 equal \F I \C r parts ; whereof AI will contain 8 ; and through q, . Xf y, z, &c. the points of division, draw planes ■ -J parallel to the base. These planes will cut K \ 3M the solid AG into 15 partial parallelopipedons, all equal to each other, because they have ^X- K equal bases and equal altitudes — equal bases, ^ '. smce every section MIKL, made parallel to A\ the base ABCD of a prism, is equal to that base (Prop. II.), equal altitudes, because the -^ altitudes are the equal divisions A:c, xy^ yz, &c. But of those 15 equal parallelopipedons, 8 are con- tained in AL ; hence the solid AG is to the soHd AL as 15 is to 8, or generally, as the altitude AE is to the ahitude AI. Again, if the ratio of AE to AI cannot be exactly expressed in numbers, it is to be shown, that notwithstanding, we shall have solid AG : solid AL : : AE : AI. For, if this proportion is not correct, suppose we have sol. AG : sol. AL ; : AE ; AO greater than AT. Divide AE into equal parts, such that each shall be less than 01 ; there will be at least one point of divisiop m, between O and I. Let P be the parallelopipedon, whose 43ase is ABCD, and altitude Am ; since the altitudes AE, Awi, are to each other as the two whole numbers, we shall have sol. AG : P : : AE : Am. But by hypothesis, we have «oZ. AG : sol. AL : : AE : AO ; therefore, f sol.AL : P : : AO : Am. But AO is greater than Am ; hence if the proportion is correct, the solid AL must be greater than P. On the contrary, how- ever, it is less : hence the fourth term of this proportion soL AG ; sol. AL : : AE : x, cannot possibly be a line greater than AI. By the same mode of reasoning, it might be shown that the fourth term cannot be less than AI ; therefore it is equal to AI ; hence rectangular parallelopipedons having the same base are to each other as their altitudes. BOOK VII. 155 PROPOSITION XII. THEOREM. Two rectangular parallelopipedons, having the same altitude, are to each other as their bases. ^ w \^ I V 1 Let the parallelopipedons j- ' E IT AG, AK, have the same al- titude AE ; then will they be to each other as their bases AC, AN. Having placed the two v Ir hHCr solids by the side of each ^ other, as the figure repre- sents, produce the plane ONKL till it meets the plane DCGH in PQ ; you will thus have a third par- JM allelopipedon AQ, which may be compared with each ^ of the parallelopipedons AG, AK. The two solids AG, AQ, having the same base AEHD are to each other as their altitudes AB, AO ; in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their altitudes AD, AM. Hence we have the two proportions, sol. AG : sol. AQ : : AB : AO, sol. AQ : sol. AK : : AD : AM. Multiplying together the corresponding terms of these propor- tions, and omitting in the result the common multiplier sol. AQ ; we shall have sol.AG : sol.AK : : ABxAD : AOxAM. But AB X AD represents the base ABCD ; and AO x AM rep- resents the base AMNO ; hence two rectangular parallelopipe- dons of the same altitude are to each other as their bases. I PROPOSITION XIII. THEOREM. Any two rectangular parallelopipedons are to each other as the products of their bases by their altitudes, that is to say, as the products of their three dimensions. 166 GEOMETRY. 2S i K e For, having placed the two I E H solids AG, AZ, so that their surfaces have the common angle BAE, produce the planes necessary for com- Va pleting the third parallelopi- y x pedon AK having the same altitude with the parallelopi- pedon AG. By the last propo- sition, we shall have sol. AG ; sol. AK ; : ABCD : AMNO. kD But the two parallelopipedons AK, AZ, having the same base -^ AMNO, are to each other as \ their altitudes AE, AX ; hence :b we have sol. AK : sol. AZ : : AE : AX. Multiplying together the corresponding terms of these propor- tions, and omitting in the result the common multiplier sol. AK ; we shall have soZ. AG : sol. AZ : : ABCDxAE : AMNO x AX. Instead of the bases ABCD and AMNO, put AB x AD and AO X AM it will give sol.AG : soLAZ : : ABxADxAE : AOxAMxAX. Hence any two rectangular parallelopipedons are to each other, &c. Scholium. We are consequently authorized to assume, as the measure of a rectangular parallelopipedon, the product of its base by its altitude, in other words, the product of its three dimensions. In order to comprehend the nature of this measurement, it is necessary to reflect, that the number of linear units in one dimension of the base multiplied by the number of hnear units in the other dimension of the base, will give the number of superficial units in the base of the parallelopipedon (Book IV. Prop. IV. Sch.). For each unit in height there are evidently as many sohd units as there are superficial units in the base. Therefore, the number of superficial units in the base multi- plied by the number of linear units in the altitude, gives the number of solid units in the parallelopipedon. If the three dimensions of another parallelopipedon are valued according to the same linear unit, and multiplied together in the same manner, the two products will be to each ofier as BOOK VIL 157 the solids, and will serve to express their relative magni- tude. The magnitude of a solid, its volume or extent, forms what is called its solidity ; and this word is exclusively employed to designate the measure of a solid ; thus we say the solidity of a rectangular parallelopipedon is equal to the product of its base by its altitude, or to the product of its three dimensions. As the cube has all its three dimensions equal, if the side is 1, the solidity will be 1 x 1 x 1 =: 1 : if the side is 2, the solidity will be 2 X 2 X 2=8 ; if the side is 3, the solidity will be 3 x 3 x 3=27 ; and so on : hence, if the sides of a series of cubes are to each other as the numbers 1, 2, 3, &c. the cubes themselves or their solidities will be as the numbers 1, 8, 27, &c. Hence it is, that in arithmetic, the cube of a number is the name given to a product which results from three factors, each equal to this number. If it were proposed to find a cube double of a given cube, the side of the required cube would have to be to that of the given one, as the cube-root of 2 is to unity. Now, by a geo- metrical construction, it is easy to find the square root of 2 ; but the cube-root of it cannot be so found, at least not by the simple operations of elementary geometry, which consist in employing nothing but straight lines, two points of which are known, and circles whose centres and radii are determined. Owing to this difficulty the problem of the duplication of the cube became celebrated among the ancient geometers, as well as that of the trisection of an angle, which is nearly of the same species. The solutions of which such problems are sus- ceptible, have however long since been discovered ; and though less simple than the constructions of elementary geometry, they are not, on that account, less rigorous or less satisfactory. PROPOSITION XIV. THEOREM. The solidity of a parallelopipedon, and generally of any prism, is equal to the product of its base by its altitude. For, in the first place, any parallelopipedon is equivalent to a rectangular parallelopipedon, having the same altitude and an equivalent base (Prop. X.). Now the solidity of the latter is equal to its base multiplied by its height ; hence the solidity of the former is, in like manner, equal to the product of its base by its altitude. In the second place, any triangular prism is half of the par- allelopipedon so constructed as to have the same altitude and a double base (Prop. VII.). But the solidity of the latter is equal 158 GEOMETRY. to its base multiplied by its altitude ; hence that of a triangular prism is also equal to the product of its base, which is half that of the parallelopipedon, multiplied into its altitude. In the third place, any prism may be divided into as many triangular prisms of the same altitude, as there are triangles capable of being formed in the polygon which constitutes its base. But the solidity of each triangular prism is equal to its base multiplied by its altitude ; and since the altitude is the same for all, it follows that the sum of all the partial prisms must be equal to the sum of all the partial triangles, which con- stitute their bases, multiplied by the common altitude. Hence the solidity of any polygonal prism, is equal to the product of its base by its altitude. Co7\ Comparing two prisms, which have the same altitude, the products of their bases by their altitudes will be as the bases simply ; hence two prisms of the same altitude are to each other as their bases. For a like reason, two prisms of the same base are to each other as their altitudes. And when neither their bases nor their altitudes are equal, their solidities will be to each other as the products of their bases and altitudes. PROPOSITION XV. THEOREM. Tivo tjnangucar pyramids, having equivalent bases and equal altitudes f are equivalent, or equal in solidity. liCt S-ABC, S-«6c, be those two pyramids ; let their equiva- lent bases ABC, abc, be situated in the same plane, and let AT be their common altitude. If they are not equivalent, let ^-abc BOOK VII. 159 be the smaller : and suppose Ka to be the altitude of a prism, which having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, &c. each less than Aa, and let k be one of those parts ; through the points of division pass planes parallel to the plane of the bases ; the corresponding sections formed by these planes in the tw^o pyra- mids w^ill be respectively equivalent, namely DEF to def, GHI io ghi, &c, (Prop. III. Cor. 2.). This being granted, upon the triangles ABC, DEF, GHI, &c. taken as bases, construct exterior prisms having for edges the parts AD, DG, GK, &c. of the edge SA ; in like manner, on bases def, ghi, klm, &c. in the second pyramid, construct inte- rior prisms, having for edges the corresponding parts of Sa. It is plain that the sum of all the exterior prisms of the pyramid S-ABC will be greater than this pyramid ; and also that the sum of all the interior prisms of the pyramid S-abc will be less than this pyramid. Hence the difference, between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids tliem- selves. Now, beginning with the bases ABC, ahcj the second exte- rior prism DEF-G is equivalent to the first interior prism def-ay because they have the same altitude k, and their bases DEF, def, are equivalent ; for like reasons, the third exterior prism GHI-K and the second interior prism ghi-d are equivalent ; the fourth exterior and the third interior ; and so on, to the last in each series. Hence all the exterior prisms of the pyramid S-ABC, excepting the first prism ABC-D, have equivalent cor- responding ones in the interior prisms of the pyramid S-abc : hence the prism ABC-D, is the difference between the sum of all the exterior prisms of the pyramid S-ABC, and the sum of the interior prisms of the pyramid S-abc. But the difference between these two sets of prisms has already been proved to be greater than that of the two pyramids ; which latter diffe- rence we supposed to be equal to the prism «-ABC : hence the prism ABC-D, must be greater than the prism a- ABC. But in reality it is less ; for they have the same base ABC, and the altitude Ax of the first is less than Aa the altitude of the second. Hence the supposed inequality between the two pyramids can- not exist ; hence the two pyramids S-ABC, S-abc, having equal altitudes and equivalent bases, are themselves equivalent. 16a GEOMETRY. PROPOSITION XVI. THEOREM. Every triangular pyramidis a third part of the triangular pt'ism having the same base and the same altitude. Let F-ABC be a triangular pyramid, ABC-DEF a triangular prism of the same base and the same altitude ; the pyramid will be equal to a third of the prism. Cut off the pyramid F-ABC from the prism, by the plane FAC ; there will remain the solid F-ACDE, which may be consi- dered as a quadrangular pyramid, whose vertex is F, and whose base is the parallelogram ACDE. Draw the diagonal CE ; and pass the plane FCE, which will cut the quadrangular pyramid into two triangular ones F-ACE,F-CDE. These two triangular pyramids have for their common altitude the perpendicular let fall from F on the plane ACDE ; they have equal bases, the triangles ACE, CDE being halves of the same parallelogram ; hence the two pyramids F-ACE, F-CDE, are equivalent (Prop. XV.). But the pyramid F-CDE and the pyramid F-ABC have equal bases ABC,DEF ; they have also the same altitude, namely, the distance between the parallel planes ABC, DEF ; hence the two pyramids are equivalent. Now the pyramid F-CDE has already been proved equivalent to F-ACE ; hence the three pyramids F-ABC, F-CDE, F-ACE, which compose the prism ABC-DEF are all equivalent. Hence the pyramid F-ABC is the third part of the prism ABC-DEF, which has the same base and the same altitude. Cor. The solidity of a triangular pyramid is equal to a third part of the product of its base by its altitude. PROPOSITION XVII. THEOREM. The solidity of every pyramid is equal to the base multiplied by a third of the altitude. BOOK VII. 161 Let S-ABCDE be a pyramid. Pass the planes SEE, SEC, through the diagonals EB, EC ; the polygonal pyramid S-ABCDE will be divided into several trian- fular pyramids all having the same altitude 10. But each of these pyramids is measured by multiplying its base ABE, BCE, or CDE, by the third part of its altitude SO (Prop. XVI. Cor.) ; hence the sum of these triangular pyra- mids, or the polygonal pyramid S-ABCDE will be measured by the sum of the triangles ABE, BCE, CDE, or the polygon ABODE, multiplied by one third of SO ; hence every pyramid is mea- sured by a third part of the product of its base by its altitude. Cor. 1. Every pyramid is the third part of the prism which has the same base and the same altitude. Cor. 2. Two pyramids having the same altitude are to each other as their bases. Cor. 3. Two pyramids havmg equivalent bases are to each other as their altitudes. Cor. 4. Pyramids are to each other as the products of their bases by their altitudes. Scholium. The solidity of any polyedral body may be com- puted, by dividing the body into pyramids ; and this division may be accomplished in various ways. One of the simplest is to make all the planes of division pass through the vertex of one solid angle ; in that case, there will be formed as many partial pyramids as the polyedron has faces, minus those faces which form the solid angle whence the planes of division proceed. PROPOSITION XVIII. THEOREM. If a pyramid he cut by a plane parallel to its base, the fmsium that remains wlien the small pyramid is taken away, iV equi- valent to the sum of three pyramids having for their common altitude the altitude of the fi^stum, and for bases the lower base of the frustum, the upper base, and a mean proportional between the two bases. O* 21 162 GEOMETRY. Let S-ABCDE be a pyra- mid cut by the plane abcdet parallel to its base; let T-FGH be a triangular pyramid hav- ing the same altitude and an equivalent base with the pyra- mid S-ABCDE. The two bases may be regarded as situated in the same plane ; in which case, the plane ahcdy if produced, will form in the triangular pyramid a section fgh situated at the same distance above the common plane of the bases ; and therefore the section ^A will be to the section abode as the base FGH is to the base ABD (Prop. III.), and since the bases are equivalent, the sections will be so hkewise. Hence the pyramids S-abcde, T-fgh are equivalent, for their altitude is the same and their bases are equivalent. The whole pyramids S-ABCDE, T-FGH are equivalent for the same rea- son ; hence the frustums ABD-dahf FGH-^ are equivalent ; hence if the proposition can be proved in the single case of the frustum of a triangular pyramid, it will be true of every other. • I^et FGH-/i^ be the frustum of a tri- angular pyramid, having parallel bases : through the three points F, g, H, pass the plane F^H ; it will cut off from the frustum the triangular pyramid ^-FGH. This pyramid has for its base the lower base FGH of the frustum ; its altitude likewise is that of the frustum, because the vertex g lies in the plane of the up- per base fgh. This pyramid being cut off, there will remain the quadrangular pyramid g-fhUF, whose vertex is g, and base fhHF, Pass the plane j^H through the three points /, ^, H ; it will divide the quad- rangular pyramid into two triangular pyramids g-FfH, g-fW. The latter has for its base the upper base gfli of the frustum ; and for its altitude, the altitude of the frustum, because its ver- tex H lies in the lower base. Thus we already know two of the three pyramids which compose the frustum. It remains to examine the third g-FfH. Now, if ^K be drawn parallel to /F, and if we conceive a new pyramid K-FfH, having K for its vertex and FfH for its base, these two pyramids will have the same base FfH ; they will also have the sarte altitude, because their vertices g and K lie in the line ^K, parallel to F/, and consequently parallel to the BOOK TIL 163 plane of the base : hence these pyramids are equivalent. But the pyramid K-F/H may be regarded as having its vertex iu /, and thus its altitude will be the same as that of the frustum : as to its base FKH, we are now to show that this is a mean proportional between the bases FGH and fgh. Now, the tri- angles FHK,^A, have each an equal angle F=/; hence FHK : /^A : : FKxFH : fgxfh (Book IV. Prop. XXIV.) ; but because of the parallels, FK=^, hence FHK ifgh : : FH : fh. We have also, FHG : FHK : : FG : FK or fg. But the similar triangles FGH, fgh give FG-.fg-.-.FH-.fli; hence, FGH : FHK : : FHK : fgh; or the base FHK is a mean proportional between the two bases FGH, fgh. Hence the frustum of a triangular pyramid is equivalent to three pyramids whose common altitude is that of the frustum and whose bases are the lower base of the frustum, the upper base, and a mean proportional between the two bases. PROPOSITION XIX. THEOREM. Similar triangular prisms are to each other as the cubes ofthett homologous sides. Let CBD-P, chd'p, be two similar triangular prisms, of which BC, he, are homologous sides: then will the prism CBD-P be to the prism chd-p, asBC3to6c3. For, since the prisms are similar, the planes which con- tain the homologous solid an- C C B gles B and h, are similar, like placed, and equally inclined to each other (Def. 1 7.) : hence the solid angles B and h, are equal (Book VI. Prop. XXL Sch.). If these solid angles be applied to each other, the angle chd will coincide with CBD, the side ha with B A, and the prism chd-p will take the position ^cd-p. From A draw AH perpendicular to the common base of the prisms : then will the plane BAH be perpendicular to the plane of the com- 164 GEOMETRY. mon base (Book VI. Prop. XVI.). Through a, in the plane BAH, draw ah perpendicular to BH : then will ah also be per- pendicular to the base BDC (Book VI. Prop. XVII.) ; and AH, ah will be the altitudes of the two prisms. Now, because of the similar triangles ABH,aBA, and of the similar parallelograms AC, «c, we have AH : a/t : : AB : «& : : BC : he. But since the bases are similar, we have base BCD : base bed : : BC^ : be^ (Book IV. Prop. XXV.) ; hence, base BCD : base bed : : AH^ : ah\ Multiplying the antecedents by AH, and the consequents by ahy and we have base BCD x AH : base bed xah : : AiP : ah\ But the solidity of a prism is equal to the base multiplied by the altitude (Prop. XIV.) ; hence, the prism BCD-P : prism bed-p : : AH^ : ah^ : : BC^ : bc^y. or as the cubes of any other of their homologous sides. Cor. Whatever be the bases of similar prisms, the prisms will be to each other as the cubes of their homologous sides. For, since the prisms are similar, their bases will be similar polygons (Def. 17.); and these similar polygons maybe di- vided into an equal number of similar triangles, similarly placed (Book IV. Prop. XXVI.) : therefore the two prisms may be divided into an equal number of triangular prisms, having their faces similar and like placed ; and therefore, equally inclined (Book VI. Prop. XXI.) ; hence the prisms will be similar. But these triangular prisms will be to each other as the cubes of their homologous sides, which sides being proportional, the sums of the triangular prisms, that is, the polygonal prisms, will be to each other as the cubes of their homologous sides. PROPOSITION XX. THEOREM. Two similar pyramids are to eaek other as the cubes of theh^ homologous sides. BOOK VIL 165 For, since the pyramids are similar, the soHd angles at the vertices will be contained by the same number of similar planes, like placed, and equally inchned to each other (Def. 17.). Hence, the solid angles at the vertices may be made to coincide, or the two pyramids may be so placed as to have the solid angle S common. In that position, the bases ABCDE, ahcde, a/ will be parallel ; because, since the homolo- gous faces are similar, the angle Safe is equal to SAB, and Sfec to SBC ; hence the plane ABC is parallel to the plane abc (Book VI. Prop. XIIL). This being proved, let SO be the perpendicular drawn from the vertex S to the plane ABC, and o the point where this perpen- dicular meets the plane ahc: from what has already been shown, we shall have SO : So : : SA ; Sa : : AB : ah (Prop. III.) ; and consequently, iSO : iSo : : AB : ah. But the bases ABCDE, ahcde^ being similar figures, we have ABCDE : ahcde : : AB^ ; ah^ (Book IV, Prop. XXVII.). Multiply the corresponding terms of these two proportions ; there results the proportion, ABCDE xiSO : ahcde x^So : : AB^ : ah^ Now ABCDE X iSO is the solidity of the pyramid S-ABCDE, and ahcde xjSo is that of the pyramid ^-ahcde (Prop. XVII.) ; hence two similar pyramids are to each other as the cubes of their homologous sides. General Scholium^ The chief propositions of this Book relating to the solidity of polyedrons, may be exhibited in algebraical termsy and so recapitulated in the briefest manner possible. Let B represent the base of a prism ; H its altitude : the solidity of the prism will be B x H, or BH. Let B represent the base of a pyramid ; H its altitude : the solidity of the pyramid will be B x ^H, or H x ^B, or ^BH. Let H represent the altitude of the frustum of a pyramid, having parallel bases A and B ; VAB will be the mean pro- portional between those bases ; and the solidity of the frustum willbeiHx(A+B+N/AB). In fine, let P and p represent the solidities of two similar prisms or pyramids ; A and a, two homologous edges : then wo shall have ^ V : p \ I A? V a\ 166 GEOMETRY. BOOK VIII. THE THREE ROUND BODIES. Definitions, M E l^ ^^ 1. A cylinder is the solid generated by the revolution of a rectangle ABCD, conceived to turn about the immoveable side AB. In this movement, the sides AD, BC, con- tinuing always perpendicular to AB, describe equal circles DHP, CGQ, which are called the hoses of the cylinder, the side CD at the same time describing the convex surface. The immoveable line AB is called the axis of the cylinder. Every section KLM, made in the cylinder, at right angles to the axis, is a circle equal to either of the bases ; for, whilst the rectangle ABCD turns about AB, the line KI, perpen- dicular to AB, describes a circle, equal to the base, and this circle is nothing else than the section made perpendicular to the axis at the point I. Every section PQG, made through the axis, is a rectangle double of the generating rectangle ABCD. 2. A cone is the solid generated by the revolution of a right- angled triangle SAB, conceived to turn about the immoveable side SA. In this movement, the side AB describes a circle BDCE, named the base of the cone ; the hypothenuse SB describes the convex surface of the cone. The point S is named the vertex of the conCf SA the axis or the altitude^ and SB the side or the apothem. Every section HKFI, at right angles to the axis, is a circle ; every section SDE, through the axis, is an isosceles triangle, double of the generating triangle SAB. 3. If from the cone S-CDB, the cone S-FKH be cut off by a plane parallel to the base, the remaining solid CBHF is called a truncated cone, or the frustum of a cone. BOOK VIII. 167 We may conceive it to be generated by the revolution of a trapezoid ABHG, whose angles A and G are right angles, about the side AG. The immoveable line AG is called the axis or altitude of the frustum, the circles BDC, HFK, are its bases, and BH is its side, 4. Two cylinders, or two cones, are similar, when their axes are to each other as the diameters of their bases. 5. If in the circle ACD, which forms the base of a cylinder, a polygon ABODE be inscribed, a right prism, constructed on this base ABODE, and equal in altitude to the cylinder, is said to be inscribed in the cylin- der, or the cylinder to be circumscribed about the prism. The edges AF, BG, OH, &c. of the prism, being perpendicular to the plane of the base, are evidently included in the convex sur- face of the cylinder ; hence the prism and the cylinder touch one another along these edges. 6. In like manner, if ABOD is a poly- gon, circumscribed about the base of a cylinder, a right prism, constructed on this base ABOD, and equal in altitude to the cylinder, is said to be circumsjcribed about the cylinder, or the cylinder to be inscribed in the prism. Let M, N, &c. be the points of contact in the sides AB, BO, &c. ; and through the points M,N,&c. let MX, NY, &c. be drawn Ak perpendicular to the plane of the base : these perpendiculars will evidently lie both in the surface of the cylinder, and in that of the circumscribed prism ; hence they will be their lines of contact. 7. If in the circle ABODE, which forms the base of a cone, any polygon ABODE be inscribed, and from the vertices A, B, C, D, E, lines be drawn to S, the vertex of the cone, these lines may be regarded as the sides of a pyramid whose base is the polygon ABODE and vertex S. The sides of this pyramid are in the convex surface of the cone, and the pyramid is said to be inscribed in the cone. 168 GEOMETRY. 8. The sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within, called the centre. The sphere may be con- ceived to be generated by the revolution of a semicircle DAE about its diameter DE : for the surface described in this movement, by the curve DAE, will have all its points equally distant from its cen- tre C. 9. Whilst the semicircle DAE revolving round its di- ameter DE, describes the sphere ; any circular sector, as DCF orFCH, describes a solid, which is named a spherical sector. 10. The radius of a sphere is a straight line drawn from the centre to any point of the surface ; the diameter or axis is a line passing through this centre, and terminated on both sides by the surface. All the radii of a sphere are equal ; all the diameters are equal, and each double of the radius. 11. It will be shown (Prop. VII.) that every section of the sphere, made by a plane, is a circle : this granted, a great cir- cle is a section which passes through the centre ; a small circle, is one which does not pass through the centre. 12. A plane is tangent to a sphere, when their surfaces have but one point in common. 13. A zone is a portion of the surface of the sphere included between two parallel planes, which form its bases. One of these planes may be tangent to the sphere ; in which case, the zone has only a single base. 14. A spherical segment is the portion of the solid sphere, included between two parallel planes which form its bases. One of these planes may be tangent to the sphere ; in which case, the segment has only a single base. 15. The altitude of a zone or of a segment is the distance between the two parallel planes, which form the bases of the zone or segment. Note. The Cylinder, the Cone, and the Sphere, are the three round bodies treated of in the Elements of Geometry. BOOK VIII. 169 PROPOSITION I. THEOREM, The convex surface of a cylinder is equal to the circumference of its base multiplied by its altitude. Let CA be the radius of the given cylinder's base, and H its ahitude : the circumference whose radius is CA being rep- resented by circ. CA, we are to show that the convex surface of the cyhnder is equal to circ, CA Inscribe in the circle any regular polygon, BDEFGA, and construct on this polygon a right prism having its altitude equal to H, the altitude of the cylin- der : this prism will be inscribed in the cylinder. The convex surface of the prism is equal to the perimeter of the polygon, multiplied by the altitude H (Book VII. Prop. I.). Let now the arcs which subtend the sides of the polygon be continually bisected, and the number of sides of the polygon indefinitely increased : the perimeter of the polygon will then become equal to circ. CA (Book V. Prop. VIII. Cor. 2.), and the convex sur- face of the prism will coincide with the convex surface of the cylinder. But the convex surface of the prism is equal to the perimeter of its base multiplied by H, whatever be the number of sides : hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude. PROPOSITION II. THEOREM. The solidity of a cylinder is equal to the product of its base by it^ altitude. 170 GEOMETRY. Let CA be the radius of the base of the cylinder, and H the altitude. Let the circle whose radius is CA be repre- sented by area CA, it is to be proved that the solidity of the cylinder is equal to area C A x H. Inscribe in the circle any regu- lar polygon BDEFGA, and con- struct on this polygon a right prism having its altitude equal to H, the altitude of the cylinder : this prism vs^ill be inscribed in the cylinder. The solidity of the prism will be equal to the area of the polygon multiplied by the altitude H (Book VIL Prop. XIV.). Let now the number of sides of the polygon be indefinitely increased : the solidity of the new prism will still be equal to its base multiplied by its altitude. But when the number of sides of the polygon is indefinitely increased, its area becomes equal to the area CA, and its pe- rimeter coincides with circ. CA (Book V. Prop. VIII. Cor. 1. &L 2.) ; the inscribed prism then coincides with the cylinder, since their altitudes are equal, and their convex surfaces per- pendicular to the common base : hence the two solids will be equal ; therefore the solidity of a cylinder is equal to the product of its base by its altitude. Cor. 1. Cylinders of the same altitude are to each other as their bases ; and cyHnders of the same base are to each other as their altitudes. Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their diameters ; and the cylinders being similar, the diameters of their bases are to each other as the altitudes (Def 4.) ; hence the bases are as the squares of the altitudes ; hence the bases, multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. Scholium. Let R be the radius of a cylinder's base ; H the altitude : the surface of the base will be tt.R^ (Book V. Prop. XII. Cor. 2.) ; and the solidity of the cylinder will be nW x H, or^.R^.H. BOOK VIII. 171 PROPOSITION III. THEOREM. The convex surf (tee of a cone is equal to the circumference of its base, multiplied hy half its side. Let the circle ABCD be the base of a cone, S the vertex, SO the ahitude, and SA the side : then will its convex sur- face be equal to czVc.OAx ^S A. For, inscribe in the base of the cone any regular polygon ABCD, and on this polygon as a base conceive a pyramid to be constructed having S for its vertex : this pyramid will be a regular pyramid, and will be inscribed in the cone. From S, draw SG perpendicular to one of the sides of the polygon. The convex surface of the inscribed pyramid is equal to the perimeter of the polygon which forms its base, multiplied by half the slant height SG (Book VII. Prop. TV.). Let now the number of sides of the inscribed polygon be indefinitely increased ; the perimeter of the inscribed polygon will then become equal to circ. OA, the slant height SG will become equal to the side SA of the cone, and the convex surface of the pyramid to the convex surface of the cone. But whatever be the number of sides of the polygon which forms the base, the convex surface of the pyramid is equal to the perimeter of the base multiplied by half the slant height : hence the convex surface of a cone is equal to the circumference of the base multiphed by half the side. Scholium. Let L be the side of a cone, R the radius of its base ; the circumference of this base will be S^^.R, and the sur- face of the cone will be 2nR x 4^L, or nKL. PROPOSITION IV. THEOREM. The convex surface of the frustum of a cone is equal to its side multiplied hy half the sum of the circumferences of its two bases. 172 GEOMETRY. Let BIA-DE be a frustum of a cone : then will its convex surface be equal to AD x (circ.OA+circ.CVy For, inscribe in the bases of the frustums two regular polygons of the same number of sides, and having their homologous sides parallel, each to each. The lines joining the ver- tices of the homologous angles may- be regarded as the edges of the frus- tum of a regular pyramid inscribed in the frustum of the cone. The con- vex surface of the frustum of the pyramid is equal to half the sum of the perimeters of its bases multiplied by the slant height fh (Book VII. Prop. IV. Cor.). Let now the number of sides of the inscribed polygons be indefinitely increased : the perimeters of the polygons will be- come equal to the circumferences BIA, EGD ; the slant height ^h will become equal to the side AD or BE, and the surfaces of the two frustums will coincide and become the same surface. But the convex surface of the frustum of the pyramid will still be equal to half the sum of the perimeters of ihe upper and lower bases multiplied by the slant height : hence the sur- face of the frustum of a cone is equal to its side multiplied by half the sura of the circumferences of its two bases. Cor. Through /, the middle point of AD, draw ZKL paral- lel to AB, and li, Dd, parallel to CO. Then, since A/, /D, are equal, Ai, id, will also be equal (Book IV. Prop. XV. Cor. 2.) : hence, K/ is equal to ^ (O A + CD). But since the circumfe- rences of circles are to each other as their radii (Book V. Prop. XL), the circ. K/=|(aVc. 0A + arc. CD) ; therefore, the convex surface of a frustum of a cone is equal to its side multi- plied by the circumference of a section at equal distances from the two bases. Scholium. If a line AD, lying wholly on one side of the line OC, and in the same plane, make a revolution around OC, the surface described by AD will have for its measure AD x /drcAO + drcDCy ^^ ^jy ^ ^.^^ ^g- . ^^^ jj^^^ j^^ j^^ jj^^ being perpendiculars, let fall from the extremities and from the middle point of AD, on the axis OC. For, if AD and OC are produced till they meet in S, the surface described by AD is evidently the frustum of a cone #*■. BOOK Vm. 173 having AO and DC for the radii of its bases, the vertex of the whole cone being S. Hence this surface will be measured as we have said. This measure will always hold good, even when the point D falls on S, and thus forms a whole cone ; and also when the hne AD is parallel to the axis, and thus forms a cylinder. In the first case DC would be nothing ; in the second, DC would be equal to AO and to IK. PROPOSITION V. THEOREM. The solidity of a cone is equal to its base multiplied by a third of its altitude. Let SO be the altitude of a cone, OA the radius of its base, and let the area of the base be designated by area OA : it is to be proved that the solidity of the cone is equal to area OAx^SO. Inscribe in the base of the cone any regular polygon ABDEF, and join the vertices A, B, C, &c. with the vertex S of the cone : then will there be inscribed in the cone a regular pyramid having the same vertex as the cone, and hav- ing for its base the polygon ABDEF. The solidity of this pyramid is equal to its base multiplied by one third of its alti- tude (Book VII. Prop. XVII.). Let now the number of sides of the polygon be indefinitely increased : the polygon will then become equal to the circle, and the pyramid and cone will coincide and become equal. But the solidity of the pyramid is equal to its base multiplied by one third of its altitude, what- ever be the number of sides of the polygon which forms its base : hence the solidity of the cone is equal to its base multi- phed by a third of its altitude. Cor. A cone is the third of a cy finder having the same base and the same altitude ; whence it follows, 1. That cones of equal altitudes are to each other as their bases ; 2. That cones of equal bases are to each other as their altitudes ; 3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes. P* 174 GEOMETIIY. Cor. 2. The solidity of a cone is equivalent to the solidity of a pyramid having an equivalent base and the same altitu^^ (Book VII. Prop. XVIL). Scholium. Let R be the radius of a cone's base, H its alti- tude ; ^he solidity of the cone will be wR^ x ^H, or ^nWR. PROPOSITION VI. THEOREM. The solidity of the frustum of a cone is equal to the sum of the solidities of three cones whose common altitude is the altitude of the frustum, and whose bases are, the upper base of the frus- tum, the lower base of the frustum, and a mean proportional between them. Let AEB-CD be the frustum of a cone, and OP its altitude ; then will its solidity be equal to i^r X OP X (A02+DP2+ AO X DP). For, inscribe in the lower and upper bases two regular polygons having the same number of sides, and having their homologous sides parallel, each to each. Join the vertices of the homologous angles and there will then be inscribed in the frustum of the cone, the frustum of a regular pyramid. The sohdity of the frustum of the pyramid is equivalent to three pyramids having the common altitude of the frustum, and for bases, the lower base of the frustum, the upper base of the frustum, and a mean proportional between them (Book VII. Prop. XVIII.). Let now, the number of sides of the inscribed polygons be indefinitely increased : the bases of the frustum of the pyramid will then coincide with the bases of the frustum of the cone, and the two frustums will coincide and become the same solid. Since the area of a circle is equal to R^.n (Book V. Prop. XII. Cor. 2.), the expression for the solidities of the frustum will become for the first pyramid |OP x OA^tt. for the second iOP x PD^.tt for the third J OP x AO x PD.^r ; since AG X FD.n is a mean proportional between OA^.^r and PD'.tt. Hence the solidity of the frustum of the cone is measured by i^iOP X (OAH PDH AO X PD). BOOK VIII. 175 PROPOSITION VII. THEOREM. Every section of a sphere, made by a plane, is a circle. Let AMB be a section, made by a plane, in the sphere whose centre is C. From the point C, draw CO perpen- dicular to the plane AMB ; and diffe- rent lines CM, CM, to different points of the curve AMB, which terminates the section. The oblique lines CM, CM, CA, are equal, being radii of the sphere ; hence they are equally distant from the perpendicular CO (Book VL Prop. V. Cor.) ; therefore all the lines OM, OM, OB, are equal ; consequently the section AMB is a circle, whose centre is Q. Cor 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere ; hence all great circles are equal. Cor. 2. Two great circles always bisect each other ; for their common intersection, passing through the centre, is a diameter. Cor. 3. Every great circle divides the sphere and its surface into two equal parts : for, if the two hemispheres were sepa- rated and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other. Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line, perpendicular to the plane of the small circle. Cor. 5. Small circles are the less the further they lie from the centre of the sphere ; for the greater CO is, the less is the chord AB, the diameter of the small circle AMB. Cor. 6. An arc of a great circle may always be made to pass through any two given points of the surface of the sphere ; for the two given points, and the centre of the sphere make three points which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points. 176 GEOMETRY. PROPOSITION VIII. THEOREM. Evety plane perpendiculaj- to a radius at its extremity is tangent to the sphere. Let FAG be a plane perpendicular to the radius OA, at its extremity A. Any point M in this plane being as- sumed, and OM, AM, being drawn, the angle 0AM will be a right angle, and hence the distance OM will be greater than OA. Hence the point M lies without the sphere ; and as the same can be shown for every other point of the plane FAG, this plane can have no point but A common to it and the surface of the sphere ; hence it is a tangent plane (Def. 12.) Scholium. In the same way it may be shown, that two spheres have but one point in common, and therefore touch each other, when the distance between their centres is equal to the sum, or the difference of their radii ; in which case, the centres and the point of contact lie in the same straight line. PROPOSITION IX. LEMMA. If a regular semi-polygon he revolved about a line passing through the centre and the vertices of two opposite angles, the surface described by its perimeter will be equal to the axis mul- tiplied by the circumference of the inscribed circle. Let the regular semi-polygon ABCDEF, be revolved about the line AF as an axis : then will the surface described by its pe- rimeter be equal to AF multiplied by the circumference of the inscribed circle. From E and D, the extremities of one of the equal sides, let fall the perpendiculars EH, DI, on the axis AF, and from the cen- tre O draw ON perpendicular to the side DE : ON will be the radius of the inscribed circle (Book V. Prop. XL). Now, the sur- face described in the revolution by any one side of the regular polygon, as DE, has BOOK VIII. 177 been shown to be equal to DE x circ. NM (Prop. IV. Sch.). But since the triangles EDK, ONM, are similar (Book IV. Prop. XXL), ED ; EK or HI : : ON : NM, or as circ. ON ; circ. NM ; hence ED X circ. NM=HI X circ. ON ; and since the same may be shown for each of the other sides, it is plain that the surface described by the entire perimeter is equal to (FH + HH- IP + PQ + QA) X circ 0N= AF x circ. ON. Cor. The surface described by any portion of the perime- ter, as EDC, is equal to the distance between the two perpen- diculars let fall from its extremities on the axis, multiplied by the circumference of the inscribed circle. For, the surface described by DE is equal to HI x circ. ON, and the surface described by DC is equal to IP x cii'c. ON ; hence the surface described by ED + DC, is equal to (HI + IP) x circ. ON, or equal to HP x circ. ON. PROPOSITION X. THEOREM. The surface of a sphere is equal to the product of its diameter hy the circumference of a great circle. Let ABCDE be a semicircle. Inscribe in it any regular semi-polygon, and from the centre O draw OF perpendicular to one of the sides. Let the semicircle and the semi-polygon be revolved about the axis AE : the semi- circumference ABCDE will describe the surface of a sphere (Def 8.) ; and the pe- rimeter of the semi-polygon will describe a surface which has for its measure AE x circ. OF (Prop. IX.), and this will be true whatever be the number of sides of the po- lygon. But if the number of sides of the polygon be indefi- nitely increased, its perimeter will coincide with the circumfe- rence ABCDE, the perpendicular OF will become equal to OE, and the surface described by the perimeter of the semi- polygon will then be the same as that described by the semi- circumference ABCDE. Hence the surface of the sphere is equal to AE x circ. OE. Cor. Since the area of a great circle is equal to the product \ of its circumference by half the radius, or one fourth of the 23 178 GEOMETRY. V diameter (Book V. Prop. XII.), it follows that the surface of a sphere is equal to four of its great circles : that is, equal to 4^.0 A2 (Book V. Prop. XII. Cor. 2.). Scholium 1. The surface of a zone is equal to its altitude mul- tiplied by the circumference of a great circle. For, the surface described by any portion of the perimeter of the inscribed polygon, as BC + CD, is equal to EH xcirc. OF (Prop. IX. Cor.). But when the number of sides of the polygon is indefinitely increased, BC + CD, becomes the arc BCD, OF becomes equal to OA, and the surface described by BC + CD, becomes the surface of the zone described by the arc BCD; hence the sur- face of the zone is equal to EH x circ. OA. Scholium 2. When the zone has but one base, as the zone described by the arc ABC D, its surface will still be equal to the altitude AE multiplied by the circumference of a great circle. Scholium 3. Two zones, taken in the same sphere or in equal spheres, are to each other as their altitudes ; and any zone is to the surface of the sphere as the altitude of the zone is to the diameter of the sphere. PKGPOSITION XL LEMMA. If a triangle and a rectangle ^ having the same base and the same altitudCfturn together about the common base^ the solid described by the triangle will be a third of the cylinder described by the rectangle. Let ACB be the triangle, and BE the rectangle. On the axis, let fall the perpen- dicular AD : the cone described by the triangle ABD is the third part of the cylinder described by the rectan- gle AFBD (Prop. V. Cor.) ; also the cone described by the triangle ADC is the third part of the cylinder de- scribed by the rectangle ADCE ; hence the sum of the two cones, or the solid described by ABC, is the third part of the two cylinders taken together, or of the cylinder described by the rectangle BCEF. BOOK VIIL 17^ If the perpendicular AD falls without the triangle ; the solid described by ABC will, in that case, be the difference of the two cones described by ABD and ACD ; but at the same time, the cylinder de- scribed by BCEF will be the difference ^ ^ ^ of the two cylinders described by AFBD and AECD. Hence the solid, described by the revolution of the triangle, will still be a third part of the cylinder described by the revolution of the rectangle having the same base and the same altitude. Scholium. The circle of which AD is radius, has for its measure n x AD^ ; hence n x AD^ x BC measures the cylinder described by BCEF, and ^n x AD^ x BC measures the solid described by the triangle ABC. PROPOSITION XII. LEMMA. If a triangle he revolved about a line drawn at pleasure through its vertex, the solid described by the triangle will have for its measure^ the area of the triangle multiplied by two thirds of the circumference traced by the middle point of the base. Let CAB be the triangle, and CD the line about which it revolves. Produce the side AB till it meets the axis CD in D ; from the points A and B, draw AM, BN, perpendicular to the axis, and CP perpendicular to DA produced. The solid described by the tri- angle CAD is measured by ^n x AM^ X CD (Prop. XI. Sch.) ; the solid described by the triangle CBD is measured by ^n x BN^ x CD ; hence the difference of those solids, or the solid described by ABC, will have for its measure iTr(AM2— BN^) x CD. To this expression another form may be given. From I, the middle point of AB, draw IK perpendicular to CD ; and through B, draw BO parallel to CD: we shall have AM + BN=2IK (Book IV. Prop. VII.) ; and AM— BN=AO ; hence (AM + BN) X (AM— NB), or AM^— BN2=2lKx AO (Book IV. Prop. X.). Hence the measure of the solid in question is ex- pressed by f^xIKxAOxCD. GEOMETRY. JVLKJSr But CP being drawn perpendicular to AB, the triangles ABO, DCP will be similar, and give the proportion AO : CP : : AB : CD ; hence AO x CD = CP x AB ; but CP X AB is double the area of the triangle ABC ; hence we have A0xCD=2ABC; hence the solid described by the triangle ABC is also measured by |7r X ABC X IK, or which is the same thing, by ABC x Icirc. IK, circ. IK being equal to 27i^ x IK. Hence the solid described hy the revolution of the triangle A^C, has for its measure the area of this triangle multiplied by two thirds of the circumference traced by I, the middle point of the base. Cor. If the sideAC = CB, the line CI will be perpen- dicular to AB, the area ABC will be equal to AB x |CI, and the solidity ^71 x ABC x IK will become fTtxABx IKxCI. But the triangles ABO, CIK, are similar, and give the proportion AB : BO or MN : ; CI : IK; hence ABxIK=MNxCI; hence the solid described by the isosceles triangle ABC will have for its measure ^n x CP x MN. Scholium. The general solution appears to include the sup- position that AB produced will meet the axis ; but the results would be equally true, though AB were parallel to the axis. Thus, the cylinder described by AMNB p i\ B is equal to jt.AM^.MN ; the cone descri- bed by ACM is equal to i^r.AM^.CM, and the cone described by BCN to inAM^ CN. Add the first two sohds and take away the third ; we shall have the solid described by ABC equal to tt.AM^ (MN + iCM— iCN): and since CN— CM =MN, this expres- sion is reducible to tt.AM^.^MN, or Iti.CP^.MN ; which agrees with the conclusion found above. BOOK VIIL 181 PROPOSITION XIII. LEMMA. If a regular semi-polygon he revolved about a line passing through the centre and the vertices of two opposite angles, the solid described will be equivalent to a cone, having for its base the inscribed circle, and for its altitude twice the axis about which the semi-polygon is revolved. Let the semi-polygon FABG be revolved about FG : then, if OI be the radius of the inscribed circle, the solid described will be measured by ^area 01 x 2FG. For, since the polygon is regular, the triangles OFA, OAB, OBC, &c. are equal and isosceles, and all the perpendiculars let fall from O on the bases FA, AB, (fee. will be equal to 01, the radius of the inscribed circle. Now, the solid described by OAB has forits measure |0P X MN (Prop. Xll.Cor.); the solid described by the triangle OFA has for its measure fTiOP X FM, the solid described by the triangle OBC, has for its measure f tiOP x NO, and since the same may be shown for the solid described by each of the other triangles, it Ibllows that the entire solid described by the semi-polygon is mea- sured by |nOP.(FM + MN + NO + OQ + QG),or |7>0FxFG; which is also equal to \nO\^ x 2FG. But yr.OP is the area of the inscribed circle (Book V. Prop, XII. Cor. 2.) : hence the solidity is equivalent to a cone whose base is area 01, and altitude 2FG. PROPOSITION XIV. THEOREM. The solidity of a sphere is equal to its surface multiplied by a third of its radius. a 182 GEOMETRIC. Inscribe in the semicircle ABCDE a regular semi-polygon, having any number of sides, and let 01 be the radius of the circle inscribed in the polygon. If the semicircle and semi-polygon be revolved about EA, the semicircle will describe a sphere, and the semi-polygon a solid which has for its measure f^OP x EA (Prop. XIII.) ; and this will be true whatever be the number of sides of the polygon. But if the number of sides of the polygon be indefinitely increased, the E semi-polygon will become the semicircle, 01 will become equal to OA, and the solid described by the semi-polygon will become the sphere : hence the solidity of the sphere is equal to fTiOA^xEA, or by substituting 20A for EA, it becomes jn.OA^ X OA, which is also equal to 4nOA^ x ^OA. But 4n.OA^ is equal to the surface of the sphere (Prop. X. Cor.) : hence the solidity of a sphere is equal to its surface multiplied by a third of its radius. Scholium 1. The solidity of every spherical sector is equal to the zone which forms its base, multiplied by a third of the radius. For, the solid described by any portion of the regular poly- gon, as the isosceles triangle OAB, is measured by fyrOP x AF (Prop. XII. Cor.) ; and when the polygon becomes the circle, the portion OAB becomes the sector AOB, 01 becomes equal to OA, and the solid described becomes a spherical sector. But its measure then becomes equal to fyr. AO^ x AF, which is equal to 27r. AO X AF X |AO. But 27r.AO is the circumference of a great circle of the sphere (Book V. Prop. XII. Cor. 2.), which being multiplied by AF gives the surface of the zone which forms the base of the sector (Prop. X. Sch. 1.) : and the proof is equally applicable to the spherical sector described by the circular sector BOC : hence, the solidity of the spherical sector is equal to the zone which forms its base, multiplied by a third of the radius. Scholium 2. Since the surface of a sphere whose radius is R, is expressed by 47rR2 (Prop. X. Cor.), it follows that the surfaces of spheres are to each other as the squares of their radii ; and since their solidities are as their surfaces multiplied by their radii, it follows that the solidities of spheres are to each other as the cubes of their radii, or as the cubes of their diameters. BOOK VIII. 183 Scholium 3. Let R be the radius of a sphere ; its surface will be expressed by 47iR^, and its solidity by 47iR2 x ^R, or ^TiR^. If the diameter is called D, we shall have R=|D, and R^=|D^ : hence the solidity of the sphere may likewise be expressed by PROPOSITION XV. THEOREM. Tfie surface of a sphere is to the whole surface of the circum- scribed cylinder, including its bases, as 2 is to 3 ; and the so- lidities of these two bodies are to each other in the same ratio. D Let M PNQ be a great circle of the sphere ; ABCD the circumscribed square : if the semicircle PMQ and the half square PADQ are at the same time made to revolve about the diameter PQ, the semicircle will gene- M rate the sphere, while the half square will generate the cylinder circum- scribed about that sphere. . The altitude AD of the cylinder is equal to the diameter PQ ; the base of the cylinder is equal to the great circle, since its diameter AB is equal to MN ; hence, the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter (Prop. 1.). This measure is the same as that of the surface of the sphere (Prop. X.) ; hence the surface of the sphere is equal to the convex surface of the circumscribed cylinder. But the surface of the sphere is equal to four great circles ; hence the convex surface of the cylinder is also equal to four great circles : and adding the two bases, each equal to a great circle, the total surface of the circumscribed cylinder will be equal to six great circles ; hence the surface of the sphere is to the total surface of the circumscribed cylinder as 4 is to 6, or as 2 is to 3 ; which was the first branch of the Proposition. In the next place, since the base of the circumscribed cylin- der is equal to a great circle, and its altitude to the diameter, the solidity of the cylinder will be equal to a great circle mul- tiplied by its diameter (Prop. II.). But the solidity of the sphere is equal to four great circles multiplied by a third of the radius (Prop. XIV.); in other terms, to one great circle multi- plied by I of the radius, or by f of the diameter ; hence the sphere is to the circumscribed cylinder as 2 to 3, and conse- quently the solidities of these two bodies are as their surfaces. 184 GEOMETRY. * -^ Scholium. Conceive a polyedron, all of whose faces touch the sphere ; this polyedron may be considered as formed of pyramids, each having for its vertex the centre of the sphere, and for its base one of the polyedron's faces. Now it is evi- dent that all these pyramids will have the radius of the sphere for their common altitude : so that each pyramid will be equal to one face of the polyedron multiplied by a third of the radius : hence the whole polyedron will be equal to its surface multi- plied by a third of the radius of the inscribed sphere. It is therefore manifest, that the solidities of polyedrons cir- cumscribed about the sphere are to each other as the surfaces of those polyedrons. Thus the property, which we have shown to be true with regard to the circumscribed cylinder, is also true with regard to an infinite number of other bodies. We might likewise have observed that the surfaces of poly- gons, circumscribed about the circle, are to each other as their perimeters. PROPOSITION XVI. PROBLEM. If a circular segment he supposed to make a revolution about a diameter exterior to it, required the value of the solid which it describes. Let the segment BMD revolve about AC. On the axis, let fall the perpendiculars BE, DF ; from the centre C, draw CI perpendicular to the chord BD ; also draw the radii CB, CD. The solid described by the sector BCA is equal to fTr.CB^.AE (Prop. XIV. Sch. 1.) ; the solid described by the sector DCA=|7r.CB2.AF; hence the difference of these two solids, or the solid described by the sector DCB=|^.CB2.(AF— AE) = |rr.CB2.EF. But the solid described by the isosceles triangle DCB has for its mea- sure f^.CP.EF (Prop. XII. Cor.) ; hence the solid described by the segment BMD=|rr.EF.(CB^— CP). Now, in the right- angled triangle CBI, we have CB^—CP^BP^^BD^ ; hence the solid described by the segment BMD will have for its mea- sure |7..EF.iBD2, or i^.BD-.EF. Scholium. The solid described by the segment BMD is to the sphere which has BD for its diameter, as i7T.BD-.EF is to i^.BD^ or as EF to BD, BOOK VIII. 185 PROPOSITION XVII. THEOREM. Every segment of a sphere^ included between two parallel planes^ is measured by the half sum of its bases multiplied by its alti- tudet plus the solidity of a sphere whose diameter is this same altitude. Let BE, DF, be the radii of the two bases of the segment, EF its altitude, the segment being described by the revolu- tion of the circular space BMDFE about the axis FE. The solid described by the segment BMD is equal to ^Tr.BD^.EF (Prop. XVI.) ; and the truncated cone de- scribed by the trapezoid BDFE is equal toi7r.EF.(BEHDF-+BE.DF)(Prop.VI.); hence the segment of the sphere, which is the sum of those two solids, must be equal to i^.EF.(2BEH2DF2H-2BE.DF+BD?) But, drawing BO parallel to EF, we shall have DO=DF— BE, hence DO^^DF— 2DF.BE + BE« (Book IV. Prop. IX.) ; and consequently BD^^BOH D02=:EF2+ DF— 2DF . BE + BEl Put this value in place of BD^ in the expression for the value of the segment, omitting the parts which destroy each other ; we shall obtain for the solidity of the segment, i7rEF.(3BE^+3DF2+EF), an expression which may be decomposed into two parts ; the /TT.BEH^.DFx one i7r.EF.(3BEH3DF2), or EF.( ^ ) being the half sum of the bases multiplied by the altitude ; while the other iTT.EF^ represents the sphere of which EF is the diame- ter (Prop. XIV. Sch.) : hence every segment of a sphere, &c. Cor. If either of the bases is nothing, the segment in ques- tion becomes a spherical segment with a single base ; hence any spherical segment^ with a single base, is equivalent to half the cylinder having the same base and the same altitude, plus the sphere of which this altitude is the diameter. General Scholium. Let R be the radius of a cylinder's base, H its altitude ; the solidity of the cylinder will be nR^ x H, or tiR^H. Let R be the radius of a cone's base, H its altitude : the solidity of the cone will be nR^ x ^H, or ^tiR^H. Let A and B be the radii of the bases of a truncated cone, a* 24 186 GEOMETRY. H its altitude : the solidity of the truncated cone will be |7r.H. (AHB2+AB). Let R be the radius of a sphere ; its solidity will be ^ttR^. Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base : the solidity of the sector will be f ^R2H. Let P and Q be the two bases of a spherical segment, H its piQ altitude: the solidity of the segment will be ^ .H+^rr.H^ til If the spherical segment has but one base, the other being nothing, its solidity will be iPH+^TiH^. BOOK IX. OP SPHERICAL TRIANGLES AND SPHERICAL POLYGONS. Definitions, 1. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles. These arcs are named the sides of the triangle, and are always supposed to be each less than a semi-circumference. The angles, which their planes form with each other, are the angles of the triangle. 2. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 3. A spherical polygon is a portion of the surface of a sphere terminated by several arcs of great circles. 4. A lune is that portion of the surface of a sphere, which is included between two great semi-circles meeting in a common .diameter. 5. A spherical wedge or ungula is that portion of the solid sphere, which is included between the same great semi-circles, and has the lune for its base. 6. A spherical pyramid is a portion of the solid sphere, in- cluded between the planes of a solid angle whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes. 7. The pole of a circle of a sphere is a point in the surface equally distant from all the points in the circumference of this circle. It will be shown (Prop. V.) that every circle, great or 1^ small, has always two poles. BOOK IX. 187 PROPOSITION I. THEOREM. In every spherical triangle^ any side is less than the sum of the other two. Let O be the centre of the sphere, and ACB the triangle ; draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be drawn ; these planes will form a solid angle at the centre O ; and the an- gles AOB, AOC, COB, will be measured" by AB, AC, BC, the sides of the spherical triangle. But each of the three plane an- gles forming a solid angle is less than the sum of the other two (Book VI. Prop. XIX.) ; hence any side of the triangle ABC is less than the sum of the other two. PROPOSITION II. THEOREM. The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points. Let ANB be the arc of the great circle which joins the points A and B ; and without this hne, if possible, let M be a point of a shorter path between A and B. Through the point M, draw MA, MB, arcs of great circles ; and take BN=MB. By the last theorem, the arc ANB is shorter than AM + MB ; take BN^BM, from both ; there will re- main AN'*■ BOOK IX. 191 the point D, its extremity A will describe. the arc of the great circle AMB. If the arc AM were required to be produced, and nothing were given but the points A and M through which it was to pass, we should first have to determine the pole D, by the intersection of two arcs described from the points A and M as centres, with a distance equal to a quadrant ; the pole D being found, we might describe the arc AM and its prolongation, from D as a centre, and with the same distance as before. In fine, if it be required from a given point P, to let fall a perpendicular on the given arc AM ; find a point on the arc AM at a quadrant's distance from the point P, which is done by describing an arc with the point P as a pole, intersecting AM in S : S will be the point required, and is the pole with which a per- pendicular to AM may be described passing through the point P. PROPOSITION VI. THEOREM. TTie angle formed by two arcs of great circles^ is equal to the an- gle formed by the tangents of these arcs at their point of inter- section, and is measured by the arc described from this point of intersection, as a pole, and limited by the sides, produced if necessary. Let the angle BAG be formed by the two A^ arcs AB, AC ; then will it be equal to the "" angle FAG formed by the tangents AF, AG, and be measured by the arc DE, described about A as a pole. For the tangent AF, drawn in the plane of the arc AB, is perpendicular to the radius ^ AO ; and the tangent AG, drawn in the plane of the arc AC, is perpendicular to the same radius AO. Hence the angle FAG is equal to the angle contained by the planes ABO, OAC (Book VI. Def. 4.) ; which is that of HL the arcs AB, AC, and is called the angle BAC. In like manner, if the arcs AD and AE are both quadrants, the lines OD, OE, will be perpendicular to OA, and the angle DOE will still be equal to the angle of the planes AOD, AOE : hence the arc DE is the measure of the angle contained by these planes, or of the angle CAB. Cor. The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles and included between their sides : hence it is easy to make an angle of this kind equal to a given angle. im GEOMETRY. Scholium. Vertical angles, such as ACO and BCN are equal ; for either of them is still the angle formed by the two planes ACB, OCN. It is farther evident, that, in the intersection of two arcs ACB, OCN, the two adjacent angles ACO, OCB, taken together, are equal to two right angles. PROPOSITION VII. THEOREM. If from the vertices of the three angles of a spherical triangle, as poles, three arcs he described forming a second triangle, the vertices of the angles of this second triangle, will he respectively poles of the sides of the first. From the vertices A, B, C, as poles, let the arcs EF, FD, ED, be described, forming on the surface of the sphere, the triangle DFE ; then will the points D, E, and F, be respec- tively poles of the sides BC, AC, AB. For, the point A being the pole of the arc EF, the dis- tance AE is a quadrant ; the point C being the pole of the arc DE, the distance CE is like- wise a quadrant : hencie the point E is removed the length of a quadrant from each of the points A and C ; hence, it is the pole of the arc AC (Prop. V. Cor. 3.). It might be shown, by the same method, that D is the pole of the arc BC, and F that of the arc AB. Cor. Hence the triangle ABC may be described by means of DEF, as DEF is described by means of ABC. Triangles so described are called polar tmangles, or supplemental tri- angles. BOOK IX. 193 PROPOSITION VIII. THEOREM. The same supposition continuing as in the last Pi'oposition, each angle in one of the triangles, will he measured by a semicir- cumference, minus the side lying opposite to it in the other triangle. For, produce the sides AB, AC, if necessary, till they meet EF, in G and H. The point A being the pole of the arc GH, the angle A will be measured by that arc (Prop. VL). But the arc EH is a quadrant, and likewise GF, E being the pole of AH, and F of AG ; hence EH -f GF is equal to a semi- circumference. Now, EH + GF is the same as EF+ GH ; hence the arc GH, which mea- sures the angle A, is equal to a semicircumference minus the side EF. In like manner, the angle B will be measured by ^circ, — DF : the angle C, by | circ. — DE. And this property must be reciprocal in the two triangles, since each of them is described in a similar manner by means of the other. Thus we shall find the angles D, E, F, of the triangle • DEFtobe measured respectively by icirc. — BC, | circ. — AC, I circ— AB. Thus the angle D, for example, is measured by the arc MI ; but MI + BC = MC + BI=i circ. ; hence the arc MI, the measure of D, is equal to \ circ— BC ; and so of all the rest. Scholium. It must further be observed, that besides the triangle ]l)EF, three others might be formed by the intersection of the three arcs DE, EF, DF. But the proposition immediately before us is ap- plicable only to the central triangle, which is distinguished from the other three by the circumstance (see the last figure) that the two angles A and D lie on the same side of BC, the two B and E on the same side of AC, and the two C and F on the same side of AB. R 25 194 GEOMETRY. PROPOSITION IX. THEOREM If around the vertices of the two angles of a given spherical tri- angle, as poles, the circumferences of two' circles be described which shall pass through the third angle of the triangle; if then, through the other point in which these circumferences intersect and the two first angles of the triangle, the arcs of great cir- cles be drawn, the triangle thus formed will have all its parts equal to those of the given triangle. Let ABC be the given triangle, CED, DFC, the arcs described about A and B as poles ; then will the triangle ADB have all its parts equal to those of ABC. For, by construction, the side ADrr: AC, DB=BC, and AB is common ; hence these tvv^o triangles have their sides equal, each to each. We are now to show, that the angles opposite these equal sides are also equal. If the centre of the sphere is supposed to be at O, a solid angle may be conceived as formed at O by the three plane angles AOB, AOC, BOC ; likewise another solid angle may be conceived as formed by the three plane angles AOB, AOD, BOD. And because the sides of the triangle ABC are equal to those of the triangle ADB, the plane angles forming the one of these solid angles, must be equal to the plane angles forming the other, each to each. But in that case we have shown that the planes, in which the equal angles lie, are equally inclined to each other (Book VI. Prop. XXI.) ; hence all the angles of the spherical triangle DAB are respectively equal to those of the triangle CAB, namely, DAB=BAC, DBA=ABC, and ADB=ACB; hence the sides and the angles of the triangle ADB are equal to the sides and the angles of the triangle AC3. Scholium. The equality of these triangles is not, however, an absolute equality, or one of superposition ; foi* it would be impossible to apply them to each other exactly, unless they •were isosceles. The equality meant here is what we have already named an equality by symmetry ; therefore we shall call the triangles ACB, ADB, symmetrical triangles. BOOK IX. 195 PROPOSITION X. THEOREM. Tivo triangles on the same sphere^ or on equal spheres^ are equal in all their parts, when two sides and the included angle of the one are equal to two sides and the included angle of the othery each to each. Suppose the side AB=EF, the side AC =EG, and the angle BAC^FEG ; then will the two triangles be equal in all their parts. " For, the triangle EFG may be placed on the triangle ABC, or on ABD symmetrical with ABC, just as two rectilineal triangles are placed V^ upon each other, when they have an B equal angle included between equal sides. Hence all the parts of the triangle EFG will be equal to all the parts of the trian- gle ABC ; that is, besides the three parts equal by hypothesis, we shall have the side BC=FG, the angle ABC=EFG, and the angle ACB=EGF. PROPOSITION XI. THEOREM. Two triangles on the same sphere, or on equal spheres, are equal in all their parts, when two angles and the included side of the one are equal to two angles and the included side of the other, each to each. For, one of these triangles, or the triangle symmetrical with it, may be placed on the other, as is done in the corres- ponding case of rectilineal triangles (Book I. Prop. VI.). PROPOSITION XII. THEOREM. If two triangles on the same sphere, or on equal spheres, have all their sides equal, each to each, their angles will likewise be equal, each to each, the equal angles lying opposite the equal sides. 196 GEOMETRY. This truth is evident from Prop. IX, where it was shown, that with three given sides AB, AC, BC, there can only be two triangles ACB, ABD, differing as to the position of their parts, and equal as to the magnitude of those parts. Hience tho§e two triangles, having all their sides re- spectively equal in both, must either be absolutely equal, or at least symmetrically so ; in either of which cases, their corres- ponding angles must be equal, and lie opposite to equal sides. PROPOSITION XIII. THEOREM. In every isosceles spherical triangle, the angles opposite the equal sides are equal ; and conversely, if two angles of a spherical tnangle are equal, the triangle is isosceles. First. Suppose the side AB=AC; we shall have the angle C=B. For, if the arc AD be" drawn from the vertex A to the middle point D of the base, the two triangles ABD, ACD, will have all the sides of the one respectively equal to the corresponding sides of the other,- namely, AD common, BD = DC, and AB== AC : hence by the last Proposition, their an- gles will be equal; therefore, B = C. Secondly. Suppose the angle B = C ; we shall have the side AC=AB. For, if not, let AB be the greater of the two ; take BO=iAC, and draw OC. The two sides BO, BC, are equal to the two AC, BC ; the angle OBC, contained by the first two is equal to ACB contained by the second two. Hence the two triangles BOC, ACB, have all their other parts equal (Prop. X.) ; hence the angle OCBi=ABC : but by hypothesis, the angle ABC =r ACB ; hence we have OCB=ACB, which is absurd ; hence it is absurd to suppose AB different from AC ; hence the sides AB, AC, opposite to the equal angles B and C, are equal. Scholium. The same demonstration proves the anglie BAD = DAC, and the angle BDAirzADC. Hence the two last are right angles ; hence the are drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to that base, and bisects the vertical angle. BOOK IX. 197 PROPOSITION XIV. THEOREM. In any spherical trianghy the greater side is opposite the greater angle ; and conversely ^ the greater angle is opposite the greater side. Let the angle A be greater than the angle B, then will BC be greater than AC ; and con- versely, if BC is greater than AC, then will the angle A be -^^ greater than B. First. Suppose the angle A>B ; make the angle BAD=B ; then we shall have AD=DB (Prop. XIII.) : but AD + DC is greater than AC ; hence, putting DB in place of AD, we shall have DB + DC, or BC>AC. Secondly. If we suppose BC > AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC=AC ; if -BAC were less than ABC, we should then, as has justi)een shown, find BC less than six right angles, and greater than tioo. For, in the first place, every angle of a spherical triangle is less than two right angles : hence the sum of all the three is less than six right angles. Secondly, the measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (Prop. VIII.) ; hence the sum of all the three, is measured by the three semicircumferences minusih^ sum of all the sides of the polar triangle. Now this latter sum is less than a circumference (Prop, III.) ; therefore, taking it away from three semicircumferences, the remainder will be greater than one semicircumference, which is the measure of two right angles ; hence, in the second place, the sum of all the angles of a sphe- rical triangle is greater than two right angles. Cor. 1 . The sum of all the angles of a spherical triangle is not constant, like that of all the angles of a rectilineal triangle ; it varies between two right angles and six, without ever arriving at either of these limits. Two given angles therefore do not serve to determine the third.* Cor. 2. A spherical triangle may have two, or even three of its angles right angles ; also two, or even three of its angles obtuse. BOOK IX. 199 Cor, 3. If the triangle ABC is hi-rectangular, in other words, has two right angles B and C, the vertex A will be the pole of the base BC ; and the sides AB, AC, will be quadrants (Prop. V. Cor. 3.). If the angle A is also a right angle, the tri- angle ABC will be tri-rectangular ; its' angles will all be right angles, and its sides quadrants, tri-rectangular triangles make half a hemisphere, four make a hemisphere, and the tri-rectangular triangle is obviously con- tained eight times in the surface of a sphere. Scholium. In all the preceding observations, we have supposed, in conformity with (Def. 1.) that sphe- rical triangles have always each of their sides less than a semicircum- ference ; from which it follows that any one of their angles is always less than two right angles. For, if the side AB is less than a semicir- cumference, and AC is so likewise, both those arcs will require to be E produced, before they can meet in D. Now the two angles ABC, CBD, taken together, are equal to two right angles ; hence the angle ABC itself, is less than two right angles. We may observe, however, that some spherical triangles do existr in which certain of the sides are greater than a semicir- cumference, and certain of the angles greater than two right angles. Thus, if the side AC is produced so as to form a whole circumference ACE, the part which remains, after subtracting the triangle ABC from the hemisphere, is a new triangle also designated by ABC, and having AB, BC, AEDC for its sides. Here, it is plain, the side AEDC is greater than the semicir- cumferencc AED ; and at the same time, the angle B opposite to it exceeds two right angles, by the quantity CBD. The triangles whose sides and angles are so large, have been excluded by the Definition ; but the only reason was, that the solution of them, or the determmation of their parts, is always reducible to the solution of such triangles as are comprehended by the Definition. Indeed, it is evident enough, that if the sides and angles of the triangle ABC are known, it will be easy to discover the angles and sides of the triangle which bears the same name, and is the difference between a hemisphere and the former triangle. 200 GEOMETRY. PROPOSITION XVII. THEOREM. The surface of a lune is to the surface of the sphere, as the angle of this lune, is to four right angles, or as the arc which mea- sures that angle, is to the circumference. Let AMBN be a lune ; then will its surface be to the surface of the sphere as the angle NCM to four right angles, or as the arc NM to the circumference of a great circle. Suppose, in the first place, the arc MN to be to the circumference MNPQ as some one rational number is to ano- ther, as 5 to 48, for example. The cir- cumference MNPQ being divided into 48 equal parts, MN will contain 5 of them ; and if the pole A were joined with the several points of division, by as many quadrants, we should in the hemisphere AMNPQ have 48 tri- angles, all equal, because all their parts are equal. Hence the whole sphere must contain 96 of those partial triangles, the lune AMBNA will contain 10 of them ; hence the lune is to the sphere as 10 is to 96, or as 5 to 48, in other words, as the arc MN is to the circumference. If the arc MN is not commensurable with the circumference, we may still show, by a mode of reasoning frequently exem- plified already, that in that case also, the lune is to the sphere as MN is to the circumference. Cor. 1. Two lunes are to each other as their respective angles. Cor. 2. It was shown above, that the whole surface of the sphere is equal to eight tri-rectangular triangles (Prop. XVI. Cor. 3.) ; hence, if the area of one such triangle is taken for unity, the surface of the sphere will be represented by 8. This granted, if the right angle be assumed equal to 1, the surface of the lune whose angle is A will be expressed by 2A ; for, 4 : 8 : : A : 2A. Thus we have here two different unities ; one for angles, being the right angle ; the other for surfaces, being the tri-rectangu- lar spherical triangle, or the triangle whose angles are all right angles, and whose sides are quadrants. Scholium. The spherical ungula, bounded by the planes AMB, ANB, is to the whole solid sphere, as the angle. A is to BOOK IX. 201 four right angles. For, the lunes being equal, the spherical ungulas will also be equal ; hence two spherical ungulas are to each other, as the angles formed by the planes which bound them. PROPOSITION XVIII. THEOREM. T\oo symmetrical spherical triangles are equivalent. Let ABC, DEP, be two symmetri- cal triangles, that is to say, two tri- angles having their sides AB=DE, AC=DF, CB=EF, and yet incapa- ble of coinciding with each other ; / I q p / we are to show that the surface ABC *' ^"* is equal to the surface DEF. Let P be the pole of the small circle passing through the three points A, B, C ;* from this point draw the equal arcs PA, PB, PC (Prop. V.) ; at the point F, make the angle DFQrzACP, the arc FQ=r:CP ; and draw DQ, EQ. The sides DF, FQ, are equal to the sides AC, CP ; the an- gle DFQ= ACP : hence the two triangles DFQ, ACP are equal in all their parts (Prop. X.) ; hence the side DQ=:AP, and the angle DQF=APC, In the proposed triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, being equal (Prop. XII.). if the angles DFQ, ACP, which are equal by construction, be taken away from them, there will remain the angle QFE, equal to PCB. Also the sides QF, FE, are equal to the sides PC, CB ; hence the two triangles FQE, CPB, are equal in all their parts ; hence the side QE^PB, and the angle FQE = CPB. Now, the triangles DFQ, ACP, which have their sides re- spectively equal, are at the same time isosceles, and capable of <5oinciding, when applied to each other ; for having placed AB on its equal DF, the equal sides will fall on each other, and thus the two triangles will exactly coincide : hence they are equal ; and the surface DQF=:APC. ' For a like reason, the surface FQE = CPB, and the surface DQE=:APB ; hence we * The circle which passes through the three points A, B, C, or which cir- cumscribes the triangle ABC, can only be a small circle of the sphere ; for if it were a great circle, the three sides AB, BC, AC, would lie in one plane, and the triangle ABC would be reduced to one of its sides. 26 202 GEOMETRY. have DQF+FQE— DQE=APC + CPB— APB, or DFE = ABC ; hence the two symmetrical triangles ABC, DEF are equal in surface. Scholium. The poles P and Q might lie within triangles ABC, DEF: in which case it would be requisite to add the three triangles DQF, FQE, DQE, together, in or- der to make up the triangle DEF ; and in like manner, to add the three triangles APC, CPB, APB, together, in order to make up the triangle ABC : in all other respects, the de- monstration and the result would still be the same PROPOSITION XIX. THEOREM. If the circumferences of two great circles intersect each other on the surface of a hemisphere^ the sum of the opposite triangles thus formed, is equivalent to the surface of a lune whose angle ■ is equal to the angle formed by the circles. Let the circumferences AOB, COD, in- tersect on the hemisphere OACBD ; then will the opposite triangles AOC, BOD be equal to the lune whose angle is BOD. For, producing the arcs OB, OD, on the other hemisphere, till they meet in N, the arc OBN will be a semicircumfe- rence, and AOB one also ; and taking OB from both, w^e shall h'ave BN=AO. For a like reason, we have DN=CO, and BD= AC. Hence the two triangles AOC, BDN, have their three sides respectively equal ; besides they are so placed as to be symmetrical ; hence they are equal in surface (Prop. XVIII.) , but the sum of the triangles BDN, BOD, is equivalent to the lune OBNDO whose angle is BOD : hence AOC 4- BOD is equivalent to the lune whose angle is BOD. Scholium. It is likewise evident that the two spherical pyra- mids, which have the triangles AOC, BOD, for bases, are together equivalent to the spherical ungula whose angle is BOD. BOOK IX. 203 PROPOSITION XX. THEOREM. The surface of a spherical triangle is measured hy the excess of the sum of its three angles above two right angles. Let ABC be the proposed triangle : pro- duce its sides till they meet the great circle DEFG drawn at pleasure without the tri- angle. By the last Theorem, the two tri- angles ADE, AGH, are together equiva- lent to the lune whose angle is A, and which is measured by 2A (Prop. XVII. Cor. 2.). Hence we have ADE + AGH=2A ; and for a like reason, BGF + BID=2B, and CIH + CFE=2C. But the sum of those six triangles exceeds the hemisphere by twice the triangle ABC, and the hemisphere is represented by 4 ; therefore twice the triangle ABC is equal to 2A + 2B + 2C — 4 ; and consequently, once ABC=A + B + C — 2 ; hence every spherical triangle is measured by the sum of all its angles minus two right angles. Cor. 1. However many right angles there be contained in this measure, just so many tri-rectangular triangles, or eighths of the sphere, will the proposed triangle contain (Prop. XVII. Cor. 2.). If the angles, for example, are each equal to f of a right angle, the three angles will amount to 4 right angles, and the proposed triangle will be represented by 4—2 or 2 ; there- fore it will be equal to two tri-rectangular triangles, or to the fourth part of the whole surface of the sphere. Cor. 2. The spherical triangle ABC is equivalent to the lune whose angle is A + B + C •1 ; likewise the spherical pyramid, which has ABC for its base, is equivalent to the spherical ungula whose angle is 1. Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, which has ABC for its base, is compared with the tri-rectangular pyramid, and a similar proportion is found to subsist between them. The solid angle at the vertex of the pyramid, is in like manner compared with the solid angle at the vertex of the tri- rectangular pyramid. These comparisons are founded on the coincidence of the corresponding parts. If the bases of the 204 GEOMETRY. pyramids coincide, the pyramids themselves will evidently co- incide, and likewise the solid angles at their vertices. From this, some consequences are deduced. First. Two triangular spherical pyramids are to each other as their bases : and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other, as the polygons which form their bases. Second. The solid angles at the vertices of these pyramids, are also as their bases ; hence, for comparing any two solid angles, we have merely to place their vertices at the centres of two equal spheres, and the solid angles will be to each other as the spherical polygons intercepted between their planes or faces. The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other : this angle, which may be called a right solid angle, will serve as a very natural unit of measure for all other solid angles. And if so, the same number, that exhibits the area of a spherical polygon, will exhi- bit the measure of the corresponding solid angle. If the area of the polygon is f , for example, in other words, if the polygon is f of the tri-rectangular polygon, then the corresponding solid angle will also be | of the right solid angle. PROPOSITION XXI. THEOREM. The surface of a spherical polygon is measured by the sum of all its angles, minus two right angles multiplied by the numbet of sides in the polygon less two. From one of the vertices A, let diago- nals AC, AD be drawn to all the other vertices ; the polygon ABCDE will be divided into as many triangles minus two as it has sides. But the surface of each triangle is measured by the sum of all its angles minus two right angles ; and the sum of the angles in all the tri- angles is evidently the same as that of all tlie angles in the po- lygon ; hence, the surface of the polygon is equal to the sum of all its angles diminished by twice as many right angles as it has sides minus two. Scholium. Let s be the sum of all the angles in a spherical polygon, n the number of its sides ; the right angle being taken for unity, the surface of the polygon will be measured by s— 2.(n— -2), or 5— 2 71 + 4. APPENDIX. THE REGULAR POLYEDRONS. A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. There are five such polyedrons. First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five : hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles ; for six angles of such a triangle are equal to four right angles, and cannot form a solid angle (Book VI. Prop. XX.). Secondly. If the faces are squares, their angles may be ar- ranged by threes : hence results the hexaedron or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle. Thirdly. In fine, if the faces are regular pentagons, their angles likewise may be arranged by threes : the regular dode- caedron will result. We can proceed no farther : three angles of a regular hexa- gon are equal to four right angles ; three of a heptagon are greater. Hence there can only be five regular polyedrons ; three formed with equilateral triangles, one with squares, and one with pen- tagons. ^ Construction of the Tetraedron. Let ABC be the equilateral triangle which is to form one face of the tetrae- dron. At the point O, the ceotre of this triangle, erect OS perpendicular to the plane ABC ; terminate this perpendicular in S, so that AS=AB; draw SB, SC : the pyramid S-ABC will be the tetrae- dron required. For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC, are equally re- S 206 APPENDIX. m moved from the perpendicular SO, and consequently equal (Book VI. Prop. V.). One of them SA=AB ; hence the four faces of the pyramid S-ABC, are trian- gles, equal to the given triangle ABC. And the solid angles of this pyramid are all equal, because each of them is formed by three equal plane angles: hence this pyramid is a regular tetrae- dron. Construction of the Hexaedron Let ABCD be a given square. On the base ABOD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evidently be equal squares ; and its solid angles all equal, each being formed with three right angles : hence this prism is a regular hexaedron or cube. The following propositions can be easily proved. 1. Any regular polyedron may be divided into as many regular pyramids as the polyedron has faces ; the common vertex of these pyramids will be the centre of the polyedron ; and at the same time, that of the inscribed and of the circum- scribed sphere. 2. The solidity of a regular polyedron is equal to its sur- face multiplied by a third part of the radius of the inscribed sphere. 3. Two regular polyedrons of the same name, are two simi- lar solids, and their homologous dimensions are proportional ; hence the radii of the inscribed or the circumscribed spheres are to each other as the sides of the polyedrons. 4. If a regular polyedron is inscribed in a sphere, the planes drawn from the centre, through the different edges, will divide the surface of the sphere into as many spherical polygons, all equal and similar, as the polyedron has faces. PLANE TRIGONOMETRY. 207 PLANE TRIGONOMETRY. In every triangle there are six parts : three sides and three angles. These parts are so related to each other, that if a certain number of them be known or given, the remaining ones can be determined. Plane Trigonometry explains the methods of finding, by cal- culation, the unknown parts of a rectilineal triangle, when a sufficient number of the six parts are given. When three of the six parts are known, and one of them is a side, the remaining parts can always be found. If the three angles were given, it is obvious that the problem would be in- determinate, since all similar triangles would satisfy the con- ditions. It has already been shown, in the problems annexed to Book III., how rectilineal triangles are constructed by means of three given parts. But these constructions, which are called graphic methods^ though perfectly correct in theory, would give only a moderate approximation in practice, on account of the im- perfection of the instruments required in constructing them. Trigonometrical methods, on the contrary, being independent of all mechanical operations, give solutions with the utmost accuracy. They are founded upon the properties of lines called trigo- nometrical lines, which furnish a very simple mode of express- ing the relations between the sides and angles of triangles. We shall first explain the properties of those lines, and the principal formulas derived from them ; formulas which are of great use in all the branches of mathematics, and which even furnish means of improvement to algebraical analysis. We shall next apply those results to the solution of rectilineal tri- angles. DIVISION OF TKfi CIRCUMFERENCE. I. For the purposes of trigonometrical calculation, the cir- cumference of the circle is divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds. The semicircumference, or the measure of two right angles, contains 180 degrees ; the quarter of the circumference, usually denominated the quadrant, and which measures the right an- gle, contains 90 degrees. XL Degrees, minutes, and seconds, are respectively desig- 208 PLANE TRIGONOMETRY. nated by the characters : «, ', " : thus the expression 16° 6' 15" represents an arc, or an angle, of 16 degrees, 6 minutes, and 15 seconds. III. The complement of an angle, or of an arc, is what re- mains after taking that angle or that arc from 90°. Thus the complement of 25° 40' is equal to 90° — 25° 40'=64° 20^ ; and the complement of 12° 4' 32" is equal to 90° — 12^ 4' 32" = 77° 55' 28". In general, A being any angle or any arc, 90° — A is the com- plement of that angle or arc. If any arc or angle be added to its complement, the sum will be 90°. Whence it is evident that if the angle or arc is greater than 90°, its complement will be negative. Thus, the complement of 160° 34' 10" is — 70° 34' 10". In this case, the complement, taken positively, would be a quantity, which being subtracted from the given angle or arc, the remainder would be equal to 90°. The two acute angles of a right-angled triangle, are together equal to a right angle ; they are, therefore, complements of each other. IV. The supplement of an angle, or of an arc, is what re- mains after taking that angle or arc from 180°. Thus A being any angle or arc, 180° — A is its supplement. In any triangle, either angle is the supplement of the sum of the two others, since the three together make 180°. If any arc or angle be added to its supplement, the sum will be 180°. Hence if an arc or angle be greater than 180°, its supplement will be negative. Thus, the supplement of 200°. is — 20°. The supplement of any angle of a triangle, or indeed of the sum of either two angles, is always positive. GENERAL IDEAS RELATING TO THE TRIGONOMERICAL LINES V. The sine of an arc is the perpendicular let fall from one extremity of the arc, on the diameter which passes through the other extremity. Thus, MP is the sine of the arc AM, or of the angle ACM. The tangent of an arc is a line touching the arc at one extremity, and limited by the prolongation of the diameter which passes through the other extremity. Thus AT is the tangent of the arc AM, or of the angle ACM. S' N tC q"^ s p r / r \ / A A s \ J^^' ^ h^ V E PLANE TRIGONOMETRY. 209 The secant of an arc is the line drawn from the centre of the circle through one extremity of the arc and limited by the tangent drawn through the other extremity. Thus CT is the secant of the arc AM, or of the angle ACM. The versed sine of an arc, is the part of the diameter inter- cepted between one extremity of the arc and the foot of the sine. Thus, AP is the versed sine of the arc AM, or the smgle ACM. These four lines MP, AT, CT, AP, are dependent upon the arc AM, and are always determined by it and the radius ; they are thus designated : MP=sin AM, or sin ACM, ATirrtangAM, or tang ACM, CTi^secAM, or sec ACM, AP=ver-sin AM, or ver-sin ACM. VI. Having taken the arc AD equal to a quadrant, from the points M and D draw the lines MQ, DS, perpendicular to the radius CD, the one terminated by that radius, the other termi- nated by the radius CM produced ; the lines MQ, DS, and CS, will, in like manner, be the sine, tangent, and secant of the arc MD, the complement of AM. For the sake of brevity, they are called the cosine^ cotangent, and cosecant, of the arc AM, and are thus designated : MQ=:cosAM, or cos ACM, DSmcot AM, or cot ACM, CS=cosec AM, or cosec ACM. In general, A being any arc or angle, we have cos A=sin (90° — A), cot A = tang (90° — A), cosec A =: sec (90° — A). The triangle MQC is, by construction, equal to the triangle CPM ; consequently CPrrMQ : hence in the right-angled tri- angle CMP, whose hypothenuse is equal to the radius, the two sides MP, CP are the sine and cosine of the arc AM : hence, the cosine of an arc is equal to that part of the radius inter- cepted between the centre and foot of the sine. The triangles CAT, CDS, are similar to the equal triangles CPM, CQM ; hence they are similar to each other. From these principles, we shall very soon deduce the different rela- tions which exist between the lines now defined : before doing so, however, we must examine the changes which those lines undergo, when the arc to which they relate increases from zero to 180°. The angle ACD is called the first quadrant ; the angle DCB, the second quadrant ; the angle BCE, the third quadrant ; and the angle ECA, the fourth quadrant S* 27 210 PLANE TRIGONOMETRY. VII. Suppose one extrem- ity of the arc remains fixed in A, while the other extremity, marked M, runs successively throughout the whole extent of the semicircumference, from A to B in the direction ADB. When the point M is at A, or when the arc AM is zero, the three points T, M, P, are confounded with the point A ; whence it appears that the sine and tangent of an arc zero, are zero, and the cosine and secant of this same arc, are each equal to the radius. Hence if R represents the radius oi the circle, we have sin 0=0, tang 0=0, cos 0=R, secO=R. VIII. As the point M advances towards D, the sine increases, and so likewise does the tangent and the secant ; but the cosine, the cotangent, and the cosecant, diminish. When the point M is at the middle of AD, or when the arc AM is 45°, in which case it is equal to its complement MD, the sine MP is equal to the cosine MQ or CP ; and the trian- gle CMP, having become isosceles, gives the proportion MP : CM : : 1 : V2, or sm 45^ R Hence 1 : V2. R sin 45°= cos 45o=-;^=lR\/2. In this same case, the triangle CAT becomes isosceles and equal to the triangle CDS ; whence the tangent of 45° and its cotangent, are each equal to the radius, and consequently we have tang 45° = cot 4!r=R. IX. The arc AM continuing to increase, the sine increases till M arrives at D ; at which point the sine is equal to the ra- dius, and the cosine is zero. Hence we have sin 90° = R, cos 90° = ; and it may be observed, that these values are a consequence of the values already found for the sine and cosine of the arc zero ; because the complement of 90^ being zero, we have sin 90°=:cos 0°r=R, and cos 90°=zsin 0°=0. PLANE TRIGONOMETRY. 211 As to the tangent, it increases very rapidly as the point M approaches D ; and finally when this point reaches D, the tan- gent properly exists no longer, because the lines AT, CD, being parallel, cannot meet. This is expressed by saying that the tangent of 90° is infinite ; and we write tang 90°= ao The complement of 90'' being zero, we have tang O=cot 90O and cot Orrtang 90o. Hence cot 90° =0, and cot 0=ao . X. The point M continuing to advance from D towards B, the sines diminish and the cosines increase. Thus M'P' is the sine of the arc AM', and M'Q, or CP' its cosine. But the arc M'B is the supplement of AM', since AM' + M'B is equal to a semicircumference ; besides, if M'M is drawn parallel to AB, the arcs AM, BM', which are included between parallels, will evidently be equal, and likewise the perpendiculars or sines MP, M'P'. Hence, the sine of an arc or of an angle is equal to the sine of the supplement of that arc or angle. The arc or angle A has for its supplement 180° — ^A; hence generally, we have sin A = sin (ISO^ — A.) The same property might also be expressed by the equation sin (90° + B) = sin (90°-— B), B being the arc DM or its equal DM', XI. The same arcs AM, AM', which are supplements of each other, and which have equal sines, have also equal co- sines CP, CP' ; but it must be observed, that these cosines lie in different directions. The line CP which is the cosine of the arc AM, has the origin of its value at the centre C, and is esti- mated in the direction from C towards A ; while CP', the cosine of AM' has also the origin of its value at C, but is estimated in a contrary direction, from C towards B. Some notation must obviously be adopted to distinguish the one of such equal lines from the other ; and that they may both be expressed analytically, and in the same general formula, it is necessary to consider all lines which are estimated in one di- rection as positive, and those which are estimated in the con- trary direction as negative. If, therefore, the cosines which are estimated from C towards A be considered as positive, those estimated from C towards B, must be regarded as nega- tive. Hence, generally, we shall have, cos A=— cos (180°--A) that is, the cosine of an arc or angle is equal to the cosine of its supplement taken negatively. The necessity of changing the algebraic sign to correspond 212 PLANE TRIGONOMETRY. with the change of direction in the trigonometrical Hne, may be illustrated by the fol- lowing example. The versed sine AP is equal to the radius CA minus CP the cosine AM : that is, ver-sin AM=R — cos AM. Now when the arc AM be- comes AM' the versed sine AP, becomes A F, that is equal to R + CF. But this expression cannot be derived from the formula, ver-sin AMrrR — cos AM, unless we suppose the cosine AM to become negative as soon as the arc AM becomes greater than a quadrant. At the point B the cosine becomes equal to — R ; that is, cos 180°=— R. For all arcs, such as ADBN', which terminate in the third quadrant, the cosine is estimated from C towards B, and is consequently negative. At E the cosine becomes zero, and for all arcs which terminate in the fourth quadrant the cosines are estimated from C towards A, and are consequently positive. The sines of all the arcs which terminate in the first and second quadrants, are estimated above the diameter BA, while the sines of those arcs which terminate in the third and fourth quadrants are estimated below it. Hence, considering the former as positive, we must regard the latter as negative. XII. Let us now see what sign is to be given to the tangent of an arc. The tangent of the arc AM falls above the line BA, and we have already regarded the lines estimated in the direc- tion AT as positive : therefore the tangents of all arcs which terminate in the first quadrant will be positive. But the tan- gent of the arc AM', greater than 90°, is determined by the intersection of the two lines M'C and AT. These lines, how- ever, do not meet in the direction AT ; but they meet in the opposite direction AV. But since the tangents estimated in the direction AT are positive, those estimated in the direction AV must be negative : therefore, the tangents of all arcs which ter- minate in the second quadrant will he negative. When the point M' reaches the point B the tangent AV will become equal to zero : that is, tang 1 80° =0. When the point M' passes the point B, and comes into the position N', the tangent of the arc ADN'will be the line AT : PLANE TRIGONOMETRY. 213 hence, the tangents of all arcs which terminate in the third quad- rant are positive. At E the tangent becomes infinite : that is, tang 270° = 00. When the point has passed along into the fourth quadrant to N, the tangent of the arc ADN'N will be the line AV : hence, the tangents of all arcs which terminate in the fourth quadrant are negative. The cotangents are estimated from the line ED. Those which lie on the side DS are regarded as positive, and those which lie on the side DS' as negative. Hence, the cotangents are posi- tive in the first quadrant, negative in the second, positive in the third, and negative in the fourth. When the point M is at B the cotangent is infinite ; when at E it is zero : hence, cot 180°=— 00 ; cot 270° = 0. Let q stand for a quadrant ; then the following table will show the signs of the trigonometrical lines in the different quadrants. \q 2q Sq 4q Sine + + — — Cosine + — — + Tangent + — ' + — ^ Cotangent -r — -f — XIII. In trigonometry, the sines, cosines, &c. of arcs or an- gles greater than 180° do not require to be considered ; the angles of triangles, rectilineal as well as spherical, and the sides of the latter, being always comprehended between and 180°. But in various applications of trigonometry, there is fre- quently occasion to reason about arcs greater than the semi- circumference, and even about arcs containing several circum- ferences. It will therefore be necessary to find the expression of the sines and cosines of those arcs whatever be their magnitude. We generally consider the arcs as positive which are esti- mated from A in the direction ADB, and then those arcs must be regarded as negative which are estimated in the contrary direction AEB. We observe, in the first place, that two equal arcs AM, AN with contraiy algebraic signs, have equal sines MP, PN, with contrary algebraic signs ; while the cosine CP is the same for both. The equal tangents AT, AV, as well as the equal cotangents DS, DS', have also contrary algebraic signs. Hence, calling X the arc, we have in general, sin ( — x)= — sin x cos ( — ^a;)=cos x tang ( — x)= — ^tanga; cot ( — x)= — cot a; 214 PLANE TRIGONOMETRY. By considering the arc AM, and its supplement AM', and recollecting what has been said, we readily see that, sin (an arc) = sin (its supplement) cos (an arc)= — cos (its supplement) tang (an arc)= — tang (its supplement) cot (an arc) = — cot (its supplement). It is no less evident, that if one or several circumfe- rences were added to any arc AM, it would still termi- nate exactly at the point M, and the arc thus increased would have the same sine as the arc AM ; hence if C rep- resent a whole circumfe- rence or 360°, we shall have sin x= sin (C + a;)=sinx=sin (2C + a:), &c. The same observation is ap- phcable to the cosine, tan- gent, &c. Hence it appears, that whatever be the magnitude of x the proposed arc, its sine may always be expressed, with a proper sign, by the sine of an arc less than 180°. For, in the first place, we may subtract 360° from the arc x as often as they are contained in it ; and y being the remainder, we shall have sin a;=sin y. Then xiy is greater than 180°, make y = 180° -}-%, and we have sin y= — sin z. Thus all the cases are reduced to that in which the proposed arc is less than 180° ; and since we farther have sin (90° + x)=sin (90°-^x), they are likewise ultimately reducible to the case, in which the proposed arc is between zero and 90°. XIV. The cosines are always reducible to sines, by means of the formula cos A = sin (90° — A) ; or if we require it, by means of the formula cos A = sin (90° + A): and thus, if we can find the value of the sines in all possible cases, we can also find that of the cosines. Besides, as has already been shown, that the negative cosines are separated from the positive cosines by the diameter DE ; all the arcs whose extremities fall on the right side of DE, having a positive cosine, while those whose extremities fall on the left have a negative cosine. Thus from 0° to 90° the cosines are positive ; from 90° to 270° they are negative ; from 270° to 360° they again become positive ; and after a whole revolution they assume the same values as in the preceding revolution, for cos (360°+a:)=cosar. PLANE TRIGONOMETRY. 215 From these explanations, it will evidently appear, that the sines and cosines of the various arcs which are multiples of the quadrant have the following values : sm 0°=0 sin 180° =0 sin 360° =0 sin 540° =0 sin 720° = &c. sm 90° =R sin 270°=— R sin 450° =R sin 630°=— R sin 810°=R &c. cos 0°=R cos 180°=— R cos 360° =R cos 540°=— R cos 720° =R iStc. cos 90° =0 cos 270° =0 cos 450° =0 cos 630° =0 cos 810°=0 &c. And generally, k designating any whole number we shall have sin 2^.90°=0, cos (2A:+1) . 90°=0, sin (4^+1) . 90° =R, cos 4^ . 90° =R, sin (4A:— 1) . 90°=— R, cos (4A; + 2) . 90°=— R. What we have just said concerning the sines and cosines renders it unnecessary for us to enter into any particular de- tail respecting the tangents, cotangents, &c. of arcs greater than 180° ; the value of these quantities are always easily de- duced from those of the sines and cosines of the same arcs : as we shall see by the formulas, which we now proceed to explain. THEOREMS AND FORMULAS RELATING TO SINES, COSINES, TANGENTS, &c. XV. The sine of an arc is half the chord which subtends a double arc. For the radius CA, perpen- dicular to the chord MN, bi- sects this chord, and likewise the arc MAN ; hence MP, the sine of the arc MA, is half the chord MN which subtends the arc MAN, the double of MA. The chord which subtends the sixth part of the circum- ference is equal to the radius ; hence ^^^°-or sin 30°=4R, sm 12 S' D \s r/y"'^^ a^ P ^ P' \ / A Jb — \ \ N' t^ V E m other words, the sine of a third part of the right angle is equal to the half of the radius 21Q PLANE TRIGONOMETRY* S' D S N sy^ Q~^ f P' \ / ^ ^ y^ i N' \ R J^ V E XVI. The square of the sine of an arc^ together with the square of the cosine, is equal to the square of the radius ; so that in general terms we have siii2A4-cos2A=R2. This property results im- mediately from the right-an- gled triangle CMP, in which MP2+CP2=CM2. It follows that when the sine of an arc is given, its co- sine may be found, and re- ciprocally, by means of the formulas cos A = db V (R^— sin^A), and sin A = di: V (R^— cos^A) . The sign of these formulas is +, or — , because the same sine MP answers to the two arcs AM, AM', whose cosines CP, CP', are equal and have contrary signs ; and the same cosine CP answers to the two arcs AM, AN, whose signs MP, PN, are also equal, and have contrary signs. Thus, for example, having found sin 30°=iR, we may de- duce from itcos30°,orsin60°=N/(R2— iR2)=v/fR2rriRv/3. XVII. The sine and cosine of an arc A being given, it is re- quired to find the tangent, secant, cotangent, and cosecant of the same arc. The triangles CPM, CAT, CDS, being similar, we have the proportions : CP : PM : : CA : AT ; or cos A : sin A : : R : tang A= Rsin A CP : CM : : CA ; CT : or cos A : R : : R : sec A= PM : CP r : CD : DS ; or sin A : cos A : : R : cot A- PM : CM : : CD : CS ; or sin A : R : : R : cosec A- cos A _R2_ cos A RcosA sin A R2 sin A which are the four formulas required. It may also be observed, that the two last formulas might be deduced from the first two, by simply putting 90° — A instead of A. From these formulas, may be deduced the values, with their proper signs, of the tangents, secants, &c. belonging to any arc whose sine and cosine are known ; and since the progres- sive law of the sines and cosines, according to the different arcs to which they relate, has been developed already, it is unnecessary to say more of the law which regulates the tan- gents and secants. PLANE TRIGONOMETRY. 217 By means of these formulas, several results, which have already been obtained concerning the trigonometrical lines, may be confirmed. If, for example, we make A =90°, we shall have sin A=R, cos A=0 ; and consequently tang 90° = p 2 — , an expression which designates an infinite quantity ; for, the quotient of radius divided by a very small quantity, is very great, and increases as the divisor diminishes ; hence, the quo- tient of the radius divided by zero is greater than any finite quantity. The tangent being equal to R ; and cotangent to R.-^- ; cos sm it follows that tangent and cotangent will both be positive when the sine and cosine have like algebraic signs, and both negative, when the sine and cosine have contrary algebraic signs. Hence, the tangent and cotangent have the same sign in the diagonal quadrants : that is, positive in the 1st and 3d, and negative in the 2d and 4th ; results agreeing with those of Art. XII. It is also apparent, from the above formulas, that the secant has always the same algebraic sign as the cosine, and the co- secant the same as the sine. Hence, the secant is positive on the right of the vertical diameter DE, and negative on the left of it ; the cosecant is positive above the diameter BA, and neg- ative below it : that is, the secant is positive in the 1st and 4th quadrants, and negative in the 2d and 3d : the cosecant is posi- tive in the 1st and 2d, and neg8^tive in the 3d and 4th. XVIII. The. formulas of the preceding Article, combined with each other and with the equation sin ''^A + cos ^ArzR^, furnish some others worthy of attention. First we have R^ + tang^ A = R^ + ^' ^'f^ ^ =z cos-* A RMBin'A + cos^A)^_R^. ^^„^^ R=+tang^ A=sec= A, a COS -A cos- A formula which might be immediately deduced from the right- angled triangle CAT. By these formulas, or by the right-an- gled triangle CDS, we have also R'^ + cot^ Arzcosec^ A. Lastly, by taking .the product of the two formulas tang A=r RsinA 1 ,4 RcosA u * a ^ a t»o — , and cot A= — : , we have tang Ax cot A=R-, a cos A' smA TJ2 p2 ■^ formula which gives cot A=-7- Xj.^ndtangA= t—t- T>2 . We likewise have cot B= . ^ tangB T 28 218 PLANE TRIGONOMETRY. Hence cot A : cot B : : tang B : tang A ; that is, the cotan- gents of two arcs are reciprocally proportional to their tangents. The formula cot Ax tang A=R'^ might be deduced imme- diately, by comparing the similar triangles CAT, CDS, which give AT ; CA : : CD : DS, or tang A : R : : R : cot A. XIX. The sines and cosines of two arcs^ a and b, being given, it is required to find the sine and cosine of the sum or difference of these arcs. Let the radius AC=R,the arc AB=a, the arc BD=fe, and con- sequently ABD=:a + h. From the points B and D, let fall the perpendiculars BE, DF upon AC ; from the point D, draw DI per- pendicular to BC ; lastly, from the point I draw IK perpendicu- lar, and IL parallel to, AC. F' C FX KE P The similar triangles BCE, ICK, give the proportions, CB : CI : : BE : IK,or R : cos 6 : : sin« : ik='^"^^^'^- R cos a cos h. R" CB : CI : : CE : CK, or R : cos 6 : : cos a : CK=: The triangles DIL, CBE, having their sides perpendicular, each to each, are similar, and give the proportions, CB : DI : : CE : DL, or R ; sin 6 : : cos a : DL= cos a sin b. R CB : DI : : BE : IL, or R : sin b sm a IL sm a sin b. R But we have IK+DL-DF: Hence sin (a + 5), and CK— IL =: CF=;Cos (a+b). sin a cos & + sin i cos a sin {a-\-b)=: cos (a 4-?))=: R cos a cos b — sin a sin b . R The values of sin (a — b) and of cos (a — b) might be easily deduced from these two formulas ; but .they may be found directly by the same figure. For, produce the sine DI till it meets the circumference at M; then we have BM=BD=6, and MI = ID = sin b. Through the point M, draw MP perpen- dicular, and MN parallel to, AC : since MIr=DI, we have MN =IL, and IN=DL. But we have IK— lN=MP=sin (a—b), and CK + MN = CP = cos (a—b) ; hence PLANE TRIGONOMETRY. 219 . / ,^ sin a cos h — sin h cos a sin {a—h) = : g ,/ cos a cos 6 + sin «5 sin h cos (a — o) = R These are the formulas which it was required to find. The preceding demonstration may seem defective in point of generahty, since, in the figure which we have followed, the arcs a and 6, and even « + &, are supposed to be less than 90°. But first the demonstration is easily extended to the case in which a and h being less than 90°, their sum a-\-h'\s greater than 90°. Then the point F would fall on the prolongation of AC, and the only change required in the demonstration would be that of taking cos {a + h) = — CF' ; but as we should, at the same time, have CF'^FL' — CK', it would still follow that cos (a4-&) = CK' — FL', or R cos (a + 6)= cos a cos h — sin a sin h. And whatever be the values of the arcs a and h, it is easily, shown that the formulas are true : hence we may regard them as established for all arcs. We will repeat and number the formulas for the purpose of more convenient reference. . . , ,v sin a cos ft + sin h cos a . ' ^ sm (a-f &) = - -^ (1.). . , . , v" sin a cos h — sin h cos a .„ v sm {a—h)=- (2.). , , cos a cos h — sin a sin 6 , , cos (a + 6)= — : ^ (3.) -. cos a cos 6+.sin a smh . . • cos {a—b) = g^ (4.) XX. If, in the formulas of the preceding Article, we make b=a, the first and the third will give . ^ 2 sin « cos a ^ cos^ a — sin^ a 2 cos^ a — R" sm 2«= p , cos 2a= ^ = ^ formulas which enable us to find the sine and cosine of the double arc, when we know the sine and cosine of the arc itself. Reciprocally, to divide a given arc «, into two equal parts, let us, in the same formulas, put \a instead of a : we shall have sin ^^ asini^cosjfl ^^^ ^^cos^ i«-sin^ ig R R Now, cos^ \a + s\v? ifl!=R^ and cos^^a — sin^ \az=iR cos a, there results by adding and subtracting cos2 ia=iR2+xR cos a, and ^m^\a=:^W — iR cos a ; whence sin ia-= n/ GR2— iR cos ti)=\s^2W—2R Qo^a, cos |a= v/(iRHiR cos a)==iN/2R2+2Rcosa. 220 PLANE TRIGONOMETRY. If we put 2a in the place of a, we shall have, sin a= \^(iR2— iR cos 2a)=iV2K^—2R cos 2a. cos a=\/(iR2 + iR cos2a)=:iV2RH2R cos 2a. Making, in the two last formulas, a=r:45°, gives cos 2a=0, and sin 45°= ViR2^Rv/i; ind also, cos 45°=V^^=RVl, Next, make a =22° 30', which gives cos 2a=^B v^i, and we have sin 22° 30'=R (Vi — i%^i, and cos 22° 30'=Rv^(^+iv^i). XXI. If we multiply together formulas (1.) and (2.) Art. XIX. and substitute for cos^ a, R^ — sin^ a, and for cos^ 6, R^ — sin^ b ; we shall obtain, after reducing and dividing by R^ sin (a + b) sin (a — b) = sin^ a — sin^ b = (sin a + sin b) (sin a — sin b). or, sin (a — b) : sin a — sin b : : sin a + sin ft ; sin {a + b). XXII. The formulas of Art. XIX. furnish a great number of consequences ; among which it will be enough to mention those of most frequent use. By adding and subtracting we obtain the four which follow, 2 sin (a + ft) + sin (a — 6)=_-sin a cos b. R 2 • sin (a-ffe) — sin (a — ft) =^ sin ft cos a, R 2 . cos (a + ft) + cos (a — ft) = ^|5-cos a cos ft. R 2 . cos (a — ft) — cos (a + ft)=-^sin a sin ft. R' and which serve to change a product of several sines or co- sines into lineal' sines or cosines, that is, into sines and cosines multiplied only by constant quantities. XXIII. If in these formulas we put a+ft==i?, « — &=^> which p-\-Q V — q gives a—-^, ^~~2~ ' ^® ^^^^^ ^^ 2 sinp + sin ^=-p^i» \{p-^-!- sin y _ cos j^ {p — q) _ cot ^ ( jp — ^) cos ^ — cos^ sinj(j9— (/) R sin jo— sin g _ sin ^ {p — q) _ tang ^ {p — q) cosp + cosq C0sj(^— ^j . R sin ^ — sin q _ cos J {p -\- q) _• cot i (p + q) COS g— -cos ^ sin i {p + q) R cos ;? + cos g ^ cos I (p + q) cos ^ ( jo — q) _^ cot | (jp -f {?) cos (7 — cos^ sin|^(jw + ^)'sin J (^— ^) tang J (jo — q) sin p + sin q 2sin ^ (p + q) cos 4 (/? — q) cos ^ (jo — q) sin (i? + ^)"~2sini(jo-F^)cosi(jo + 5')~cos^(p + ^) sin jp — sin q 2sin^ (jo — q) cos^ (p + q) sin ^ (jp — gr) sm (jp + g)~2sin i (jp + 5') cos i (p + q) ~sin i (p + ^)" Formulas which are the expression of so many theorems. From the first, it follows that the sum of the sines of two arcs is to the difference of these sines, as the tangent of half the sum of the arcs is to the tangent of half their difference. T* 222 PLANE TRIGONOMETRY. XXV. In order likewise to develop some formulas reUtiye to tangents, let us consider the expression tang {a-\-b)= -^ — J^^ ^^ which by substituting the values of sin (a+b) and cos (cf + ft), we shall find / . i-\ R (sin a cos 5 + sin 6 cos a) cos a cos b — sm o sm a ,, , . cos a tane a , . " cos b tang b Now we have sm a= ^ , and.sm b= p — ^- : substitute these values, dividing all the terms by cos cf cos b; we shall have 7 . i.\ K^ (tang a + tang b) tang {a + h)= "^ ^ \ / 5 ^ W — tang a tang b which is the value of the tangent of the sum of two arcs, ex- pressed by the tangents of each of these arcs. For the tangent of their difference, we should in like manner find / \\ R^ (tang a-^tang b) *^"§ (^~^)=RH- tang « tang &. ' Suppose b=a; for the duplication of the arcs, we shall have the formula 2 R2 tang a ^^"g2^=R^_tang^a- Suppose b=2a; for their triplication, we shall have the for- mula . tang3«^ g;ft^"g^ + ^^"g^/) ; & R2_tang a tang 2 a in which, substituting the value of tang 2 a, we shall have ^ R'^— 3tang2«. XXVI. Scholium, The radius R iDeing entirely arbitrai-y, is generally taken equal to 1, in which case it does not appear in the trigonometrical formulas. For example the expression for the tangent of twice an arc when R=l, becomes, ^ ^ 2 tang a tang 2 a= 5_— 1 — tang^ a- If we have an analytical formula calculated to the radius of 1, and wish to apply it to another circle in which the radius is R, we must multiply each term by such a power of R as will make all the terms homogenious : that is, so that each shall contain the same number of literal factors. PLANE TRIGONOMETRY. 22S CONSTRUCTION AND DESCRIPTION OF THE TABLES. XXVII. If the radius of a circle is taken equal to 1, and the lengths of the lines representing the sines, cosines, tangents, cotangents, &c. for every minute of the quadrant be calculated, and written in a table, this would be a table of natural sines, cosines, <&c. XXVIII. If such a table were known, it would be easjr to calculate a table of sines, &c. to any other radius ; since, in different circles, the sines, cosines, &c. of arcs containing the same number of degrees, are to each other as their radii. XXIX. If the trigonometrical lines themselves were used, it would be necessary, in the calculations, to perform the opera- tions of multiplication and division. To avoid so tedious a method of calculation, we use the logarithms of the sines, co- sines, &c. ; so that the tables in common use show the values of the logarithms of the sines, cosines, tangents, cotangents, &c, for each degree and minute of the quadrant, calculated to a given radius. This radius is 10,000,000,000, and consequently its logarithm is 10. XXX. Let us glance for a moment at one of the methods of calculating a table of natural sines. The radius. of a circle being 1, the circumference is known to be 3.14159265358979. This being divided successively, by 180 and 60, or at once by 10800, gives .0002908882086657, for the arc of 1 minute. Of so small an arc the sine, chord, and arc, differ almost imperceptibly from the ratio of equality ; so that the first ten of the preceding figures, that is, .0002908882 may be regarded as the «ine of 1' ; and in fact the sine given in the tables which run to seven places of figures is .0002909. By Art. XVI. we have for any arc, cos= \/(l — sin^). This theorem gives, in the present case, cos 1' = . 9999999577. Then by Art. XXIT. we shall have 2 cos I'xsin 1'— sin 0'=:sin 2' = .00058 17764 .2 cos I'xsin 2'~sin l'=sin 3' = .0008726646 2 cos r X sin 3'-— sin 2' = sin 4' = .00 11635526 2 cos I'xsin 4'— sin 3' = sin 5' = .00 14544407 2 cos r X sin 5'— sin 4'=sin 6' =.001 7453284 &;c. &c. &LC. Thus may the work be continued to any extent, the whole diflficulty consisting in the multiplication of each successive re- sult by the quantity 2 cos 1' = 1.9999999154. 224 PLANE TRIGONOMETRY. Or, the sines of 1 ' and 2' being determined, the work might be continued thus (Art. XXI.) : sin r : sin 2' — sin 1' : : sin 2' + sin 1' : sin 3' sin 2' : sin 3' — sin 1' : : sin 3' + sin 1' : sin 4' sin 3' : sin 4' — sin T : : sin 4' + sin 1' : sin 5' sin 4' : sin 5' — sin 1' : : sin 5' + sin 1' : sin 6' .&c. &c. &c. In hke manner, the computer might proceed for the sines of degrees, &c. thus : • sin 1° : sin 2°— sin 1° : : sin 2° + sin 1° : sin 3° sin 2° : sin 3°-^sin 1° :: sin 3° + sin 1° : sin 4° sin 3° : sin 4° — sin T : : sin 4° + sin 1° : sin 5= &c. &c. &c. Above 45° the process may be considerably simpUfied by the theorem for the. tangents of the sums and differences of arcs. For, when the radius is unity, the tangent of 45° is also unity, and tan (a + b) will be denoted thus : tan(45° + 6)=i+-^. ^ ^ ^ 1 — tan b And this, again, may be still further simplified in practice. Th§ secants and cosecants may be found from the cosines and sines. TABLE OF LOGARITHMS. XXXI. If the logarithms of all the numbers between r and any given number, be calculated and arranged in a tabular form, such table is called a table of logarithms. The table annexed shows the logarithms of all numbers between 1 and 10,000. The first column, on the left of each page of the table, is the column of numbers, and is designated by the letter N ; the deci- mal part of the logarithms of these numbers is placed directly opposite them, and on the same horizontal line. The characteristic of the logarithm, or the part which stands to the left of the decimal point, is always known, being 1 less than the places of integer figures in the given number, and there- fore it is not written in the table of logarithms. Thus, for all numbers between 1 and 10, the characteristic is : for num- bers between 10 and 100 it is 1, between 100 and 1000 it is 2, «Sz:c. PLANE TRIGONOMETRY. 225 PROBLEM. To find from the table the logarithm of any number, . CASE I. When the number is less than 100. Look on the first page of the table of logarithms, along the columns of numbers under N, until the number is found ; the number directly opposite it, in the column designated Log., is the logarithm sought. ^ . CASE II. When the number is greater than 100, and less than 10,000. Find, in the column of numbers, the three first figures of the given number. Then, pass across the page, in a horizontal line, into the columns marked 0, 1,2, 3, 4, &c., until you come to the column which is designated by the fourth figure of the given number : to the four figures so found, tw^o figures taken from the column marked 0, are to be prefixed. If the four figures found, stand opposite to a row of six figures in the column marked 0, the two figures from this column, which are to be prefixed to the four before found, are the first two on the left hand ; but, if the four figures stand opposite a line of only four figures, you are then to ascend the column, till you come to the line of six. figures : the two figures at the left hand are to be prefixed, and then the decimal part of the logarithm is obtained. To this, the characteristic of the logarithm is to be prefixed, which is always one less than the places of integer figures in the given number. Thus, the logarithm of 1 122 is 3.049993. In several of the columns, designated 0, 1, 2, 3, &c., small dots are found. Where this occurs, a cipher must be written for each of these dots, a«d the two figures which are to be pre- fixed, from the first column, are then found in the horizontal line directly below. Thus, the log. of 2 188 is 3.340047, the two dots being changed into two ciphers, and the 34 from tlie column 0, prefixed. The two figures from the colum 0, must also be taken from the line below, if any dots shall have been passed over, in passing along the horizontal line : thus, the loga- rithm of 3098 is 3.491081, the 49 from the column being taken from the line 310, 29 226 PLANE TRIGONOMETRY. CASE III. When the number exceeds 10,000, or consists of five or more places of figures. Consider all the figures after the fomrth from the left hand, as ciphers. Find, from th.e table, the logarithm of the first four places, and prefix a characteristic which shall be one less than the number of places including the ciphers. Take from the last column on the right of the page, marked D, the number on the same horizontal line with the logarithm, and multiply this num- ber by the numbers that*have been considered as ciphers: then, cut off from the rlgflt hand as many places for decimals as there are figures in the multiplier, and add the product, so obtained, to the first logarithm : this sum will be the logarithm sought. Let it be required to find the logarithm of 672887. The log. of 672800 is found, on the 1 1th page of the table, to be 5.827886, after prefixing the characteristic 5. The corresponding num- ber in the column D is 65, which being multiplied by 87, the figures regarded as ciphers, gives 5655 ; then, pointing off two places for decimals, the number to be added is 56.55. This number being added to 5.827886, gives 5.827942 for the loga- rithm of 672887 ; the decimal part .55, being omitted. This method of finding the logarithms of numbers, from the table, supposes that the logarithms are proportional to their respective numbers, which is not rigorously true. In the exam- ple, the logarithm of 672800 is 5.827886 ; the logarithm of 672900, a number greater by 100, 5.827951 : the difference of the logarithms is 65. Now, as 100, the difference of the numbers, is to 65, the difference of their logarithms, so is 87, the diffe- rence between the given number .and the least of the numbers used, to the difference of their logarithm's, which is 56.55 : this difference being added to 5.827886, the logarithm of the less number, gives 5.827942 for the logarithm of 672887. The use of the column of differences is therefore manifest. When, however, the decimal part which is to be omitted ex- ceeds .5, we come nearer to the true result by increasing the next figure to the left by 1 ; and this will be done in all the calculations which follow. Thus, the difference to be added, was nearer 57 than 56 ; hence it would have been more exact to have added the former number. The logarithm of a vulgar fraction is 6qual to the loga- rithm of the numerator, minus the logarithm of the denom- PLANE TRIGONOMETRY. 227 inator. The logarithm of a decimal fraction is found, hy con- sidering it as a whole number j and then prefixing to the decimal part of its logaiHthm a negative chai-acteristic, greater hy unity than the number of ciphers between the decimal point and the first significant place of figures. Thus, the logarithm of .0412, is 2^614897. PROBLEM. To find from the table, a number answering to a given logarithm. XXXII Search, in the column of logarithms, for the decimal part of the given logarithm, and if it be exactly found, set down the corresponding number. Then, if the characteristic of the given logarithm be positive,point off, from the left of the number found, one place more for whole numbers than there are units in the characteristic of the given logarithm, and treat the other places as decimals ; this will give the number sought. If the characteristic of the given logarithm be 0, there will be one place of whole numbers ; if it be — 1, the number will be entirely decimal ; if it be — 2, there will be one cipher be- tween the decimal point and the first significant figure ; if it be — 3, there will be two, &c. The number whose logarithm is 1.492481 is found in page 5, and is 31.08. But if the decimal part of the logarithm cannot be exactly found in the table, take the number answering to the nearest less logarithm ; take also from the table the corresponding dif- ference in the column D : then, subtract this less logarithm from the given logarithm ; and having annexed a sufficient number of ciphers to the remainder, divide it by the difference taken from the column D, and annex the quotient to the number an- swering to the less logarithm : this gives the required number, nearly. This rule, like the one for finding the logarithm of a number when the places exceed four, supposes the numbers to be proportional to* their corresponding logarithms. Ex. 1. Find the number answering to the logarithm 1.532708. Here, The given logarithm, is . . _ 1.532708 Next less logarithm of 34,09, is - - 1.532627 Their difference is -- - - - 81 And the tabular difference is 128 : hence 128) 81.00 (63 which being annexed to 34,09, gives 34.0963 for the number answering to the logarithm 1.532708. • liUS PLANE TRIGONOMETRY. Ex, 2. Required the number answering to tiie logarithm 3.233568. The given logarithm is 3.233568 The next less tabular logarithm of 1712, is 3.233504 Diff.= 64 Tab. Diff.=253) 64.00 (^5 Hence the number sought is 1712.25, marking four places of integers for the characteristic 3. TABLE OF LOGARITHMIC SINES. XXXIII. In this table are arranged the logarithms of the numerical values of the sines, cosines, tangents, and cotangents, of all the arcs or angles of the quadrant, divided to minutes, and calculated for a radius of 10,000,000,000. The logarithm of this radius is 10. In the first and last horizontal line, of each page, are written the degrees whose logarithmic sines, &:c, are expressed on the page. The vertical columns on the left and right, are columns of minutes. CASE L To find, in the table, the logarithmic sine, cosine, tangent, or co- tangent of any given arc or angle. 1. If the angle be less than 45°, look in the first horizontal line of the different pages, until the number of degrees be found ; then descend along the column of minutes, on the left of the page, till you reach the number showing the minutes ; then pass along the horizontal line till you come into the column designated, sine, cosine, tangent, or cotangent, as the case may be : the number so indicated, is the logarithm nought. Thus, the sine, cosine, tangent, and cotangent of 19° 55', are found on page 37, opposite 55, and are, respectively, 9.532312, 9.973215, 9.559097, 10.440903. 2. If the angle be greater than 45°, search along the bottom line of the different pages, till the number of degrees are found ; then ascend along the column of minutes, on the right hand side of the page, till you reach the number expressing the mi- nutes ; then pass along the horizontal line into the columns designated tang., cotang., sine, cosine, as the case may be ; the number so pointed out is the logarithm required. I PLANE TRIGONOMETRY. 229 It will be seen, that the column designated sine at the top of the page, is designated cosine at the bottom ; the one desig- nated tang., by cotang., and the one designated cotang., by tang. The angle found by taking the degrees at the top of the page, and the minutes from the first vertical column on the left, is the complement of the angle, found by taking the corresponding degrees at the bottom of the page, and the minutes traced up in the right hand column to the same horizontal line. This being apparent, the reason is manifest, why the columns desig- nated sine, cosine, tang., and cotang., when the degrees are pointed out at the top of the page, and the minutes counted downwards, ought to be changed, respectively, into cosine, sine, cotang., and tang., when the degrees are shown at the bottom of the page, and the minutes counted upwards. If the angle be greater than 90°, we have only to subtract it from 180°, and take the sine, cosine, tangent, or cotangent of the remainder. The secants and cosecants are omitted in the table, being easily found from the cosines and sines. For, sec. = ; or, taking the logarithms, log. sec.=r2 COS. log. R — log. cos.=20-^log. COS. ; that is, the logarithmic secant is found by suhstracting the logarithmic cosine from 20. And R^ • cosec . = , or log. cosec.=2 log. R — log. sine =20 — log. sine sine ; that is, the logarithmic cosecant is found by subtracting the logarithmic sine from 20. It has been shown that R^mtang. x cotang. ; therefore, 2 log. R:i=log. tang. + log. cotang.; or 20=log. tang. + log. cotang. The column of the table, next to the column of sines, and on the right of it, is designated by the letter D. This column is calculated in the following manner. Opening the table at any page, as 42, the sine of 24° is found to be 9.609313 ; of 24° r, 9.609597 : their difference is 284 ; this being divided by 60, the number of seconds in a minute, gives 4.73, which is entered in the column D, omitting the decimal point. Now, supposing the increase of the logarithmic sine to be propor- tional to the increase of the arc, and it is nearly so for 60", it follows, that 473 (the last two places being regarded as deci- mals) is the increase of the sine for 1". Similarly, if the arc be 24° 20', the increase of the sine for 1", is 465, the last two places being decimals. The same remarks are equally apph- cable in respect of the column D, after the column cosine, and of the column D, between the tangents and cotangents. The column D, between the tangents and cotangents, answers U 23a PLANE TRIGONOMETRY, ^ to either of these columns ; since of the same arc, the lug. tang. + log. cotang— 20. Therefore, having two arcs, a and 6, log. tang 6 + log. cotang 6:=:log. tang a + log. cotang q,\ or, log. tang h — log. tang a=log. cotang h — log. cotang a. Now, if it were required to find the logarithmic sine of an arc expressed in degrees, minutes, and seconds, we hatve only to find the degrees and minutes as before ; then multiply the corresponding tabular number by the seconds, cut off two places to the right hand for decimals, and then add the product to the number first found, for the sine of the given arc. Thus, if we wish the sine of 40° 26' 28". The sine 40° 26' . - . - 9.811952 Tabular difference =:= 247 Number of seconds rr: 28 Product = Q9.16, to be added -= 69.16 Gives for the sine of 40° 26' 28" =: 9.812021.16 The tangent of an arc, in which there are seconds, is found in a manner entirely similar. In regard to the cosine and co-' tangent, it must be remembered, that they increase while the arcs decrease, and decrease while the arcs are increased, con- sequently, the proportional numbers found for the seconds must be subtracted, not added. Ex, To find the cosine 3° 40' 40". Cosine 3° 40^ 9.999110 Tabular difference == 13 Number of seconds = 40 Product zr: 5.20, which being subtracted i= 5.20 Gives for the cosine of 3° 40' 40" 9.999104.80 CASE II. To find the degrees, minutes , and seconds answering to any given logarithmic sine, cosine, tangent, or cotangent. Search in the table, and in the proper column, until the num- ber be found ; the degrees are shown either at the top or bot- tom of the page, and the minutes in the side columns, either at the left or right. But if the number cannot be exactly found in the table, take the degrees and minutes answering to the nearest less logarithm, the logarithm itself, and also the corresponding tabular difference. Subtract the logarithm taken, from the PLANE TRIGONOMETRY. 23 1 given logarithm, annex two ciphers, and then divide the re- mainder by the tabulai* difference : the quotient is seconds, and is to be connected with the degrees and minutes before found ; to be added for the sine and tangent, and subtracted for the cosine and cotangent. Ex. 1. To find the arc answering to the sine 9.880054 Sine 49° 20', next less in the table, 9.879963 Tab. Diff. 181)9100(50" Hence the arc 49° 20' 50" corresponds to the given sine 9.880054. Ex. 2. To find tJie arc corresponding to cotang. 10.008688f. Cotang 44° 26', next less in the table 10,008591 Tab. Dift 421)9700(23" Hence, 44° 26' — 23" =44° 25' 37" is the arc corresponding to the given cotangent 10.008688. r PRINCIPLES FOR THE SOLUTION OF RECTIUNEAL TRI- ANGLES. THEOREM I. In every right angled triangle, radius is to the sine ^f either of the acute angles, as the hypothenuse to the opposite side . and raditis is to the cosine of either qfihe acute ^mgks, as the hypothenuse to the adjacent side. Let ABC be the proposed tri- ^ angle, right-angled at A : from the point C as a centre, with a v ^y^ radius CD equal to the radius of ^fT^ the tables, describe the arc DE, ^y^ \ which will measure the angle C ■; ^^ \ on CD let fall the perpendicular r\^^ 1 \ EF, which will be the sine of the ^ :B*D -flL angle C, and CF will be its co- sine. The triangles CBA, CEF, are similar, and give the pro- portion, CE : EF : : CB : BA : hence H :«inC : : BC : BA. 232 PLANE TRIGONOMETRY. ■M' -f. But we also have, -'^ «^ CE : CF : ; CB : CA : hence R : cos C : : CB ; CA. Cor. If the radius R=l, we shall have, AB=CB sin C, and CA = CB cos C. Hence, in every right angled triangle, the perpendicular is equal to the hypothenuse multiplied by the sine of the angle at the base ; and the base is equal to the hypothenuse multiplied, by the cosine of the angle at the base ; the radius being equal to unity. THEOREM II. r^*> %im< In every right angled triangle, radius is to the tangent of ei- ther of the acute angles, as the side adjacent to the side op- posite. Let CAB be the proposed tri- " angle. With any radius, as CD, de- scribe the arc DE, and draw the tangent DG. From the similar triangles CDG,CAB, we shall have, CD : DG :: CA : AB : hence, ^ R : tang C : : CA : AB. ; Cor. 1. If the radius R=l, AB=CA tang C. Hence, the perpendicular of a right angled triangle is equal to the base multiplied by the tangent of the angle at the base, the radius being unity. Cor. 2. Since the tangent of an arc is equal to the cotangent of its complement (Art. VI.), the cotangent of B may be sub- stituted in the proportion for tang C, which will give R : cot B ; : CA : AB. THEOREM III. In every rectilineal triangle, the sines of the angles are to each other as the opposite sides. PLANE TRIGONOMETRY. 233 Let ABC be the proposed triangle ; AD the perpendicular, let fail from the vertex A on the opposite side BC : there may be two cases. First. If the perpendicular falls within -^ the triangle ABC, the right-angled triangles ABD, ACD, will give, R : sin B : : AB : AD. R : sin C : : AC : AD. In these two propositions, the extremes are equal ; hence, sin C ; sin B : : AB : AC. Secondly. If the perpendicular falls without the triangle ABC, the right- angled triangles ABD, ACD, will still give the proportions, R : sin ABD : : AB : AD, R:sinC : : AC : AD; from which we derive sin C : sin ABD : : AB : AC. But the angle ABD is the supplement of ABC, or B ; hence siji ABD = sin B ; hence we still have sinC;sinB::AB:AC. THEOREM ly. In every rectilineal triangle, the cosine of either of the angles is equal to radius multiplied by the sum of the squares of the sides adjacent to the angle, minus the square of the side opposite^ divided by twice the rectangle of the adjacent sides. Let ABC be a triangle : then will AB2+BC2— AC2 s cos B=- 2AB X BC. First. If the perpendicular falls within the triangle, we shall have AC2=:AB2+ BC^— 2BC X BD (Book IV. Prop. XII.); , j.^. AB2+BC2--AC2 ^ , . ,, . , , , , , . , hence BD= ^^^ . But m the right-angled triangle D ABD, we have 2BC R cos B : : AB U* 30 BD; 234 PLANE TRIGONOMETRY. „ RxBD , ... , , .«^ hence, cos 13= ^ , or by substituting the value of BD, ^ „ AB2+BC2— AC2 cosB=.Rx ^^^^^^ Secondly. If the perpendicular falls without the triangle, we shall have AC2=:ABHBC2+2BCxBD; hence ^^ AC2— AB2— BC2 ^^=" 2BC • But in the right-angled triangle BAD, RxBD D B C we still have cos ABD= — ~^u~ '•> ^^^ t^© angle ABD being supplemental to ABC, or B, we have T5 Auri RxBD cos B= — cos ABD=: -^^ . hence by substituting the value of BD, we shall again have ^ „ AB2+BC2— AC^ cosB^Rx ^ABxBC ' Scholium. Let A, B, C, be the three angles of any triangle ; a, hy c, the sides respectively opposite them : by the theorem, we shall have cos B=R x - — ^ . And the same principle, when applied to each of the other two angles, will, in like man- &2^c2_^2 a^-^y^-^ ner give cos A=R x ^r '> and cos C=R x — ^"T — ' Either of these formulas may readily be reduced to one in which the computation can be made by logarithms. Recurring to the formula R^ — R cos A=2sin^ J A (Art. XXIII.), or 2sin^^A=R^ — RcosA, and substituting for cosA, we shall have 2sinHA=:R^-R^x^^^^' _ R^x26c-R^(&Hc^-~-a^ a^-^^-~c^-i-26c 26c ^ 26c = R.x±:|=i)Wx(?±^-4M^=^). Hence 26c 26c sin JA=Rv/(fcth^g^±f=^). For the sake of brevity, put J (a4-6+c)=j9, or fl+6+c=2j9; we have a + 6 — c=2p — ^2c, a + c — h^s=2p'^2b ; hence PLANE TRIGONOMETRY. 235 THEOREM V. In every rectilineal triangle, the sum of two sides is to their diffe- rence as the tangent of half the sum of the angles opposite those sides, to the tangent of half their difference. For, AB : BC : : sin C : sin A (Theo- rem III,). Hence, AB + BC : AB— BC : : sin C + sin A ; sin C — sin A. But sin sinC + sin A : sin C — sin A : : tang — - — ; tang ^^ (Art. XXIV.) ; hence, AB+BC : AB— BC : : tang ^±^ : tang "^, which is the property we had to demonstrate. With the aid of these five theorems we can solve all the cases of rectilineal trigonometry. Scholium. The required part should always be found from the given parts ; so that if an error is made in any part of the work, it may not affect the correctness of that which follows. SOLUTION OF RECTILINEAL TRIANGLES BY MEANS OF LOGARITHMS. It has already been remarked, that in order to abridge the calculations which are necessary to find the unknown parts of a triangle, we use the logarithms of the parts instead of the parts themselves. Since the addition of logarithms answers to the multiplica- tion of their corresponding numbers, and their subtraction to the division of their numbers ; it follows, that the logarithm of the fourth term of a proportion will be equal to the sum of the logarithms of the second and third terms, diminished by the logarithm of the first term. Instead, however, of subtracting the logarithm of the first term from the sum of the logarithms of the second and third terms, it is more convenient to use the arithmetical complement of the first term. The arithmetical complement of a logarithm is the number which remains after subtracting the logarithm from 10. Thus 10—9.274687 = 0.725313: hence, 0.725313 is the arithmetical complement of 9,274687, 236 PLANE TRIGONOMETRY. It is now to be shown that, the difference between two loga- rithms is truly found, by adding to the first logarithm the arith- metical complement of the logarithm to be subtracted, and dimin- ishing their sum by 10. Let a = the first logarithm. b = the logarithm to be subtracted. c = 10 — &=the arithmetical complement o(b. Now, the difference between the two logarithms will be expressed by a — b. But from the equation c=10 — b, we have c — 10= — b ; hence if we substitute for — b its value, we shall have a — b=a-\-c — 10, which agrees with the enunciation. When we wish the arithmetical complement of a logarithm, we may write it directly from the tables, by subtracting the left hand figure from 9, then proceeding to the right, subtract each figure from 9, till we reach the last significant figure, which must be taken from 10 : this will be the same as taking the logarithm from 10. Ex. From 3.274107 take 2.104729. Common method. By ar.-comp. 3.274107 3.274107 2.104729 ar.-comp. 7.895271 DifF. 1.169378 sum 1.169378 after re- jecting the 10. We therefore have, for all the proportions of trigonometry, the following RULE. Add together the arithmetical complement of the logarithm of the the first term, the logarithm of the second term, and the loga- rithm of the tliirdterm, and their sum after rejecting 10, will be the logarithm of the fourth term. And if any expression occurs in which the arithmetical complement is twice used, 20 must be rejected from the sum. PLANE TRIGONOMETRY. 237 SOLUTION OF RIGHT ANGLED TRIANGLES. Let A be the right angle of the proposed right angled triangle, B and C the other two angles ; let a be the hypothenuse, h the side opposite the angle B, c the side opposite the angle C. Here we must consider that the ^^ c" two angles C and B are complements of each other ; and that consequently, according to the different cases, we are entitled to assume sin C=cos B, sin B=cos C, and likewise tang B= cot C, tang C=cot B. This being fixed, the unknown parts of a right angled triangle may be found by the first two theo- rems ; or if two of the sides are given, by means of the pro- perty, that the square of the hypothenuse is equal to the sum of the squares of the other two sides. EXAMPLES. Ex. 1. In the right angled triangle BCA, there are given the hypothenuse «=250, and the side 6=240 ; required the other parts. R : sin B : : a : 6 (Theorem I.), or, « : 6 ; : R : sin B. When logarithms are used, it is most convenient to write the proportion thus. As hyp. a - 250 - ar.-comp. log. - 7.602060 To side 6 - 240 2.380211 So is R ----------- 10.000000 To sin B - 73° 44' 23" (after rejecting 10) 9.982271 But the angle C=90°— B=90°— 73° 44' 23"= 16° 15' 37". or, C might be found by the proportion, As hyp. a - 250 - ar.-comp. log. - 7.G02060 To side 6 - 240 2.380211 So is R - 10.000000 To cos C - 16° 15' 37" ------ 9.982271 To find the side c, we say. As R - - ar. comp. log. - 0.000000 To tang. C 16° 15' 37" - - - - 9.464889 So is side 6 240 - - - - 2.380211 To side c 70.0003 - - - 1.845100 238 PLANE TRIGONOMETRY. Or the side c might be found from the equation For, c2=e{2-~i2=(^_,, J) X (a—ft) : hence, 2 log. c=log. (« + ?>) + log. (a — 6), or log. c=Jlog. (aH-6)+J]og. (a— !^) « + 6=250 + 240=490 log. 2.690196 a— 6=250— 240=10 - - 1.000000 2 ) 3.690196 Log. c 70 - - - - - - 1.845098 ^ JBa;. 2. In the right angled triangle BCA, there are given, side 6=384 yards, and the angle B=53° 8' : required the other parts. To find the third side c. R : tang B : : c : 6 (Theorem II.) or, tang B : R : : 6 : c. Hence, As tang B 53° 8' ar.-comp. log. 9.875010 Is to R 10.000000 So is side 6 384 2.584331 To side c 287.965 .... 2.459341 Note. When the logarithm whose arithmetical complement is to be used, exceeds 10, take the arithmetical complement with reference to 20 and reject 20 from the sum. To find the hypothenuse a, R : sin B : : a : 6 (Theorem I.). Hence, As sin B 53° 8' ar. comp. log. 0.096892 Is to R 10.000000 ^ ^^ So is side 6 384 - - - . 2.584331 To hyp. a 479.979 - - - 2.681223 Ex, 3. In the right angled triangle BAG, there are given, side c=195, angle B=47° 55', required the other parts. Arts. Angle C=42° 05', a=290.953, 6=215.937. SOLUTION OF RECTILINEAL TRIANGLES IN GENERAL. Let A, B, C be the three angles of a proposed rectilineal tri- angle ; a, 6, c, the sides which are respectively opposite them ; the different problems which may occur in determining three of these quantities by means of the other three, will all be redu- cible to the four following cases. PLANE TRIGONOMETRY. 239 CASE I. Given a side and two angles of a triangle, to find the remaining parts. First, subtract the sum of the two angles from two right an- gles, the remainder will be the third angle. The remaining sides can then be found by Theorem III. I. In the triangle ABC, there are given the angle A =58° 07', the angle B=22° 37', and the side c=408 yards: required the remaining angle and the two other sides. To the angle A - - - - - =58° 07' Add the angle B =22° 37' Their sum - - - - - . =80° 44' taken from 180° leaves the angle C - =99° 16'. This angle being greater than 90° its sine is found by taking that of its supplement 80° 44'. As sine C Is to sine A So is side c So side a As sine C Is to sine B So is side c To side b To find the side a. 99° 16' ar.-comp. log. 58° 07' . . - - 408 351.024 - . . - To find the side b. 99° 16' ar.-comp. log. 22° 37' - - . . 408 . - - - - 158.976 - - - 0.005705 9.928972 2.610660 2.545367 0.005705 9.584968 2.610660 2.201333 2. In a triangle ABC, there are given the angle A =38° 25' B=57° 42', and the side C=400: required the remaining parts. Ans, Angle C=83° 53', side a =249.974, side ^>=:340.04. CAS? II. Criven two sides of a triangle, and an angle opposite one of them, to find the third side and the two remaining angles. 240 PLANE TRIGONOMETRY. 7.931814 2.334454 9.584968 9.851236 1. In the triangle ABC, there are given side AC =216, BC = 117, and the angle A =22° 37', to find the remaining parts. Describe the triangles ACB, ACB', as in Prob. XI. Book III. Then find the angle B by- Theorem III. As side B'C or BC 117 ar.-comp. log. Is to side AC 216 So is sine A 22° 37' To sine B' 45° 13' 55" or ABC 134° 46' 05" Add to each A 22° 37' 00 " 22° 37' 00" Take their sum 67° 50' 55" 157° 23' 05" From 180° 00' 00" 180° 00' 00" Rem. ACB' 112° 09' 05" ACB=22° 36' 55" To find the side AB or AB'. As sine A 22° 37' ar.-comp. log. Is to sine ACB' 112° 09' 05" -^ ,, i^ So is side B'C 117 - , - To side AB' 281.785 - - - The ambiguity in this, and similar examples, arises in con- sequence of the first proportion being true for both the trian- gles ACB, ACB'. As long as the two triangles exist, the am- biguity will continue. But if the side CB, opposite the given angle, be greater than AC, the arc BB' will cut the line ABB', on the same side of the point A, but in one point, and then there will be but one triangle answering the conditions. If the side CB be equal to the perpendicular Cd, the arc BB' will be tangent to ABB', and in this case also, there will be but one triangle. When CB is less than the perpendicular Cd, the arc BB' will not intersect the base ABB', and in that case there will be no triangle, or the conditions are impossible. 2. Given two sides of a triangle 50 and 40 respectively, and the angle opposite the latter equal to 32° : required the remain- ing parts of the triangle. Ans, If the angle opposite the side 50 be acute, it is equal to 41° 28' 59", the third angle is then equal to 106° 31' 01", and the third side to 72.368. If the angle opposite the side 50 be obtuse, it is equal to 138° 31' 01", the third angle to 9° 28' 59", and the remaining side to 12.436. 0.415032 9.966700 2.068186 2.449918 PLANE TRIGONOMETRY. 241 CASE III. Given two sides of a triangle^ with their included angle, to find the third side and the two remaining angles. Let ABC be a triangle, B the given angle, and c and a the given sides. Knowing the angle B, v^^e shall like- wise know the sum of the other two an- gles C + A=180°— B, and their half sum 1- (C + A)=90— JB. We shall next A h 'C compute the half difference of these two angles by the propor- tion (Theorem V.), c-f a ; c — a : : tang J (C + A) or cot J B : tang J (C — A,) in which we consider c>fl and consequently C>A. Having found the half difference, by adding it to the half sum \ (C + A), we shall have the greater angle C ; and by subtract- ing it from the half-sum, we shall have the smaller angle A. For, C and A being any two quantities, we have always, C=rHC + A)+J(C— A) A=:ri(C+A)-4(C-A). Knowing the angles C and A to find the third side h, we have the proportion. sin A : sin B : : a : 6 Ex, 1. In the triangle ABC, leta=450, c=540, and the in- cluded angle Bur 80° : required the remaining parts. c + air=990,c— a=90, 180°— B = 100° = C + A. Asc + a 990 ar.-comp. log. 7.004365 Is toe— a 90 1.954243 So is tang J (C + A) 50° . - - 10.076187 To tang \ (C— A) 6° 11' - - - 9.034795 Hence, 50° + 6° ll'=:56° ir=C ; and 50°— 6° ir=:43° 49' =A. To find the third side h. As sine A 43° 49' ar.-comp. log. 0.159672 Is to sine B 80° 9.993351 So is side a 450 - - - - - 2.65321 3 To side h 640.082 ... - 2.806236 Ex. 2. Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 04': required the other parts. ^715. Angles, 33° 34' 39 ", 18° 21' 21", side 2400. X31 242 PLANE TRIGONOMETRY. CASE IV. Given the three sides of a triangle, to find the angles. We have from Theorem IV. the formula, sin i A=R V^(1S(£E5) i„^hich p represents the half sum of the three sides. Hence, srn^A=k {P-^\lP-') , or 2 log. sin JA=2 log. R + log. (p — 6) + log. {p--c) — log. c — log. h. Ex. 1. In a triangle ABC, let b=40, c=34, and a^25 : required the angles. Here jo= ^ =49.5, jo — &=^9.5, and /?— c=I5.5. 2 Log. R - 20.000000 log. ip—b) 9.5 0.977724 log. (p^c) 15.5 - - - - - 1.190332 — log. c 34 ar.-comp. - - 8.468521 — log. h 40 ar.-comp. - - 8.397940 2 sin J A 19.034517 log. sin J A 19° 12' ^9" - - - 9-517258 Angle A=38° 25' 18". In a similar manner we find the angle B=83° 53' 18" and the angle C=57° 41' 24". Ex. 2. What are the angles of a plane triangle whose sides are, a=60, 6=50, and c=40? ^715. 41° 24' 34", 55° 46' 16" and 82° 49' 10". APPLICATIONS. Suppose the height of a building AB were required, the foot of it being accessible. PLANE TRIGONOMETRY. 243 On the ground which we suppose to be horizontal or very nearly so, measure a base AD, neither very great nor very small in comparison with the altitude AB ; then at D place the foot of the circle, or what- ever be the instrument, with which we are to measure the angle BCE formed by the hori- zontal line CE parallel to AD, A. D and by the visual ray direct it to the summit of the building- Suppose we find AD or CE 1=67.84 yards, and the angle BCE=4lo 04' : in order to find BE, we shall have to solve the right angled triangle BCE, in which the angle C and the adjacent side CE are known. To find the side EB. As R ar.-comp. - 0.000000 Is to tang. C 41° 04' 9.940183 So is EC 67.84 1.831486 ToEB 59.111 1.771669 Hence, EB=59.111 yards. To EB add the height of the instrument, which we will suppose to be 1.12 yards, we shall then have the required height AB=60.231 yards. If, in the same triangle BCE it were required to find the hypothenuse, form the proportion As cos C 41° 04' ar.-comp. - - log. 0.122660 Is to R 10.000000 So is CE 67.84 .-.--.--- 1.831486 To CB 89.98 - - 1.954146 Note. If only the summit B of the building or place whose height is required were visible, we should determine the dis- tance BE by the method shown in the following example ; this distance and the given angle BCE are suflicient for solv- ing the right angled triangle BCE, whose side, increased by the height of the instrument, will be the height required. 244 PLANE TRIGONOMETRY 2. To find upon the ground the distance of the point A from an inaccessible object B, we must measure a base AD, and the two adjacent angles BAD, ADB. Sup- pose we have found AD= 588.45 yards, BAD = 103° 55' 55", and BDA = 36^ 04'; we shall thence get the third angle ABD=40° 05", and to obtain AB, we shall form the proportion As sine ABD 40° 05" Is to sin BDA 36° 04' So is AD 588.45 ToAB - - 538.943 ar.-comp. log. 0.191920 9.769913 2.769710 2.731543 If for another inaccessible object C, we have found the an- gles CAD =35° 15', ADC = 119° 32', we shall in like manner find the distance AC = 1201.744 yards. 3. To find the distance between two inaccessible objects B and C, we determine AB and AC as in the last example ; we shall, at the same time, have the included angle BAC=BAD — DAC. Suppose AB has been found equal to 538.818 yards, AC = 1201.744 yards, and the angle BAC=68° 40' 55"; to get BC, we must resolve the triangle BAC, in which are known two sides and the included angle. AsAC + AB 1740.562 ar.-comp. log.- 6.759311 Is to AC— AB 662.926 2.821465 B-t-C So is tang.— -- 55° 39' 32" 10.165449 To tang. B— C 29° 08' 19" 9.746225 Hence - - - But we have - B— C 2 B + C =29° 08' 19' = 55° 39' 32' Hence B =84° 47' 51" and C =26° 31' 13" I PLANE TRIGONOMETRY. 245 Now, to find the distance BC make the proportion, As sine B 84° 47' 51" ar.-comp. - log. - 0.001793 Is to sine A 68° 40' 55 ' 9.969218 So is AC 1201.744 8.079811 To BC 1124.145 3.050822 4. Wanting to know the distance between two inaccessible objects which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and found to be, of the nearest, 57° ; of the most remote, 25° 30' I required the distance between them. Ans. 173.656 feet. 6. In order to find the distance between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distance of a third point C from each, was measured, viz. CA=588 feet and CB =672 feet, and also the contained angle ACB=55° 40': requi- red the distance AB. Ans. 592.967 feet. 6. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51° : then measuring in a direct line 180 feet farther from the hill, the angle of ele- vation of the top of the tower was 33° 45' : required the height of the tower. Ans. 83.9983 feet. 7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen, at a distance from each other equal to 200 yards, from the former of which A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE equal to 200 yards, and the following angles were taken, viz. AFC=83° ACF=i 54° 31', ACD=53° 30', BDC=156° 25', BDE=54° 30', and BED =88° 30' : required the distance AB. Ans. 345.46 yards. 8. From a station P there can be seen three objects, A, B and C, whose distances from each other are known, viz. AB= 800, AC = 600, and BC— 400 yards. There are also measured the horizontal angles, A?C=33° 45', BPC=22° 30'. It is re- quired, from these data, to determine the three distances PA, PC and PB. Ans. PA=710.193, PC = 1042.522, PB = 934.291 yards. X* 246 SPHERICAL TRIGONOMETRY. SPHERICAL TRIGONOMETRY. I. It has already been shown that a spherical triangle is formed by the arcs of three great circles- intersecting each other on the surface of a sphere, (Book IX. Def 1). Hence, every spherical triangle has six parts : the sides and three angles. Spherical Trigonometry explains the methods of determin- ing, by calculation, the unknown sides and angles of a spheri- cal triangle when any three of the six parts are given. II. Any two parts of a spherical triangle are said to be of the same species when they are both less or both greater than 90<* ; and they are of different species when one is less and the other greater than 90°. III. Let ABC be a spherical triangle, and O the centre of the sphere. Let the sides of the tri- angle be designated by letters corresponding to their opposite angles: that is, the side opposite the angle A by «, the side oppo- site B by 6, and the side opposite C by c. Then the angle COB will be represented by a, the an- gle CO A by h and the angle BOA by c. The angles of the spherical triangle will be equal to the angles included between the planes which determine its sides (Book IX. Prop. VI.). From any point A, of the edge OA, draw AD perpendicular to the plane COB. From D draw DH perpendicular to OB, and DK perpendicular to OC ; and draw AH and AK : the last lines will be respectively perpendicular to OB and OC, (Book VI. Prop. VI.) The angle DHA will be equal to the angle B of the spheri- cal triangle, and the angle DKA to the angle C. The two right angled triangles OKA, ADK, will give the proportions R : sin AOK : : OA : AK, or, R x AK= OA sin b. R : sin AKD : : AK : AD, or, R x ADrrAK sin C. Hence, R^ x AD = AO sin h sin C, by substituting for AK its value taken from the first equation. SPHERICAL TRIGONOMETRY. 247 In like manner the triangles AHO, ADH, right angled at H and D, give R : sin c : : AO : AH, or R x AH=:AO sin c R : sin B : : AH : AD, or Rx AD=AH sin B. Hence, R^x AD=AO sin a sin C. Equating this with the value of R^ x AD, before found, and di- viding by AO, we have • -o. • • T» sin C sin c sin sin C=sin c sin B, or - — 1^=- — r ( 1 ) sin B sinh ^ ' or, sin B : sin C : : sin 5 : c that is, The sines of the angles of a spherical triangle are to each other as the sines of their opposite sides, IV. From K draw KE perpendicular OB, and from D draw DF parallel to OB. Then will the angle DKFr=COB=a, since each is the complement of the angle EKO. In the right angled triangle OAH, we have R : cos c : : OA : OH ; hence AO cos c=RxOH=RxOE + R.DF. In the right-angled triangle OKE R : cos a : : OK : OE, or RxOEzrrOK cos a. But in the right angled triangle OKLA. R : cos 6 : : OA : OK, or, R x OK=OA cos 6. TT w rwn i^ A <^os a cos b Hence RxOE=OA. p In the right-angled triangle KFD R : sin a : KD : DF, or R x DF=KD sin a. But in the right angled triangles OAK, ADK, we have R : sin 5 : : OA : AK, or Rx AK=OA sin b R : cos K : AK : KD, or RxKD-AK cos C , „^ OA sin b cos C hence KD= ^ , and _ _ ^ OA sin a sin b cos C RxDF= o2 : therefore -. . OA cos a cos b AO sin a sin b cos C OA cos c= ^ + ^2 » or R* cos c=R cos a cos 6 + sin a sin b cos C. 248 SPHERICAL TRIGONOMETRY. Similar equations may be deduced for each of the other sides. Hence, generally, R2 cos a=R cos b cos c + sin b sin c cos A. ) R2 cos b='R cos a cos c + sin a sin c cos B. > (2.) R2 cos c=R cos b cos a + sin b sin c cos B. ) That is, radius square into the cosine of either side of a spheH- cal triangle is equal to radius into the rectangle of the cosines of the two other sides plus the rectangle of the sines of those sides into the cosine of their included angle. V. Each of the formulas designated (2) involves the three sides of the triangle together vv^ith one of the angles. These formulas are used to determine the angles when the three sides are known. It is necessary, however, to put them under an- other form to adapt them to logarithmic computation. Taking the first equation, we have . R^ cos a — R cos b cos c cos A= : — r-- sm sm c Adding R to each member, we have R2 cos a + R sin b sin c — R cos b cos c R + cos A: sin b sin c But, R+cos A^ ^ cQ^'?A (^rt. XXIII.), and R sin b sin c — R cos b cos c=— R^ cos (b + c) (Art, XIX.) ; , 2 cos^^A _ R2 (cos a — cos (b + c)) nence, ■!-» — = t » z^ K sm sm c ^ ^ sini (u + b + c) sini (b+c-a) sm sm c ^ ' Putting 5 =a+Z> + c, we shall have |5=i(a+6 + c) and ^s — a=^(b + c — a) ; hence cos i j.^^^^HsWS^EK'^ ^ ^ sm sm c cos \ B^R y/"^" ^ ( " ) ^^". (^'!=S ^ (3.) sm a sm c cos 1 r _TJ4. /^^" i ( g ) sin (^ s—c) ^i.-KV sin a sin 6 J SPHERICAL TRIGONOMETRY. 249 Had we subtracted each member of the first equation from R, instead of adding, we should, by making similar reductions, have found sm sm sin _ . / sin i(a 4- b—c) sin -^ (a + c~b) "^ sm c J 15 — K V gjjj ^gjjj ^ ^ ~ sin a sin b >(4.) Putting 5=a4-6 + c, we shall have J« — rt=J(&4-c — a), ^s — b=h (a + €—b), and ^s — c=^{a-^b — c) hence, sin 1A=R^ /«^" a^-^) si" a^-^) sin 6 sin c sin ^R = R ^/ ^nrgg^I^^n" (j^g^ I (5.) sm a sm c sin y.^R^/ s"^ (^^— ^) si" (i^— ») ^ sin a sin b VI. We may deduce the value of the side of a triangle in terms of the three angles by applying equations (4.), to the polar triangle. Thus, if «', b', c\ A', B', C', represent the sides and angles of the polar triangle, we shall have A=180°— a', B = 180°— 6', C=180°— c' ; a=l80°—A\ 6=180°— B', and c=180°— C (Book IX. Prop. VII.) : hence, omitting the ', since the equa- tions are applicable to any triangle, we shall have cos }.=R^/ ^^^I^ + B-C) '^^^ i (A + C-B) ' sin B sin C cos i h=n^ Ao^ i (A + B-C) cos i (B + C-A) sin A sin C cos I c=Ry/ cos i (A+C-B) cos j (B+C— A) sin A sin B. 32 he-) 250 SPHERICAL TRIGONOMETRY. Putting S=A + B + C, we shall have JS~A =i{A + B— A), ^S— B =i (A + C— B) and JS— C=KA+B— C), hence cos J^a=R\/ '="^ (tS-C) cos gS-B) sin B sin C cos JL&=R^ A°« (JS-C) cos gS-A) sin A sin C ^ /cos(iS-B)cosaS-A) COS|C=R V : r— sm A sm B M^O VII. If we apply equations (2.) to the polar triangle, we shall have — R^ cos A'=R cos B' cos C — sin B' sin C cos a\ Or, omitting the ', since the equation is applicable to any tri- angle, we have the three symmetrical equations, R^.cos A=sin B sin C cos a — ^R cos B cos C R^.cos B=sin A sin C cos b-^R cos A cos C W8.) R^.cos C=sin A sin B cos c — R cos A cos b) That is, radius square into the cosine of either angle of a sphe- rical triangle^ is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus radius into the rectangle of their cosines. VIII. All the formulas necessary for the solution of spheri- cal triangles, may be deduced from equations marked (2.). If we substitute for cos h in the third equation, its value taken from the second, and substitute for cos^ a its value R^ — sin^ a, and then divide by the common factor R.sin «, we shall have R.cos c sin a:zi^\n c cos a cos B + R.sin h cos C T, . X- /, \ • . , sin B sin c But equation (1.) gives sm h= . > ; sm C hence, by substitution, R cos c sin a=sin c cos a cos B-f R. Dividing by sin c, we have cos c R - — sin a=cos a cos B + R sm c sin B cos C sin c sin C sin B cos C sin C SPHERICAL TRIGONOMETRY. 251 Bat, ^=~ (Art. XVII.). sin R Tlierefore, cot c sin a=cos a cos B + cot C «in B. Hence, we may write the three symmetrical equations, cot a sin b=cos b cos C + cot A sin C -x cot b sin c=cos c cos A + cot B sin A > (9.) cot c sin azrcos a cos B + cot C sin B / That is, in every spherical triangle, the cotangent of one of the sides into the sine of a second side, is equal to the cosine of the se- cond side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle. IX. We shall terminate these formulas by demonstrating Napier^ s Analogies, which serve to simplify several cases in the solution of spherical triangles. If from the first equations (2.) cos c be eliminated, there will result, after a little reduction, R cos A sin c=R cos a sin b — cos C sin a cos b. By a simple permutation, this gives R cos B sin c=R cos b sin a — cos C sin b cos a. Hence by adding these two equations, and reducing, we shall have sin c (cos A+cos B)=(R — cos C) sin (a+&) ^ . sin c sin « sin ft ^ ^^ . But smce -: — ri=- — K=- — ^, we shall have sm C sm A sm B sin c (sin A + sin B) = sin C (sin a + sin b), and sin c (sin A — sin B)=siri C (sin a — sin b), . Dividing these two equations successively by the preceding one ; we shall have sin A + sin B _ sin C sin a + sin b cosA+cosB~R — cos C * sin (a+b) sin A — sin B_ sin C sin a — sin b cos A+cos B~R — cos C ' sin (a+bj' 252 SPHERICAL TRIGONOMETRY. And reducing these by the formulas in Articles XXIII. and XXIV., there will result tangi(A+B)=cotJC.5S^ii^ ° ^ ^ ^ COS J(a + fe) tang J (A— B) =cot W.-^-Vt r(. ^^^ ^ ^ sin^(a + 6) Hence, two sides a and i with the included angle C being given, the two other angles A and B may be found by the analogies, cosJ(a+6) : cos^(a — b) : : cot |^ C : tangJ(A + B) sin ^ {a-\-b) : sin J {a — b) : : cot ^ C : tang J (A — B). If these same analogies are applied to the polar triangle of ABC, we shall have to put 180°— A', 180°— B', 180°— «', 180^— 6', 180° — c', instead of a, 6, A, B, C, respectively; and for the result, we shall have after omitting the ', these two analogies, cosJ(A + B) : cosJ(A — B) : : tangle : tangJ(rt-{-6) sinJ(A + B) : sin^(A— B) : : tangjc : tangj(« — 6), by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These foiir proportions are known by the name of Napier's Analogies, X. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solu- tions corresponding to the two results in Case II. of rectilineal triangles. It is also plain that this ambiguity will extend itself to the corresponding case of the polar triangle, that is, to the case in which there are given two angles and a side opposite one of them. In every case we shall avoid all false solutions by recollecting, 1st. That every angle, and every side of a spherical triangle is less than 180°. 2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally. NAPIER'S CIRCULAR PARTS. XI. Besides the analogies of Napier already demonstrated, that Geometer also invented rules for the solution of all the cases of right angled spherical triangles. SPHERICAL TRIGONOMETRY. In every right angled spherical triangle BAG, there are six parts : three sides and three angles. If we omit the consideration of the right angle, which is always known, there will be five remain- ing parts, two of which must be given before the others can be determined. The circular parts, as they are called, are the two sides c and 6, about the right angle, the complements of the oblique angles B and G, and the complement of the hypothenuse a. Hence there are five circular parts. The right angle A not being a circular part, is supposed not to separate the circular parts c and &, so that these parts are considered as adjacent to each other. If any two parts of the triangle be given, their corresponding circular parts will also be known, and • these together with a required part, will make three parts under consideration. Now, these three parts will all lie together, or one of them will be sepa- rated from both of the others. For example, if B and c were given, and a required, the three parts considered would lie together. But if B and G were given, and b required, the parts would not lie together ; for, B would be separated from C by the part a, and from b by the part c. In either case B is the middle part. Hence, when there are three of the circular parts under consideration, the middle part is that one of them to which both of the others are adjacent, or from which both of them are separated. In the former case the parts are said to be adjacent^ and in the latter case the parts are said to be opposite. This being premised, we are now to prove the following rules for the solution of right angled spherical triangles, which it must be remembered apply to the circular parts, as already defined. 1st. Radius into the sine of the middle part is equal to the rect- angle of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part is equal to the red- angle of the cosines of the opposite parts. These rules are proved by assuming each of the five circu- lar parts, in succession, as the middle part, and by taking the extremes first opposite, then adjacent. Having thus fixed the three parts which are to be considered, take that one of the general equations for oblique angled triangles, which shall con- tain the three corresponding parts of the triangle, together with the right angle : then make A = 90°, and after making the reduc- tions corresponding to this supposition, the resulting equation will prove the rule for that particular case. i. - y 'k 254 SPHERICAL TRIGONOMETRY. For example, let comp. a be the middle part and the ex- tremes opposite. The equation to be applied in this case must contain a, b, c, and A. The first of equations (2.) contains these four quantities : hence R2 cos fl=R cos b cos c + sin 6 sin c cos A. If A = 90° cos A=:0 ; hence R cos a=cos b cos c ; that is, radius into the sine of the middle part, (which is the complement of a,) is equal to the rectangle of the cosines of the opposite parts. Suppose now that the complement of a were the middle part and the ex- tremes adjacent. The equation to be appUed must contain the four quan- tities fl, B, C, and A. It is the first of equations (8.). c R^ cos A=sin B sin C cos a — R cos B cos C. Making A =90°, we have sin B sin C cos a=R cos B cos C, or R cos a=cot B cot C ; that is, radius into the sine of the middle part is equal to the rectangle of the tangent of the complement of B into the tan- gent of the complement of C, that is, to the rectangle of the tangents of the adjacent circular parts. Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under consideration will then be the perpendicular b and the angle C. The equation to be applied must contain the four parts A, B, C, and 6 : it is the second of equations (8.), R^ cos B=sin A sin C cos b — R cos A cos C. Making A =90°, we have, after dividing by R, R cos B=sin C cos 6. Let comp. B be still the middle part and the extremes adja- cent. The equation to be applied must then contain the four four parts a, B, c, and A. It is similar to equations (7.). cot a sin c=cos c cos B + cot A sin B. But if A =90°, cot A=0 ; hence, cot a sin crrcos c cos B ; or, R cos Brrrcot a tang c. SPHERICAL TRIGONOMETRY. 255 And by pursuing the same method of demonstration when each circular part is made the middle part, we obtain the five fol- lowing equations, which embrace all the cases. R cos a=:cos b cos c=cot B cot C R cos Birrcos b sin C==cot a tang c R cos C = cos csinB=:cot <2 tang 6 R sin 6=sin A CA till CD is equal to 90°, and con- / \ ceive the arc of a great circle to be y^ \ drawn through B and D. Then C ^y^ \ will be the pole of the arc BD, and ^^^^.-^ _--J A the angle C will be measured by B-^I!. .^^ — \ BD (Book IX. Prop. VI.), and the \.... ^ \b angles CBD and D will be right an- *'J^*---^ / gles. Now before the remaining "l) parts of the quadrantal triangle can be found, at least two parts must be given in addition to the side BC = 90° ; in which case two parts of the right angled tri- angle BDA, together with the right angle, become known. Hence the conditions which enable us to determine one of these triangles, will enable us also to determine the other. 3. In the quadrantal triangle BCA, there are given CB=r90°, the angle C=42° 12', and the angle A=115° 20' : required' the remaining parts. Having produced CA to D, making CD =90° and drawn the arc BD, there will then be given in the right angled triangle BAD, the side fl=C=42° 12', and the angle BAD=180°— BAC = 180°— 115° 20' = 64°40',to find the remaining parts. To find the side d. The side a will be the middle part, and the extremes oppo« site: hence, R sin <2 = sin A sin d. As sin A 64° 40' ar.-comp. log. 0.043911 Isto R 10.000000 So is sin a 42° 12' . - - - 9.827189 To sin d 48° 00' 15" - - - 9.871100 To find the angle B. The angle A will correspond to the middle part, and the ex- tremes will be opposite : hence R cos A=sin B cos «. As cos a 42° 12' ar.-comp. log. 0.130296 Isto R - - - - - - 10.000000 So is cos A 64° 40' .... 9.631326 To sin B 35° 16' 53" - - - 9^761622 SPHERICAL TRIGONOMETRY. 269 To find the side b. The side b will be the middle part, and the extremes adja- cent : hence, R sin b=coi A tang a. As R - ar.-comp. log. 0.000000 Is to cot A 64° 40' - - . - 9.675237 So is tang a 42° 12' - - - - 9.957485 To sin b 25° 25' 14" - - - 9.632722 Hence, CA=90°— 6=90°— 25° 25' 14" =64° 34' 46" CBAr=90°— ABD=90°— 35° 16' 53 '=54° 43 07" BA^d - - . - =48° 00' 15". 4. In the right angled triangle BAC, right angled at A, there are given a=115° 25', and c=60° 59' i^-equired the remaining parts. ( B=148° 56' 45" Ans. ) C= 75° 30' 33" ( 6=152° 13' 50". 6. In the right angled spherical triangle BAC, right angled at A, there are given c=116° 30' 43", and 6=29° 41' 32" : re- quired the remaining parts. ( C=:103° 52' 46" Ans, ) B= 57° 28' 22" (a =112° 47' 58". 6. In a quadrantal triangle, there are given the quadrantal side ==90°, an adjacent side =115° 09', and the included angle = 115° 55' ; required the remaining parts. side, 113° 18' 19" ^w*- ^ o^^io. S 117° 33' 52" angles, ^^j^ 4Q. q^.. SOLUTION OF OBLIQUE ANGLED TRIANGLES BY LOGARITHMS. There are six cases which occur in the solution of oblique angled spherical triangles. 1. Having given two sides, and an angle opposite one of them. 2. Having given two angles, and a side opposite one of them. 3. Having given the three sides of a triangle, to find the angles. 260 SPHERICAL TRIGONOMETRY. 4. Having given the three angles of a triangle, to find the sides. 5. Having given two sides and the included angle. 6. Having given two angles and the included side. CASE I. Cfiven two sides , and an angle opposite one of them, to find the re- maining parts. For this case we employ equation {!.) ; As sin a : sin 6 : : sin A : sin B. • Ex. 1. Given the side a=44° 13' 45", &z=84° 14' 29" and the angle A=32° 26' 07" : required the remaining parts. To find the angle B. As sin a 44° 13' 45" ar.-comp. log. 0.156427 Is to sin b 84° 14' 29" - - - 9.997803 So is sin A 32° 26' 07" - - - 9.729445 To sin B 49° 54' 38" or sin B' 130° 5' 22" 9.883685 Since the sine of an arc is the same as the sine of its supple- ment, there will be two angles corresponding to the logarithmic sine 9.883685 and these angles will be supplements of each other. It does not follow however that both of them will satisfy all the other conditions of the question. If they do, there will be two triangles ACB', ACB ; if not, there will be but one. To determine the circumstances under which this ambiguity arises, we will consider the 2d of equations (2.). R^ cos b=R cos a cos c+sin a sin c cos B. from which we obtain _, R^ cos b — R cos 0! cos c cos B = : -. . sm a sm c Now if cos b be greater than cos a or cos c, we shall have R^ cos 6>R cos a cos c, or the sign of the second member of the equation will depend on that of cos b. Hence cos B and cos b will have the same SPHERICAL TRIGONOMETRY. 261 sign, or B and b will be of the same species, and there will be but one triangle. But when cos fe>cos a, sin ft < sin a : hence, If the sine of the side opposite the required angle be less than the sine of the other given side, there will be but one triangle. If however, sin ft > sin a, the cos ft will be less than cos a, and it is plain that such a value may then be given to c as to render R^cos ftR cos A cos C, and hence the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B >cos A the sin Bsin A, the cos B will be less than cos A, and it is plain that such a value may then be given to cos C, as to render R^cos B Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle there will be two solutions. Let us first suppose the side b to be less than 90°, or equal to 79° 12' 10". If now, we let fall from the angle C a perpendicular on the base BA, the triangle will be divided into two right angled tri- angles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules we find, Angle C=130°54'26" Side c=119°03'26". If we take the side 6=100° 47' 50", we shall find Angle C=156° 15' 04" Side c=:152° 14' 18". 264 SPHERICAL TRIGONOMETRY. Ex, 2. In a spherical triangle ABC there are given A=103'' 59' 57", B=46° 18' 7", and a=42° 8' 48" ; required the remain- ing parts. There will but one triangle, since sin B1 3581 4248 4913 5578 6241 6904 7565 •8226 8885 3047 3714 4381 5046 5711 6374 7036 7698 8358 9017 9676 0333 0989 1645 2299 2952 3605 4256 4906 5556 6204 6852 7499 8144 8789 9432 ..75 0717 1358 1998 3114 3781 4447 5113 5777 6440 7102 7764 8424 9083 9741 0399 1055 1710 2364 3018 3670 4321 4971 5621 3181 3S48 4514 5179 5843 6506 7169 7830 8490 9149 819544 820201 0858 1514 2168 2822 3474 4126 4776 5426 9807 0464 1120 1775 2430 3083 3735 4386 5036 5686 6334 6981 7628 8273 8918 9561 .204 0845 1486 2126 2764 3402 4039 4675 5310 5944 6577 7210 7841 8471 9873 0530 1186 1841 2495 3148 3800 4451 5101 5751 6399 7046 7692 8338 8982 9625 .268 0909 1550 2189 9939 0595 1251 1906 2560 3213 3865 4516 5166 5815 6464 7111 7757 8402 9046 9690 .332 0973 1614 2253 2892 3530 4166 4802 5437 6071 6704 7336 7967 8597 9227 9855 0482 1109 1735 2360 2983 3606 4229 4850 826075 6723 7369 8015 8660 9304 9947 830589 1230 1870 832509 3147 3784 4421 5056 5691 6324 6957 7588 8219 838849 9478 840106 0733 1359 1985 2609 3233 3855 4477 6269 6917 7563 8209 8853 9497 .139 0781 1422 2062 2700 3338 3975 4611 5247 5881 6514 7146 7778 8408 6528 7175 7821 8467 9111 9754 •396 1037 1678 2317 2956 3593 4230 4866 5500 6134 6767 7399 8030 8660 2637 3275 3912 4548 5183 5817 6451 7083 7715 8345 2828 3466 4103 4739 5373 6007 6641 7273 7904 8534 9164 9792 0420 1046 1672 2297 2921 3544 4166 4788 3020 3657 4294 4929 5564 6197 6830 7402 8093 8723 9352 9981 0608 1234 1860 2484 3108 3731 4353 4974 3083 3721 4357 4993 5627 6261 6894 7525 8156 8786 8912 9541 0169 0796 1422 2047 2672 3295 3918 4539 8975 9604 0232 0859 1485 2110 2734 3357 3980 4601 9038 9667 0294 0921 1547 2172 2796 3420 4042 4664 9101 9729 0357 0984 1610 2235 2859 3482 4104 4726 9289 9918 0545 1172 1797 2422 3046 3669 4291 4912 9415 ..43 0671 1297 1922 2547 3170 3793 4415 5036 N. 1 I 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 D. 1 12 A TABLE OP LOGARITHMS FROM 1 TO 10,000 N. 1 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 |D.l 700 845098 5160. 5222 5284 5346 1 5408 5470 5532 5594 5656 62 1 701 6718 5780 5842 5904 5966 6028 6090 6151 6213 6275 62 702 6337 6399 6461 6523 6585 6646 6708 6770 6832 6894 62 703 6955 7017 7079 7141 7202 7264 7320 7388 7449 7511 62 704 7573 7634 7696 7758 7819 7881 7943 8004 8066 8128 62 705 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 706 8805 8866 8928 8989 9051 9112 9174 9235 9297 9358 61 707 9419 9481 9542 9604 9665 9726 9788 9849 9911 9972 61 708 850033 0095 0156 0217 0279 0340 0401 0462 0524 0585 61 709 710 0646 0707 1320 0769 1381 0830 0891 0952 1564 1014 1625 1075 1136 1747 1197 1809 61 61 851258 1442 1503 1686 711 1870 1931 1992 2053 2114 2175 2236 2297 2358 2419 61 712 2480 2541 2602 2663 2724 2785 2846 2907 2968 3029 61 713 3090 3150 3211 3272 3333 3394 3455 3516 3577 3637 61 714 3698 3759 3820 3881 3941 4002 4003 4124 4185 4245 61 715 4306 4367 4428 4488 4549 4610 4670 4731 4792 4852 61 716 4913 4974 5034 5095 5156 5216 5277 5337 5398 5459 61 717 5519 5580 5640 5701 5761 5822 5882 5943 6003 6064 61 718 6124 6185 6245 6306 6366 6427 6487 6548 6608 6668 60 719 720 6729 857332 6789 7393 6850 6910 7513 6970 7574 7031 7634 7091 7152 7212 7272 7875 60 60 7453 7694 7755 7815 721 7935 7995 8056 8116 8176 8236 8297 8357 8417 8477 60 722 8537 8597 8657 8718 8778 8838 8898 8958 9018 9078 60 723 9138 9198 9258 9318 9379 9439 9499 9559 9619 9679 60 724 9739 9799 9859 9918 9978 ..38 ..98 .158 .218 .278 60 725 860338 0398 0458 0518 0578 0637 0697 0757 0817 0877 60 726 0937 0996 1056 1116 1176 1236 1295 1355 1415 1475 60 727 1534 1594 1654 1714 1773 1833 1893 1952 2012 2072 60 728 2131 2191 2251 2310 2370 2430 2489 2549 2608 2668 60 729 730 2728 2787 3382 2847 2906 2966 3025 3085 3144 3739 3204 3263 60 59 863323 3442 3501 3561 3620 3080 3799 3858 731 3917 3977 4036 4096 4155 4214 4274 4333 4392 4452 59 732 4511 4570 4630 4689 4748 4808 4867 4926 4985 5045 59 733 5104 5163 5222 5282 5341 5400 5459 5519 5578 5637 59 734 5696 5755 5814 5874 5933 5992 6051 6110 6169 6228 59 735 6287 6346 6405 6465 6524 6583 6642 6701 6760 6819 59 736 6878 6937 6996 7055 7114 7173 7232 7291 7350 7409 59 737 7467 7526 7585 7644 7703 7762 7821 7880 7939 7998 59 738 8056 8115 8174 8233 8292 8350 8409 8468 8527 8586 59 739 740 8644 8703 9290 8762 8821 8879 9466 8938 9525 8997 9584 9056 9114 9173 59 59 869232 9349 9408 9642 9701 9760 741 9818 9877 9935 9994 ..53 .111 .170 .228 .287 .345 59 742 870404 0462 0521 0579 0638 0696 0755 0813 0872 0930 68 743 0989 1047 1106 1164 1223 1281 1339 1398 1456 1515 58 744 . 1573 1631 1690 1748 1806 1865 1923 1981 2040 2098 68 745 2156 2215 2273 2331 2389 2448 2506 2564 2622 2681 68 746 2739 2797 2855 2913 2972 3030 3088 3146 3204 3262 68 747 3321 3379 3437 3495 3553 3611 3669 3727 3785 3844 68 748 3902 3960 4018 4076 4134 4192 4250 4308 4366 4424 58 749 750 4482 4540 4598 4656 4714 5293 4772 5351 4830 4888 4945 5003 5582 58 68 875061 5119 5177 5235 5409 5466 5524 751 5640 5698 5756 5813 5871 5929 5987 6045 6102 6160 68 752 6218 6276 6333 6391 6449 6507 6564 6622 6680 6737 68 753 6795 6353 6910 6908 7026 7083 7141 7199 7256 7314 68 754 7371 7429 7487 7544 7602 7659 7717 7774 7832 7889 68 755 7947 8004 8062 8119 8177 8234 8292 8349 8407 8464 67 756 8522 8579 8637 8694 8752 8809 8866 8924 8981 9039 67 757 9096 9153 9211 9268 9325 9383 9440 9497 9555 9612 67 758 9669 9726 9784 9841 9898 9956 ..13 ..70 .127 .185 67 759 880242 0299 0356 0413 0471 0528 0585 0642 0699 0756 67 N. |ll2|3|4|5|6|7|8|9!D. 1 A TABLE OF LOGAEITHBIS FROM 1 TO 10,000 13 N. |l|2|3|4|5|6|7|8|9|D. 1 760 880814 0871.0928,0985 1042 1099, 1156 1213 1271 1328 57 1 761 1385 1442 1499 1556 1613 1670 1727 1784 1841 1898 57 762 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 67 763 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 57 764 3093 3150 3207 3264 3321 3377 3434 3491 3548 3605 57 765 3661 3718 3775 3832 3888 3945 4002 4059 4115 4172 57 766 4229 4285 4342 4399 4455 4512 4569 4625 4682 4739 57 767 4795 4852 4909 4965 5022 5078 5135 5192 5248 5305 57 768 5361 5418 5474 5531 5587 5644 5700 5757 5813 5870 57 769 770 5926 5983 6039 6096 6152 6716 6209 6265 6321 6378 6434 56 56 886491 6547 6604 6660 6773 6829 6885 6942 6998 771 7054 7111 7167 7223 7280 7336 7392 7449 7505 7561 56 772 7617 7674 7730 7786 7842 7898 7955 8011 8067 8123 56 773 8179 8236 8292 8348 8404 8460 8516 8573 8629 8685 56 774 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 56 775 9302 9358 9414 9470 9526 9582 9638 9694 9750 9806 56 776 9862 9918 9974 ..30 ..86 .141 .197 .253 .309 .365 56 777 890421 0477 0533 0589 0645 0700 0756 0812 0868 0924 56 T78 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 56 779 780 1537 1593 2150 1049 2206 1705 2262 1760 2317 1816 2373 1872 1928 1983 2039 2595 56 56 892095 2429 2484 2540 781 2651 2707 2762 2818 2873 2929 2985 3040 3096 3151 56 7*2 3207 3262 3318 3373 3429 3484 3540 3595 3651 3706 56 783 3762 3817 3873 3928 3984 4039 4094 4150 4205 4261 55 784 4316 4371 4427 4482 4538 4593 4648 4704 4759 4814 55 785 4870 4925 4980 5036 5091 5146 5201 5257 5312 5367 55 786 5423 5478 5533 5588 5044 5699 5754 5809 5864 5920 55 787 5975 6030 6085 6140 6195 6251 6306 6361 6416 6471 65 788 6526 6581 6636 0692 6747 6802 6857 6912 6967 7022 65 789 790 7077 7132 7187 7737 7242 7792 7297 7352 7407 7462 7517 7572 55 55 897627 7682 7847 7902 7957 8012 8067 8122 791 8176 8231 8286 8341 8396 8451 8506 8561 8615 8670 65 792 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 66 793 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 55 794 9821 9875 9930 9985 ..39 ..94 .149 .203 .258 .312 56 795 900367 0422 0476 0531 0586 0640 0695 0749 0804 0859 65 796 0913 0968 1022 1077 1131 1186 1240 1295 1349 1404 65 797 1458 1513 1567 1622 1676 1731 1785 1840 1894 1948 54 798 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 54 799 800 2547 2601 2655 3199 2710 3253 2764 3307 2818 2873 2927 3470 2981 3524 3036 64 54 903090 3144 3361 3416 3578 801 3633 3687 3741 3795 3849 3904 3958 4012 4066 4120 64 802 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 803 4716 4770 4824 4878 4932 4986 5040 5094 5148 5202 54 804 5256 5310 5364 6418 5472 5526 5580 5634 5688 5742 54 805 5796 5850 5904 5958 6012 6066 6119 6173 6227 6281 54 806 6335 6389 6443 6497 6551 6604 66.58 6712 6766 6820 54 807 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 64 808 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 809 810 7949 908485 8002 8539 8056 8592 8110 8163 8217 8753 8270 8324 8378 8914 8431 8967 54 54 8646 8699 8807 8860 811 9021 9074 9128 9181 9235 9289 9342 9396 9449 9503 54 812 9556 9610 9663 9716 9770 9823 9877 9930 9984 ..37 53 813 910091 0144 0197 0251 0304 0358 0411 0464 0518 0571 53 814 0624 0678 0731 0784 0838 0891 0944 0998 1051 1104 53 815 1158 1211 1264 1317 1371 1424 1477 1530 1584 1637 53 816 1690 1743 1797 1850 1903 1956 2009 2063 2116 2169 53 817 2222 2275 2328 2381 2435 2488 2541 2594 2647 2700 53 818 2753 2806 2859 2913 2966 3019 3072 3125 3178 3231 53 819 3284 3337 3390 34431 3496 3549 3602 3655 3708 3761 63 N. i 1 FT" I 3 1 4 5 1 6 1 7 1 8 1 9 D.t Bb 14 A TABLE OP LOGARITHMS FROM 1 TO 10,000. N. 1 |i!2|3|4|5|6|7|8|9|D. j 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 913814 4343 4872 5400 5927 6454 6980 7506 8030 8555 3867 4396 4925 5453 5980 6507 7033 7558 8083 8607 3920 4449 4977 5505 6033 6559 7085 7611 &135 8659 9183 9706 0228 0749 1270 1790 2310 2829 3348 3865 4383 4899 5415 5931 6445 6959 7473 7986 8498 9010 3973 4502 503O 5558 6085 6612 7138 7663 8188 8712 4026 4555 .5083 5611 6138 6664 7190 7716 8240 8764 9287 9810 0332 0853 1374 1894 2414 2933 3451 3969 4079 4608 5136 5664 6191 6717 7243 7768 8293 8816 4132 4660 5189 5716 6243 6770 7295 7820 8345 8869 4184 4713 5241 5769 6296 6822 734« 7873 8397 8921 9444 9967 0489 1010 1530 2050 2570 3089 3607 4124 4237 4766 5294 5822 6349 6875 7400 7925 8450 8973 9496 ..19 0541 1062 1582 2102 2622 3140 3658 4176 4290 4819 5347 5875 6401 6927 7453 7978 8502 9026 9549 ..71 0593 1114 1634 2154 2674 3192 3710 4228 53 53 53 63 53 53 53 52 52 52 52 52 52 52 52 62 52 52 62 52 52 52 56 51 51 51 61 51 61 51 51 51 51 51 61 51 51 61 51 51 50 60 50 50 50 50 50 50 50 50 50 50 60 60 60 50 60 49 49 49 919078 9601 920123 0645 1166 1686 2206 2725 3244 3762 9130 9653 0176 0697 1218 1738 2258 2777 3296 3814 4331 484S 5354 5879 6394 6908 742Z 7935 8447 8959 9235 9758 0280 0801 1322 1842 2362 2881 3399 3917 4434 4951 5467 5982 6497 7011 7524 8037 8549 9061 9572 ..83 0592 1102 1610 2118 2626 3133 3639 4145 9340 9862 0384 090G 1426 1946 2466 2985 3503 4021 9392 9914 0436 0958 1478 1998 2518 3037 3555 4072 924279 4796 5312 5828 6342 6857 7370 7883 8396 8908 4486 .5003 5518 6034 6548 7062 7576 8088 8601 9112 4538 5054 5570 6085 6600 7114 7627 8140 8652 9163 4589 5106 5621 6137 6651 7165 7678 8191 8703 9215 9725 .236 0745 1254 1763 2271 2778 3285 3791 4296 4641 5157 5673 6188 6702 7216 7730 8242 8754 9266 4693 5209 5725 6240 6754 7268 7781 8293 8805 9317 4744 5261 5776 6291 6805 7319 7832 8345 8857 9368 9879 .389 0898 1407 1915 2423 2930 3437 3943 4448 4953 5457 5960 6463 6966 7468 7969 8470 8970 9469 9968 0467 0964 1462 1958 2455 2950 3445 3939 4433 929419 9930 930440 0949 1458 1966 2474 2981 3487 3993 9470 9981 0491 1000 1509 3017 2524 3031 3538 4044 9521 ..32 0542 1051 1560 2068 2575 3082 3589 4094 4599 5104 5608 6111 6614 7117 7618 8119 8620 9120 9619 0118 0616 1114 1611 2107 2603 3099 3593 4088 9623 .134 0643 1153 1661 2169 2677 3183 3690 4195 4700 5205 5709 6212 6715 7217 7718 8219 8720 9220 9719 0218 0716 1213 1710 2207 2702 3198 3692 4186 9674 .185 0694 1204 1712 2220 2727 3234 3740 4246 4751 5255 5759 6262- 6765 7267 7769 8269 8770 9270 9776 .287 0796 1305 1814 2322 2829 3335 3841 4347 4852 5356 5860 6363 6865 7367 7869 8370 8870 9369 9869 0367 0865 1362 1859 2355 2851 3346 3841 4335 9827 .338 0847 1356 1865 2372 2879 3386 3892 4397 934498 6003 5507 6011 6514 7016 7518 8019 8520 9020 939519 940018 0516 1014 1511 2008 2504 3000 3495 3989 4549 5054 5558 6061 6564 7066 7568 8069 8570 9070 9569 0068 0566 1064 1561 2058 2554 3049 3544 4038 4650 5154 5658 0162 6665 7167 7668 8169 8670 9170 9669 0168 0666 1163 1660 2157 2653 3148 3643 4137 4801 5306 5809 6313 6815 7317 7819 8320 8820 9320 9819 0317 0815 1313 1809 2306 2801 3297 3791 4285 4902 5406 5910 6413 6916 7418 7910 8420 8920 9419 9918 0417 0915 1412 1909 2405 2901 3396 3890 4384 9769 0267 0765 1263 1760 2256 2752 3247 3742 4236 N. 1 1 1 1 2 1 3 1 4 1 5 1 6 t 7 1 8 ' 9 1 D. 1 w: »: A TABLE Of LOGARITHMS FROM 1 TO 10,000. 15 F" lll2l3|4|5|6}7|8|9|D. 1 880 944483 45321 4581 46311 4680 4729 4779 48281 48771 4927 49 1 881 4976 5025 5074 5124 5173 5222 5272 5321 5370 5419 49 1 882 6469 5518 5567 5616 5665 5715 5764 5813 5862 5912 49 883 5961 6010 6059 6108 6157 6207 6256 6306 6354 6403 49 884 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 49 885 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 886 7434 7483 7532 7581 7630 7679 7728 7VVV 7826 7875 49 887 7924 7973 8022 8070 8119 8168 8217 8266 8315 8364 49 888 8413 8462 8511 8560 8609 8657 8706 8756 8804 8853 49 889 890 8902 8951 8999 9048 9097 9146 9634 9195 9683 9244 9731 9292 9780 9341 49 49 949390 9439 9488 9536 9585 9829 891 9878 9926 9975 ..24 ..73 .121 .170 .219 .267 .316 49 892 950365 0414 0462 0511 0560 0608 0657 0706 0764 0803 49 893 0851 0900 0949 0997 1046 1095 1143 1192 1240 12«9 49 894 1338 1386 1435 1483 1532 1580 1629 1677 1726 1776 49 8<*5 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 48 896 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 897 2792 2841 2889 2938 2986 3034 3083 3131 3180 3228 48 898 3276 3325 3373 3421 3470 3518 3566 3615 3663 3711 48 899 900 3760 3808 4291 3856 3905 4387 3953 4001 4049 4532 4098 4580 4146 4628 4194 4677 48 48 954243 4339 4435 4484 901 4725 4773 4821 4869 4918 4966 6014 5062 5110 .5168 48 902 6207 5255 5303 5351 5399 5447 6495 5543 5592 5640 48 903 5688 5736 5784 5832 .5880 5928 .5976 6024 ^073 6120 48 904 6168 6216 6265 6313 6361 6409 6457 6605 6553 6601 48 905 6649 6697 6745 6793 6840 6888 6936 6984 7032 7^80 48 906 7128 7176 7224 7272 7320 7368 7416 7464 751.2 7559 48 907 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 48 908 8086 8134 8181 8229 8277 8325 8373 8421 8468 8516 48 909 910 8564 8612 9089 8659 9137 8707 8755 8803 8850 8898 9375 8946 8994 9471 48 48 959041 9185 9232 9280 9328 9423 911 9518 9566 9614 9661 9709 9757 9804 9852 9900 9947 48 912 9995 ..42 ..90 .138 -185 ,233 .280 .328 .376 .423 48 913 960471 0518 0566 0613 0661 0709 0756 .0804 0851 0899 48 914 0946 0994 1041 1089 1136 1184 1231 1279 1326 1374 47 916 1421 1469 1516 1563 1611 1658 1706 17.53 1801 1848 47 916 1895 1943 1990 2038 208512132 2180 2227 2275 2322 47 917 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 918 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 47 919 920 3316 3363 3410 3882 3457 3504 3977 3552 4024 3599 3646 4118 3693 4165 3741 4212 47 47 963788 3835 3929 4071 921 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 47 922 4731 4778 4825 4872 4919 4966 5013 5061 5108 5155 47 923 5202 5249 5296 5343 5390 5437 5484 5531 5578 5625 47 924 5672 5719 5766 5813 5800 5907 5,954 6001 6048 6095 47 925 6142 6189 6236 6283 6329 6376 64?3 6470 6517 6564 47 926 6611 6658 6705 6752 6799 6845 6892 6939 6986 7033 47 927 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 928 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 929 930 8016 8062 8109 8576 8156 8623 8203 8670 8249 8716 8296 8343 8810 8390 8856 8436 47 47 968483 8530 8763 8903 931 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 932 9416 9463 9509 9556 9602 9649 9695 9742 9789 9835 47 933 9882 9928 9975 ..21 ..68 .114 .161 .207 .254 .300 47 934 970347 0393 0440 0486 0533 0579 0626 0B72 0719 0765 46 935 0812 0858 0904 0951 0997 1044 1090 1137 1183 1229 46 936 1276 1322 1369 1415 1461 1508 1554 1601 1647 1693 46 937 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 46 938 2203 2249 3295 2342 2388 2434 2481 2527 2573 2619 46 939 26661 2712' 2758> 2804' 2851' 2897' 2943' 2989' 3035' 3082' 46 "nT 1 lll2|3l4l5|6|7|8|9|D. 16 A TABLE OP LOGARITHMS FK03I 1 TO 10,000 . N. 1 |l|2|3|4|5|6|7t8l9|D. 1 940 973128 3174 3220 3266 3313 3359! 3405 3451 3497 3543 46 941 3590 3636 3682 3728 3774 3820 3866 3913 3959 4005 46 94a 4051 4097 4143 4189 4235 4281 4327 4374 4420 4466 46 943 4512 4558 4604 4650 4696 4742 4788 4834 4880 4926 46 944 4972 5018 5064 5110 5156 5202 5248 5294 5340 .5386 46 945 5432 5478 6524 5570 5616 5662 5707 5753 5799 5845 46 946 5891 5937 5983 6029 6075 6121 6167 6212 62.58 6.304 46 947 6350 6396 6442 6488 6533 6579 6625 6671 6717 6763 46 948 6808 6854 6900 6946 6992 7037 7083 71291 7175 7220 46 949 950 7266 7312 7358 7403 7449 7495 7541 7586 7632 8089 7678 81.35 46 46 977724 7769 7815 7861 7906 7:^52 7998 8043 951 8181 8226 8272 8317 8363 8409 8454 8.500 8546 8591 46 952 8637 8683 8728 8774 8819 8865 8911 8956 9002 9047 46 953 9093 9138 9184 9230 9275 9.321 9366 9412 9457 9503 46 954 9548 9594 9639 9685 9730 9776 9821 9867 9912 99.58 46 955 980003 0049 0094 0140 0185 0231 0276 0322 0367 0412 45 956 0458 0503 0549 0594 0640 0685 0730 0776 0821 0867 45 957 0912 0957 1003 1048 1093 1139 1184 1229 1275 1320 45 958 1366 1411 1456 1501 1547 1.592 1637 1683 1728 1773 45 959 960 1819 1864 2316 1909 2362 1954 2000 2045 2090 2135 2181 2226 45 45 982271 2407 2452 2497 2543 2588 26.33 2678 961 2723 2769 2814 2859 2904 2949 2994 3040 3085 3130 45 962 3175 3220 ^265 3310 3356 3401 3446 3491 3536 3581 45 963 3626 3671 3716 3762 3807 3852 3897 3942 3987 4032 45 964 4077 4122 4167 4212 4257 4302 4347 4392 4437 4-1:82 45 965 4527 4572 4617 4662 4707 4752 4797 4842 4887 4932 45 966 4977 5022 5067 5112 5157 5202 5247 5292 53,37 5382 45 967 5426 5471 5516 5561 5606 .5651 5696 5741 5786 5830 45 968 5875 5920 5965 6010 6055 6100 6144 6189 6234 6279 45 969 970 6324 6369 6413 6458 6.503 6951 6548 6593 6637 7085 6682 6727 45 986772 6817 6861 6906 6996 7040 7130 7175 45 971 7219 7264 7309 7353 7398 7443 7488 7.532 7577 7622 45 972 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 973 8113 8157 8202 8247 8291 8336 8381 8425 8470 8514 45 974 8559 8604 8648 8693 8737 8782 8826 8871 8916 8960 45 975 9005 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 976 9450 9494 9539 9583 9628 9672 9717 9761 9806 9850 44 977 9895 9939 9983 ..28 ..72 .117 .161 .206 .2.50 .294 44 978 990339 0383 0428 0472 0516 0561 0605 0650 0694 0738 44 979 0783 0827 0871 0916 13.59 0960 1004 1049 1093 1137 1,580 1182 1625 44 44 991226 1270 1315 1403 1448 1492 1 .536 981 1669 1713 1758 1802 1846 1890 1935 1979 2023 2067 44 982 2111 2156 2200 2244 2288 2.333 2377 2421 2465 2.509 44 983 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 44 984 2995 3039 3083 3127 3172 3216 3260 3304 3348 3392 44 985 3436 3480 3524 3568 3613 3657 3701 3745 3789 3833 44 986 3877 3921 3965 4009 4053 4097 4141 4185 4229 4273 44 987 4317 4361 4405 4449 4493 4.537 4581 4625 4669 4713 44 988 4757 4801 4845 4889 4933 4977 .5021 .5065 5108 5152 44 989 990 5196 995635 5240 5679 5284 5328 5767 5372 5416 5460 5504 5942 5547 5591 44 44 5723 5811 5854 .5898 5986 6030 991 6074 6117 6161 6205 6249 6293 6337 6380 6424 6468 44 992 6512 6555 6599 6643 6687 6731 6774 6818 6862 6906 44 993 6949 6993 7037 7080 7124 7168 7212 7255 7299 7343 44 994 7386 7430 7474 7517 7.561 7605 7648 7692 7736 7779 44 995 7823 7867 7910 7954 7998 8041 8085 8129 8172 8216 44 996 8259 8303 8347 8390 8434 8477 8,521 8564 8608 8652 44 997 8695 8739 8782 8826 8869 8913 89.56 9000 9043 9087 44 998 9131 9174 9218 926] 9305 9348 9392 9435 9479 9.522 44 999 9565 9609 96521 9696 97391 9783 9826 98701 9913 9957 43 N. 1 |l|2|3|4|6|6|7|8|9lD. 1 A TABLE OP LOGARITHMIC SINES AND TANGENTS, FOR EVERY DEGREE AND MINUTE OP THE QUADRANT. N.B. The minutes in the left-hand column of each page, increasing downwards, belong to the degrees at the top ,• and those increasing upwards, in the right-hand column, belong to the degrees below. l]b» 18 (0 L>egi *Ce.) A TABLE OF LOGARITHMIC 17 1 Sine 1 D. 1 Cosine | D. 1 Tang. D. 1 Cotang. 1 1 =F 0.000000 10.000000 0.000000 liiiiniie. 60 1 6.463726 601717 000000 00 6,463726 501717 13.5.36274 69 2 764756 293485 000000 00 764756 293483 235244 58 3 940847 208231 000000 00 940847 208231 059153 57 4 7.065786 161517 000000 00 7.065786 161517 12.934214 56 5 162696 131968 000000 00 162696 131969 837304 55 6 241877 111575 9.999999 01 241878 111578 758122 54 7 308824 96653 999999 01 308825 996.53 691175 53 8 366816 85254 999999 01 366817 85254 633183 52 9 417968 76263 9.99999 01 417970 76263 582030 51 10 11 463725 7.505118 68988 999998 01 01 463727 68988 636273 50 49 62981 9.999998 7.505120 62981 12.494880 12 542906 57936 999997 01 542909 57933 457091 48 13 577668 53641 999997 01 577672 53642 422328 47 14 609853 49938 999996 01 609857 49939 390143 46 15 639816 46714 999996 01 639820 46715 360180 45 16 667845 43881 999995 01 667849 43882 332151 44 17 694173 41372 999995 01 694179 41373 305821 43 18 718997 39135 999994 01 719003 39136 280997 42- 19 742477 37127 999993 01 742484 37128 257cl6 41 20 21 764754 35315 999993 9.999992 01 01 764761 35136 235239 40 39 7.785943 33672 7.785951 33673 12.214049 22 806146 32175 999991 01 806155 32176 19.3845 38 23 825451 30805 999990 01 825460 30806 174540 37 24 843934 29547 999989 02 843944 29549 156056 36 25 861662 28388 999988 02 861674 28390 138326 35 26 878695 27317 999988 02 878708 27318 121292 34 27 895085 26323 999987 02 895099 26325 104901 33 28 910879 25399 999986 02 910894 25401 089106 32 29 926119 24538 999985 02 926134 24540 073866 31 30 31 940842 23733 22980 999983 9.999982 02 02 940858 7.955100 23735 22981 059142 30 29 7.955082 12.044900 32 968870 22273 999981 02 968889 22275 031111 28 33 982233 21608 999980 02 982253 21610 017747 27 oM 995198 20981 999979 02 995219 20983 004781 26 35 8.007787 20390 999977 02 8.007809 20392 11.992191 25 36 020021 19831 999976 02 020045 19833 979955 24 37 031919 19302 999975 02 031945 19305 968055 23 38 043501 18801 999973 02 043527 18803 956473 22 39 054781 18325 999972 02 054809 18327 945191 21 40 41 065776 17872 17441 999971 9.999969 02 02 065806 8.076531 17874 17444 934194 20 19 8.076500 11.923469 42 086965 17031 999968 02 086997 17034 913003 18 43 097183 16639 999966 02 097217 16042 902783 17 44 107167 16265 999964 03 107202 16268 892797 16 45 116926 15908 999963 03 116963 15910 883037 15 46 126471 15566 999961 03 126510 15568 873490 14 47 135810 15238 999959 03 135851 15241 864149 13 48 144953 14924 999958 03 144996 14927 855004 12 49 153907 14622 999956 03 153952 14627 846048 11 5: 51 162681 14333 999954 03 03 162727 14336 837273 11.828672 10 9 8.171280 14054 9.999952 8.171328 14057 52 179713 13786 999950 03 179703 13790 820237 8 53 187985 13529 999948 03 188036 13532 811964 7 64 196102 13280 999946 03 196156 13284 803844 6 55 204070 13041 999944 3 204126 13044 795874 5 56 211895 12810 999942 4 211953 12814 788047 4 57 219581 12587 999940 04 219641 12590 780359 3 68 227134 12372 999938 04 227195 12376 772805 2 59 234557 12164 999936 04 234621 12168 765379 1 60 241855 11963 999934 04 241921 11967 758079 n Cosine | Sine 1 Colang. 1 Tang. 1 M. | 89 Degrees. SINES AND TANGE^'TS. (1 DeglGG.] 19 M. Sine ) D. Cosine 1 D. Tang. 1 D. Cotang. 1 "0" 8,241855 11963 9.999934 04 8.241921 11967 11.758079 60 1 249033 11768 999932 04 249102 11772 750898 59 2 256094 11580 999929 04 256165 11584 743835 58 3 263042 11398 999927 04 263115 11402 736885 57 4 269881 11221 999925 04 269956 11225 7300441 56 5 276614 11050 999922 04 276691 11054 723309 55 % 283243 10883 999920 04 283323 10887 716677 54 7 289773 10721 999918 04 28985G 10726 710144 53 8 296207 10565 999915 04 296292 10570 703708 52 9 302546 10413 999913 04 302634 10418 697366 51 10 11 308794 10266 10122 999910 9.999907 04 04 308884 10270 691116 50 49 8,314954 8.315046 10126 11.684954 12 321027 9982 999905 04 321122 9987 678878 48 13 327016 9847 999902 04 327114 9851 672886 47 14 332924 9714 999899 05 333025 9719 666975 46 15 338753 9586 999897 05 338S56J 9590 661144 45 16 344504 9460 999894 05 3446] 0| 9465 655390 44 17 350181 9338 999891 05 350289 9343 6,49711 43 18 355783 9219 999888 05 355895 9224 644105 42 19 ' 361315 9103 999885 05 361430 9108 638570 41 20 21 366777 8990 999882 9.999879 05 05 366895 8995 633105 11.627708 40 39 8,372171 8880 8.372292 8885 22 377499 8772 999876 05 377622 8777 622378 38 23 382762 8667 999873 05 3828891 8672 617111 37 24 387962 8564 999870 05 388092 8570 611908 36 25 393101 8464 999867 05 393234 8470 608766 35 26 398179 8366 999864 05 398315 8371 601685 34 27 403199 8271 999861 05 403338 8276 596662 33 ^8 408161 8177 999858 05 408304 8182 591696 32 ^9 413068 8086 999854 05 413213 8091 586787 31 30 31 417919 8.422717 7996 999851 9.999848 06 06 418068 8.422869 8002 581932 11.577131 30 29 7909 7914 32 427462 7823 999844 06 427618 7830 572382 28 33 432156 7740 999841 06 432315 7745 567685 27 34 486800 7657 999838 06 436962 7663 563038 26 35 441394 7577 999834 06 441560 7583 558440 25 36 4;4594i 7499 999831 06 446110 7505 553890 24 37 450440 7422 999827 06 450613 7428 549387 23 38 454893 7346 999823 06 455070 7352 544930 22 39 459301 7273 999820 06 459481 7279 540519 21 40 41 463665 8.467985 7200 999816 9.999812 06 06 463849 8.468172 7206 536151 20 19 7129 7135 11.531828 42 472263 7060 999809 06 472454 7066 527546 18 43 476498 6991 999805 06 476693 6998 523307 17 44 480693 6924 99980] 06 480892 6931 5191.08 16 45 484848 6859 999797 07 485050 6865 514950 15 46 488963 6794 999793 07 489170 6801 510830 14 47 493040 6731 999790 07 493250 6738 506750 13 48 497078 6669 999788 07 497293 6676 502707 12 49 501080 6608 999782 07 501298 6615 498702 11 50 51 505045 6548 999778 07 07 505267 8.509200 6555 6496 494733 10 9 8.508974 6489 9.999774 11.490800 52 512867 6431 999769 07 513098 6439 486902 8 53 516726 6375 999765 07 516961 6382 483039 7 54 520551 6319 999761 07 5207P9 6326 479210 6 55 524343 6264 999757 07 5245fc> 5 C272 475414 5 56 528102 6211 999753 07 528349 6218 471651 4 67 631828 6158 999748 07 532080 6165 467920 3 58 535523 6106 999744 07 635779 6113 464221 2 59 539186 6055 999740 07 539447 6062 460553 1 60 542819 6004 999735 07 543084 6012 456916 B Cosine j Sine j Cotang. 1 Tang. jAf.j rib Degrees so (2 Degrees.) a TABLE OF LOGARITHMIC T Sine j D. (Cosine j D. T;uiii. D. Cotanc. 1 8.542819 6004 9.999735 07 8.. 543084 6012 1 1.456916 GO 1 546422 5965 999731 07 546691 5962 453309 59 2 549995 6906 999726 07 .550268 6914 449732 58 3 553539 6858 999722 08 553817 5866 446183 57 4 557054 5811 999717 08 557336 5819 442664 56 5 560540 6766 999713 08 560828 6773 439172 .55 6 563999 5719 999708 08 564291 5727 436709 54 7 567431 5674 999704 08 567727 5682 432273 53 8 570836 5630 999699 08 571137 56.38 4288631 52 | 9 574214 5587 999694 08 574520 5595 426480 51 10 11 577566 8.580892 5544 5602 999689 9.999086 08 08 577877 5552 422123 50 49 8.581208 5510 11.418792 12 584193 5460 999680 08 584514 5468 416486 48 13 587469 5419 999675 08 587795 5427 412205 47 14 690721 .5379 999670 08 591061 5387 408949 46 15 593948 5339 999665 08 594283 5347 405717 45 16 597152 5300 999660 08 697492 6308 402508 44 17 600332 5261 999656 08 600677 6270 399323 43 18 603489 5223 9996.50 08 603839 5232 396161 42 19 606623 6186 999645 09 606978 5194 393022 41 20 31 609734 5149 999640 9.999035 09 09 610094 8.613189 51.58 389906 40 39 8.612823 •5112 5121 11.386811 22 615891 5076 999629 09 616262 5085 383738 38 23 618937 5041 999624 09 619313 5050 380687 37 24 621962 5006 999619 09 622343 .5016 377657 36 25 624965 4972 999614 09 626352 4981 374648 35 26 627948 4938 999608 09 628340 4947 371660 34 27 630911 4904 999603 09 631308 4913 368692 33 28 633854 4871 999597 09 634256 4880 365744 32 29 636776 4839 999592 09 637184 4848 362816 31 30 31 639680 8.642563 4806 999586 9.999681 09 09 640093 4816 4784 359907 30 4776 8.642982 11. 3570181 29 1 32 645428 4743 999576 09 645863 4753 364147 28 33 648274 4712 999570 09 648704 4722 351290 27 34 651102 4682 999664 09 651537 4691 348463 26 35 653911 4652 999558 10 654352 4661 345648 25 36 656702 4622 999563 10 067149 4631 342851 24 37 659475 4592 999547 10 659928 4602 340072 23 38 662230 4563 999541 10 662689 4573 837311 22 39 664968 4535 999536 10 665433 4544 334567 21 40 667689 4606 999629 10 668160 4526 331840 20 41 8.670393 4479 9.999524 10 8.670870 4488 11.329130 19 42 673080 4451 999518 10 673563 4461 326437 18 43 675761 4424 999512 10 676239 4434 323761 17 44 678405 4397 999506 10 678900 4417 321100 16 45 681043 4370 999500 10 681544 4380 318456 15 46 683666 4344 999493 10 684172 4364 316828 14 47 686272 4318 999487 10 686784 4328 313216 13 48 688803 4292 999481 10 689381 4303 310619 12 49 691438 4267 999475 10 69-1963 4277 308037 11 50 51 693993 8.696543 4242 9994G9 9.990463 10 11 694529 8.697081 4252 305471 11.302919 10 9 4217 4228 52 699073 4192 999456 11 699617 4203 300383 8 53 701689 4168 999450 11 702139 4179 297861 7 54 704090 4144 999443 11 704646 4155 295354 6 55 706577 4121 999437 11 707140 4132 292860 5 56 709049 4097 999431 11 709618 4108 29i>382 4 57 711507 4074 999424 11 712083 4085 287917 3 58 713962 4051 999418 11 714534 4062 285466 2 59 716383 4029 999411 11 716972 4040 283028 1 60 718800 4006 999404 11 71939G 4017 280604' 01 L Cosine | Sine 1 (Jotaiic. 1 Tang. |m7| 87 Degrees. »♦ SINES AND TANGENTS. (3 DcgrceS.) 21 M. Sine D. Cosine | D. 'J'ans;. 1 D. Cotanir. | | 8.718800 4006 9.999404 11 8.719396 4017 11.280604 60 1 721204 3984 999398 11 721806 3995 278194 59 2 723595 3962 999391 11 724204 3974 275796 58 3 725972 3941 99^384 11 726588 3952 273412 57 4 728337 3919 999378 11 728959 3930 271041 56 5 730688 3898 999371 11 731317 3909 268683 55 6 733027 3877 999364 12 733663 3889 266337 54 7 735354 3857 999357 12 735996 3868 264004 53 8 737667 3836 999350 12 738317 3848 261683 52 9 739969 3816 999343 12 740626 3827 259374 51 10 11 742259 8.744536 3796 999336 12 12 742922 3807 257078 50 49 3776 9.999329 8.745207 3787 11.254793 12 746802 3756 999322 12 747479 3768 252521 48 13 749055 3737 999315 12 749740 3749 250260 47 14 751297 3717 999308 12 751989 3729 248011 46 15 753528 3698 999301 12 754227 3710 245773 45 16 755747 3679 999294 12 756453 3692 243547 44 17 757955 3661 999286 12 758668 3673 241332 43 18 760151 3642 999279 12 760872 3655 239128 42 19 762337 3624 999272 12 763065 3636 236935 41 20 21 764511 8.766675 3606 999265 9.999257 12 12 765246 3618 234754 40 39 3588 8.767417 3600 11.232583 22 768828 3570 9992.50 13 769578 3583 230422 38 23 770970 3553 999242 13 771727 3565 228273 37 24 773101 3.535 999235 13 773866 3548 226134 36 25 775223 3518 999227 13 775995 3531 224005 35 26 777333 3501 999220 13 778114 3514 221886 34 27 779434 3484 999212 13 780222 3497 219778 33 28 781524 3467 999205 13 782320 3480 217680 32 29 783605 3451 999197 13 784408 3464 215592 31 30 31 785675 .3431 999189 13 13 786486 3447 213514 30 29 8.787736 3418 9.999181 8.788554 3431 11.211446 32 789787 3402 999174 13 790613 3414 209387 28 33 791828 3386 999166 13 792662 3399 207338 27 34 793859 3370 999158 13 794701 3383 205299 26 35 795881 3354 999150 13 796731 3368 203269 25 36 797894 3339 999142 13 798752 3352 201248 24 37 799897 3323 999134 13 800763 3337 199237 23 38 801892 3308 999126 13 802765 3322 197235 22 39 40 41 S03876 805852 3293 3278 999118 13 804758 3307 3292 195242 193258 21 20 19 999110 13 13 806742 8.807819 3263 9.999102 8.808717 3278 11.191283 42 809777 3249 999094 14 810683 3262 189317 18 43 811726 3234 999086 14 812641 3248 187359 17 44 813667 3219 999077 14 814589 3233 185411 16 45 815599 3205 999069 14 816529 3219 183471 15 46 817522 3191 999061 14 818461 3205 181539 14 47 819436 3177 999053 14 820384 3191 179616 13 48 821343 3163 999044 14 822298 3177 177702 12 49 823240 3149 999036 14 824205 3163 175795 11 50 51 825130 8.827011 3135 999027 14 14 826103 3150 3136 173897 10 9 3122 9.999019 8.827992 11,172008 52 828884 3108 999010 14 829874 3123 170126 8 53 830749 3095 999002 14 831748 3110 168252 7 54 832607 3082 998993 14 833613 3096 166387 6 55 834456 3069 998984 14 835471 3083 164529 5 56 836297 3056 998976 14 837321 3070 162679 4 57 838130 3043 998967 15 839163 3057 160837 3 68 839956 3030 998958 15 840998 3045 159002 2 59 841774 3017 998950 15 842825 3032 167175 1 60 843585 3000 998941 15 8446441 3019 155356 U Cosine Sine 1 Cotang. 1 1 Tang. 1 M. 86 Degrees. >¥•' 22 (4 Degrees.) a TABLE OF LOGARITHMIC T| Sine D. 1 Cosine | D. | Tang. 1 D. | Cotang. 1 1 8.843585 3005 9.998941 15 8.844644 3019 11.1553661 60 1 845387 2992 998932 16 846465 3007 153645 69 2 847183 2980 998923 15 848260 2996 151740 68 3 848971 2P67 998914 15 850067 2982 149943 57 4 850751 2956 998906 15 851846 2970 148154 56 5 862525 2943 998896 16 853628 2968 146372 55 6 854291 2931 998887 15 866403 2946 144597 54 7 856049 2919 998878 15 867171 2935 142829 .53 8 857801 2907 998869 15 868932 2923 141068 .52 1 9 859546 2896 998860 15 860686 2911 139314 51 10 11 861283 8.863014 288 1 998861 9.998841 16 16 862433 2900 2888 137567 11.135827 60 49 2873 8.864173 12 864738 2861 998832 15 866906 2877 134094 48 13 866456 2860 998823 16 867632 2866 132368 47 14 868165 2839 998813 16 869351 2854 130649 46 15 869868 2828 998804 16 871064 2843 128936 45 16 871565 2817 998796 16 872770 2832 127230 44 17 873255 2806 998785 16 874469 2821 125531 43 18 874938 2795 998776 16 876162 2811 123838 42 19 876615 2786 998766 16 877849 2800 122161 41 20 21 878285 2773 998767 16 16 879629 2789 120471 40 39 8.879949 2763 9.998747 8.881202 2779 11.118798 22 881607 2762 998738 16 882869 2768 117131 38 23 883258 2742 998728 16 884630 2758 115470 37 24 884903 2731 998718 16 886186 2747 113815 36 25 886542 2721 998708 16 887833 2737 112167 35 26 888174 2711 998699 16 889476 2727 110624 34 27 889801 2700 998089 16 891112 2717 108888 33 28 891421 2690 998679 16 892742 2707 107258 32 29 893035 2680 998669 17 894366 2697 105634 31 30 31 894643 2670 2660 998659 17 17 895984 2687 104016 30 29 8.896246 9.998649 8.897696 2677 11.102404 32 897842 2661 998639 17 899203 2667 100797 28 33 899432 2641 998629 17 900803 2658 099197 27 34 901017 2631 998619 17 902398 2648 097692 26 35 902696 2622 998609 17 903987 2638 096013 25 36 904169 2612 998599 17 905570 2629 094430 24 37 905736 2603 998589 17 907147 2620 092863 23 38 907297 2593 998578 17 908719 2610 091281 22 39 40 41 908863 910404 2584 998668 17 910286 911846 8.913401 2601 2592 089716 088164 21 20 T9 2575 998558 IT 17 8.911949 2666 9.998548 2683 11.086699 42 913488 2556 998537 17 914961 2674 085049 18 43 915022 2547 998527 17 916495 2665 083.506 17 44 916660 2538 998516 18 918034 2566 081966 16 45 918073 2629 998506 18 919668 2547 080432 15 46 919691 2620 998496 18 921096 2638 078904 14 47 921103 2612 998486 18 922619 2630 077381 13 48 922610 2603 998474 18 924136 2621 075864 12 49 924112 2494 998464 18 925649 2512 074351 11 50 61 926609 8.927100 2486 24*77 998453 18 18 927166 8.928668 2603 072844 10 9 9.998442 2495 11.071342 52 928587 2469 998431 18 930166 2486 069845 8 53 930068 2460 998421 18 931647 2478 068363 7 54 931544 2462 998410 18 933134 2470 066866 6 55 933015 2443 998399 18 934616 2461 065384 5 56 934481 2436 998388 18 936093 2453 U63907 4 57 935942 2427 998377 18 937566 2445 062435 3 58 937398 2419 998366 18 939032 2437 060968 2 59 93885C 2411 998355 18 940494 2430 069506 1 60 940296 2403 998344 18 941962 2421 068048 1 Cosine 1 1 Sine 1 1 Cotang. 1 1 Tang. |M.| 85 Degrees. SINES AND TANGENTS. (5 Degrees.) 23 M Sine D. Cosine 1 D. 1 Tang. 1 D. 1 Ootang;. | | 8.940296 2403 9.998344 19 8.941952 2421 11.058048 60 1 941738 2394 998333 19 943404 2413 056596 59 2 943174 2387 998322 19 944S52 2405 055148 58 3 944606 2379 998311 19 946295 2397 053705 57 4 946034 2371 998300 19 947734 2390 052266 56 ?i 947456 2363 998289 19 949168 2382 050832 55 h 948874 2355 998277 19 950597 2374 049403 54 V 950287 2348 99826G 19 952021 2366 047979 53 p 951696 2340 998255 19 953441 2360 046559 52 c 953100 2332 998243 19 954856 2351 045144 51 10 ll' 954499 8.955894 2325 998232 9.998220 19 19 956267 8.957674 2344 043733 50 49 2317 2337 11.042326 15 957284 2310 998209 19 959075 2329 040925 48 U- 958670 2302 998197 19 960473 2323 039527 47 14 960052 2295 998186 19 961866 2314 038134 46 If' 961429 2288 998174 19 963255 2307 036745 45 If. 962801 2280 998163 19 964639 2300 035361 44 ir 964170 2273 998151 19 966019 2293 033981 43 18 965534 2266 998139 20 967394 2286 032606 42 19 966893 2259 998128 20 968766 2279 0312.34 41 20 21 968249 8.969600 2252 998116 9.998104 20 20 9701.33 8.971496 2271 029867 40 39 2244 2265 11.028504 22 970947 2238 998092 20 972855 2257 027145 38 23 972289 2231 998080 20 974209 2251 025791 37 24 973628 2224 998068 20 975560 2244 024440 36 25 974962 2217 998056 20 976906 2237 023094 35 26 976293 2210 998044 20 978248 2230 021752 34 27 977619 2203 998032 20 979586 2223 020414 33 28 978941 2197 998020 20 980921 2217 019079 32 29 980259 2190 998008 20 982251 2210 017749 31 30 31 981573 8.982883 2183 997996 9.997984 20 20 983577 ,2204 016423 30 29 2177 8.984899 2197 11.015101 32 984189 2170 997972 20 986217 2191 013783 28 33 985491 2163 997959 20 987532 2184 012468 27 34 986789 2157 997947 20 988842 2178 011158 26 35 988083 2150 997935 21 990149 2171 009851 25 36 989374 2144 997922 21 991451 2165 008549 24 37 990660 2138 997910 21 992750 21.58 007250 23 38 991943 2131 997897 21 994045 2152 005955 22 39 993222 2125 997885 21 995337 2146 004663 21 40 41 994497 8.995768 2119 2112 997872 9.997860 21 21 996624 2140 003376 20 19 8.997908 21,34 11.002092 42 997036 2106 997847 21 999188 2127 000812 18 43 998299 2100 997835 21 9.000465 2121 10.999535 17 44 999560 2094 997822 21 001738 2115 998262 16 45 9.000816 2087 997809 21 003007 2109 996993 15 46 002069 2082 997797 21 004272 2103 995728 14 47 003318 2076 997781 21 005534 2097 994466 13 48 004563 2070 997771 21 006792 2091 993208 12 49 005805 2064 997758 21 008047 2085 991953 11 50 51 007044 9.008278 2058 997745 21 21 009298 9.010546 2080 990702 10.989454 9 2052 9.997732 2074 52 009510 2046 997719 21 011790 2068 988210 8 53 010737 2040 997706 21 013031 2062 986969 7 54 011962 2034 997693 22 014268 2056 985732 6 55 013182 2029 997680 22 015502 2051 984498 5 56 014400 2023 997667 22 016V32 2045 983268 4 57 015613 2017 997654 22 017959 2040 982041 3 58 016824 2012 997641 22 019183 2033 980817 2 59 018031 2006 997628 22 020403 2028 979597 1 60 019235 2000 997614 22 021620 2023 978380 ^ Cosine | 1 Sine 1 1 Cotang. 1 1 Tang. |M. I 84 Degrees. 24 (6 Degrees.) a FABLE OF LOGABITHMIC ;^ Sine D. Cosine | 1). Tang. D. 1 Cotang. 1 1 9.019235 200p 1995 9.997614 22 9.021620 2023 10.978380 60 1 020435 997601 22 022834 2017 977166 59 2 021632 1989 997588 22 024044 2011 975956 58 3 022825 1984 997574 22 025251 2006 974749 57 4 024016 1978 997561 22 026455 2000 973545 56 5 025203 1973 997547 22 027655 1995 972345 55 6 026386 1967 997534 23 028852 1990 971148 54 7 027567 1962 997520 23 030046 1985 969954 53 8 028744 1957 997507 23 031237 1979 968763 52 9 029918 1951 997493 23 032425 1974 e'67575 51 10 11 031089 1947 997480 23 23 033609 9.034791 1969 966391 50 49 9.032257 1941 9.997466 1964 10.965209 12 033421 1936 997452 23 035969 1958 964031 48 13 034582 1930 997439 23 0.37144 1953 962856 47 14 035741 1925 997425 23 038316 1948 961684 46 15 036896 1920 997411 23 039485 1943 960515 45 16 038048 1915 997397 23 040651 1938 959349 44 17 039197 1910 997383 23 041813 1933 9.58187 43 18 040342 1905 997369 23 042973 1928 957027 42 19 041485 1899 997355 23 044130 1923 955870 41 20 21 042625 1894 997341 9.997327 23 24 045284 9.046434 1918 9.54716 40 39 9.043762 1889 1913 10.9,53566 22 044895 1884 997313 24 047582 1908 952418 38 23 046026 1879 997299 24 048727 1903 951273 37 24 047154 1875 997285 24 049869 1898 950131 36 25 048279 1870 997271 24 051008 1893 948992 35 26 049400 1865 997257 24 052144 1889 947856 34 27 050519 1860 997242 24 053277 1884 946723 33 28 051635 1855 997228 24 054407 1879 945593 32 29 052749 1850 997214 24 055535 1874 944465 31 30 053859 1845 997199 24 056659 1870 943341 30 31 054966 1841 9.997185 24 9.057781 1865 10.942219 29 32 056071 1836 997170 24 058900 1869 941100 28 33 0.57172 1831 997156 24 060016 1855 939984 27 34 058271 1827 997141 24 061130 1851 938870 26 35 059307 1822 997127 24 062240 1846 937760 25 36 060460 1817 997112 24 063348 1842 936652 24 37 061551 1813 997098 24 064453 1837 935547 23 38 002639 1808 997083 25 065556 1833 934444 22 39 063724 1804 997068 25 066655 1828 933345 21 40 41 064806 9.065885 1799 997053 25 25 067752 1824 932248 20 19 1794 9.997039 9.068846 1819 10.931154 42 066962 1790 997024 25 069938 1815 930062 18 43 068036 1786 997009 25 071027 1810 928973 17 44 069107 1781 996994 25 072113 1806 927887 16 45 070176 1777 996979 25 073197 1802 926803 15 46 071242 1772 996964 25 074278 1797 925722 14 47 072306 1768 996949 25 075356 1793 924644 13 48 073366 1763 996934 25 076432 1789 923568 12 49 074424 1759 996919 25 077505 1784 922495 11 50 51 075480 9.076.533 1755 996904 25 25 078576 1780 1776 921424 10 9 1 1750 9.996889 9.079644 10.920356 52 077583 1746 996874 25 1 080710 1772 919290 8 53 078031 1742 996858 25 081773 1767 9 J 8227 7 54 079676 1738 996843 25 082833 1763 917167 6 55 080719 1733 996828 25 083891 1759 916109 5 56 081759 i 1729 996812 26 084947 1755 915053 4 57 08279V 1 1725 996797 26 086000 1751 914000 3 58 083832 1721 996782 26 087050 1747 912950 2 59 084864 1717 996766 26 088098 1 1743 911902 1 60 085894 1713 996751 26 089144 ! 1738 9108.56 Cosine Sine 1 1 «;o.au{^. 1 1 Tang. 1 M. | 83 Degrees. siivEs AND TANGLKTs. ^7 Dcgrees.) 25 M. Sine D. Cosine 1 D. Tang. 1 D. Cotang. ( 1 9.085894 1713 9.996751 26 9.089144 1738 10.910850 -60 1 086922 1709 996735 26 090187 1734 909813 59 2 087947 1704 996720 26 091228 1730 908772 58 3 088970 1700 996704 26 092266 1727 907734 57 4 089990 1696 996688 26 093302 1722 906698 56 6 091008 1692 996673 26 094336 1719 905664 55 6 092024 1688 996657 26 095367 1715 9046'}3 54 7 093037 1684 996641 26 096395 1711 903605 53 8 094047 1680 996625 26 097422 1707 902578 52 9 095056 1676 996610 26 098446 1703 901554 51 10 11 096062 1673 996594 9.996578 26 27 099468 9.100487 1699 900532 50 49 9.097065 1668 1695 10.899513 12 098066 1665 996562 27 101504 1691 898496 48 13 099065 1661 996546 27 102519 1687 897481 47 14 J00062 1657 996530 27 103532 1684 896468 46 15 101056 1653 996514 27 104542 1680 895458 45 16 102048 1649 996498 27 105550 1676 894450 44 17 103037 1645 996482 27 106556 1672 893444 43 18 104025 1641 996465 27 107559 1669 892441 42 19 105010 1638 996449 27 108560 1665 891440 41 20 21 105992 1634 1630 996433 27 27 109559 1661 890441 40 39 9.106973 9.996417 9.110556 1658 10.889444 22 107951 1627 990400 27 111551 1654 888449 38 23' 108927 1023 996384 27 112543 1650 887457 37 24 109901 1619 996368 27 113533 1646 886467 36 25 110873 1616 996351 27 114521 1643 885479 35 26 111842 1612 996335 27 115507 1639 884493 34 27 112809 1608 996318 27 116491 1636 883509 33 28 113774 1605 996302 28 117472 1632 882528 32 29 114737 1601 996285 28 118452 1629 881548 31 30 31 115698 1597 1594 996269 9.996252 28 28 119429 1625 880571 10.879.596 30 29 9.116656 9.120404 1622 32 117613 1590 996235 28 121377 1618 878623 28 33 118567 1587 996219 28 122348 1615 877652 27 34 119519 1583 996202 28 123317 1611 876683 26 35 120469 1580 996185 28 124284 1607 875716 25 36 121417 1576 996168 28 125249 1604 874751 24 37 122362 1573 996151 28 126211 1601 873789 23 38 123306 1569 996134 28 127172 1597 872828 22 39 124248 1566 996117 28 128130 1594 871870 21 40 41 125187 1562 996100 9.996083 28 29 129087 9.130041 1591 870913 20 19 9.126125 1559 1587 10.8699.59 42 127060 1556 996066 29 130994 1584 869006 18 43 127993 1552 996049 29 131944 1581 868056 17 44 128925 1549 996032 29 132893 1577 867107 16 45 129854 1545 996015 29 1338.39 1574 866161 15 46 130781 1542 995998 29 1.34784 1571 865216 14 47 131706 1539 995980 29 135726 1567 864274 13 48 . 132630 1535 995963 29 136667 1564 863333 12 49 133551 1532 995946 29 137605 1561 862395 11 50 51 134470 9.135387 1529 1525 995928 9.995911 29 29 138542 1558 861458 10 9 9.139476 1555 10.860524 52 136303 1522 99.5894 29 140409 1561 859591 8 53 137216 1519 995876 29 141340 1548 858660 7 54 138128 1516 995859 29 142269 1545 857731 6 55 139037 1512 995841 29 143196 1542 856804 6 56 139944 1509 995823 29 144121 1539 855879 4 57 140850 1506 995806 29 145044 1535 854956 3 58 141754 1503 995788 29 145966 1532 854034 2 59 142655 1500 995771 29 146885 1.529 1526 853115 1 60 143555 1496 995753 29 147803 852197 3 Cosine { 1 bine | roiai.t. 1 .! Tung |M;| Cc Degreei. 26 (8 Degrees.; a table op logarithmic Ti Sine D. 1 Cosine | D. | Tan^j. 1 D. 1 Ct)tajig. 1 1 9.143556 1496 9.996753 30- 9.147803 1526 10.8621971 60 1 1 144453 1493 995735 30 148718 1523 851282 59 2 145349 1490 995717 30 149632 1520 850368 68 3 146243 1487 995699 30 150544 1517 849466 67 4 147136 1484 995681 30 161454 1514 848546 66 5 148026 1481 995664 30 152363 1511 847637 55 G 148915 1478 995646 30 153^9 1508 846731 54 7 149802 1475 995628 30 164174 1505 845826 53 8 150686 1472 995610 3a 155077 1502 844923 52 9 151569 1469 995591 30 155978 1499 844022 21 10 11 152451 1466 995573 30 30 156877 9.157775 1496 843123 50 49 9 153330 1463 9.995655 1493 10.842225 12 154208 1460 995537 30 158671 1490 841329 48 13 155083 1467 995519 30 159565 1487 840435 47 14 155957 1454 995501 31 160467 1484 839543 46 15 156830 1451 995482 31 161347 1481 838653 46 16 167700 1448 996464 31 162236 1479 837764 44 17 158569 1445 995446 31 163123 1476 836877 43 18 169435 1442 995427 31 164008 1473 835992 42 19 100301 1439 996409 31 164892 1470 835108 41 20 21 161164 9.162026 1436 1433 996390 31 31 165774 1467 834226 40 39 9.996372 9.166654 1464 10.833346 22 162885 1430 995353 31 167532 1461 832468 38 23 163743 1427 99.5334 31 168409 1468 831591 37 24 164600 1424 995316 31 169284 1466 830716 36 25 165454 1422 995297 31 170157 1453 829843 35 26 166307 1419 995278 31 171029 1460 828971 34 27 167159 1416 995260 31 171899 1447 828101 33 28 168008 1413 995241 32 172767 1444 827233 32 29 168856 1410 995222 32 173634 1443 826366 31 30 31 169702 9.170547 1407 995203 32 32 174499 1439 825501 30 29 1405 9.996184 9,175362 1436 10.824638 32 171389 1402 995165 32 176224 1433 823776 28 33 172230 1399 995146 32 177084 1431 822916 27 34 173070 1396 995127 32 177942 1428 822058 26 35 173908 1394 996108 32 178799 1426 821201 25 36 174744 1391 996089 32 179665 1423 820346 24 37 175578 1388 995070 32 180508 1420 819492 23 38 176411 1386 996051 32 -181360 1417 818640 22 39 177242 1383 995032 32 182211 1416 817789 21 40 41 178072 1380 1377 995013 32 32 183059 1412 816941 20 19 9.178900 9.994993 9.183907 1409 10.816093 42 179726 1374 994974 32 184752 1407 815248 18 43 180551 1372 994955 32 185597 1404 814403 17 44 181374 1369 994935 32 186439 1402 813561 16 45 182196 1366 994916 33 187280 1399 812720 15 46 183016 1364 994896 33 188120 1396 811880 14 47 183834 1361 994877 33 188958 1393 811042 13 48 184651 1369 994857 33 189794 1391 810206 12 49 185466 1356 994838 33 190629 1389 809371 11 50 51 186280 1353 994818 33 33 191462 9.192294 1386 1384 808538 10.807706 10 9 9.187092 1351 9.994798 52 187903 1348 994779 33 193124 1381 80C876 8 53 188712 1346 994759 33 193953 1379 80604? 7 54 189519 1343 994739 33 194780 1376 805220 6 55 190325 1341 994719 33 195606 1374 804394 6 56 191130 1338 994700 33 196430 1371 803570 4 57 191933 1336 994680 33 197253 1369 802747 3 58 192734 1333 994660 33 198074 1366 801926 2 59 193534 1330 994640 33 198894 1364 801106 7 60 194332 1328 994620 33 199713 1361 800287 1 Cosine 1 1 «'"« 1 1 Cotang. 1 Tang. \ M. 81 Degreea. SINES AND TANGENTS. (9 Degrees.) 27 M. Sine j D. 1 Cosine | D. Tang. D. Cotang. 1 9.194332 1328 9.994620 33 9,199713 1361 10.800287 60 1 1951S9 1326 994600 33 200629 1359 799471 59 2 195926 1323 994580 33 201345 1356 798655 5S 3 196719 1321 994560 34 202159 1354 797841 57 4 197511 1318 994540 34 202971 1352 797029 56 5 198302 1316 994519 34 203782 1349 796218 55 6 199091 1313 994499 34 204592 1347 796408 64 7 199879 1311 994479 34 205400 1345 794600 53 8 200666 1308 994459 34 206207 1342 793793 52 9 201461 1306 994438 34 2^7013 1340 792987 51 10 11 202234 1304 1301 994418 34 34 207817 1338 792183 50 49 9,203017 9.994397 9,208619 1335 10,791381 12 203797 1299 994377 34 209420 1333 790580 48 13 204577 1296 994357 34 210220 1331 789780 47 14 205354 1294 994336 34 211018 1328 788982 46 15 206131 1292 994316 34 211815 1326 788185 45 16 206906 1289 994295 34 212611 1324 787389 44 17 207679 1287 994274 36 213405 1321 786595 43 18 208452 1285 994254 35 214198 1319 785802 42 19 209222 1282 994233 35 214989 1317 785011 41 20 21 209992 9,210760 1280 994212 35 35 215780 9,216668 1315 784220 40 39 1278 9,994191 1312 10,783432 22 211526 1275 994171 35 217366 1310 782644 38 23 212291 1273 994160 36 218142 1308 781858 37 24 213055 1271 994129 35 218926 1305 781074 36 25 213818 1268 994108 35 219710 1303 780290 35 26 214579 1266 994087 35 220492 1301 779608 34 27 215338 1264 994066 35 221272 1299 778728 33 28 216097 1261 994045 35 222052 1297 , 777948 32 ^ ^16854 1259 994024 35 •222830 1294 777170 31 30 31 217609 1257 1255 994003 35 35 223606 129,2 776394 30 29 9-218363 9.993981 9.224382 1290 10.775618 32 219116 1253 993960 35 225156 1288 774844 28 33 -215868 1250 993939 35 226929 1286 774071 27 34 22D618 1248 993918 35 226700 1284 773300 26 35 221367 1246 993896 36 227471 1281 772529 25 36 222115 1244 993875 36 228239 1279 771761 24 37 222861 1242 993854 36 229007 1277 770993 23 38 223606 1239 993S32 36 229773 1275 770227 22 39 224349 1237 993811 36 230539 1273 769461 21 40 41 225092 1235 993789 36 36 231302 1271 768698 20 19 9.225833 1233 9.993768 9-232065 1269 10.767935 42 226573 1231 993746 36 232826 1267 767174 18 43 227311 1228 993725 36 233586 1265 766414 17 44 228048 1220 993703 36 234345 1262 766665 16 45 228784 1224 993681 36 235103 1260 764897 15 46 229518 1222 993660 36 235859 1258 764141 14 47 230252 1220 993638 36 236614 1256 763386 13 48 230984 1218 993616 36 237368 1254 762632 12 49 231714 1216 993594 37 238120 1252 761880 11 50 51 232444 1214 993572 37 37 238872 1250 76112S 10.760378 10 9 9.233172 1212 9.993550 .9.239622 1248 52 233899 1209 993528 37 240371 1246 769629 8 63 234625 1207 993506 37 241118 1244 758882 7 54 235349 1205 993484 37 241865 1242 758135 6 55 236073 1203 993462 37 242610 1240 757390 5 56 236795 1201 993440 37 243354 1238 756646 4 57 237515 1199 993418 37 244097 1236 766903 3 58 238235 1197 993396 37 244839 1234 7.55161 2 59 238958 1195 993374 37 246678 1232 754421 1 60 230670 1193 993351 37 246319 1230 753681 Ll Cosine i Sine I Cotang. Tang. M. | 30 Degrees. 58 (10 Degr ees. ) A TABLE OP LOGARITHMIC^ M. Sine 1 D. Cosine 1 D. Tang. 1 D. 1 Cotang. 1 1 9.239670 1193 9.993351 37 9.246319 1230 10.753681 60 1 240386 1191 993329 37 247057 1228 752943 59 2 241101 1189 993307 37 247794 1226 752206 68 3 241814 1187 993285 37 248530 1224 751470 57 4 242526 1185 9S3262 37 249264 1222 750736 56 6 243237 1183 993240 37 249998 1220 760002 55 6 243947 1181 993217 38 2507.30 1218 749270 54 7 244656 1179 993195 38 261461 1217 748639 53 8 245363 1177 993172 38 252191 1215 747809 52 9 246069 1175 993149 38 252920 1213 747080 51 10 11 246775 1173 1171 993127 38 38 263648 1211 746362 50 49 9.247478 9.993104 9.254374 1209 10.746626 12 248181 1169 993081 38 256100 1207 744900 48 13 248883 1167 993059 38 266824 1205 744176 47 14 249583 1165 993036 38 266547 1203 743463 46 15 250282 1163 993013 38 257269 1201 742731 45 16 250980 1161 992990 38 2.57990 1200 742010 44 17 251677 1159 992967 38 258710 1198 741290 43 18 252373 1158 992944 38 259429 1196 740571 42 19 253067 1156 992921 38 260146 1194 739854 41 20 21 253761 9.254453 1154 1152 992898 38 38 260863 9.261578 1192 739137 40 39 9.992875 1190 10.738422 22 255144 1160 992852 38 262292 1189 737708 38 23 255834 1148 992829 39 263005 1187 736995 37 24 266523 1146 992806 39 263717 1185 736283 36 25 257211 1144 992783 39 264428 1183 735572 35 26 257898 1142 992759 39 265138 1181 734862 34 27 258583 1141 993736 39 266847 1179 734153 33 28 259268 1139 992713 39 266565 1178 733445 32 29 259951 1137 992690 39 267261 1176 732739 31 30 31 260633 1135 1133 992666 39 39 267967 1174 732033 30 29 9.261314 9.992643 9.268671 1172 10.731329 32 261994 1131 992619 39 269376 1170 730625 28 33 262673 1130 992596 39 270077 1169 729923 27 34 263351 1128 992572 39 270779 1167 729221 26 35 264027 1126 992549 39 271479 1165 728521 25 36 264703 1124 992526 39 272178 1164 727822 24 37 265377 1122 992501 39 272876 1162 727124 23 38 266051 1120 992478 40 273573 1160 726427 •22 39 266723 1119 992464 40 274269 1158 726731 21 40 41 267395 9.268065 1117 992430 40 40 274964 1167 726036 20 19 1115 9.992406 9.275658 1155 10.724342 42 268734 1113 992382 40 276351 1153 723649 18 43 269402 1111 992359 40 277043 1151 722967 17 44 270069 1110 992335 40 277734 1150 722266 16 45 270735 1108 992311 40 278424 1148 721676 15 46 271400 1106 992287 40 279113 1147 720887 14 47 272064 1105 992263 40 279801 1145 720199 13 48 272726 1103 992239 40 280488 1143 719512 12 49 273388 1101 992214 40 281174 1141 718826 11 50 51 274049 9.274708 1099 992190 40 40 281858 9.282642 1140 718142 10 9 1098 9.992106 1138 10.717458 52 275367 1096 992142 40 283225 1136 716775 8 53 276024 1094 992117 41 283907 1135 716093 7 64 276681 1092 992093 41 284588 1133 715412 6 55 277337 1091 992069 41 285268 1131 714732 5 56 277991 1089 992044 41 286947 1130 714053 4 57 278644 1087 992020 41 286624 1128 713376 3 58 279297 1086 991996 41 287301 1126 712699 2 59 279948 1084 991971 41 287977 1125 712023 1 60 280599 1082 991947 41 288662 1123 711348 1 Cosine | Sine . 1 1 Colang. 1 1 Tang. 1 M. j 79 Degrees, SINES AND TANGENTS (11 Degrees. ) 29 M. 1 Sine 1 D. 1 Co^ne 1 D. 1 Tang. 1 D. 1 Cotang. 1 1 T 9.280599 1082 9.991947 41 9.288652 1123 10.7113481 60 1 1 281248 1081 991922 41 289326 1123 710674 59 2 281897 1079 991897 41 289999 1120 710001 58 3 282544 1077 991873 41 290671 1118 709329 57 4 283190 1076 991848 41 291342 1117 708658 56 5 283836 1074 991823 41 292013 1115 707987 55 & 284480 1072 991799 41 292682 1114 707318 54 7 285124 1071 991774 42 293350 1112 706650 53 8 285766 1069 991749 42 294017 1111 705983 52 9 286408 1067 991724 42 294684 1109 705316 51 10 11 287048 9,287687 1066 991699 9.991674 42 42 295349 1107 704651 50 49 1064 9.296013 1106 10.703987 12 288326 1063 991649 42 296677 1104 703323 48 13 288964 1061 991624 42 297339 1103 702661 47 14 289600 1059 991599 42 298001 1101 701999 46 15 290236 1058 991574 42 298662 1100 701338 45 16 290870 1056 991549 42 299322 1098 700678 44 17 291504 1054 991524 42 299980 1096 700020 43 18 293137 1053' 991498 42 300638 1095 699362 42 19 292768 1051 991473 42 301295 1093 698705 41 20 293399 1050 991448 42 301951 1092 698049 40 21 9.294029 1048 9.991422 42 9.302607 1090 10.697393 39 22 294658 1046 991397 42 303261 1089 696739 38 23 295280 1045 991372 43 303914 1087 696086 37 24 295913 1043 991346 43 304567 1086 695433 36 25 296539 1042 991321 43 305218 1084 694782 35 26 297164 1040 991295 43 305869 1083 694131 34 27 297788 1039 991270 43 306519 1081 693481 33 28 298412 1037 991244 43 307168 1080 692832 32 29 299034 1036 991218 43 307815 1078 692185 31 30 31 299655 9.300276 1034 991193 9.991167 43 43 308463 9.309109 1077 691537 30 29 1032 1075 10.690891 32 300895 1031 991141 43 309754 1074 690246 28 33 301514 1029 991115 43 310398 1073 689602 27 34 302132 1028 991090 43 811042 1071 688958 26 35 302748 1026 991064 43 311685 1070 688315 25 36 303364 1025 991038 43 312327 1068 687673 24 37 303979 1023 991012 43 312967 1067 687033 23 38 304593 1022 990986 43 313608 1005 686392 22 39 305207 1020 990960 43 314247 1064 685753 21 40 305819 1019 990934 44 314885 1062 685115 20 41 9.806430 1017 9.990908 44 9.315523 1061 10.684477 19 42 307041 1016 990882 44 316159 1060 683841 18 43 307650 1014 990855 44 316795 1058 683205 17 44 308259 1013 990829 44 317430 1057 683570 16 45 308867 1011 990803 44 318064 1055 681936 li 46 309474 1010 990777 44 318697 1054 681303 47 310080 1008 990750 44 319329 1053 680671 13 48 310685 1007 990724 44 319961 1051 680039 12 49 311289 1005 990697 44 320592 1050 679408 11 50 311893 1004 990671 44 321222 1048 678778 10 51 9.312495 1003 9.990644144 9.321851] 1047 10.678149 9 62 313097 1001 990618144 322479 1045 677521 8 53 313698 1000 990591 44 323106 1044 676894 7 54 314297 •998 9905651 44 323733 1043 676267 6 55 314897 997 990538: 44 324358! 1041 675642 5 56 315495 996 9905 ir 45 324983 1040 675017 4 57 316092 994 990485 45 325607; 1039 6743931 3 58 316689 993 990458 45 326231' 1037 6737691 2 59 317284 991 990431 45 326853 1036 673147! 1 60 317879 990 990404 45 327475 1035 672525* Cosine | Bine | Uotaiig. j Tang, j Cc 78 Degrees. 30 (12 Degrees.) a TABLE OF LOGAEITHMIC M. 1 Sine 1 D. 1 Cosine 1 D. 1 Tang. 1 D. 1 Cota.ig. 1 1 T 9.317879 990 9.990404 45 9.327474 1035 10.672526 60 1 318473 988 990378 45 328095 1033 671905 59 2 319066 987 990351 45 328715 1032 671285 58 3 319658 986 990324 45 329334 1030 670666 57 4 320249 984 990297 45 329953 1029 670047 56 5 320840 983 990270 45 330570 1028 669430 55 6 ,321430 982 990243 45 331187 1026 668813 54 7 322019 980 990215 45 331803 1025 668197 53 8 322607 979 990188 45 332418 1024 667582 52 9 323194 977 990161 45 333033 1023 666967 51 10 11 323780 976 990134 45 46 333646 1021 666354 50 49 9.324366 975 9.990107 9.334259 1020 10.665741 12 324950 973 990079 46 334871 1019 665129 48 13 325534 972 990052 46 335482 1017 664518 47 14 326117 970 990025 46 336093 1016 663907 46 15 326700 969 98-^997 46 336702 1015 663298 45- 16 327281 968 98997a 46 337311 1013 662689 44 17 327862 966 989942 46 337919 1012 662081 43 18 328442 965 989915 46 338527 1011 661473 42 19 329021 964 989887 46 339133 1010 660867 41 20 21 329599 962 961 989860 46 46 339739 9.340344 1008 1007 660261 40 39 9.330176 9.989832 10.659656 22 330753 960 989804 46 340948 1006 659052 38 23 331329 958 989777 46 341552 1004 658448 37 24 331903 957 989749 47 342155 1003 657845 36 25 332478 956 989721 47 342757 1002 657243 35 26 333051 954 989693 47 343358 1000 656642 34 27 333624 953 989665 47 843958 999 656042 33 28 334195 952 989637 47 344558 998 655442 32 29 334766 950 989609 47 345157 997 654843 31 30 31 335337 949 989582 9.989553 47 47 345755 9.346353 996 654245 30 29 9.335906 948 994 10.653647 32 336475 946 989525 47 346949 993 653051 28 33 337043 945 989497 47 347545 992 652455 27 34 337610 944 989469 47 348141 991 651859 26 35 338176 943 989441 47 348735 990 651265 650671 25 36 338742 941 989413 47 349329 988 24 37 339306 940 989384 47 349922 987 650078 23 38 339871 939 989356 47 350514 986 649480 22 39 340434 937 989328 47 351106 985 648894 21 40 41 340996 936 989300 9.989271 47 47 351697 983 982 648303 20 19 9.341558 935 9.352287 10.647713 42 342119 934 989243 47 352876 981 647124 18 43 342679 932 989214 47 353465 980 646535 17 44 343239 931 989186 47 354053 979 645947 16 45 343797 930 989157 47 354640 977 645360 15 46 344355 929 989128 48 355227 976 644773 14 47 344912 927 989100 48 355813 975 644187 13 48 345469 926 989071 48 356398 974 643602 12 49 346024 925 989042 48 356982 973 643018 11 50 51 346579 924 989014 9.988985 48 48 357566 9.358149 971 970 642434 10 9 9.347134 922 10.641851 52 347687 921 988956 48 358731 969 641269 8 53 348240 920 988927 48 359313 968 640687 7 54 348792 919 988898 48 359893 967 640107 6 55 349343 917 988869 48 360474 966 639526 5 56 349893 916 988840 48 361053 965 638947 4 57 350443 915 988811 49 361632 963 638368 3 58 350992 914 988782 49 362210 962 637790 2 59 351540 913 988753 49 362787 961 637213 1 60 352088 911 988724 49 363364 960 636636 l: Cosine 1 Sine 1 Colaiig. Tang 1 M. 1 77 Degrees. SINES AND TANGENTS. (13 Degrees.) 31 J4. 1 Sine D. 1 Cosine 1 I). 1 Tanp. 1 D. 1 Ootang. 1 1 ^ 9.3520S8 911 9.988724 49 9.363364 960 10.636636 60 1 352635 910 988695 49 363940 959 636060 59 2 353181 909 988666 49 364515 958 635485 58 3 353726 908 988636 49 365090 957 634910 57 4 354271 907 988607 49 365664 955 6343361 56 I 5 354815 905 988578 49 366237 954 633763 55 6 355358 904 988548 49 366810 953 633190 54 7 355901 903 988519 49 367382 952 632618 53 8 356443 902 988489 49 367953 951 632047 52 9 356984 901 988460 49 368524 950 631476 51 10 11 357524 899 988430 9.988401 49 49 369094 949 630906 50 49 9.358064 898 9.369663 948 10.630337 12 358603 897 988371 49 370232 946 629768 48 13 359141 896 988342 49 370799 945 629201 47 14 359678 895 988312 50 371367 944 628633 46 15 360215 893 988282 50 371933 943 623067 45 16 360752 892 988252 50 372499 942 627501 44 17 361287 891 988223 50 373064 941 626936 43 18 361822 890 988193 /50 373629 940 626371 42 19 362356 889 988163 50 374193 939 625807 41 20 21 362889 9.363422 888 988133 50 50 374756 938 625244 40 39 887 9.988103 9.375319 937 10.624681 22 363954 885 988073 50 375881 935 624119 38 23 364485 884 988043 50 376442 934 623558 37 24 365016 883 988013 50 377003 933 622997 36 25 365546 882 987983 50 377563 932 622437 35 26 366875 881 987953 50 378122 931 621878 34 27 366604 880 987922 50 378681 930 621319 33 28 367131 879 987892 50 379239 929 620761 32 29 367659 877 987862 50 379797 928 620203 31 30 31 368185 876 987832 9.987801 51 51 380354 927 619046 30 29 9.368711 875 9.380910 926 10.619090 32 369236 874 987771 51 381466 925 618534 28 33 369761 873 987740 51 382020 924 617980 27 34 370285 872 987710 51 382575 923 617425 26 35 370808 871 987679 51 383129 922 616871 25 36 371330 870 987649 51 383682 921 616318 24 37 .371852 869 987618 51 384234 920 615766 23 38 372373 867 987588 51 384786 919 615214 22 39 372894 866 987557 51 385337 918 614663 21 40 41 373414 865 987526 9.987496 51 51 385888 917 614112 20 19 9.373933 864 9.386438 915 10.613562 42 374452 863 987465 51 386987 914 613013 18 43 374970 862 987434 51 387536 913 612464 17 44 375487 861 987403 52 388084 912 611916 16 45 376003 860 987372 52 388631 911 611369 15 46 376519 859 987341 52 389178 910 610822 14 47 377035 858 997310 52 389724 909 610276 13 48 377549 857 987279 52 390270 908 609730 12 49 378003 856 987248 52 390815 907 609185 11 50 51 378577 854 987217 9.987186 52 52 391360 906 608640 10.608097 10 9 9.379089 853 9.391903 905 52 379601 852 987155 52 392447 904 607553 8 63 380113 851 987124 52 392989 903 607011 7 54 380624 850 987092 52 393531 902 606469 6 55 381134 849 987061 52 394073 901 . 605927 5 56 381643 848 987030 52 394614 900 605386 4 67 382152 847 986998 52 395154 899 604846 3 58 382661 846 986967 52 395694 898 604306 2 59 383168 845 986936 52 396233 897 603767 1 60 383675 844 986904 52 396771 896 603229 ^ Cosine 1 1 Sine 1 1 Cotang. 1 Tang. |M. 1 76 Degrees. 32 (14 Degrees.; a TABLE OF LOGARITHMIC pn Sine 1 D. Cosine | D. Tang. 1 D. 1 Cotang. 1 1 9.383675 844 9.986904 52 9.396771 896 10.603229 60 1 384182 843 986873 53 397309 896 602691 59 2 384687 842 986841 53 397846 895 602154 58 3 385192 841 986809 53 398383 894 601617 57 4 385697 840 986778 53 398919 893 601081 66 5 386201 839 986746 53 399455 892 600545 55 6 386704 838 986714 53 399990 891 600010 54 7 387207 83T 986683 53 400524 890 599476 53 8 387709 836 986651 53 401058 889 598942 52 9 388210 835 986619 53 401591 888 598409 51 10 11 388711 834 986587 53 53 402124 887 597876 50 49 9.389211 833 9.986555 9.402656 886 10.597344 12 389711 832 986523 53 403187 885 696813 48 13 390210 831 986491 53 403718 884 596282 47 14 390708 830 986459 53 404249 883 695751 46 15 391206 828 986427 53 404778 882 595222 45 16 391703 827 986395 53 405308 881 694692 44 17 392199 826 986363 54 405836 880 594164 43 18 392695 825 986331 54 406364 879 593636 42 19 393191 824 986299 54 406892 878 593108 41 20 21 393685 823 986266 9.986234 54 54 407419 877 592581 40 39 9.394179 822 9.407945 876 10.592055 22 394673 821 986202 54 408471 875 591529 38 23 395166 820 986169 54 408997 874 691003 37 24 395658 819 986137 54 409521 874 690479 36 25 396150 816 986104 54 410045 873 689955 35 26 396641 817 986072 54 410569 872 589431 34 27 3971-32 817 986039 54 411092 871 588908 33 28 397621 816 986007 54 411615 870 588385 32 29 398111 815 985974 54 412137 869 587863 31 30 31 398600 814 985942 9.985909 54 55 412658 868 687342 30 29 9.399088 813 9.413179 867 10.586821 32 399575 812 - 985876 55 413699 866 586301 28 33 400062 811 985843 55 414219 865 585781 27 34 400549 810 985811 55 414738 864 585262 26 35 401035 809 985778 55 41.5257 864 584743 26 36 401520 808 . 985745 55 415775 863 684225 24 37 402005 807 985712 55 416293 862 5837-07 23 38 402489 806 985679 55 416810 861 583190 22 39 402972 805 985646 55 417326 860 682674 21 40 41 403455 804 985613 55 55 417842 859 582158 20 19 9.403938 803 9.985580 9.418358 858 10.581642 42 404420 802 985547 55 418873 857 581127 18 43 404901 801 985514 55 419387 856 580613 17 44 405382 800 985480 55 419901 855 580099 16 45 405862 799 985447 55 420415 855 579585 15 46 406341 798 985414 56 420927 854 579073 14 47 406820 .797 985380 56 421440 853 678560 13 48 407299 796 985347 56 421952 852 678048 12 49 407777 795 985314 56 422463 851 677537 11 50 51 408254 794 985280 9.985247 56 56 422974 850 677026 10 9 9.408731 794 9.423484 849 10.576516 52 409207 793 985213 56 423993 848 576007 8 53 409682 792 985180 56 424503 848 675497 7 54 410157 791 985146 56 425011 847 674989 6 55 410632 790 985113 56 425519 846 674481 5 56 411106 789 985079 56 426027 845 673973 4 57 411579 788 985045 56 426534 844 573466 3 58 412052 787 985011 56 427041 843 572959 2 59 412524 786 984978 56 427547 843 672453 1 ^ 412996 785 984944! 56 428052 842 571948 ^ Cosine Sine 1 j Uoiang. 1 Tang =,M.| 75 Degrees. SINES AND TANGENTS. (15 Degrees.) 33 M. 1 Sine 1 D. 1 Cosine 1 D. 1 Tang. 1 D. 1 Cotang. 1 1 ^ 9.412996 785 9.984944 57 9.428052 842 10.571948 60 1 413467 784 984910 57 428557 841 571443 59 2 413938 783 984876 57 429062 840 670938 58 3 414408 783 984842 57 429566 839 670434 57 4 414878 782 984808 57 430070 838 569930 56 6 415347 781 984774 57 430573 838 569427 55 6 415815 780 984740 57 431075 837 668926 54 7 416283 779 994706 57 431677 836 568423 53 8 416751 778 984672 57 432079 835 667921 62 9 417217 777 984637 57 432680 834 567420 51 10 11 417684 9. 41-8150 770 984603 9.984569 57 57 433080 833 566920 50 49 775 9.433580 832 10.. 566420 12 418615 774 984535 57 434080 832 565920 48 13 419079 773 984500 57 434579 831 565421 47 14 419544 773 984466 57 435078 830 564922 46 16 420007 772 984432 58 435576 829 564424 45 16 420470 771 984397 58 436073 828 563927 44 17 420933 770 984363 58 436570 828 563430 43 18 421395 769 984328 58 437067 827 662933 42 19 421857 768 984294 58 437563 826 662437 41 20 21 422318 9 422778 767 984259 58 58 438059 9.438554 825 824 661941 40 39 767 9.984224 10.561446 22 423238 766 984190 58 439048 823 560952 38 23 423697 765 984155 58 439543 823 560457 37 24 424156 764 984120 58 440036 822 569964 36 25 424615 763 984085 58 440529 821 559471 36 26 425073 762 984050 58 44102a 820 568978 34 27 425530 761 984015 58 441514 819 558486 33 28 425987 760 983981 58 442006 819 557994 32 29 426443 760 983946 58 442497 818 567503 31 30 31 426899 9.427354 759 758 983911 9.983875 58 58 442988 817 657012 30 29 9.443479 816 10.656521 32 427809 757 983840 59 443968 816 566032 28 33 428263 756 983805 59 444458 815 555542 27 34 428717 755 983770 59 444947 814 555053 26 35 429170 754 983735 59 446435 813 554566 25 36 429623 753 983700 59 445923 812 5.54077 24 37 430075 752 983664 59 446411 812 653589 23 38 430527 752 983629 59 446898 811 553102 22 39 430978 751 983594 59 447384 810 652616 21 40 41 431429 750 983558 59 59 447870 9.448356 809 552130 20 19 9.431879 749 9.983523 809 10.561644 42 432329 749 9S3487 59 448841 808 651159 18 43 432778 748 983452 69 449326 807 650674 17 44 433226 747 9834161 59 449810 806 550190 16 45 433675 746 983381 59 450294 806 549706 16 46 434122 745 983345 59 450777 805 549223 14 47 434569 744 983309 59 451260 804 648740 13 48 435016 744 983273 60 451743 803 548257 12 49 435462 743 983238 60 452225 802 647775 11 60 51 435908 9.436353 742 983202 60. 60 452706 802 647294 10.546813 10 9 741 9.983166 9.453187 801 52 436798 740 983130 60 463668 800 646332 8 53 437242 740 983094 60 454148 799 645862 7 54 437686 739 983058 60 454628 799 545372 6 55 438129 738 98.3022 60 455107 798 544893 6 56 438572 737 982986 60 456686 797 544414 4 57 439014 736 982950 60 456064 796 543936 3 58 439456 736 982914 60 456542 796 543468 2 59 439897 735 982878 60 457019 795 542981 1 60 440338 734 982842 60 457496 794 642504 1 1 Cosine j" 1 Sine 1 j Colang. 1 1 Tang. 1 M. | 74 Degrees. E 34 (16 Degrees.) a TABLE OP LOGARITHMIC "m" Sine D. 1 Cosine | d:i Tang. D. Cotang. 1 1 "01 9.440338 734 9.982842 60 9.457496 794 10.542504 60 1 440778 733 982805 60 457973 793 542027 59 2 441218 732 982769 61 458449 793 541551 68 3 441658 731 982733 61 458925 792 541075 57 4 442096 731 982696 61 459400 791 540600 56 6 442535 730 982660 61 459875 790 540125 55 6 442973 729 982624 61 460349 790 539651 54 7 443410 728 982587 61 460823 789 539177 53 8 443847 727 982551 61 461297 788 538703 52 9 444284 727 982514 61 461770 788 538230 fl 10 11 444720 726 982477 61 61 462242 9.462714 787 786 537758 50 49 9.445155 725 9.982441 10.537286 IS 445590 724 982404 61 463186 785 536814 48 13 446025 723 982367 61 463658 785 536342 47 14 446459 723 982331 61 464129 784 535871 46 15 446893 722 982294 61 464599 783 535401 45 16 447326 721 982257 61 465069 783 634931 44 17 447759 720 982220 62 465539 782 534461 43 18 448191 720 982183 62 466008 781 633992 42 19 448623 719 982146 62 466476 780 533524 41 20 21 449054 718 982109 62 62 466945 780 533055 10.532587 40 39 9.449485 717 9.982072 9.467413 779 22 449915 716 982035 62 467880 778 532120 38 T3 450346 716 981998 62 468347 778 531'653 37 24 450775 715 981961 62 468814 777 531186 36 25 451204 714 981924 62 469280 776 530720 35 26 451632 713 981886 6-^ 409r4tJ 775 530254 34 27 462060 713 981849 62 470211 775 529789 33 28 452488 712 981812 62 470676 774 529324 32 29 452915 711 981774 62 471141 773 528859 31 30 31 453342 710 981737 9.981699 62 63 471605 9.472068 773 528395 30 29 9.453768 710 772 10.527932 32 454194 709 981662 63 472532 771 527468 28 33 454619 708 981625 63 472995 771 527005 27 34 455044 707 981587 63 473457 770 526543 26 35 455469 707 981549 63 473919 769 626081 26 36 455893 706 981512 63 474381 769 525619 24 37 456316 705 981474 63 474842 768 525158 23 38 456739 704 981436 63 475303 767 524697 22 39 457162 704 981399 63 475763 767 524237 21 40 457584 703 981361 63 476223 766 623777 20 41 9.458006 702 9.981323 63 9.476683 765 10.523317 19 42 458427 701 981285 63 477142 765 622858 18 43 458848 701 981247 63 477601 764 522399 17 44 459268 700 981209 63 478059 763 521941 16 45 459688 699 981171 63 478517 763 521483 15 46 460108 698 981133 64 478975 762 521025 14 47 460527 698 981095 64 479432 761 520568 13 48 460946 697 981057 64 479889 761 620111 12 49 461364 696 981019 64 480345 760 519655 11 50 51 461782 695 980981 64 64 480801 759 519199 10.518743 10 9 9.462199 695 9.980942 9.481257 759 52 462616 694 980904 64 481712 758 618288 8 53 463032 693 980866 64 482167 757 517833 7 54 463448 693 980827 64 482621 757 517379 6 55 463864 692 980789 64 483075 756 516925 5 56 464279 691 980750 64 483529 755 516471 4 57 464694 690 980712 64 483982 755 616018 3 58 465108 690 980673 64 484435 754 515566 2 59 465522 689 980635 64 484887 753 615113 1 60 465935 688 980596 64 485339 753 514661 Cosine 1 1 Sine 1 1 Colang. 1 Tang. 1 M. | 73 Degrees. SINES AND TANGENTS. (^17 Degrees.) 35 M. Sine D. Cosine 1 D. 1 Tang. 1 D. Cotang. 1 1 9.465935 688 9.980596 64 9.485339 765 10.614661 60 1 466348 688 980558 64 485791 •£62 514209 59 2 466761 687 980519 65 486242 761 613768 68 3 467173 686 980480 65 486693 751 513307 67 4 467585 685 980442 65 487143 750 612857 56 5 467996 685 980403 65 487593 749 512407 55 6 466407 684 980364 65 488043 749 511957 64 7 468817 683 980325 65 488492 748 511508 53 8 469227 683 980286 66 488941 747 511059 52 9 469637 682 980247 65 489390 747 510610 51 10 11 470046 681 980208 9.980169 65 65 489838 9.490286 746 610162 10.609714 60 49 9.470455 680 746 12 470863 680 980130 65 490733 746 509267 48 13 471271 679 980091 65 491180 744 508820 47 14 471679 678 980052 65 491627 744 608373 46 15 472086 678 980012 65 492073 743 507927 45 16 472492 677 979973 65 492519 743 607481 44 17 472898 676 979934 66 492965 742, 507035 43 18 473304 676 979895 66 493410 741 506590 42 19 473710 675 979856 66 493854 740 506143 41 20 21 474115 674 979816 9.979776 66 66 494299 740 505701 10.505257 40 39 9.474519 674 9.494743 740 22 474923 673 979737 66 496186 739 604814 38 23 475327 672 979697 66 495630 738 604370 37 24 475730 672 979658 66 496073 737 603927 36 25 476133 671 979618 66 490515 737 503485 35 26 476536 670 979579 66 496957 736 503043 34 27 476938 669 979539 66 497399 736 502601 33 28 477340 669 979499 66 497841 735 602159 32 29 477741 668 979459 66 498282 734 501718 31 30 31 478142 667 979420 9.979380 66 66 498722 9.499163 734 733 501278 30 29 9.478542 667 10.500837 32 478942 666 979340 66 499603 733 600397 28 33 *'479342 665 979300 67 500042 732 499958 27 34 479741 665 979260 67 500481 731 499519 26 35 480140 664 979220 67 600920 731 499080 26 36 480539 663 979180 67 501359 730 498641 24 37 480937 663 979140 67 501797 730 498203 23 38 481334 662 979100 67 502236 729 497765 22 39 481731 661 979059 67 502672 728 497328 21 40 41 482128 661 979019 9.978979 67 67 503109 728 496891 10.496454 20 19 9.482525 660 9.503546 727 42 482921 659 978939 67 503982 727 496018 18 43 483316 659 978898 67 604418 726 495582 17 44 483712 658 978868 67 504854 725 495146 16 45 484107 657 978817 67 505289 726 494711 15 46 484501 657 978777 67 505724 724 494276 14 47 484895 656 978736 67 506159 724 493841 13 48 485289 655 978696 68 506693 723 493407 12 49 485682 655 978655 68 607027 722 492973 11 60 51 486075 9.486467 654 978615 9.978574 68 68 507460 9.507893 722 721 492540 10 9 653 10.492107 62 486860 653 978533 68 608326 721 491674 8 53 487251 652 978493 68 608759 720 491241 7 64 487643 651 978452 68 509191 719 490809 6 65 488034 651 978411 68 509622 719 490378 6 56 488424 650 978370 68 510064 718 489946 4 57 488814 650 978329 68 610485 718 489516 3 68 489204 649 978288 68 510916 717 . 489084 2 59 489693 648 978247 68 511346 716 488664 1 60 489982 648 978206 68 611776 716 488224 1 _l Cosine 1 1 Sine 1 1 Cotang. 1 'Jang. 1 M.~ 72 Degrees. 30 (1 B Degr ees.) A TABLE OF LOGARITHMIC "m" Sine D. 1 Cosine 1 I). Tang. D. { Cotang. 1 9.489982 648 9. 9782061 68 9.511776 716 10.488224 60 1 490371 648 978165 68 512206 716 487794 59 2 490759 647 978124 68 512635 715 487365 68 3 491147 646 978083 69 513064 714 486936 57 4 491535 646 978042 69 513493 714 486507 56 5 491922 645 978001 69 513921 713 486079 56 6 492308 644 977959 69 614349 713 485651 64 7 492695 644 977918 69 514777 712 485223 53 8 493081 643 977877 6y 515204 712 484796 52 9 493466 642 977835 69 515631 711 484369 51 10 11 493851 642 977794 69 69 510057 9.516484 710 483943 10.483516 50 49 9.494236 641 9.977752 710 12 494621 641 977711 69 616910 709 483090 48 13 495005 640 977669 69 517335 709 482665 47 14 495388 639 977628 69 517761 708 482239 46 15 495772 639 977586 69 518185 708 481815 46 16 496154 638 977544 70 618610 707 481390 44 17 496537 637 977503 70 519034 706 480966 43 18 496919 637 977461 70 519458 706 480542 42 19 497301 636 977419 70 619882 705 480118 41 20 21 497682 636 977377 ■70 70 620305 705 479695 10.479272 40 39 9.498064 635 9.977335 9.520728 704 22 498444 634 977293 70 621151 703 478849 38 23 498825 634 977251 70 621673 703 478427 37 24 499204 633 977209 70 621996 703 478005 36 25 499584 632 977167 70 522417 702 477583 35 26 499963 632 977125 70 522838 702 477162 34 27 500342 631 977083 70 523259 701 476741 33 28 .500721 631 977041 70 523680 701 476320 32 29 501099 630 976999 70 524100 700 475900 31 30 31 501476 629 976957 70 70 524520 699 475480 10.475061 30 29 9.501854 629 9.976914 9.624939 699 32 502231 628 976872 71 625359 698 474641 28 33 502607 628 976830 71 625778 698 4742^2 27 34 502984 627 976787 71 526197 697 473803 26 35 503360 026 976745 71 626615 697 473385 25 36 503735 626 976702 71 527033 696 472967 24 37 504110 625 976660 71 527451 696 472549 23 38 504485 625 976617 71 527868 695 472132 22 39 504860 624 976574 71 528285 695 471715 21 40 41 605234 623 976532 71 71 628702 694 693 471298 20 19 9.505608 623 y. 976489 9.629119 0.470881 42 505981 622 976446 71 629535 693 470465 18 43 506354 622 976404 71 529950 693 470050 17 44 506727 621 976361 71 530366 692 469634 16 45 507099 620 976318 71 630781 691 469219 16 46 507471 620 976275 71 531196 691 468804 14 47 507843 619 976232 72 531611 690 468389 13 48 508214 619 976189 72 632025 690 467975 12 49 508585 618 976146 72 532439 689 467561 11 50 61 508956 9.509326 618 976103 9.976060 72 72 532853 689 467147 10.466734 10 9 617 9.533266 688 52 509696 616 976017 72 533679 688 466321 8 53 510065 616 976974 72 634092 687 465908 7 54 510434 615 976930 72 534504 687 465496 6 55 510803 615 975887 72 534916 686 465084 6 56 511172 614 975844 72 535328 686 464672 4 57 511540 613 975800 72 635739 685 464261 3 58 511907 613 975757 72 636150 685 463850 2 59 612275 612 975714 72 536561 684 463439 1 60 512642 612 975670 72 536972 684 463028 ~ Cosine Sine 1 Cotang. Tang. |M.| 71 Degrees. SINES AND TANGENTS. (19 Degrees.) 37 M. 1 Siae 1 D. 1 Cosine | D. 1 Tang. 1 D. 1 Cotang. 1 1 ~o" 9.512642 612 9.975670 73 9.536972 684 10.463028 60 1 513009 611 975627 73 637382 683 462618 59 2 513375 611. 975583 73 537792 683 462208 58 3 513741 610 975539 73 538202 682 461798 67 4 514107 609 975496 73 538611 682 461389 66 5 514472 609 975452 73 539020 681 460980 56 6 514837 608 975408 73 539429 681 460571 54 7 515202 608 9r5365 73 539837 680 460163 53 8 515566 607 975321 73 .540245 680 459755 62 9 516930 607 975277 73 540653 679 459347 61 10 11 516294 606 605 975233 73 73 541061 679 458939 10.458532 50 49 9.516657 9.975189 9.541468 678 12 517020 605 975145 73 541875 678 458125 48 13 517382 604 975101 73 .542281 677 457719 47 14 517745 604 975057 73 542688 677 457312 46 15 618107 603 975013 73 543094 676 456906 45 16 518468 603 974969 74 543499 676 456501 44 17 518829 602 974925 74 543906 675 456095 43 18 619190 601 974880 74 544310 675 455690 42 19 519551 601 974836 74 544715 674 455285 41 20 21 519911 600 974792 9.974748 74 74 545119 674 673 454881 10.464476 40 39 9.520271 600 9.545524 22 520631 699 974703 74 545928 673 454072 38 23 520990 599 974659 74 546331 672 453669 37 24 65J1349 598 974614 74 546735 672 453265 36 25 5^1707 598 974570 74 547138 671 452862 35 26 522066 597 974525 74 547540 671 452460 34 27 522424 596 974481 74 547943 670 452057 33 28 522781 696 974436 74 548345 670 451655 32^ 29 523138 595 974391 74 548747 669 451253 31 30 31 523495 595 974347 9.974302 76 75 549149 9.549560 669 450851 30 29 9.523852 594 668 10.4504.50 32 - 524208 694 974257 75 ^549951 668 450049 28 33 524564 593 974212 75 550352 667 449648 27 34 ^24920 593 974167 75 650752 667 449248 26 35 62B275 592 974122 75 651152 666 448848 25 36 6256.30 591 974077 75 551652 666 448448 24 37 525984 591 974032 75 651952 665 448048 23 i 38 526339 590 973987 75 55235-1 665 447649 22 39 526693 590 973942 75 552750 665 447250 21 40 41 527046 589 973897 75 75 553149 9.653548 664 664 446851 20 19 9.527400 589 9.973852 10.446452 42 527753 588 973807 75 553946 663 446054 18 43 528105 588 973761 75 654344 663 445056 17 44 528458 587 973716 76 554741 662 445259 16 45 528810 587 973671 76 555139 662 444801 15 46 629161 586 973625 76' 555536 681 444464 14 47 529513 586 973580 76 555933 661 444067 13 48 529864 585 973535 76 556329 660 443671 12 49 .^^302 15 585 973489 76 656725 660 443275 11 50 51 .'530565 584 973444 76 76 557121 659 442879 10 9 9.530915 684 9.973398 9.557517 659 10.442483 52 531265 583 973352 76 557913 659 442087 8 63 531614 582 973307 76 558308 658 441692 7 54 531963 582 973261 76 558702 658 441298 6 55 532312 581 973215 76 559097 657 440903 6 56 632661 581 973169 76 559491 657 440609 4 57 633009 580 973124 76 559885 656 440115 3 58 533357 580 973078 76 560279 656 439721 2 59 533704 579 973032 77 560673 655 439327 1 60 534052 578 972986 77 561066 655 438934 n Cosine Sine 1 Cotang. 1 Tang. 1 M. | Dd 70 Degrees. 'M 38 (20 Degrees.) a TABLE OF LOGARITHMIC M. 1 Sine D. 1 Cosine | D. Tang. 1 D. Cotang. 1 1 "o" 9.534052 578 9.9729861 77 9.561066 655 10.438934 60 1 534399 577 972940 77 561459 654 438541 59 2 534745 677 972894 77 561851 654 438149 58 3 535092 577 972848 77 562244 653 437756 57 4 535438 676 972802 77 662636 653 437364 56 5 535783 576 972755 77 663028 653 436972 55 6 536129 575 972709 77 563419 652 436681 54 7 536474 574 972663 77 563811 652 436189 53 < 8 536818 574 972617 77 564202 651 435798 62 9 537163 573 972570 77 564592 651 435408 51 10 11 537507 9.537851 573 972524 9.972478 77 77 564983 650 435017 50 49 572 9.565373 650 10.434627 12 538194 672 972431 78 565763 649 434237 48 13 538538 571 972385 78 566153 649 433847 47 14 638880 571 972338 78 566542 649 433458 46- 15 53^9223 570 972291 78 566932 648 433068 45 16 539565 570 972245 78 567320 648 432680 44 17 639907 569 972198 78 567709 647 432291 43 18 540249 669 972151 78 568098 647 431902 42 19 540590 568 972105 79 568486 646 431514 41 20 21 540931 9.541272 .568 972058 78 78 568873 646 431127 40 39 567 9.972011 9.569261 64o 10.430739 22 541613 567 971964 78 569648 645 430352 38 23 541953 566 971917 78 570035 645 429965 37 24 542293 566 971870 78 570422 644 42957& 36 25 542632 565 971823 78 570809 644 429191 35 26 542971 565 971776 78 671195 643 428805 34 27 543310 564 971729 79 571581 643 428419 33 28 543649 564 971682 79 571967 642 428033 32 29 543987 563 971635 79 572352 642 427648 31 30 544325 563 971588 79 572738 642 427262 30 31 9.544663 562 9.971540 79 9.573123 641 10.426877 29 32 545000 562 971493 79 573507 641 426493 28 33 545338 561 971446 79 573892 640 426108 27 34 545674 561 971398 79 574276 640 425724 26 35 546011 560 971.%1 79 574660 630 425340 25 36 546347 660 971303 79 576044 639 424956 24 37 546683 559 &71256 79 575427 639 424573 23 38 547019 559 971208 79 575810 638 424190 22 39 547354 558 971161 79 576193 638 423807 21 40 41 547689 658 971113 9.971066 79 80 576576 637 423424 20 19 9.548024 557 9.576958 637 10.423041 42 548359 557 971018 80 577341 636 422659 18 43 548693 656 970970 80 577723 636 422277 17 44 549027 556 970922 80 578104 636 421896 16 45 549360 655 970874 80 578486 635 421514 15 46 549693 655 970827 80 578867 635 421133 14 47 550026 554 970779 80 579248 634 420752 13 48 550359 554 970731 80 579629 634 420371 12 49 550692 553 970683 80 580009 634 419991 11 50 61 551024 9.551356 553 970635 9.970586 80 80 580389 633 419611 10.419231 10 9 - 552 9.580769 633 52 551687 652 970538 80 581149 632 418851 8 63 552018 552 970490 80 581528 632 418472 7 54 552349 551 970442 80 581907 632 418093 6 55 552680 551 970394 80 582286 631 417714 5 56 553010 560 970345 81 582665 631 417335 4 57 5.'>3341 550 970297 81 583043 630 416957 3 58 553670 649 970249 81 683422 630 416578 2 59 554000 649 970200 81 583800 629 416200 1 60 654329 548 970152 81 584177 629 415823 1 Cosine 1 Sine 1 Ootaug. 1 Tang. ■mT Degrees. •^ •LY- SINES AND TANGENTS. (21 Degrees.) 39 T\ Sine 1 D. 1 Cosine | D. \ Tang. 1 D. 1 Cotang. 1 1 : 9.554329 548 9,970152 81 9.584177 629 10.415823 60 1 554658 548 970103 81 584555 629 415445 59 2 554987 647 970055 81' 584932 628 415068 58 3 555315 547 970006 81 585309 628 414691 57 4 555643 546 969957 81 585686 627 414314 56 6 555971 546 969909 81 586062 627 413938 55 6 556299 545 969860 81 586439 627 413.561 54 7 656626 545 969811 81 586815 626 413185 53 8 556953 644 969762 8i 587190 626 412810 52 9 557280 544 969714 81 587566 625 41-2434 51 10 11 557606 543 969665 81 82 587941 625 625 • 412059 10.411684 50. 49 9.557932 543 9.969616 9.588316. 12 558258 543 969667 82 588691 624 411309 48 13 658583 542 969518 82 589066 624 410934 47 14 558909 642 969469 82 589440 623 410560 46 15 569234 641 969420 82 589814 623 410186 45 t 16 569568 641 969370 82 590188 623 409812 44 17 56988a 640 969321 82 690562 622 409438 43 18 660207 540 969272 82 590935 622 409065 42 19 660531 539 969223 82 591308 622 408692 41 20 21 660865 639 969173 82 82 591681 621 408319 40! 39 9.561178 638 9.969124 9.592054 621 10.407946 22 661501 638 969075 82 592426 620 407574 38 > 23 661824 637 969025 82 592798 620 407202 37' 24 662146 63? 968976 82 693170 619 406829 36^ 25 662468 536 968926 83 593542 619 406458 35 26 662790 536 9688.77 88 693914 618 406086 34 27 663112 536 968827 83 594285 618 405715 33 28 663433 535 968777 83 594656 618 405344 32- 29 663756 535 968728 83 595027 617 404973 31 30 31 664075 534 968678 9.968628 83 83 695398 617 404602 30 29 9,564396 534 9.595768 617 10.404232 32 664716 533 968578 83 596138 616 403862 28 33 665036 .633 968528 83 596508 616 403492 27. 34 605356 '532 968479 83 596878 616 403122 26^ 35 565676 632 968429 83 597247 615 402753 25 36 665996 631 968379 83 597616 615 402384 24 37 666314 631 968329 83 697985 615 402015 23 38 566632 531 968278 83 5(98354 614 401646 22^ 39 666951 630 968228 84 598722 614 401278 21 40 41 567269 630 968178 9.908128 84 «4 599091 613 400909 20 19 9.567587 629 9.599459 613 10.400541 42 667904 529 968078 84 599827 613 400173 18 43 568222 528 968027 84 600194 612 399806 17 44 668539 528 967977 84 6005612 612 i399438 16 45 568866 '.528 !967927 84 600929 611 399071 15 46 5691:72 ,527 -967870 84 601296 611 398704 14 47 669488 ,527 967826 84 601602 611 •398338 13 48 569804 526 96777S 84 60202*] 610 397971 12 49 670120 526 967725 84 602395 610 397605 11 50 51 57043S 625 967674 9.967624 84 84 60^61 9.603127 610 609 397239 10.396873 10 9 9.570751 525 52 571066 524 96757? f 84 603492 609 396507 8 53 571361 524 967525 85 60385€ 609 396142 7 54 571 69£ ) 623 96747] 85 60422S 008 395777 6 55 57200t > 623 •96742] 85 60458€ 608 395412 5 56 57232C i 623 96737( ) 85 60495S 607 395047 4 57 672636 ) 622 9673 U 85 605317 607 394683 3 58 67295( ) 622 •967^6^ 1 85 605685 607 394318 2 59 673261 J 621 96721" r 85 60604t 606 393954 1 60 67357f ) 621 9671061 85 60641C 606 393590 1 Cosine 1 1 Sine 1 1 Cotang. 1 1 Tang. 1 M. fi8 Degrees- 40 (22 Degrees.) a table op logarithmic M. I Sine 1 D. Cosine | D. 1 Tang. D. Cotang. ' ^ 9.573575 521 9.967166 85 9.606410 606 10.393590 "6(f 1 573888 520 967115 85 606773 606 393227 59 2 574200 520 967064 85 607137 605 392863 58 3 574512 519 967013 85 607500 605 392500 57 4 574824 519 966961 85 607863 604 392137 56 5 675136 619 966910 85 608225 604 391775 55 6 575447 518 966859 85 608588 604 391412 54 7 575758 518 966808 85 608950 603 391050 53 8 676069 517 966756 86 609312 603 390688 52 9 576379 517 966705 86 609674 603 390326 51 10 11 576689 516 966663 9.966602 86 86 610036 9.610397 602 389964 10.389603 60 49 9.576999 616 602 12 577309 616 966550 86 610769 602 389241 48 13 577618 515 966499 86 611120 601 388880 47 14 577927 516 966447 86 611480 601 388520 46 15 578236 514 966395 86 611841 601 388169 45 16 578545 514 966344 86 612201 600 387799 44 17 578853 513 966292 86 612561 600 387439 43 18 579162 513 966240 86 612921 600 387079 42 19 579470 513 966188 86 613281 599 386719 41 20 21 579777 512 966136 9 966085 86 87 613641 599 386359 40 39 9.680085 612 9.614000 598 10.386000 22 580392 511 966033 87 614359 598 385641 38 23 680699 511 965981 87 614718 598 385282 37 24 581005 511 965928 87 616077 597 384923 36 25 581312 610 965876 87 615435 597 384565 35 26 681618 510 965824 87 615793 597 384207 34 27 581924 509 965772 87 616151 596 383849 33 28 582229 509 965720 87 616509 696 383491 32 29 582535 509 965668 87 616867 596 383133 31 30 31 582840 508 965615 87 87 617224 9 617582 595 382776 10.382418 30 29 9.683145 608 9.965563 595 32 583449 607 965511 87 617939 595 382061 28 33 583754 607 965458 87 618295 594 381705 27 34 584058 506 965406 87 618662 594 381348 HQ 35 584361 606 965353 88 619008 594 380992 26 36 584665 •^06 965301 88 619364 693 380636 24 37 684968 505 965248 88 619721 593 380279 23 38 585272 605 966195 88 620076 593 379924 22 39 585674 504 965143 88 620432 592 379568 21 40 41 585877 504 965090 9.965037 88 88 620787 9.6211412 592 592 379213 10.378858 20 19 9.586179 603 42 586482 503 964984 88 621497 591 378503 18 43 586783 603 964931 88 621852 591 378148 17 44 587085 502 964879 88 622207 590 3VVV93 16 45 587386 602 964826 88 622561 690 377439 15 46 587688 501 964773 88 622915 590 377085 14 47 587989 501 964719 88 623269 589 376731 13 48 588289 601 964666 89 623623 589 376377 12 49 588590 500 964613 89 623976 589 376024 11 50 588890 600 964560 89 624330 588 375670 10 51 9.689190 499 9.964507 89 9.624683 588 10.375317 9 52 589489 499 964454 89 625036 588 374964 8 53 589789 499 964400 89 625388 587 374612 7 54 5900S8 498 964347 89 625741 587 374259 6 55 590387 498 964294 89 626093 587 373907 5 56 690686 497 964240 89 626446 586 373655 4 67 590984 497 964187 89 626797 586 373203 3 58 591282 497 964133 89 627149 586 372851 2 59 591580 496 964080 89 627501 585 372499 1 60 691878 496 964026' 89 627852 585 372148 ! Cosine 1 Sine 1 Couing. 1 Tang. 1 M. | C 7D(*^ b'rees. SINES AND TANGENTS. (23 Degrees.) 41 M. 1 Sine 1 D- 1 Cosine | D. 1 Tang. 1 D. Cotang, r* "^ 9.59187fi 496 9,964026 89 9.627852 585 10.372148 -60 1 692176 495 963972 89 628203 585 371797 59 2 592473 495 963919 89 628554 585 371446 58 3 69«770 495 963865 90 6289§5 584 371095 67 4 693067 4Q4 963811 90 629255 684 370745 56 5 593363 494 963757 90 €29606 583 370394 55 6 593659 493 963704 90 629956 683 370044 54 7 693955 493 963650 90 630306 583 369694 53 8 594251 493 963596 90 630656 583 369344 52 9 594547 492 963542 90 8-31005 582 368995 51 10 11 694842 492 963488 90 90 631355 682 368645 60, 49 9.595137 491 9.963434 9.631704 582 10.368296 12 695432 491 963379 90 632053 581 367947 48 13 695727 491 963325 90 632401 581 367599 47^ 14 596021 490 963271 90 632750 581 367250 46 15 696315 490 963217 90 €33098 580 366902 45 16 696609 489 968163 90 €33447 580 366553 44 17 696903 489 &63108 91 €33796 580 366205 43 18 697196 489 963054 91 €34143 579 365857 42 19 597490 488 962999 91 €34490 579 365510 41 20 21 597783 488 962945 9.962890 91 91 €34838 579 365162 40 .39 9,698075 487 9.635185 578 10.364815 22 598368 487 962836 91 635632 578 364468 38 23 698660 487 962781 91 635879 578 364121 37 24 698952 486 962727 91 638226 577 363774 36 25 599244 486 962672 91 636572 577 363428 35 26 599536 485 962617 91 636919 677 363081 34 27 599827 485 962562 91 637265 577 362735 33 28 600118 485 962508 91 637611 676 362389 32 29 €00409 484 962453 91 637956 676 362044 31 30 31 600700 9.600990 484 962398 92 92 638302 576 361698 30' 29 484 9.962343 9.638647 575 10.361353 32 601280 483 962288 92 638992 ;575 • 361008 28 33 601570 483 962238 92 639337 575 360663 27 34 601860 482 962178 92 «89682 574 360318 26 35 602150 482 962123 92 €40027 574 359973 25 36 602439 482 962067 92 €40371 574 359629 24 37 60^728 481 962012 92 6407ie 573 359284 23 38 603017 481 961957 92 €41060 673 358940 22 39 60>3305 481 961902 92 641404 573 358596 21 40 41 603594 480 961846 92 92 641747 572 358253 10.357909 20 19 9.603882 480 9.961791 9.642091 572 42 604170 479 -961735 92 642434 572 357566 18 45 604457 479 961680 92 642777 572 357223 17 44 604745 479 961624 93 643120 671 356^80 16-- 45 605032 478 961569 93 643463 571 35G537 15 46 605319 478 961513 93 043806 571 356194 14 47 605606 478 961458 93 644148 570 355S52 13 48 605892 477 961402 93 644490 670 355510 12 49 606179 477 961346 93 644832 570 355168 11 50 51 606465 9.606751 476 961290 93 93 645174 569 354826 10.364484 10 9 476 9.961235 9.645516 569 52 607036 476 961178 93 €45857 569 354143 8 53 607322 475 961123 93 646199 669 353801 7 54 607607 475 961067 93 646540 568 353460 6 55 607892 474 961J011 93 646861 568 353119 6 56 608177 474 960955 93 64725^ 568 352778 4 57 608461 474 960899 93 6476€2 567 352438 3 58 608745 473 960843 94 647903 567 352097 59 609029 473 960786 94 648243 567 351757 T 60 609313 478 960730 94 648583 566 351417 1 Cosine | 1 Sine 1 1 Colanp. 1 1 I'ane. | M. | D4 Degrees. F 42 (2 i Degrees.') a TABLE OF LOGABITHMIC pn Sine D. Cosine 1 D. Tang. 1 D. Cotang. 1 9.609313 473 9.960730 94 9.648583 566 10.351417 w 1 609597 472 960674 94 648923 566 351077 59 2 609880 472 960618 94 649263 566 350737 58 3 610164 472 960561 94 649602 566 350398 57 4 610447 471 960505 94 649942 565 350068 56 5 610729 471 960448 94 650281 565 349719 55 6 611012 470 960392 94 650620 565 349380 54 7 611294 470 960335 94 650959 564 349041 53 8 611576 470 960279 94 651297 564 348703 52 9 611858 469 960222 94 651636 564 348364 51 10 11 612140 469 960165 94 95 651974 563 348026 50 49 9.612421 469 9.960109 9.652312 563 10.347688 12 612702 468 960052 95 652650 563 347350 48 13 612983 468 959995 95 652988 563 347012 47 14 613264 467 959938 95 653326 562 346674 46 15 613545 467 959882 95 653663 562 346337 45 16 613825 467 959825 95 654000 562 346000 44 17 614105 466 959768 95 654337 561 345663 43 18 614385 466 959711 95 654674 561 345326 42 19 614665 466 959654 95 655011 561 344989 41 20 21 614944 465 959596 95 95 655348 561 344652 40 39 9.615223 465 9.959539 9.655684 560 10.344316 22 615502 465 959482 95 656020 560 343980 38 23 615781 464 959425 95 656356 560 343644 37 24 616060 464 959368 95 656692 559 343308 36 25 616338 464 959310 96 657028 659 342972 35 26 616616 463 959253 96 657364 559 342636 34 27 616894 463 959195 96 657699 559 342301 33 28 617172 462 959138 96 658034 558 341966 32 29 617450 462 959081 96 658369 558 341631 31 30 31 617727 462 959023 96 96 658704 558 341296 30 ,29 9.618004 461 9.958965 9.659039 558 10.340961 32 618281 461 958908 96 659373 557 340627 28 33 618558 461 958850 96 659708 .557 340292 27 34 618834 460 958792 96 660042 557 339958 26 35 619110 460 958734 96 660376 557 339624 25 36 619386 460 958677 96 660710 556 339290 24 37 619662 459 958619 96 661043 556 338957 23 38 619938 459 958561 96 661377 556 338623 22 39 620213 459 958503 97 661710 555 338290 21 40 41 620488 458 958445 97 97 662043 555 337957 10.337624 20 19 9.620763 • 458 9.9.58387 9.662376 655 42 621038 457 958329 97 662709 554 337291 18 43 621313 457 958271 97 663042 554 336958 17 4-1 621587 457 958213 97 663375 554 336625 16 45 621861 456 958154 97 663707 554 336293 15 46 622135 456 958096 97 664039 553 335961 14 47 622409 456 958038 97 664371 653 335629 13 48 622682 455 957979 97 664703 653 335297 12 49 622956 455 957921 97 665035 553 334965 11 50 51 623229 455 957863 9.957804 97 9V 665366 9.665697 552 334634 10.334303 10 9 9.623502 454 652 52 623774 454 957746 98 666029 652 333971 8 53 624047 454 957687 98 666360 651 333640 7 54 624319 453 957628 98 666691 551 333309 6 55 624591 453 957570 98 667021 551 332979 5 56 624863 453 957511 98 667352 551 332648 4 57 625135 452 957452 98 667682 550 332318 3 58 625406 452 957393 98 668013 550 331987 2 69 625677 452 957335 98 668343 550 831657 1 60 625948 451 9572761 98 668672 550 331328 1 Cosine I Sine 1 1 CotaiiEt. 1 Tang. 1 M. | 65 Degr ees. SIXES AND TANGENTS . (25 Degrees.) 43 F Sine D. Cosine j D. 1 Tang. 1 D. 1 Cotang. i j 9.625948 451 9.957276 98 9.668673 550 10.3313271 601 1 626219 451 957217 98 669002 549 330998 59 3 626490 451 957158 98 669332 549 330608 58 3 626760 450 957099 98 669661 549 330339 57 4 627030 450 957040 98 669991 548 330009 56 5 627300 450 956981 98 670320 548 329680 55 6 627570 449 956921 99 670649 548 329351 54 7 627840 449 956862 99 670977 548 329023 53 8 628109 449 956803 99 671306 547 328694 52 9 628378 448 956744 99 671634 547 328366 51 10 11 628647 448 956684 99 99 671963 547 328037 10.327709 50 49 9.628916 447 9.95662.'> 9.672291 547 12 629185 447 956566 99 672619 546 327381 48 13 629453 447 956506 99 672947 546 327053 47 14 629721 446 956447 99 673274 546 326726 46 15 629989 446 956387 99 673602 546 326398 45 16 630257 446 956327 99 673929 545 326071 44 17. 630524 446 956268 99 674257 545 325743 43 18 630792 445 956208 100 674584 545 325416 42 19 631059 445 956148 100 674910 544 325090 41 20 21 631326 445 956089 100 100 675237 544 324763 10.324436 40 39 9.631593 444 9.956029 9.675564 544 22 631859 444 955969 100 675890 544 324110 38 23 632125 444 955909 100 676216 543 323784 37 24 632392 443 955849 100 676543 543 323457 36 25 632658 443 955789 100 676869 543 323131 35 26 632923 443 955729 100 677194 543 322806 34 27 633189 442 955669 100 677520 542 322480 33 28 633454 442 955609 100 677846 542 322154 32 29 633719 442 955548 100 678171 542 321829 31 30 633984 441 955488 100 678496 542 321504 30 31 9.634249 ^ 441 9.955428 101 9.678821 541 10.321179 29 32 634514 440 -955368 101 679146 541 320854 28 33 6,34778 440 955307 101 679471 541 320529 37 34 635042 440 955247 101 679795 541 320205 26 35 635306 439 955186 101 680120 540 319880 25 36 635570 439 955126 101 680444 540 319556 24 37 635834 439 955065 101 680768 540 319232 23 38 636097 438 955005 101 681092 540 318908 22 39 636360 438 954944 101 68L416 539 318584 21 40 41 636623 438 954883 9 954823 101 101 681740 50 539 318260 20 19 9.636886 437 9.682063 10.317937 42 637148 437 954762 101 682387 539 317613 18 43 637411 437 954701 101 682710 538 317290 17 44 637673 437 954640 101 683033 538 316967 16 45 637935 436 954579 101 683356 538 316644 15 46 638197 436 954518 102 683679 538 316321 14 47 638458 436 954457 102 684001 537 31.5999 13 48 638720 435 954396 102 684324 537 315676 12 49 638981 435 954335 102 684646 537 31.5354 11 50 639242 435 954274 102 684968 537 315032 10 51 9.639503 434 9.954213 102 9.686290 536 10.314710 9 52 639764 434 954152 102 685612 536 314388 8 53 640024 434 954090 102 685934 536 314066 7 54 640284 433 954029 102 686255 536 313745 6 55 640544 433 953968 102 686577 C35 313423 5 56 640804 433 953906 102 686898 635 313102 4 57 641064 432 953845 102 687219 635 312781 3 58 641324 432 953783 102 687540 535 312460 2 59 641584 432 953722 103 687861 534 312139 1 ^ 641842 431 9536601 103 688182 534 311818 Cosine 1 Sine 1 , Cotang. Tang. j M. a^ IDeg ees. 44 ^^26 Degrees.) a TABLE OF LOGARITHMIC "mT Sine D. Cosine 1 D. T.n.. D. Cotang. 9.641842 431 9.953660 103 9.688182 534 10.311818 "60 1 642101 431 953599 103 688502 534 311498 59 2 642360 431 953537 103 688823 534 311177 58 3 642618 430 953475 103 689143 533 310857 57 4 642877 430 953413 103 689463 533 310537 56 5 643135 430 953352 103 689783 533 310217 55 6 643393 430 953290 103 690103 533 309897 54 7 643650 429 953228 103 690423 533 309577 53 8 643908 429 953166 103 690742 532 309258 52 9 644165 429 953104 103 691062 532 308938 51 10 11 644423 428 953042 103 104 691381 532 308619 10.308300 50 49 9.644680 428 9.952980 9.691700 531 12 644936 428 952918 104 692019 531 307981 48 13 645193 427 952855 104 692338 531 307662 47 14 645450 427 952793 104 692656 531 307344 46 15 645706 427 952731 104 692975 531 307025 45 16 645962 426 952669 104 693293 530 306707 44 17 646218 426 952606 104 693612 530 306388 43 18 646474 426 952544 104 693930 530 306070 42 19 646729 425 952481 104 694248 530 305752 41 20 21 646984 425 952419 9.952356 104 694566 529 305434 40 39 9.647240 425 9.694883 529 10.3'05117 22 647494 424 952294 104 695201 529 304799 38 23 647749 424 952231 104 695518 529 304482 37 24 648004 424 952168 105 695836 529 304164 36 25 648258 424 952106 105 696153 528 303847 35 36 648512 423 952043 105 696470 528 303530 34 27 648766 423 951980 105 696787 528 303213 33 28 649020 423 951917 105 697103 528 302897 32 29 649274 422 951854 105 697420 527 302580 31 30 31 649527 422 951791 105 105 697736 527 302264 30 29 9.649781 422 9.951728 07698053 527 10.301947 32 650034 422 951665 105 698369 527 .301631 28 33 650287 421 951602 105 698685 526 301315 27 34 650539 421 951539 105 699001 526 300999 26 35 650792 421 951476 105 699316 526 300684 25 36 651044 420 951412 105 699632 526 300368 24 37 651297 420 951349 106 699947 526 300053 23 38 651549 420 951286 106 700263 525 299737 22 39 651800 419 951222 106 700578 525 299422 21 40 41 652052 9.652304 419 951159 106 106 700893 525 299107 10.298792 20 19 419 9.951096 9.701208 524 524 42 652555 418 951032 106 701523 298477 18 43 652806 418 9509G8 106 701837 524 298163 17 44 653057 418 950905 106 702152 524 297848 16 45 653308 418 950841 106 702466 524 297534 15 46 653558 417- 950778 106 702780 523 297220 14 47 653808 417 950714 106 703095 523 296905 13 48 654059 417 950650 106 703409 523 296591 12 49 654309 416 950586 106 703723 523 296277 11 50 51 654558 416 950522 9.950458 107 107 704036 522 295964 10.295650 10 9 9.654808 416 9.704350 522 52 655058 416 950394 107 704663 522 295337 8 53 655307 415 950330 107 704977 522 295023 7 .54 655556 415 950266 107 705290 522 294710 6 55 655805 415 950202 107 705603 521 294397 5 56 656054 414 950138 107 705916 521 294084 4 57 656302 414 950074 107 706228 521 293772 3 58 656551 414 950010 107 706541 521 293459 2 59 656799 413 940945 107 706854 521 293146 1 60 657047 413 949881 107 707166 520 292834 Cosine Sine 1 Cotan?. 1 Tang. 1 M. | G3 Degrees. SINES AND TANGENTS. (27 Degrees.) 45 M. Sine D. Cosine | D. Tang. 1 D. 1 Cotang. 1 1 "o" {>. 657047 413 9.949881 107 9.707166 520 10.292834 60 1 657295 413 949816 107 707478 620 292622 59 2 657542 412 949752 107 707790 620 292210 58 3 657790 412 949688 108 708102 520 291898 57 4 658037 412 949623 108 708414 519 291.586 56 5 658284 412 949558 108 708726 519 291274 56 6 658531 411 949494 108 709037 619 290963 54 7 658778 411 949429 108 709349 619 290651 53 8 669025 411 949364 108 709660 619 290340 52 9 659271 410 949300 108 709971 618 290029 51 10 11 659517 410 949235 108 108 710282 518 289718 50 49 9.659763 410 9.949170 9.710593 518 10.289407 12 660009 409 949105 108 710904 518 289096 48 13 660255 409 949040 108 711215 518 288785 47 14 660501 409 948975 108 711525 . 517 288475 46 15 660746 409 948910 108 711836 617 288164 45 16 660991 408 948845 108 712146 517 287864 44 17 661236 408 948780 109 712456 517 287544 43 18 661481 408 948715 109 712766 516 287234 42 19 661726 407 948650 109 713076 516 286924 41 20 21 661970 9.662214 407 407 948584 109 109 713386 516 286614 10.286304 40 39 9.948519 9.713696 516 22 662459 407 948454 109 714005 516 285995 38 23 662703 406 948388 109 714314 515 285686 37 24 662946 406 948323 109 714624 616 285376 36 25 663190 406 948257 109 714933 616 28506? 35 26 663433 405 948192 109 715242 616 284768 34 27 663677 405 948126 109 716551 614 284449 33 28 663920 405 948060 109 715860 614 284140 32 29 664163 405 947995 110 716168 614 283832 31 30 31 664406 404 947929 110 716477 614 283523 30 29 9.664648 404 9.947863 9.716785 614 10.283216 32 664891 404 947797 110 717093 513 282907 28 33 665133 403 947731 110 717401 513 282599 27 34. 665375 403 947665 110 717709 613 282291 26 35 665617 403 947600 110 718017 513 281983 25 36 665859 402 947533 no 718326 513 281675 24 37 666100 402 947467 110 718633 512 281367 23 38 666342 402 947401 110 718940 612 281060 22 39 666583 402 947335 no 719248 512 280762 21 40 41 666824 401 947269 no no 719566 612 280446 20 19 9.667066 401 9.947203 9.719862 612 10.280138 42 667306 401 947136 111 720169 611 279831 18 43 667546 401 947070 111 720476 611 279624 17 44 667786 400 947004 111 720783 511 279217 16 45 668027 400 946937 111 721089 611 278911 15 46 668267 400 946871 111 721396 611 278604 14 47 668506 399 946804 111 721702 610 278298 13 48 668746 399 946738 111 722009 510 277991 12 49 668986 399 946671 111 722316 510 277686 11 60 51 669225 399 946604 9.946538 111 111 722621 510 277379 10 9 9.669464 398 9.722927 610 10.277073 52 669703 398 946471 111 723232 609 276768 8 53 669942 398 946404 111 723538 609 276462 7 54 670181 397 946337 111 723844 509 276156 6 65 670419 397 946270 112 724149 609 275851 5 66 670658 397 946203 112 724454 609 275546 4 57 670896 897 946136 112 724759 508 275241 3 58 671134 396 946069 112 725065 608 274935 2 59 671372 396 946002 112 726369 508 274631 1 60 671609 396 945935 112 725674 508 274326 n Cosine Sine 1 i Cotang. Tang. |M.| Degrees. 46 (28 Degrees.) a TABLE OF LOGARITHMIC M. 1 Sine D. I Cosine 1 D. 1 Taut?. 1 n. Cotang. 1 1 ~~0 9.671609 396 9.945935 112 9.725674 508 10.274323 60 1 671847 395 945868 112 725979 608 274021 69 2 672084 395 945800 112 726284 607 273716 58 3 672321 395 945733 112 726688 507 273412 67 4 672558 395 945666 112 726892 507 273108 66 6 672795 394 945598 112 727197 607 272803 55 6 673032 394 945531 112 727501 507 272499 54 7 €73268 394 945464 113 727806 506 272195 53 8 673505 394 945396 113 728109 506 271891 52 9 673741 393 945328 113 728412 606 271588 51 10 673977 393 945261 113 728716 606 271284 50 11 9.674213 393 9.945193 113 9.729020 606 10.270980 49 12 674448 392 946125 113 729323 506 270677 48 13 674684 • 392 946058 113 729626 605 270374 47 14 674919 392 944990 113 729929 506 270071 46 15 675155 392 944922 113 730233 605 269767 46 16 675390 391 944864 113 730535 505 269465 44 17 675624 391 944786 113 730838 504 269162 43 18 675859 391 944718 113 731141 504 268859 42 19 676094 391 944660 113 731444 604 268566 41 20 21 676328 9.676562 390 944582 114 114 731746 504 268254 40 39 390 9.944514 9.732048 604 10.267952 22 676796 390 944446 114 732351 503 267649 38 23 677030 390 944377 114 732653 603 26731:7 37 24 677264 389 944309 114 732955 503 267045 36 25 677498 389 944241 114 733257 503 266743 36 26 677731 389 944] 72 114 733568 603 266442 34 27 677964 388 944104 114 733860 502 266140 33 28 678197 388 944036 114 734162 602 266838 32 29 678430 388 943967 114 734463 502 265537 31 30 31 678683 388 943899 114 114 734764 502 265236 10.264934 30 29 9 678895 387 9.943830 9.735066 502 32 679128 387 943761 114 735367 502 264633 28 33 679360 387 943693 116 73.5668 601 264332 27 34 679592 387 943624 115 735969 601 264031 26 35 679824 386 943656 115 736269 501 263731 25 36 680056 386 943486 116 736670 501 263430 24 37 680288 386 943417 115 736871 501 263129 23 38 680519 385 943348 115 737171 500 262829 22 39 680750 385 943279 116 737471 500 262529 21 40 41 680982 386 943210 116 115 737771 9.738071 500 262229 20 19 9.681213 385 9.943141 600 10.261929 42 681443 384 943072 115 738371 500 261629 18 43 681674 384 943003 116 738671 499 261329 17 44 681905 384 942934 116 738971 499 261029 16 45 682135 384 942864 115 739271 499 260729 16 46 682365 383 942795 116 739670 499 260430 14 47 682596 383 942726 116 739870 499 260130 13 48 682825 383 942656 116 740169 499 259831 12 49 683055 383 942587 116 740468 498 259532 11 50 61 683284 382 942517 116 116 740767 498 259233 10 9 9.683514 382 9.942448 9.741066 498 10.258934 S2 683743 382 942378 116 741365 498 258635 8 63 683972 382 942308 116 741664 498 268336 7 64 684201 381 942239 116 741962 49? 268038 6 55 684430 381 942169 116 742261 497 257739 5 56 684658 381 942099 116 742559 497 257441 4 57 684887 380 942029 116 742858 497 267142 3 58 685115 380 941959 il6 743156 497 256844 2 59 685343 380 941889 117 743454 497 256546 1 60 685571 380 941819 117 743752 496 256248 Li Cosine 1 Sine 1 1 Colang. 1 Tang. 1 M. | 61 Degrees. SINES AND TANGENTS . ^29 E legrees •) 47 JL. Sine 1 D. Cosine 1 D. | Tang. 1 D. Cotang. 1 1 9.685571 380 9.941819 117 9.743752 496 10.256248 60 1 685799 379 941749 117 744050 496 255950 59 2 686027 379 941679 117 744348 496 255652 58 3 686254 379 941609 117 744645 496 255355 57 4 686482 379 941539 117 744943 496 255057 56 5 686709 378 941469 117 745240 496 254760 55 6 686936 378 941398 117 745538 495 254462 54 7 687163 378 941328 117 745835 495 254165 53 8 687389 878 941258 117 746132 495 253868 52 9 687616 377 941187 117 746429 495 253571 51 10 11 687843 9.688069 377 377 .941117 9.941046 117 118 746726 495 253274 10.252977 50 49 9.747023 494 12 688295 377 940975 118 747319 494 252681 48 13 688521 376 940905 118 747616 494 252384 47 14 688747 376 940834 118 747913 494 252087 46 15 688972 376 940763 118 748209 494 251791 45 16 689198 376 940693 118 748505 493 251495 44 17 689423 375 940622 118 748801 493 251199 43 18 689648 375 940551 118 749097 493 250903 42 19 689873 375 940480 118 749393 493 250607 41 20 21 690098 9.690323 375 374 940409 9.940338 118 118 749689 493 250311 40 39 9.749985 493 10.250015 32 690548 374 940267 118 750281 492 249719 38 23 690772 374 940196 118 750576 492 249424 37 24 690996 374 940125 119 750872 492 249128 36 25 691220 373 940054 119 751167 492 248833 35 26 691444 373 939982 119 751462 492 248538 34 27 691668 373 939911 119 751757 492 248243 33 28 691892 373 939840 119 752052 491 247948 32 29 692115 372 939768 119 752347 491 247653 31 30 31 692339 372 939697 9.939625 119 119 752642 491 247358 ,30 29 9.692562 372 9.752937 491 10.247063 32 692785 371 939554 119 753231 491 246769 28 33 693008 371 939482 119 753526 491 246474 27 34 693231 371 939410 119 758820 490 246180 26 35 693453 371 939339 119 754115 490 245885 25 36 693676 370 939267 120 754409 490 245591 24 37 693898 370 939195 120 754703 490 245297 23 38 694120 370 939123 120 754997 490 245003 22 39 694342 370 939052 120 755291 490 244709 21 40 41 694564 9.694786 369 938980 120 120 755585 489 244415 20 19 369 9.938908 9.755878 489 10.244122 42 695007 309 938636 120 756172 489 243828 18 43 695229 369 938763 120 756465 489 243535 17 44 695450 368 938691 120 756759 489 ^43241 16 45 695671 368 938619 120 757052 489 242948 15 46 695892 368 938547 120 757345 488 242655 14 47 696113 368 938475 120 757638 488 242362 13 48 696334 367 938402 121 757931 488 242069 12 49 696554 367 938330 121 758224 488 241776 11 50 51 696775 367 938258 121 121 758517 488 241483 10 9 9.696995 367 9.938185 9.758810 488 10.241190 52 697215 366 938113 121 759102 487 240898 8 53 697435 366 938040 121 759395 487 240605 7 54 697654 366 937967 121 759687 487 240313 6 55 697874 366 937895 121 759979 487 240021 5 56 698094 365 937822 121 . 760272 487 239728 4 57 698313 365 937749 121 760564 487 239436 3 58 698532 365 937676 121 760856 • 486 239144 2 59 698751 365 937604 121 761148 486 238852 1 60 698970 364 937531 121 761439 486 238561 I Cosine 1 ■ 1 Sine 1 1 Cotang. 1 Tang. |M.| Degrees. 48 (30 Degr ees.) A TABLE OF LOGARITHMIC T| Sine 1 D. 1 Cosine 1 D. | 'J'ansr. | D. 1 Cotang. 1 j 9.698970 364 9.937531 121 9.761439 486 10.238561 60 1 699189 364 937458 122 761731 486 238269 59 2 699407 364 937386 122 762023 486 237977 58 3 699626 364 937312 122 762314 486 237686 57 4 699844 363 937238 122 762606 485 237394 56 5 700062 363 937166 122 762897 485 237103 55 6 700280 363 937092 122 763188 485 236812 64 7 700498 363 937019 122 763479 486 236521 53 8 700716 363 936946 122 763770 485 236230 52 9 700933 362 936872 122 764061 485 236939 51 10 11 701151 9.701368 362 936799 9.936725 122 122 764362 484 235648 50 49 362 9.764643 484 10.235367 12 701585 362 936662 123 764933 484 236067 48 13 701802 361 936578 123 766224 484 234776 47 14 702019 361 936506 123 765614 484 234486 46 15 702236 361 936431 123 765805 484 234196 45 16 702452 361 936357 123 766096 484 233905 44 17 702669 360 936284 123 766385 483 233016 43 18 702885 360 936210 123 766676 483 233326 42 19 703101 360 936136 123 766965 483 233036 41 20 21 703317 360 936002 9.935988 123 123 767255 483 232746 40 39 9.703533 359 9.767645 483 10.232456 22 703749 359 935914 123 767834 483 232166 38 23 703964 359 935840 123 768124 482 231876 37 24 704179 359 935766 124 768413 482 231587 36 25 704395 359 935692 124 768703 482 231297 35 26 704610 358 935618 124 768992 482 231008 34 27 704825 358 935543 124 769281 482 230719 33 28 705040 358 935469 124 769570 482 230430 32 29 705254 358 935396 124 769860 481 230140 31 30 31 705469 9.705683 357 935320 9.935246 •124 124 770148 481 229852 30 29 357 9.770437 481 10.229563 32 705898 357 935171 124 770726 481 229274 28 33 706112 357 936097 124 771016 481 228985 27 34 706326 .356 936022 124 771303 481 228697 26 35 706539 356 934948 124 771592 481 228408 25 36 706753 356 934873 124 771880 480 228120 24 37 706967 356 934798 125 772168 480 227832 23 38 707180 355 934723 125 772467 480 227543 22 39 707393 355 934649 126 772745 480 227265 21 40 41 707606 9.707819 355 355 934574 9.934499 125 125 773033 480 226967 10.226679 20 19 9.773321 480 42 708032 354 934424 126 773608 479 226392 18 43 708245 354 934349 125 773896 479 226104 17 44 708458 354 934274 125 774184 479 225816 16 45 708670 354 934199 125 774471 479 226529 15 46 708882 353 934123 125 774759 479 225241 14 47 709094 353 934048 126 776046 479 224954 13 48 709306 353 933973 126 775333 479 224667 12 49 709518 353 933898 126 775621 478 2243S'9 11 50 51 709730 353 933822 9.933747 126 126 776908 478 224092 10 9 9 709941 362 9.776195 478 10.223805 52 710153 352 933671 126 776482 478 223518 8 63 710364 362 933596 126 776769 478 223231 7 54 710575 362 933520 126 777055 478 222945 6 55 710786 351 933446 126 777342 478 222668 5 56 710997 351 933369 126 777628 477 222372 4 57 711208 351 93329,1 126 777915 477 222086 3 58 711419 351 933217 126 778201 477 221799 2 59 711629 360 9331411 126 778487 477 221512 1 60 711839 350 933066' 12C 778774 477 221226 0| 1 Cosine 1 1 SU.e 1 1 Cotang. 1 1 Tiing. 1 M. 1 59 Degrees. SINES AND TANGENTS . (31 L ) agrees 49 M. Sine D. Cosine 1 D. Tansr. D. Cotang. 1 1 "T 9 711839 350 9.933066 126 9.778774 477 10.221226 60 1 712050 350 932990 127 779060 477 220940 59 2 712260 350 932914 127 779346 176 220654 58 3 712469 349 932838 127 779632 476 220368 57 4 712679 849 932762 127 779918 476 220082 56 5 712889 349 932685 127 780203 476 219797 55 6 713098 349 932609 127 780489 476 219511 54 7 713308 349 932533 127 780775 476 219225 53 8 713517 348 932457 127 781060 476 218940 52 9 713726 348 932380 127 781346 475 218654 51 10 11 713935 348 932304 9.932228 127 127 781631 9.781916 475 218369 50 49 9.714144 348 475 10.218084 12 714352 347 932151 127 782201 475 217799 48 13 714561 347 932075 128 782486 475 217514 47 14 714769 347 931998 128 782771 475 217229 46 15 714978 347 931921 128 783056 475 216944 45 16 715186 347 931845 128 783341 475 216659 44 17 715394 346 931768 128 783626 474 216374 43 18 715602 346 931691 128 783910 474 216090 42 19 715809 346 931614 128 784195 474 215805 41 20 21 716017 9.716224 346 931537 9.931460 128 128 784479 474 215521 40 39 345 9.784764 474 10.215236 22 716432 345 931383 128 785048 474 214952 38 23 716639 345 931306 128 785332 473 214668 37 24 716846 345 931229 129 785616 473 214384 36 25 717053 345 931152 129 785900 473 214100 35 26 717259 344 931075 129 786184 473 213816 34 27 717466 344 930998 129 786468 473 213532 33 28 717673 844 930921 129 786752 473 213248 32 29 717879 344 930843 129 787036 473 212964 31 30 31 718085 343 930766 129 129 787319 472 212681 30 29 9.718291 343 9.930688 9.787603 472 10.212397 32 718497 343 930611 129 787886 472 212114 28 33 718703 343 930533 129 788170 472 211830 27 34 718909 343 930456 129 788453 472 211547 26 35 719114 342 930378 129 788736 472 211264 25 36 719320 342 930300 130 789019 472 210981 24 37 719525 342 930223 130 789302 471 210698 23 38 719730 342 930145 130 789585 471 210415 22 39 719935 341 930067 130 789868 471 210132 21 40 41 720140 9.720345 341 929989 9.929911 130 130 790151 471 209849 20 19 341 9.790433 471 10.209567 42 720549 341 929833 130 790716 471 209284 18 43 720754 340 929755 130 790999 471 209001 17 U 720958 340 929677 130 791281 471 208719 16 45 721162 340 929599 130 791563 470 208437 15 46 721366 340 929521 130 791846 470 208154 14 47* 721570 340 929442 130 792128 470 207872 13 48 721774 339 929364 131 792410 470 207590 12 49 721978 339 929286 131 792692 470 207308 11 50 51 722181 9.722385 339 339 929207 131 131 792974 470 207026 10 9 9.929129 9.793256 470 10.206744 52 722588 339 929050 131 793538 469 206462 8 53 722791 338 928972 131 793819 469 206181 7 54 722994 338 928893 131 794101 469 205899 6 55 723197 338 928815 131 794383 469, 205617 6 56 723400 338 928736 131 794664 469 205336 4 57 723603 337 928657 131 794945 469 205055 3 58 723805 337 928578 131 795227 469 204773 2 59 724007 337 928499 131 795508 468 204492 1 60 724210 337 928420 131 795789 468 204211 Lj Cosine 1 Sine 1 Cotang. 1 Tang. m Be 53 Degrees G 60 (32 Degrees.) a table of Logarithmic M. 1 Sine 1 D. 1 Cosine | D. 1 Tang. 1 ■■■- 1 Cotang. 1 1 ~ 9.724210 337 9.928420 132 9.795789 \ 468 10.2042111 60| 1 724412 337 928342 132 796070 468 203930 59 2 724614 336 928263 132 796351 468 203649 58 3 724816 336 928183 132 796632 468 203368 57 4 725017 336 928104 132 796913 468 203087 56 5 725219 336 928025 132 797194 468 202806 55 6 725420 335 927946 132 797475 468 202525 54 7 725622 335 927867 132 797755 468 202245 53 8 725823 335 927787 132 798036 467 201964 52 9 726024 335 927708 132 798316 467 201684 51 10 11 726225 335 927629 132 132 798596 9.798877 467 467 201404 50 49 9.726426 334 9.927549 10.201123 12 726626 334 927470 133 799157 467 200843 48 13 726827 334 927390 133 799437 467 200563 47 14 727027 334 927310 133 799717 467 200283 46 15 727228 334 927231 133 799997 466 200003 45 16 727428 333 927151 133 800277 466 199723 44 17 727628 333 927071 133 800557 466 199443 43 18 727828 333 926991 133 800836 466 199164 42 19 728027 333 926911 133 801116 466 198884 41 20 21 728227 333 926831 133 133 801396 466 198604 40 39 9.728427 332 9.926751 9.801675 466 10.198325 22 728626 332 926671 133 801955 466 198045 38 23 728825 332 926591 133 802234 465 197766 37 24 729024 332 926511 134 802513 465 197487 36 25 729223 331 926431 134 802792 465 197208 35 26 729422 331 926351 134 803072 465 196928 34 27 729621 331 926270 134 803351 465 196649 33 28 729820 331 926190 134 803630 465 196370 32 29 730018 330 926110 134 803908 465 196092 31 30 31 730216 9.730415 330 926029 134 134 804187 465 195813 30 29 330 9.925949 9.804466 464 10.195534 32 730613 330 925868 134 804745 464 195255 28 33 730811 330 925788 134 805023 464 194977 27 34 731009 329 925707 134 805302 464 194698 26 35 731206 329 925626 134 805580 464 194420 25 36 731404 329 925545 135 805859 464 194141 24 37 731602 329 925465 135 806137 464 193863 23 38 731799 329 925384 135 806415 463 193585 22 39 731996 328 925303 135 806693 463 193307 21 40 41 732193 328 925222 9.925141 135 135 806971 9.807249 463 193029 20 19 9.732390 328 463 10.192751 42 732587 328 925060 135 807527 463 192473 18 43 732784 328 924979 135 807805 463 192195 17 44 732980 327 924897 135 808083 463 191917 16 45 733177 327 924816 135 808361 463 191639 15 46 733373 327 924735 136 808638 462 191362 14 47 733569 327 924654 136 808916 462 191084 13 48 733765 327 924572 136 809193 462 190807 12 49 733961 326 924491 136 809471 462 190529 11 50 51 734157 326 924409 136 136 809748 462 190252 10 9 9.734353 326 9.924328 9.810025 462 10.189975 52 734549 326 924246 136 810302 462 189698! 81 53 734744 325 924164 136 810580 462 189420J 7| 64 734939 325 924083 136 810857 462 189143 6 55 735135 325 924001 136 811134 461 188866 5 66 735330 325 923919 136 811410 461 188590 4 57 735525 325 923837 136 811687 461 188313 3 58 735719 324 923755 137 811964 461 1880.36 2 59 735914 324 923673 137 812241 461 187759 1 60 736109 324 923591 137 812517 461 187483 1 " Cosine 1 Sine 1 1 Cotang. j Tang. 1 "W 57 Degrees. ■^ SINES AND TANGEKTS . (33 Degrees 51 M. Sine D. Cosine | D. Tang. D. Cotang. 1 9.736109 324 9.923591 137 9.812517 461 10.187482 60 1 736303 324 923509 137 812794 461 187206 59 2 736498 324 923427 137 81.3070 461 186930 68 3 736692 323 923.345 137 813.347 460 186653 57 4 736886 323 923263 137 813623 460 186377 66 ft 737080 323 923181 137 813899 460 186101 55 6 737274 323 923098 137 814175 460 185825 64 7 737467 323 923016 137 814452 460 185548 63 8 737661 322 922933 137 814728 460 185272 62 9 737855 322 922851 137 815004 460 184996 61 10 11 738048 322 922768 9.922686 138 138 815279 460 184721 60 49 9.738241 322 9.815555 459 10.184445 12 738434 322 922603 138 81,5831 459 184169 48 13 738627 321 922520 138 816107 459 183893 47 14 738820 321 922438 138 816382 459 183618 46 15 739013 321 922355 138 816658 459 183342 46 16 739206 321 922272 138 816933 459 183067 44 17 739398 321 922189 138 817209 459 182791 43 18 739590 320 922106 138 817484 459 182516 42 19 739783 320 922023 138 817759 459 182241 41 20 21 739975 320 921940 138 139 818035 458 181965 40 39 9.740167 320 9.921857 9.818310 458 10.181690 22 740359 320 921774 139 818585 458 181415 38 23 740550 319 921691 139 818860 458 181140 37 24 740742 319 921607 1.39 819135 458 180865 36 25 740934 319 921524 139 819410 458 180590 36 26 741125 319 921441 139 819684 458 180316 34 27 741316 319 921357 139 819959 458 180041 33 28 741508 318 921274 139 820234 458 179766 32 29 741699 318 921190 139 820508 457 179492 31 30 31 741889 9.742080 318 921107 9.921023 139 139 820783 457 179217 30 29 318 9,821057 457 10.178943 32 742271 318 920939 140 821332 457 178668 28 33 742462 317 920856 140 821606 457 178394 27 34 742652 317 920772 140 821880 457 178120 26 35 742842 317 920688 140 822154 457 177846 25 36 743033 317 920604 140 822429 457 177571 24 37 743223 317 920520 140 822703 457 177297 23 38 743413 316 920436 140 822977 456 177023 22 39 743002 316 920352 140 823250 456 176750 21 40 743792 316 920268 140 823524 456 176476 20 41 9.743982 316 9.920184 140 9.823798 456 10.176202 19 42 744171 316 920099 140 824072 456 175928 18 43 744361 315 920015 140 824345 456 175655 17 44 744550 315 919931 141 824619 456 176381 16 45 744739 315 919846 141 824893 456 175107 15 46 744928 315 919762 141 825166 456 174834 14 47 745117 315 919677 141 825439 455 174561 13 48 745306 314 919593 141 825713 455 174287 12 49 745494 314 919508 141 825986 455 174014 11 60 745683 814 919424 141 826259 455 173741 10 51 9.745871 314 9.919339 141 9.826532 455 10.173468 9 62 74G059 314 919254 141 826805 455 173195 8 53 746248 313 919169 141 827078 455 172922 7 54 746436 313 919085 141 827351 455 172649 6 55 746624 313 919000 141 827624 455 172370 5 56 746812 313 918915 142 827897 454 172103 4 57 746999 313 918830 142 828170 454 171830 3 58 747187 312 918745 142 828442 454 171558 2 59 747374 312 918659 142 828715 454 171285 1 60 747562 312 9185741 142 828987 454 171013 n Cosine Sine 1 Colaiig. j Tang. 1 M. 56 Degrees. 52 (34 Degrees.) a TABLE OP LOGARITHMIC M. 1 Sine 1 D. 1 Cosine | D. 1 Tang. 1 D. 1 Cotang. 1 1 9.747562 312 9.918574 142 9.828987 454 I0.17l\0l3 60 1 747749 312 918489 142 829260 454 170740 59 2 747936 312 918404 142 829532 454 170468 58 3 748123 311 918318 142 829805 454 170195 57 4 748310 311 9182.33 142 830077 454 169923 56 5 748497 311 918147 142 830349 453 169651 55 6 748683 311 918062 142 830621 453 •169379 54 7 .748870 311 917976 143 830893 453 169107 53 8 749056 310 917891 143 831 1&5 453 168835 52 9 749243 310 ' 917805 143 831437 453 168563 51 10 11 749429 9.749615 . 310 310 917719 143 143 831709 453 168291 50 49 9.917634 9.831981 453 10.168019 12 749801 310 917548 143 832253 453 167747 48 13 749987 309 917462 143 832525 453 167475 47 14 750172 309 947376 143 832796 453 167204 46 15 750358 309 - 9ri7290 143 833068 452 166932 45 16 750543 309 917204 143 833339 452 166661 44 17 750729 .309 9iril8 144 833611 452 166389 43 18 750914 308 917032 144 833882 452 166118 42 19 751099 308 916946 144 834154 452 165846 41 20 21 751284 9.751469 308 308 916859 9.916773 144 144 834425 4.52 165575 40 39 9.834696 452 10.165304 22 751654 308 916687 144 834967 452 165033 38 23 751839 308 916600 144 835238 452 164762 37 24 752023 307 916514 144 835509 452 164491 36 25 752208 307 916427 144 835780 451 164220 35 26 752392 307 916341 144 838051 451 163949 34 27 752576 307 916254 144 836322 451 163678 33 28 752760 307 916167 145 836593 451 163407 32 29 752944 306 916081 145 836864 451 163136 31 30 31 753128 306 915994 145 145 837134 9.837405 451 451 162866 30 29 9.753312 306 9.915907 10.162595 32 753495 306 915820 145 837675 451 162325 28 33 753679 306 915733 145 837946 451 162054 27 34 753862 305 915646 145 838216 451 161784 26 35 754046 305 915559 145 838487 450 161513 25 36 754229 305 915472 145 838757 450 161243 24 37 754412 305 915385 145 839027 450 160973 23 38 754595 305 915297 145 839297 450 160703 22 39 754778 304 915210 145 839568 450 160432 21 40 41 754960 304 915123 9.915035 146 146 839838 450 160162 20 19 9.755143 304 ,9.840108 450 10.159892 42 755346 '?c304 ~9 14948 146 840378 450 159622 18 43 755508 '-304 914860 146 840647 450 159353 17 44 755690 304 914773 146 840917 449 159083 16 45 755872 303 914685 146 841187 449 158813 15 46 756054 303 914598 146 841457 449 1.58543 14 47 756236 303 914510 146 841726 449 158274 13 48 756418 303 914422 146 841996 449 158004 12 49 756600 303 914334 146 842266 449 157734 11 50 51 756782 302 914246 9.914158 147 147 842535 9.842805 449 157465 10 9 9.756963 302 449 10.157195 52 757144 302 914070 147 843074 449 156926 8 53 757326 302 913982 147 843343 449 156657 7 54 757507 302 913894 147 843612 449 156388 6 55 757688 301 913806 147 843882 448 156118 5 56 757869 301 913718 147 844151 448 155849 4 57 758050 301 913630 147 844420 448 155580 3 58 758230 301 913541 147 844689 448 15.5311 2 59 758411 301 913453 147 844958 448 155042 1 60 758591 301 91.3365 147 845227 448 154773 Cosine 1 1 Sine 1 1 Colang. 1 tang. M.| 55 Degrees. SINES AND TANGENTS (35 Degrees. ) 63 M. Sine 1 D. Cosine 1 P. . Tanc. D. Cotang. 1 1 "Q- 9.758591 301 9.913365 147 9.845227 448 10.154773 60 1 758772 300 913276 147 845496 448 154504 59 2 758952 300 913187 148 845764 448 154236 58 3 759132 300 913099 148 846033 448 153967 57 4 759312 300 913010 148 846302 448 153698 56 5 759492 300 912922 148 846570 447 153430 55 6 759672 299 912833 148 846839 447 153161 54 7 759852 299 912744 148 847107 447 152893 53 8 760031 299 912655 148 847376 447 152624 52 9 760211 299 912566 148 847644 447 152356 51 10 11 760390 299 912477 9.912388 148 148 847913 9.848181 447 152087 50 49 9.760569 298 447 10.151819 12 760748 298 912299 149 848449 447 151551 48 13 760927 298 912210 149 848717 447 151283 47 14 761106 298 912121 149 848986 447 151014 46 15 761285 298 912031 149 849254 447 150746 45 16 761464 298 911942 149 849522 447 150478 44 17 761642 297 911853 149 849790 446 150210 43 18 761821 297 911763 149 850058 446 149942 42 19 761999 297 911674 149 850325 446 149675 41 20 21 762177 297 911584 9.911495 149 149 850593 446 149407 40 39 9.762356 297 9.850861 446 10,149139 22 762534 296 911405 149 851129 446 148871 38 23 762712 296 911315 150 851396 446 148604 37 24 762889 296 911226 150 851664 446 148336 36 26 763067 296 911136 150 851931 446 148069 35 26 763245 296 911046 150 852199 446 147801 34 27 763422 296 910956 150 852466 446 147534 33 28 763600 295 910866 150 852733 445 147267 32 29 763777 295 910776 150 853001 445 146999 31 30 31 763954 295 910686 150 150 853268 445 146732 10,146465 30 29 9.764131 295 9.910596 9.853535 445 32 764308 295 910506 150 853802 445 146198 28 33 764485 294 910415 150 854069 445 145931 27 34 764662 294 910325 151 854336 445 145664 26 35 764838 294 910235 151 854603 445 145397 25 36 765015 294 910144 151 854870 445 145130 24 37 765191 294 910054 151 855137 445 144863 23 38 765367 294 909963 151 855404 445 144596 22 39 765544 293 909873 151 855671 444 144329 21 40 41 785720 293 909782 9.909691 151 151 855938 444 144062 20 19 9.765896 293 9.856204 444 10,143796 42 766072 293 909601 151 856471 444 143529 18 43 766247 293 909510 151 856737 444 143263 17 44 766423 293 909419 151 857004 444 142996 16 45 766598 292 909328 152 857270 444 142730 15 46 766774 292 909237 152 857537 444 142463 14 47 766949 292 909146 152 857803 444 142197 13 48 767124 292 909055 152 858069 444 141931 12 49 767300 292 908964 152 858336 444 141664 11 50 51 767475 291 ^08873 9.908781 152 152 858602 443 141398 10 ~9 9.767649 291 9,858868 443 10.141132 52 767824 291 908690 152 859134 ■ 443 140866 8 53 767999 291 908599 152 859400 443 140600 7 54 768173 291 908507 152 859666 443 140334 6 55 768348 290 908416 153 859932 443 140068 5 56 768522 290 908324 153 860198 443 139802 4 57 768697 290 908233 153 860464 443 139536 3 68 768871 290 908141 153 860730 443 139270 2 59 769045 290 908049 153 860995 443 139005 1 60 769219 290 907958' 153 861261 443 138739 y Cosine „ 1 Sine 1 [' Uotang. 1 Tang. 1 M. | Ee 54 Degrees. 64 (36 Degrees.) a TABLE OF LouARITHMIC M. Sine • 1 D. Cosine | I). Turn. D. Cotansr. | | "o" 9.7692J9 290 9.907958 153 9.861261 443 10.1387.39 60 1 769393 289 907866 153 861527 443 138473 59 2 769566 289 90V VV4 153 861792 442 138208 58 3 769740 289 907682 153 862058 442 137942 57 4 769913 289 907590 153 862323 442 137677 56 5 770087 289 907498 153 862589 442 137411 55 6 770260 288 907406 153 862854 442 137146 54 7 770433 288 907314 154 863119 442 136S81 53 8 770606 288 907222 154 863385 442 136615 52 9 770779 288 907129 154 863650 442 136350 51 10 11 770952 288 907037 154 154 863915 442 136085 50 49 9.771125 288 9 906945 9.864180 442 10.135820 12 771298 287 906852 154 864445 442 135555 48 13 771470 287 906760 154 864710 442 135290 47 14 771643 287 906667 154 864975 441 135025 46 15 771815 287 906575 154 865240 441 134760 45 16 771987 287 906482 154 865505 441 134495 44 17 772159 287 906389 155 865770 441 134230 43 18 772331 286 906296 155 866035 441 133965 42 19 772503 286 906204 155 866300 441 133700 41 20 21 772675 286 906111 155 155 866561 441 133436 40 39 9.772847 286 9.906018 9.866829 441 10.133171 22 773018 286 905925 155 867094 441 132906 38 23 77.*? 190 286 905832 155 867358 441 132642 37 24 773361 285 905739 155 867623 441 132377 36 25 773533 285 905645 155 867887 441 132113 35 26 773704 285 905552 155 868152 440 131848 34 27 773875 285 905459 155 868416 440 131584 33 28 774046 285 905366 156 868680 440 131320 32 29 774217 285 905272 156 868945 440 131055 31 30 31 774388 284 905179 156 156 869209 440 130791 10.130527 30 29 9.774558 284 9.905085 9.869473 440 32 774729 284 904992 156 869737 440 130263 28 33 774899 284 904898 156 870001 440 129999 27 34 775070 284 904804 156 870265 440 129735 26 35 775240 284 904711 156 870529 440 129471 25 36 775410 283 904617 156 870793 440 129207 24 37 775580 283 904523 156 871057 440 128943 23 38 775750 283 904429 157 871321 440 128679 22 39 775920 283 904335 157 871.585 440 128415 21 40 41 776090 283 904241 157 157 871849 439 128151 20 19 9.776259 283 9.904147 9.872112 439 10.127888 42 776429 282 904053 157 872376 439 127624 18 43 776598 282 903959 157 872640 439 127360 17 44 776768 282 903864 157 872903 439 127097 16 45 776937 282 903770 157 873167 439 126833 15 46 777106 282 903676 157 873430 439 1265/0 14 47 777275 281 903581 157 873694 439 126306 13 48 777444 281 903487 157 873957 439 126043 12 49 777613 281 903392 158 874220 439 125780 11 50 51 777781 281 281 903298 158 158 874484 439 125516 10 9 9.777950 9.903203 9.874747 439 10.12.52.53 52 778119 281 903108 158 875010 439 124990 8 53 778287 280 903014 158 875273 438 124727 7 54 778455 280 902919 158 875536 438 124464 6 55 778624 280 902824 158 • 875800 438 124200 6 56 778792 280 902729 158 876063 438 12,3937 4 57 778960 280 902634 158 876326 438 123674 3 58 779128 280 902539 159 876589 438 123411 2 59 779295 279 902444 159 876851 438 123149 1 eo_ 779463 279 9023491 159 877114 438 122888 Ctwine 1 Sine 1 Cotang. 1 1 Tang. |M. 1 53 Degrees. SINES AND TANGENTS. (37 Degrees.) 55 M. Sine D. 1 Cosine 1 D. Tans?. D. CotariE. 1 1 "T 9.779463 279 9.902349 159 9. 877114 438 10.122886 60 1 779631 279 902253 159 877377 438 122623 59 2 779798 279 902158 159 877640 438 122360 58 3 779966 279 902063 159 877903 438 122097 57 4 780133 279 901967 159 878165 438 121835 56 5 780300 278 901872 159 878428 438 121572 55 6 780467 278 901776 159 878691 438 121309 54 7 780634 278 901681 1.59 878953 437 121047 53 8 780801 278 901585 159 879216 437 120784 52 9 780968 278 901490 159 879478 437 120522 51 10 11 781134 278 901394 160 160 879741 9.880003 437 120259 50 49 9.781301 277 9.901298 437 10.119997 12 781468 277 901202 160 880265 437 119735 48 13 781634 277 901106 160 880528 437 119472 47 14 781800 277 901010 160 880790 437 119210 46 15 781966 277 900914 100 881052 437 118948 45 16 782132 277 900818 16-) 881314 437 118686 44 17 782298 276 900722 160 881576 437 118424 43 18 782464 276 900626 160 881839 437 118161 42 19 782630 276 900529 160 882101 437 117899 41 20 21 782796 276 900433 9.900337 161 161 882363 436 117637 40 39 9.782961 276 9.882625 436 10.117375 22 783127 276 900240 161 882887 436 117113 38 23 783292 275 900144 161 883148 436 116852 37 24 783458 275 900047 161 883410 436 116590 36 25 783623 275 899951 161 883672 436 116328 35 26 783788 275 899854 161 883934 436 116066 34 27 783953 275 899757 161 884196 436 115804 33 28 - 784118 275 899660 161 884457 436 115543 32 29 784282 274 899564 161 884719 436 115281 31 30 31 784447 274 899467 9.899370 162 162 884980 430 11.5020 30 29 9.784612 274 9.885242 436 10.114758 32 784776 274 899273 162 885503 436 114497 28 33 784941 274 899176 162 885765 436 114235 27 34 785105 274 899078 162 886026 436 113974 26 35 785269 273 898981 162 886288 436 113712 25 36 785433 273 898884 162 886549 435 113451 24 37 785597 273 898787 162 886810 435 113190 23 38 785761 273 898689 162 887072 435 112928 22 39 785925 273 898592 162 887333 435 112667 21 40 41 786089 273 898494 9.898397 163 163 887594 435 112406 20 19 9.786252 272 9.887855 435 10.112145 42 786416 272 898299 163 888116 435 111884 18 43 786579 272 898202 163 888377 435 111623 17 44 786742 272 898104 163 888639 435 . 111361 16 45 786906 272 898006 163 888900 435 lUlOO 15 46 787069 272 897908 163 889160 435 110840 14 47 787232 271 897810 163 889421 435 110579 13 48 787395 271 897712 163 889682 435 110318 12 49 787557 271 897614 163 889943 435 110057 11 60 51 787720 271 897516 9.897418 163 164 890204 434 109796 10 9 9.787883 271 9.890465 434 10.109535 52 788045 271 897320 164 890725 434 109275 8 53 788208 271 897222 164 890986 434 109014 7 54 788370 270 897123 164 891247 434 108753 6 55 788532 270 897025 164 891507 434 108493 5 56 788694 270 896926 164 891768 434 108232 4 57 788856 270 896828 164 892028 434 107972 3 58 789018 270 890729 164 892289 434 107711 2 59 789180 270 896631 164 892549 434 107451 1 60 789342 269 896532 164 892810 434 107190 ~ Cosine Sine 1 Colang. 1 Tang. 1 1 Degrees 66 (38 Degrees.) a TABLE OF LOGABITHMTC M. 1 Sine D. 1 Cosine | D. Tang. D. CotanEj. 1 9.789342 269 9.896532 164 9.892810 434 10.107190,60 1 789504 269 896433 165 893070 434 106930 59 2 789665 269 896335 165 893331 434 106669 58 3 789827 269 896236 165 893591 434 106409 57 4 789988 269 896137 165 893851 434 106149 56 5 790149 269 896038 165 894111 434 105889 55 6 790310 268 895939 165 894371 434 105629 54 7 790471 268 895840 165 894632 433 105368 53 8 790632 268 895741 165 894892 433 105108 52 9 790793 268 895641 165 895152 433 104848 51 10 11 790954 268 895542 165 166 895412 9.895672 433 104588 50 49 9.791115 268 9.895443 433 10,104328 12 791275 267 895343 166 895932 433 104068 48 13 791436 267 895244 166 896192 433 103808 47 14 791596 267 895145 166 896452 433 103548 46 15 791757 267 895045 166 896712 433 103288 45 16 791917 267 894945 166 896971 433 103029 44 17 792077 267 894846 166 897231 433 102769 43 18 792237 266 894746 166 897491 433 102509 42 19 792397 266 894646 166 897751 433 102249 41 20 21 792557 266 894546 9.894446 166 167 898010 433 101990 40 39 9.792716 266 9.898270 433 10.101730 22 792876 266 894346 167 898530 433 101470 38 23 793035 266 894246 167 898789 433 101211 37 24 793195 265 894146 167 899049 432 100951 36 25 793354 265 894046 167 899308 432 100692 35 26 793514 265 893946 167 899568 432 100432 34 27 793673 265 893846 167 899827 432 100173 33 28 793832 205 893745 167 900086 432 099914 32 29 793991 265 893645 167 900346 432 099654 31 30 31 794150 264 893544 167 168 900605 432 099395 30 29 9.794308 264 9.893444 9.900864 432 10.099136 32 794467 264 893343 168 901124 432 098S76 28 33 794626 264 893243 168 901383 432 098617 27 34 794784 264 893142 168 901642 432 098358 26 35 794942 264 893041 168 901901 432 098099 25 36 795101 264 892940 168 902160 432 097840 24 37 795259 263 892839 168 902419 432 097581 23 38 795417 263 892739 168 902679 432 097321 22 39 795575 263 892638 168 902938 432 097062 21 40 41 795733 263 892536 168 169 903197 431 096803 20 19 9.795891 263 9.892435 9.903455 431 10.096545 42 796049 263 892334 169 903714 431 096286 18 43 796206 263 892233 169 903973 431 096027 17 44 796364 262 892132 169 904232 431 095768 16 45 796521 262 892030 169 904491 431 095509 15 46 796679 262 891929 169 904750 431 095250 14 47 796836 262 891827 169 905008 431 094992 13 48 796993 262 891726 169 905267 431 094733 12 49 797150 261 891634 169 905526 431 . 094474 11 50 51 797307 261 891523 9.891421 170 170 905784 9.906043 431 094216 10 9 9.797464 261 431 10.093957 52 797621 261 891319 170 906302 431 093698 8 53 797777 261 891217 170 906560 431 093440 7 54 797934 261 891115 170 906819 431 093181 6 55 798091 261 891013 170 907077 431 092923 5 56 798247 261 890911 170 907336 431 092664 4 57 798403 260 890809 170 907594 431 092406 3 58 798560 260 890707 170 907852 431 092148 2 59 798716 260 890605 170 908111 430 091889 1 60 798872 260 890503 170 908369 430 091631 ~ Cosine Sine 1 1 Coiang. 1 Tang. |M. 1 Si Degrees. SINES AND TANGENTS. (39 I )egrees 67 M. i Sine 1 I>. 1 Cosine j D. ! Tang. D. 1 Cotang. 1 1 9.798872 260 9.890503 170 9.908369 430 10.091631 60 1 799028 260 890400 171 908628 430 091372 59 2 799184 260 890298 171 908886 430 091114 58 3 799339 259 890195 171 909144 430 090856 57 4 799495 259 890093 171 909402 430 090598 56 5 799651 259 889990 171 909660 430 090340 55 6 799806 259 889888 171 909918 430 090082 54 7 799962 259 889785 171 910177 430 089823 53 8 800117 259 889682 171 910435 430 089565 52 9 800272 258 889579 171 910693 430 089307 51 10 11 800427 258 889477 9.889374 171 172 910951 430 089049 50 49 9.800582 258 9.911209 430 10.088791 12 800737 258 889271 172 91 1467 430 088533 48 13 800892 258 889168 172 911724 430 088276 47 14 801047 258 889064 172 911982 430 088018 46 .15 801201 258 888961 172 912240 430 087760 45 16 801356 257 888858 172 A912498 430 087502 44 17 801511 257 888755 172 912756 430 087244 43 18 801665 257 888651 172 913014 429 086986 42 19 801819 257 888548 172 913271 429 086729 41 20 801973 257 888444 173 913529 429 086471 40 21 9.802128 257 9.888341 173 9.913787 429 10.086213 39 22 802282 256 888237 173 914044 429 085956 38 23 802436 256 888134 173 9143,02 429 085698 37 24 802589 256 888030 173 914560 429 085440 36 25 802743 256 887926 173 '914817 429 085183 35 26 802897 256 887822 173 915075 429 084925 34 27 803050 256 887718 173 915332 429 084G68 33 28 803204 256 887614 173 91,5590 429 084410 32 29 803357 255 887510 173 915847 423 084153 31 30 80.3511 255 887406 174 916104 429 083896 30 31 9.803664 255 9.887302 174 9.916362 429 10.083638 29 32 803817 255 887198 174 916619 429 083.381 28 33 803970 255 887093 174 916877 429 083123 27 34 804123 2.55 886989 174 917134 429 082866 26 35 804276 254 886885 174 917391 429 082609 26 36 804428 254 886780 174 917648 429 082352 24 27 804581 254 886676 174 917905 429 082095 23 38 804734 254 886571 174 918163 428 081837 22 39 804886 254 886466 174 918420 420 08^.580 21 40 41 805039 254 886362 9.88.6257 175 175 918677 428 081323 10.081066 20 19 9.805191 254 9.918934 ■^"M 42 805343 253 886152 175 919191 42«^ 080809 18 43 805495 253 886047 175 919448 428 080552 17 44 805647 253 885942 175 919705 428 080295 16 45 805799 253 885837 175 919962 428 080038 15 46 805951 253 885732 175 920219 428 079781 14 47 806103 253 885627 175 920476 428 ' 079524 13 48 800254 253 8855S2 175 92073S 428 079267 12 49 806406 252 885416 175 920990 428 079010 11 50 51 806557 9.806709 252 885311 9.88.5205 176 176 921247 428 078753 10 9 252 9.921503 428 JO. 078497 52 806860 252 885100 176 921760 428 078240 8 53 807011 252 884994 176 922017 428 077983 7 54 807163 252 884889 176 922274 428 077726 6 55 807314 252 884783 176 922530 428 077470 5 56 807465 251 884677 176 922787 428 077213 4 57 807615 251 884572 176 923044 428 076956 3 58 807766 251 884466 176 923300 428 076700 2 59 807917 251 884360 176 923557 427 076443 1 60 808067 251 884254 177 923813 427 076187 _J[ Cosine 1 Sine 1 Colang. 1 Tanj,'. T 50 Degrees. H 58 (40 Degr ees.') A TABLE OF LOGAEITHMIC M. 1 Sine 1 D. 1 Cosine ] D. Tang. D. Cotanp. 1 1 "T "9.808067 251 9.884254 177 9.923813 427 10.076187 60 1 808218 251 884148 177 924070 427 07593U 59 2 808368 251 884042 177 924327 427 075673 58 3 808519 250 883936 177 924583 427 075417 57 4 808669 250 883829 177 924840 427 075180 56 5 808819 250 883723 177 925096 427 074904 55 6 808969 250 883617 177 925352 427 074648 54 7 809119 250 883510 177 925609 427 07439] 53 8 809269 250 883404 177 925865 427 074135 52 9 809419 249 883297 178 926122 427 073878 51 10 11 809569 9.809718 249 883191 9.883084 178 178 926378 427 073622 50 49 249 9.926634 427 10.073366 12 809868 249 . 882977 178 925890 427 073110 48 13 810017 249 882871 178 927147 427 072853 47 14 810167 249 882764 178 927403 427 072597 46 15 810316 248 882657 178 927659 427 072341 45 16 810465 248 882550 178 927915 427 072085 44 17 810614 248 882443 178 928171 427 071829 43 18 810763 248 882336 179 928427 427 071573 42 19 810912 248 882229 179 928683 427 071317 41 20 21 811061 248 882121 9.882014 179 179 928940 9.929196 427 427 071060 10.070804 40 39 9.811210 248 22 811358 247 881907 179 929452 427 070548 38 23 811507 247 881799 179 929708 427 070292 37 24 811655 247 881692 179 929964 426 070036 36 25 811804 247 881584 179 930220 426 069780 35 26 811952 247 881477 179 930475 426 069525 34 27 812100 247 881369 179 930731 426 069269 33 28 812248 247 881261 180 930987 426 069013 32 29 812396 246 8811.53 180 931243 426 068757 31 30 31 812544 246 881046 9.880938 180 180 931499 9.931755 426 068501 10.068245 30 29 9.812692 246 426 32 « 12840 240 880830 180 932010 426 067990 28 33 812988 246 880722 180 932266 426 067734 27 34 813135 246 880613 180 932522 426 067478 26 35 813283 246 880505 180 932778 426 067222 25 36 813430 245 880397 180 933033 426 066967 24 37 813578 245 880289 181 933289 426 066711 23 38 813725 245 880180 181 933545 426 06.6455 22 39 813872 245 880072 181 933800 426 066200 21 40 41 814019 9.814166 245 f45 879963 9.8798,55 181 181 934056 426 065944 20 19 9.934311 426 10.065689 42 814313 879746 181 934567 426 065433 18 43 814460 244 879637 181 934823 426 065177 17 44 814607 244 879529 181 935078 426 064922 16 45 814753 244 879420 181 935333 426 064667 15 46 814900 244 879311 181 935589 426 064411 14 47 815046 244 879202 182 935844 426 064156 13 48 815193 244 879093 183 936100 426 060900 IS 49 815339 244 878984 182 936355 426 063645 11 50 51 815485 243 878875 9.878766 182 182 936610 426 063390 10 9 9.815631 243 9.9.36866 425 10.063134 52 815778 243 878656 182 937121 425 062879 8 53 815924 243 878547 182 937376 425 062624 7 54 816069 243 878438 182 937632 425 062368 6 55 816215 243 878328 182 937887 425 062113 6 56 816361 243 878219 183 938142 425 061858 4 57 816507 242 878109 183 938398 425 061602 3 58 816652 242 877999 183 938653 425 061347 2 59 816798 242 877890 183 938908 425 061092 1 60 816943 242 877780! 183 939163 425 060837 Cosine 1 Sine 1 Cotang, Tang. M. | 49 Degrees. SINES AND TANGENTS . (41 Degrees.) 69 M. Sine 1 D. Cosine | D. Tann. D. Cotanfj. 1 9.816943 24^ 9.877780 183 9.939163 425 10.060837 60 1 817088 242 877670 183 939418 425 060682 59 2 817233 242 877560 183 939673 425 060327 58 3 817379 242 877460 183 939928 426 060072 67 4 817524 241 877340 183 940183 425 059817 56 6 817668 241 877230 184 940438 425 059562 65 6 817813 241 877120 184 940694 425 059306 54 7 817958 241 877010 184 940949 425 059051 53 8 818103 241 876899 184 941204 425 058796 52 9 818247 241 876789 184 941468 426 058542 51 10 11 818392 241 876678 184 184 941714 426 058286 50 49 9.818636 240 9.876568 9.941968 425 10.058032 12 818681 240 876457 184 942223 425 057777 48 13 818826 240 876.347 184 942478 425 057522 47 14 818969 240 876236 185 942733 425 057267 46 15 819113 240 876125 186 942988 426 057012 45 16 819267 240 876014 186 943243 425 056767 44 17 819401 240 876904 186 943498 425 066502 43 18 819545 239 876793 185 943752 426 056248 42 19 819689 239 875682 185 944007 425 055993 41 20 21 819832 239 875571 9.875469 186 185 944262 425 055738 10.055483 40 39 9.819976 239 9.944517 425 22 820120 239 875348 186 944771 424 055229 38 23 820263 239 876237 186 946026 424 054974 37 24 820406 239 875126 186 945281 424 054719 36 25 820650 238 875014 186 945536 424 054465 35 26 820693 238 . 874903 186 945790 424 064210 34 27 820836 238 874791 186 946045 424 053956 33 28 820979 238 874680 186 946299 424 063701 32 29 821122 238 874568 186 946564 424 053446 31 30 31 821266 238 874456 9.874344 186 186 946808 424 053192 10.052937 30 29 9.821407 238 9.947063 424 32 821650 238 874232 187 947318 424 062682 28 33 821693 237 874121 187 947672 424 052428 27 34 821835 237 874009 187 947826 424 0.52174 26 35 821977 237 873896 187 948081 424 051919 25 36 822120 237 873784 187 948336 424 051664 24 37 822262 237 873672 187 948590 424 051410 23 38 822404 237 873560 187 948844 424 051156 22 39 822546 237 873448 187 949099 424 050901 21 40 41 822688 236 873336 187 187 949353 9.949607 424 050647 20 19 9.822830 236 9.873223 424 10.050393 42 822972 236 873110 188 949862 424 0,50138 18 43 823114 236 872998 188 950116 424 049884 17 44 823255 236 872885 188 950370 424 049630 16 45 823397 236 872772 188 950626 424 049375 15 46 823539 236 872659 188 950879 424 049121 14 47 823680 235 872647 188 951133 424 048867 13 48 823821 235 872434 188 951388 424 048612 12 49 823963 236 872321 188 951642 424 048358 11 50 51 824104 9.824245 235 872208 9.872096 188 189 961896 424 048104 10 9 236 9.962150 424 10.047850 52 824386 235 871981 189 952405 424 047695 8 53 824527 235 871868 189 952659 424 047341 7 54 824668 234 871755 189 962913 424 047087 6 56 824808 234 871641 189 953167 423 046833 6 56 824949 234 871628 189 953421 423 046579 4 57 825090 234 871414 189 953675 423 046325 3 58 825230 234 871301 189 953929 423 046071 2 69 826371 234 871187 189 964183 423 046817 1 60 825511 234 871073 190 954437 423 045563 1 Cosine " «ine 1 Cotaiig. Tang. 1 M. | 48 Degrees. 60 (4i I Degrees.) a TABLE OF LOGARITHMIC M. Sine D. Cosine 1 D. Tanc D. Cotanp. 1 1 9.825511 234 9.871073 190 9.954437 423 10.046663 60 1 825651 233 870960 190 954691 423 046309 69 2 825791 233 870846 ]90 954945 423 045055 58 3 825931 233 870732 ]90 955200 423 044800 67 4 826071 233 870618 190 955454 423 044646 66 5 826211 233 870504 190 955707 423 044293 55 6 826351 233 870390 190 955961 423 044039 54 7 826491 233 870276 190 956215 423 043785 53 8 826631 233 870161 190 966469 423 043531 52 9 826770 232 870047 191 966723 423 043277 51 10 11 826910 232 869933 9.869818 191 191 966977 9.967231 423 043023 10.042769 50 49 9.827049 232 423 12 827189 232 869704 191 957485 423 042515 48 13 827328 232 869589 191 957739 423 042261 47 14 827467 232 869474 191 957993 423 042007 46 15 827600 232 869360 191 958246 423 041764 45 16 827745 232 869245 191 958600 423 041600 44 17 827884 231 869130 191 9587.54 423 041246 43 18 828023 231 869015 192 969008 423 040992 42 19 828162 231 868900 192 959202 423 040738 41 20 21 828301 231 868785 192 192 959516 9.959769 423 423 040484 10.040231 40 39 9.828439 231 9.868670 22 828578 231 868555 192 960023 423 039977 38 23 828716 231 868440 192 960277 423 039723 37 24 828855 230 868324 192 960531 423 039469 36* 25 828993 230 868209 192 960784 423 039216 35 26 829131 230 868093 192 961038 423 038962 34 27 829269 230 867978 193 961291 423 038709 33 28 829407 230 867862 193 961545 423 038455 32 29 829545 2.30 867747 193 961799 423 038201 31 30 31 829683 230 867631 9.867515 193 193 962052 423 037948 10.037694 30 29 9.829821 229 9.962306 423 32 829959 229 867399 193 962560 423 037440 28 33 830097 229 867283 193 962813 423 037187 27 34 830234 229 867167 193 963067 423 036933 26 35 830372 229 867051 193 963320 423 036680 25 36 830509 229 866935 194 963674 423 036426 24 37 830646 229 866819 194 963827 423 036173 23 38 830784 229 . 866703 194 964081 423 036919 22 39 830921 228 866586 194 964335 423 036665 21 40 41 831058 228 866470 194 194 964688 422 036412 10.036158 20 19 9.831195 228 9.866353 9.964842 422 42 831332 228 866237 194 965095 422 034905 18 43 831469 228 866120 194 965349 422 034651 17 44 831606 228 866004 195 965602 422 034398 16 45 831742 228 865887 195 966856 422 034145 15 46 831879 228 865770 195 966109 422 0.33891 14 47 832015 227 865653 195 966362 422 033638 13 48 832162 227 865636 195 966616 422 033384 12 49 832288 227 86.5419 196 966869 422 033131 11 50 51 832426 227 865302 9.865185 195 196 967123 422 032877 10 9 9.832561 227 9.967376 422 10.032624 52 832697 227 865068 195 967629 422 032371 8 53 832833 227 864950 195 967883 422 032117 7 54 832969 226 864833 196 968136 422 031864 6 55 833105 226 864716 196 968389 422 031611 5 56 833241 226 864598 196 968643 422 031357 4 67 833377 226 864481 196 968896 422 031104 3 58 833512 226 864363 196 969149 422 030861 2 59 833648 226 864245 196 969403 422 030597 1 60 833783 226 864127' 196 969656 422 030344 1 Cosine 1 Sine 1 Cotang. P Tang. 1 M. 47 Degrees. ^m " -"'i^" SINES AND TANGENTS . (43 Degrees.) 61 M. Sine D. Cosine | D. TanR. D. Cotang. 1 1 9.833783 226 9.864127 196 9.969656 422 10.030344 *60 1 833919 225 864010 196 969909 422 030091 59 2 834054 225 863892 197 970162 422 029838 58 3 834189 225 863774 197 970416 422 029584 57 4 834325 225 863656 197 970669 422 029331 56 5 834460 225 863538 197 970922 422 029078 55 6 834595 225 863419 197 971175 422 028825 54 7 834730 225 863301 197 971429 422 028571 53 8 834865 225 863183 197 971682 422 028318 52 9 834999 224 863064 197 971935 422 028065 51 10 11 835134 224 862946 198 198 972188 9.972441 422 422 027812 10.027559 50 49 9.835269 224 9.862827 12 835403 224 862709 198 972694 422 027306 48 13 835538 224 862590 198 972948 422 027052 47 14 835672 224 862471 198 973201 422 026799 46 15 835807 224 862353 198 973454 422 026546 45 16 835941 224 862234 198 973707 422 026293 44 17 836075 223 862115 198 973960 422 026040 43 18 836209 223 861996 198 974213 422 025787 42 19 836343 223 861877 198 974466 422 025534 41 20 21 836477 223 861758 199 199 974719 422 025281 10.025027 40 39 9.836611 223 9.861638 9.974973 422 22 836745 223 861519 199 975226 422 024774 38 23 836878 223 861400 199 975479 422 024521 37 24 837012 222 861280 199 975732 422 024268 36 25 837146 222 861161 199 975985 422 024015 35 26 837279 222 861041 199 976238 422 023762 34 27 837412 222 860922 199 976491 422 023509 33 28 837546 222 860802 199 976744 422 023256 32 29 837679 222 860682 200 970997 422 023003 31 30 31 837812 222 860562 9.860442 200 200 977250 9.977503 422 022750 30 29 9.837945 222 422 10.022497 32 838078 221 860322 200 977756 422 022244 28 33 838211 221 860202 200 978009 422 021991 27 34 838344 221 860082 200 978262 422 021738 26 35 838477 221 859962 200 978515 422 021485 25 36 838610 221 859842 200 978768 422 021232 24 37 838742 221 859721 201 979021 422 020979 23 38 838875 221 859601 201 979274 422 020726 22 39 839007 221 859480 201 979527 422 020473 21 40 41 839140 220 859360 9.859239 201 201 979780 422 020220 20 19 9.839272 220 9.980033 422 10.019967 42 839404 220 859119 201 980286 422 019714 18 43 839536 220 858998 201 980538 422 019462 17 44 839668 220 858877 201 980791 421 019209 16 45 839800 220 858756 202 981044 421 018956 15 40 839932 220 858635 202 981297 421 018703 14 47 840064 219 858514 202 981550 421 018450 13 48 840196 219 858393 202 981803 421 018197 12 49 840328 219 858272 202 982056 421 017944 11 50 51 840459 219 858151 9.858029 202 202 982309 421 017691 10.017438 10 9 9.840591 219 9.982562 421 52 840722 219 857908 202 982814 421 017186 8 53 840854 219 857786 202 983067 421 016933 7 54 840985 219 857665 203 983320 421 016680 6 55 841116 218 857543 203 983573 421 016427 5 56 841247 218 857422 203 983826 421 016174 4 57 841378 218 857300 203 984079 421 015921 3 58 841509 218 857178 203 984331 421 015669 2 59 841640 218 857056 203 984584 421 015416 1 60 841771 218 856934 203 984837 421 0151631 1 Cosine 1 1 Sine 1 Cotaiig. 1 Tang. 1 M. 46 Degrees. Hh 62 (44 Degrees.) a TABLE OP LOGARITHMIC nn" Sine 1 D. 1 Cosine | D. | Tuns. 1 D. | Cotansr. | f 9.841771 218 9.856934 203} 9.984837 421 10.015163 60 1 841902 218 856812 203| 985090 421 014910 59 2 842033 218 856690 204 985343 421 014657 58 3 842163 217 856568 204 985596 421 014404 57 4 842294 217 856446 204 985848 421 014152 56 5 842424 217 856323 204 986101 421 013899 55 6 842555 217 85620 1 204 986354 421 013646 54 7 842685 217 856078 204 980607 421 013393 53 8 842815 217 855956 204 986800 421 013140 52 9 842946 217 855833 204 987112 421 012888 51 10 11 843076 9.843206 217 85.5711 205 205 987365 9^.987618 421 012635 50 49 216 9.855588 421 10.012382 12 843336 216 855465 205 987871 421 012129 48 13 843466 843595 216 855342 205 988123 421 0'11877 47 14 216 855219 205 988376 421 0-11624 46 15 843725 216 855096 205 988629 421 011371 45 16 843855 216 854973 205 988882; 421 011118 44 17 843984 216 854850 205 989134 421 010866 43 18 844114 215 854727 206 989387 421 010613 42 19 844243 215 854603 206 989640 421 010360 41 20 21 844372 215 854180 9.854356 206 206 989893 9^.990145 421 010107 40 39 9.844502 2i5 421 10.009855 22 844631 215 854233 206 990398 421 009602 38 23 844760 215 854109 206 990651 421 009349 37 24 844889 215 853986 206 990903 421 009097 36 25 845018 215 853862 206 991156 421 008844 35 26 845147 215 853738 206 991409 421 008591 34 27 845276- 214 853614 207 991662 421 008338 33 28 845405 214 853490 207 991914 421 008086 32 29 845533 214 853366 207 992167 421 007833 31 30 31 845662 9.845790 214 853242 9.853118 207 207 992420 421 0075SO 30 29 214 9.992672 421 10.007328 32 845919 214 852994 207 992925 421 007075 28 33 846047 214 852869 207 993178 421 006823 27 34 846175 214 852745 207 993430 421 006570 26 35 846304 214 852620 207 993683 421 006317 25 36 846432 213 852496 208 993936 421 006064 24 37 846560 213 852371 208 994189 421 005811 23 38 846688 213 852247 208 994441 421 005559 22 39 840816 213 852122 208 994694 421 005306 21 40 846944 213 851997 208 994947 421 005053 20 41 9.847071 213 9.851872 208 9.995199 421 10.004801 19 42 847199 213 851747 208 995452 421 004548 18 43 847327 213 851622 208 995705 421 004295 17 44 847454 212 851497 209 995957 421 004043 16 45 847582 212 851372 209 996210 421 003790 15 46 847709 212 851246 209 996463 421 003537 14 47 847836 212 851121 209 996715 421 003285 13 48 847964 212 850996 209 996968 421 003032 12 49 848091 212 850870 209 997221 421 002779 11 50 848218 212 850745 209 997473 421 002527 10 51 9.848345 212 9.850619 209 9.997726 421 10.002274 9 52 848472 211 850493 210 997979 421 002021 8 53 848599 211 850368 210 998231 421 001769 7 54 848726 211 850242 210 998484 421 001516 6 55 848852 211 850116 210 9987371 421 001263 5 56 848979 211 849990 210 998989 421 001011 4 57 849106 211 849864 210 999242! 421 000758 3 58 843232 211 849738 210 999495 421 000505 2 59 849359 211 849611 210 999748 421 000253 1 Igo 849485 211 849485 210 10.000000 421 OOOOOOi Cosine I aine 1 1 Co;'aiig. 1 1 Tang. 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