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 Complete trigonometry, 
 
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Cornell University 
 Library 
 
 The original of this book is in 
 the Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924004627471 
 
COMPLETE TRIGONOMETRY 
 
 BY 
 
 WEBSTER WELLS, S.B. 
 
 PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
 INSTITUTE OF TECHNOLOGY 
 
 REVISED EDITION 
 
 D C HEATH & CO., PUBLISHEES 
 BOSTON NEW YOEK CHICAGO 
 
Copyright, 1900 and 1911, 
 By WEBSTER WELLS 
 
 1g6 
 
PREFACE. 
 
 The present volume is a revision of the author's Essen- 
 tials of Trigonometry. 
 
 In preparing the new edition, many improvements have 
 heen effected ; the attention of teachers is specially invited 
 to the following features of the work : 
 
 1. The proofs of the functions of 120°, 135°, 150°, etc.; 
 § 27. 
 
 2. The proofs of the functions of (— A), and (90° +4), 
 in terms of those of A; §§28, 29. 
 
 3. The general demonstration of the formulae 
 
 tan x — , and sinrx + cos 2 a; = 1, 
 
 cos a; 
 
 in §§ 34 and 36; the four cases being considered together. 
 
 4. The general demonstrations of the formulae 
 
 cos x 
 sin x' 
 
 cot x = — , sec 2 a; = 1 + tan 2 a;, and cse 2 * = 1 + cot 2 a; 
 
 sin 1? ' 
 
 in §§ 35 and 36. 
 
 5. The proofs of the formulae for sin (a; + y) and cos (a; + y), 
 when x and y are acute ; the two cases when x + y is acute 
 or obtuse being considered together ; § 38. 
 
 6. The proofs of the formulae for tan i x and cot £ x; § 45. 
 
 7. The new arrangement of examples in § 46. 
 
 8 The introduction of trigonometric equations in §§ 54, 
 55, and 56. 
 
 iii 
 
iv Preface. 
 
 9. The solution of right triangles by Natural Functions; 
 see Ex. 1, page 76. 
 
 The new work contains a great many more examples than 
 the old ; they have been selected with great care, and most 
 of them are new. It is not expected that every class will 
 solve all the examples; they are sufficiently numerous to 
 furnish a variety in successive years. 
 
 Attention is specially invited to the sets in §§ 96, 114, 157, 
 and 160. 
 
 In § 112 will be found a set of miscellaneous examples in 
 the solution of plane oblique triangles, and in § 155 a set in 
 spherical oblique. 
 
 In the Appendix to the Plane Trigonometry there will be 
 found a discussion of the principles of Parallel, Middle 
 Latitude, and Traverse Sailing, with problems. 
 
 The results have been worked out by aid of the author's 
 New Four Place Logarithmic Tables, which contain also 
 
 Tables of Natural Functions. 
 
 i 
 
 The present revised edition has been made with the 
 co-operation of Professor N. R. George of the Massachusetts 
 Institute of Technology, to whom are due the author's 
 thanks for the preparation of § § 32, 46, and 51 to 56 inclu- 
 sive, as well as for improvements in other parts of the text. 
 
 WEBSTER WELLS. 
 Massachusetts Institute op 
 Technology, 1911. 
 
CONTENTS. 
 
 PLANE TRIGONOMETRY. 
 
 PAGl 
 
 L Trigonometric Functions of Acute Angles 1 
 
 IL Trigonometric Functions of Angles in General 7 
 
 HI. General Formulae 26 
 
 IV. Miscellaneous Theorems 38 
 
 V. Logarithms 60 
 
 Properties of Logarithms ..... 62 
 
 Applications 68 
 
 VI. Solution op Right Triangles .... 75 
 
 Formulae for the Area of a Right Triangle . . 82 
 
 "VTI. General Properties of Triangles ... 85 
 
 Formulae for the Area of an Oblique Triangle . 90 
 
 VTLT. Solution of Oblique Triangles .... 92 
 
 Appendix to Plane Trigonometrt . . . 106 
 
 Parallel Sailing 106 
 
 Middle Latitude Sailing 107 
 
 Traverse Sailing . 109 
 
 v 
 
vi Contents 
 
 SPHERICAL TRIGONOMETRY. 
 
 PAGE 
 
 IX. Geometrical Principles Ill 
 
 X. Right Spherical Triangles 115 
 
 Solution of Right Spherical Triangles . . . 122 
 
 XI. Oblique Spherical Triangles .... 132 
 
 General Properties of Spherical Triangles . . 132 
 
 Napier's Analogies 138 
 
 Solution of Oblique Spherical Triangles . . 141 
 
 XII. Applications 153 
 
 Formulae : 
 
 Plane Trigonometry - 158 
 
 Spherical Trigonometry 161 
 
 Answers 
 
 Use op the Tables 
 
PLANE TRIGONOMETRY. 
 
 oXKo 
 
 I. TRIGONOMETRIC FUNCTIONS OP ACUTE 
 ANGLES. 
 
 1. Trigonometry treats of the properties and measure- 
 ment of angles and triangles. 
 
 In Plane Trigonometry, we consider plane figures only. 
 
 2. Definitions of the Trigonometric Functions of Acute 
 Angles. 
 
 Let BAC be any acute angle. 
 
 From any point in either side, as B, draw line BO per- 
 pendicular to AC, forming right triangle ABC. 
 
 We then have the following definitions, applicable to 
 either of the acute angles A or B: 
 
 In any right triangle, 
 
 The sine of either acute angle is the ratio of the opposite 
 side to the hypotenuse. 
 
 The cosine is the ratio of the adjacent side to the hypotenuse. 
 
 The tangent is the ratio of the opposite side to the adjacent 
 
 side. 
 
 The cotangent is the ratio of the adjacent side to the oppo- 
 site side. 
 
Plane Trigonometry. 
 
 The secant is the ratio of the hypotenuse to the adjacent side. 
 The cosecant is the ratio of the hypotenuse to the opposite 
 
 We also have the following definitions : 
 
 The versed sine of an angle is 1 minus the cosine of the 
 angle. 
 
 The coversed sine is 1 minus the sine. 
 
 The eight ratios defined above are called the Trigono- 
 metric Functions of the angle. 
 
 Representing sides BC, GA, and AB, by a, b, and c, 
 respectively, and employing the usual abbreviations, we 
 have: 
 
 ■ A a 
 
 sin A = — 
 c 
 
 tan A = r - 
 
 
 
 . c 
 
 secA = T - 
 
 b 
 
 vers A = 1 - 
 
 b 
 
 c 
 
 A ° 
 
 cos A = — 
 c 
 
 cot A = — 
 a 
 
 A C 
 
 csc A = — 
 a 
 
 covers A = 1 - 
 
 a 
 c 
 
 sin B = — 
 c 
 
 tanZ? = -- 
 a 
 
 r. c 
 
 sec_B = — 
 
 a 
 
 vers B = 1 - 
 
 a 
 c 
 
 cos B = -- 
 
 c 
 
 cotB = ^. 
 
 
 
 T> C 
 
 b 
 
 covers B = l- 
 
 b 
 c 
 
 3. It is important to observe that the values of the trigo- 
 nometric functions depend solely on the magnitude of the 
 angle, and are entirely independent of the lengths of the 
 sides of the right triangle which contains it. 
 
 For let B and B' be any two points in side AD of angle 
 DAE, and draw lines BC and B'G perpendicular to AE. 
 
Trigonometric Functions. 
 
 Then, by the definitions of § 2, 
 
 ■ . SO , • , B'C 
 
 svo.A = , and smJ = 
 
 AB AB' 
 
 But right triangles ABC and AB'C are similar, since 
 they have angle A common. 
 "Whence, by Geometry, 
 
 BG = B'C 
 AB AB'' 
 
 Thus the two values found for sin A are equal. 
 The same may be proved true of each of the remaining 
 functions. 
 
 4. We have from § 2, 
 
 sin A = cos B. sec A = esc B. 
 
 tan A = cot B. vers A = covers B. 
 
 But B is the complement of A. 
 
 Hence, the sine, tangent, secant, and versed sine of any 
 acute angle are, respectively, the cosine, cotangent, cosecant, 
 and coversed sine of the complement of the angle. 
 
 & From § 2, sin A = - = cos B, and cos A = - = sin B. 
 c c 
 
 Whence, 
 
 a = e sin A = c cos B, and 6 = c sin B = c cos A. 
 
 That is, in any rigJU triangle, either side about the right 
 angle is equal to the sine of the opposite angle, or the cosine 
 of the adjacent angle, multiplied by the hypotenuse. 
 
 6. To find the Values of the Other Seven Functions of an 
 Acute Angle, when the Value of Any One is Given. 
 
 1. Given esc A = 3 ; find the values of the remaining 
 functions of A. 
 
Plane Trigonometry. 
 B 
 
 2V3 
 
 We may write the equation cso A= — 
 
 Since the cosecant is the hypotenuse divided by the opposite side, 
 we may regard A as one of the acute angles of right triangle ABC. 
 in which hypotenuse AB = 3, and opposite side BO = 1. 
 
 By Geometry, AC = VAB 2 -BC 2 = V§~^~1 = V8 = 2 V2. 
 Then by the definitions of § 2, 
 
 sin^l ; 
 
 tan A = - 
 
 sec A = - 
 
 3 
 
 2a/2 
 
 2VI 
 
 A 2V2 
 
 cos A = ?-LZ.. 
 3 
 
 cot A = 2 V2. 
 covers ^ = 1 — = — 
 
 vers -4 = 1 — 
 
 2v5 
 
 3 3 
 
 2 
 2. Given vers -4 = ^ ; find the value of cot A. 
 5 
 
 Since vers A = 1 — cos A, we have cos A = 1 — - = -. 
 
 5 5 
 
 Then, in right triangle ABC, we take adjacent side AC =3, and 
 hypotenuse AB = 5. 
 
 Whence, SO = VaB 2 - AC 1 = V25 - 9 = Via = 4. 
 
 Q 
 
 Then, by definition, cot A = — 
 
Trigonometric Functions. 
 
 EXAMPLES. 
 
 In each, of the following, find the values of the remaining 
 functions : 
 
 3 
 
 3. sin A = -• 6. esc A = 7. 9. sec A = x. 
 
 o 
 
 4. vers-4 = — • 7. cos^l = ^- 10. tan.4 = l- 
 
 13 14 x 
 
 7 ° a 
 
 5. cotA = — - 8. covers ,1 = — • 11. sin-4=— 
 
 24 17 6 
 
 3 
 
 12. Given cot A = -; find sin A 
 
 13. Given esc A = - ; find cos A. 
 
 40 
 
 14. Given sec A = 5 : find cot A. 
 
 °1 
 
 15. Given eos^l = — ; find esc A. 
 
 ■29 
 
 16. Given tan J. = ^ ~ ; find sec .4. 
 
 o 
 
 17. Given sinJ. = ^; find vers A. 
 
 7. Functions of 45° 
 
 Let ABC be an isosceles right triangle, C being the right 
 angle, and sides AC and BC being each equal to 1. 
 
 Then, Z.4=4o°, and AB=^AC i +BC*=Vl+l=^/2. 
 
Plane Trigonometry. 
 
 Whence, by definition, 
 
 sin 45° = — =iV2. 
 V2 2 
 
 cos 45° = — = iV2. 
 V2 
 
 covers 
 
 sec45°=V2. 
 csc45°=V2. 
 
 vers45° = l-i-V2 = 
 
 45° = l-iV2 = 
 
 2-V2 
 
 2 
 2-V2 
 
 Let ABD be an equilateral triangle having each side equal 
 to 2, and draw line AC perpendicular to BD. 
 
 By Geometry, BC= \ BD = 1, Z J3^C = | Z J5AD = 30°. 
 
 Also, ^C = VZB 2 _i?G 2 = V4^1=V3. 
 Then from right triangle ABC, by definition, 
 
 cos 30° =^= sin 60°. 
 
 _ 
 
 sin 30° = i=cos60°. 
 
 tan 30° = -±- = \ VB = cot 60°. cot 30° = V3 = tan 60°. 
 
 V3 
 
 sec 30° = -^ = | V3 = esc 60°. esc 30° = 2 = sec 60°. 
 
 vers 30° = 1 - 20. = covers 60°. 
 covers 30° = 1 - \ = 1 = vers 60°. 
 
Trigonometric Functions. 
 
 II. TRIGONOMETRIC FUNCTIONS OP 
 ANGLES IN GENERAL. 
 
 9. In Geometry, we are, as a rule, concerned with angles 
 Less than two right angles ; but in Trigonometry it is con- 
 venient to consider them as unrestricted in magnitude. 
 
 Let be the centre, and XX' and YY' a pair of perpen- 
 dicular diameters, of circle AY' ; OY being above, and Y 
 below, XX', when OX is horizontal and extends to the 
 right, and OX' to the left, of 0. 
 
 Let radius OA (Fig. 1) start from the position OX, and 
 revolve about point as a pivot towards the position OY. 
 
 When OA coincides with OY, it has generated an angle 
 of 90°; when it coincides with OX', of 180°; with OY', 
 of 270°; with OX, its first position, of 360°; with OY again, 
 of 450° ; and so on. 
 
 Hence, a meaning may be attached to a positive angle of 
 any number of degrees. 
 
 10. We may also conceive of a negative angle of any 
 number of degrees. 
 
 Thus, if a positive angle indicates revolution from the 
 position OX towards OF, a negative angle, may be taken as 
 indicating revolution from the position OX in the opposite 
 direction, towards OY 1 - 
 
8 Plane Trigonometry. 
 
 Thus, if radius OA' (Fig. 2) starts from the position OX, 
 and revolves about point as a pivot towards the position 
 OY', when it coincides with OY', it lias generated an angle 
 of -90°; when it coincides with OX', of -180°; with OY, 
 of -270°; and so on. 
 
 Note. It is immaterial which direction we consider the positive 
 direction of rotation ; but having at the outset adopted a certain direc- 
 tion as positive, our subsequent operations must be in accordance. 
 
 11. In generating a positive or negative angle of any 
 number of degrees, the line from which the rotation is sup- 
 posed to commence is called the initial line of the angle, and 
 the final position of the rotating radius the terminal line. 
 
 12. To designate an angle, we always write first the letter 
 at the extremity of the initial line. 
 
 Thus, in designating the angle formed by the lines OX 
 and OA, if we regard OX as the initial line, we should call 
 it XOA ; and if we regard OA as the initial line, we should 
 call it AOX. 
 
 13. There are always two angles less than 360° in abso- 
 lute value, one positive and the other negative, having the 
 same initial and terminal line. 
 
 Thus there are formed by OX and OA' (Fig. 2) the posi- 
 tive angle XOA between 270° and 360°, and the negative 
 angle XOA' between 0° and - 90°. 
 
 We shall distinguish between these angles by referring to 
 them as "the positive angle XOA'," and "the negative 
 angle XOA'," respectively. 
 
 14. Rectangular Co-ordinates. 
 
 Let XX' and YY' be two straight lines intersecting at 
 right angles at O ; the letters being arranged as in the fig- 
 ures of § 9. 
 
 Let Pj be any point in the plane of XX' and YY', and 
 draw line F^M perpendicular to XX'. 
 
Trigonometric Functions. 9 
 
 Then 031 and MJP are called the rectangular co-ordinates 
 of P 1 ; OM is called the abscissa, and JfjPthe ordinate. 
 
 P a (-D,a) 
 
 a. 
 
 a 
 P,(Hb,-a) 
 
 Pj (b,.a) 
 
 a 
 
 P^.-a) 
 
 The lines of reference, XX' and 7F, are called the axis 
 of X and the axis of Y, respectively, and is called the 
 origin. 
 
 It is customary to express the fact that the abscissa of a 
 point is b, and its ordinate a, by saying that for the point 
 in question x = b and y = a ; or, more concisely, we may 
 refer to the point as " the point (b, a)," where the first term 
 in the parenthesis is understood to be the abscissa, and the 
 second term the ordinate. 
 
 15. If, in the figure of § 14, M and N~ be points on OX 
 and OX', respectively, such that OM = ON=b, and lines 
 P^i and PoP 3 be drawn through M and N, respectively, 
 perpendicular to XX', making MP X = NP 2 = NP Z = MP t = a, 
 each of the points P u P 2 , P 3 , and P 4 will have its abscissa 
 equal to b, and its ordinate equal to a. 
 
 To avoid this ambiguity, abscissas measured to the right 
 of are considered positive, and to the left, negative; and 
 ordinates measured above XX' are considered positive, and 
 below, negative. 
 
 Then the co-ordinates of the points will be as follows : 
 
 P„ (6, a); P 2 , (- b, a); P* (- b, -a); P„ (6, - o). 
 
io Plane Trigonometry. 
 
 Note 1. It is understood, in the above convention with regard to 
 signs, that the figure is so placed that OX is horizontal, and extends 
 to the right from 0. 
 
 Note 2. In all the figures of the present chapter, the small letters 
 are understood as denoting the lengths of the lines to which they are 
 attached, without regard to their algebraic signs ; hence they always 
 represent positive quantities. 
 
 16. If a point lies upon XX', its ordinate is zero; and 
 if it lies upon TY', its abscissa is zero. 
 
 17. General Definitions of the Functions. 
 
 We will now give general definitions of the trigonometric 
 functions, applicable to any angle whatever. 
 
 Construct axes in such a way that the initial line of the 
 angle shall be the positive direction of the axis of X, and 
 the vertex the origin. 
 
 From any point in the terminal line drop a perpendicular 
 to the axis of X, and find the co-ordinates of this point. 
 
 Then, the sine of the angle is the ratio of the ordinate of the 
 point to its distance from the origin. 
 
 Tlie cosine is the ratio of the abscissa to the distance. 
 The tangent is the ratio of the ordinate to the abscissa. 
 The cotangent is the ratio of the abscissa to the ordinate. 
 The secant is the ratio of the distance to the abscissa. 
 The cosecant is the ratio of the distance to the ordinate. 
 
 Note. The above definitions include those of § 2. 
 The definitions of the versed sine and co versed sine, given in § 2, 
 are sufficiently general to apply to any angle whatever. 
 
 18. We will now apply the definitions of § 17 in the fol- 
 lowing figures. 
 
 In each case, we construct axes in such a way that the 
 initial line of the angle shall be the positive direction of the 
 axis of X, and the vertex the origin. 
 
Trigonometric Functions. 
 L Let XOPi be any a.ncrle between 90° and 1S0°. 
 
 r 
 
 11 
 
 Let P 2 be any point on the terminal line, and draw P^V 
 perpendicular to AW: let MR, = a. OM=b, and OP t = c 
 Then the co-ordinates of P 2 are (_— b, a). 
 'Whence, by definition, 
 
 sin XOP 3 
 
 a 
 
 c 
 
 tanXOP 2 =-^- = -r 
 — b o 
 
 — 
 
 cos X0P s 
 
 -6 
 
 c 
 
 cot X0P 2 = — =--• 
 a a 
 
 esc X0P s 
 
 c 
 a 
 
 IL Let XOP 4 be any angle between ISO and 270° 
 
 Let P s be any point on the terminal line, and draw PjJtf 
 perpendicular to XX' ; letJfP s = «, OM=b, and OP s = <i 
 Then the co-ordinates of P s are (— 6, — a). 
 
12 Plane Trigonometry. 
 
 Whence, by definition, 
 
 sin XOP s = 
 
 — a a 
 
 tan XOP 3 = — ! = £• 
 — o o 
 
 sec XOP 3 = -^- = -r 
 
 cos XOP 4 = — - = 
 
 cot XOP 3 = — = -■ 
 — a a 
 
 esc XOP 3 = 
 
 — a a 
 
 -b b 
 
 III. Let XOPi be any angle between 270° and 360°. 
 Y 
 
 Let P, be any point in the terminal line, and draw MP A 
 perpendicular to XX' ; let J/P 4 = a, 031= b, and OP* = o 
 Then the co-ordinates of P 4 are (&, — a). 
 Whence, by definition, 
 
 sin XOP 4 = — = --• 
 c c 
 
 cos XOP 4 
 
 _6_ 
 
 c 
 
 tan XOP 4 = — = -?• 
 
 
 
 cotX0P 4 = 
 
 b _ b 
 — a a 
 
 sec XOP t = r 
 
 
 
 csc XOP 4 = 
 
 c c 
 — a a 
 
 19. It is evident that the terminal lines of any two 
 angles which differ by a multiple of 360° are coincident, 
 and hence the trigonometric functions of two such angles 
 are identical. 
 
Trigonometric Functions. 
 
 »3 
 
 Thus, the functions of 50°, 410°, 770°, -310°, etc., are 
 identical. 
 
 20. If the initial line of an angle coincides with OX, and 
 its terminal line lies between OX and Y, the angle is said 
 to be in the first quadrant; if the terminal line lies between 
 OY and OX 1 , the angle is said to be in the second quadrant; 
 if between OX' and OY', in the third quadrant; if between 
 O Y' and OX, in the fourth quadrant. 
 
 Thus, any positive angle between 0° and 90°, or 360° and 
 450°, or any negative angle between — 270° and — 360°, is 
 in the first quadrant ; any positive angle between 90° and 
 180°, or 450° and 540°, or any negative angle between — 180° 
 and — 270°, is in the second quadrant. 
 
 21. It follows from the definitions of § 17 that, for any 
 angle in the first quadrant, all the functions are positive. 
 
 It is also evident by inspection of the results of § 18 that : 
 
 In the second quadrant, the sine and cosecant are positive, 
 and the cosine, tangent, cotangent, and secant are negative. 
 
 In the third quadrant, the tangent and cotangent are positive, 
 and the sine, cosine, secant, and cosecant are negative. 
 
 In the fourth quadrant, the cosine and secant are positive, 
 and the sine, tangent, cotangent, and cosecant are negative. 
 
 It is customary to express the above in tabular form, as 
 follows : 
 
 Functions. 
 
 First 
 Quad. 
 
 Second 
 Quad. 
 
 Third 
 Quad. 
 
 Fourth 
 Quad. 
 
 Sine and cosecant . . . 
 Cosine and secant . . . 
 Tangent and cotangent . 
 
 + 
 + 
 + 
 
 + 
 
 + 
 
 + 
 
14 Plane Trigonometry. 
 
 22. Functions of 0° and 360°. 
 Y 
 
 a P(a,0)^ 
 
 The terminal line of 0° coincides with the initial line OX 
 Let P be a point on OX such that OP = a. 
 Then by § 16, the co-ordinates of P are (a, 0). 
 Whence by definition, 
 
 sin 0° = - = 0. 
 a 
 
 tan0° = - = 0. 
 a 
 
 sec0°=- = l. 
 a 
 
 cos 0° = - = 1. 
 a 
 
 cot 0° = ^ = oo. 
 
 escO° = - = oo. 
 
 
 
 By § 19, the functions of 360° are the same as those of 0°, 
 
 23. Functions of 90°. 
 
 Y 
 
 P (0,o) 
 
 a 
 
 .90° 
 
 For the angle 90°, OY is the terminal line. 
 Let P be a point on OY, such that OP= a. 
 Then the co-ordinates of P are (0, a). 
 Whence by definition, 
 
 sin 90° = - = 1. 
 a 
 
 tan 90° = - = oo. 
 
 sec 90° = ^ = oo. 
 
 o 
 
 > 
 
 X 
 0. 
 
 cos 90° = - = 0. 
 a 
 
 cot 90° = - = 0. 
 a 
 
 esc 90° = - 
 
Trigonometric Functions. 
 
 »5 
 
 24. Functions of 180° 
 
 «£3 
 
 180° 
 
 PC-ti.ot 
 
 o 
 
 For the angle ISO". OX' is the terminal line. 
 Let Pbe a point on OX', such that OP = a. 
 Then the co-ordinates of P are (— a, 0). 
 "Whence by definition, 
 
 sin ISO" = 
 
 = 0. 
 
 cos ISO- = = — 1. 
 
 a 
 
 tan ISO* = — = 0. 
 — ti 
 
 cotlSO i = -^=x. 
 
 sec ISO* =— = -1. 
 — a 
 
 cse ISO 3 = l 4 = oc. 
 
 
 2£x Functions of 270' 
 
 »:o 
 
 P(P,-a) 
 
 For the angle 2 70-. OT" is the terminal line. 
 Ler Pbe a point on OT'. such that OP= a. 
 Then the co-ordinates of P are (0, — a). 
 
i6 
 
 Plane Trigonometry. 
 
 Whence by definition, 
 
 sin 270° = — = -1. 
 a 
 
 tan 270°: 
 
 ao. 
 
 sec 270°=- =oo 
 
 
 cos 270°=- =0. 
 a 
 
 cot 270° = —-=0. 
 
 — a 
 
 esc 270° = — = -1. 
 
 — a 
 
 Note. No absolute meaning can be attached to- such a result as 
 cot 0° = oo ; it merely signifies that as an angle approaches 0°, its co- 
 tangent increases without limit. 
 
 A similar interpretation must be given to the equations esc 0° = ao, 
 tan 90° = co, etc. 
 
 26. Given the value of one function of an angle, to find the 
 values of the remaining functions. (Compare § 6.) 
 3 
 
 1. Given sin^l = — -; required the values of the remain- 
 ing functions of A. 
 
 The example may be solved by a method similar to that of § 6 ; 
 since the sine is the ratio of the ordinate to the distance, we may 
 regard the point of reference as having its ordinate equal to — 3, and 
 its distance equal to 5. 
 
 There are two points, P and P', which are 3 units below the axis 
 of X, and distant 5 units from O. 
 
 PC-4,-3) 
 
 P'(4,-3) 
 
 There are then two angles, XOP and XOP', in the third and fourth 
 quadrants, respectively, either of which may be the angle A. 
 
 Now, OM = OM' = ^OP^-PM 2 = V25 -9 = 4. 
 
 Then co-ordinates of Pare (- 4, — 3"), and of P', (4, — 3). 
 
Trigonometric Functions. 
 Whence by definition : 
 
 17 
 
 Angle. 
 
 Cot- 
 
 Tan 
 
 Cot. 
 
 See. 
 
 Csc 
 
 XOP 
 XOP" 
 
 4 
 
 6 
 
 4 
 5 
 
 3 
 
 4 
 3 
 4 
 
 4 
 3 
 4 
 3 
 
 5 
 
 4 
 5 
 
 4 
 
 5 
 3 
 
 
 3 
 
 Thus the two solutions to the problem are : 
 
 cosJ.=t4> tan^ = ±5, cot^l = ±i, sec^l = =F 7 , csc^ = -|; 
 5 4 o 4 o 
 
 where the upper signs refer to XOP, and the lower signs to XOP 1 . 
 
 2. Given cot-l = 3; required the values of the remain- 
 ing functions of A. 
 
 The equation may be written in the farms 
 
 rt q 
 
 cot A = -. or coti = 
 
 1 -1 
 
 We may then regard the point of reference as having its abscissa 
 equal to 3 and its ordinate equal to 1, or as having its abscissa equal 
 to — 3 and its ordinate equal to — 1. 
 
 There are t iro angles, XOP and XOP', in the first and third quad- 
 rants, respectively, either of which satisfies the given condition. 
 
 rt_Jfl 
 
 1 
 
 
 J 
 
 3 (M) 
 
 1 
 
 3 ( 
 
 
 
 — X 
 
 
 
 
 3 1 
 
 I 
 
 /tC 
 
 
 
 
 P'(-3,D 
 
 I 
 
 Then, 0P=0P ="*■ 01? + PjT = \ 9 + 1 =Via 
 
1 8 Plane Trigonometry. 
 
 Whence by definition : 
 
 Angle. 
 
 Sin. 
 
 Cos. 
 
 Tan. 
 
 Sec. 
 
 Cse. 
 
 XOP 
 XOP' 
 
 1 
 
 VIO 
 
 1 
 
 vTo 
 
 3 
 
 VIo 
 
 3 
 
 Vio. 
 
 1 
 3 
 
 1 
 3 
 
 VEo 
 
 3 
 
 VIo 
 3 
 
 VIo 
 -VIo 
 
 Thus the two solutions are : 
 
 1 3 1 
 
 sin A = ± — — , cos A = ± , tan A = -, 
 
 VIO VIO 3 
 
 sec ^ = ±^52, esc ^ = ± VlO. 
 
 o 
 
 Note. It must be clearly borne in mind, in examples like the 
 above, that the "distance" is always positive. 
 
 EXAMPLES. 
 In each of the following, find the values of the remaining 
 functions : 
 
 3. 
 
 A i» 
 
 sec^4 = -« 
 4 
 
 7. 
 
 A 60 
 
 csc .4 = 
 
 7 
 
 11. 
 
 tan A = — 7. 
 
 4. 
 
 cot^ = -^- 
 5 
 
 8. 
 
 tan .4 = — -• 
 40 
 
 12. 
 
 csc A = 3. 
 
 5. 
 
 sin^f 
 
 9. 
 
 7 
 
 sec A = 
 
 2 
 
 13. 
 
 b 
 
 6. 
 
 a 21 
 cos A = — — 
 29 
 
 10. 
 
 sin A = 
 
 5 
 
 14. 
 
 cot -4 = *. 
 
 27. Functions of 120°, 135°, 150°, etc. 
 
 P(-l,V3) 
 
Trigonometric Functions. 
 
 »9 
 
 Let OPJf be a r.cb.:_ triage :v.»vir.c OP f OX. and PM 
 eqr.xl to .. 1. and \ ?. rv svv\ tive'.v, and _ POJlf = 60'. 
 
 Tr.ev. _ XOP = l:X v \ and coordinates of Pate ^—1. % o\ 
 
 Wr.ev.oe by definition, 
 sir. r.\v = Al. 
 
 tar. t •.»-" = — % 3. 
 
 
 sa> 
 
 • w- _ _ •> 
 
 050 1L\>' = -== = 
 A 3 
 
 ■\ o. 
 
 T" "ike :r.a;r.ier may tv t»rcvtv. : r.c rev.-.aiv.hic v.«I".:es cir^u 
 in the t-.'O-a-ir.c cable, which are left as exercises feat the 
 
 Ji**j»*f- 
 
 >j -■;- 
 
 155* 
 
 
 saje 
 
 — j x 
 
 — j\ : 
 
 — j\ : 
 
 — j\ : 
 
 — i 
 
 CV*. 
 
 J^ - 
 
 i 
 
 i 
 
 Tint. 
 
 — \ 5 
 -1 
 
 * 1 
 
 \ -> 
 -\? 
 -1 
 
 — ;\ s 
 
 o*. 
 
 — ,N -- 
 -1 
 
 — \ , ; 
 
 \ s 
 
 1 
 
 -1 
 
 _2 
 — \ i 
 
 _ j 
 — \i 
 
 -\ i 
 
 W : 
 
 28. Fractions of (— -1 in terms of those of J, 
 
 s.:. - Jf = — sir. A. cos — -4>= eoe-l, v 
 
 tan^— -T = — t*n_4. eoc'— JP — — cot A. 
 
 S;V *— A * = S*?C -1. CSC v — -T = — CSC A. 
 
 & 
 
20 
 
 Plane Trigonometry. 
 
 There may be four cases : A in the first quadrant (Fig. 1), 
 A in the second quadrant (Fig. 2). A in the third quadrant, 
 (Fig. 3), or A in the fourth quadrant (Fig. 4). 
 
 X-*- 
 
 M 
 P' 
 
 Y' 
 
 Fig. 1. 
 
 Fio. 3. 
 
 In each figure, let the positive angle XOP represent the 
 angle A, and the negative angle XOP' the angle — A. 
 
 Draw PM perpendicular to XX', and produce it to meet 
 OP' at P'. 
 
 In right triangles OPM and OP'M, side OM is common, 
 aaiZPOM^ZP'OM. 
 
 Then the triangles are equal, and PM=FM an<3 
 OP = OP l . 
 
 Therefore, in each figure, 
 
 abscissa P' — abscissa P, 
 
 ordinate P' — ~ ordinate P, 
 
 and distance P' = distance P. 
 
Trigonometric Functions. 
 
 Then. 
 
 sin 
 
 (-A) = 
 
 ord. P 
 
 tan (— A) = 
 cot(— J) = 
 see(—A) = 
 esc (— A) 
 
 dist.P' 
 P 
 
 ord. P 
 
 dist.P 
 abs. P 
 
 = — sin A. 
 
 dist.P' 
 
 ord. P 
 abs. P' ' 
 
 abs. P 
 ord./" 
 
 dist. P 
 abs. F ' 
 
 dist. P' 
 
 dist.P 
 ord.P 
 
 abs.P 
 
 abs.P 
 ~ ord. P 
 
 dist.P 
 
 = cos A. 
 
 tan A 
 
 abs.P 
 dist.P 
 
 = — cotA. 
 sec A. 
 
 ord. P' 
 
 ord.P 
 
 = — csc .4, 
 
 29. Functions of (90° + A) in terms of those of A 
 To prove the formulae 
 
 sin (90° + ^)= cos 4 
 tan(90° + ^l) = -cot.4, 
 sec (90° + A) = - esc A, 
 
 for any value of A. 
 PC 
 
 cos (90° + A) = - sin A,] 
 cot (90° + A)=- tan A, 
 esc (90° + ^)= sec A,. 
 
 21 
 
 (2) 
 
22 
 
 Plane Trigonometry. 
 
 Fig. S. 
 
 There may be four cases : A in the first quadrant (Fig. 1), 
 A in the second quadrant (Fig. 2), A in the third quadrant 
 (Fig. 3), or A in the fourth quadrant (Fig. 4). 
 
 In each figure, let the positive angle XOP represent the 
 angle A, and the positive angle XOP the angle 90° 4- A. 
 
 Take OP 1 = OP, and draw PM and P'M' perpendicular 
 to XX'. 
 
 Since OP is perpendicular to OP, and OM to P'M', 
 
 ZPOM=ZOPM'. 
 
 Then right triangles 0PM and OP'M have the hypote- 
 nuse and an acute angle of one equal, respectively, to the 
 hypotenuse and an acute angle of the other, and are equal. 
 
 Whence, MP = OM' and 0M= M'P 1 . 
 
 Therefore, in each figure, 
 
 ordinate P = abscissa P, 
 
 abscissa P = — ordinate P, 
 
 and distance P' = distance P. 
 
 ord. P' abs. P 
 
 Then, sin (90° + A) = 
 cos (90° + A) 
 
 dist. P 
 
 abs. P' 
 dist. P' 
 
 dist. P 
 
 ord. P 
 dist. P" 
 
 = cos -4. 
 
 ■ sin A 
 
Trigonometric Functions. 23 
 
 tan (90° + ^)=°^' = -^ = - cot A. 
 v J abs. P ord. P 
 
 cot (90° + A) = 2^g = - ^P = _ tan A 
 v ' ord. P' abs. P 
 
 see (90° + ^) = ^ p ' = -^P = - cse A. 
 K J abs. P' ord. P 
 
 «„„ cnno , a\ dist. P dist. P „„ j 
 
 csc (90 + A) = — — — = — — — = sec A. 
 v ' ord.P abs. P 
 
 30. Tbe results of § 29 may be stated as follows : 
 
 The sine, cosine, tangent, cotangent, secant, and cosecant of 
 any angle are equal, respectively, to the cosine, minus the sine, 
 minus the cotangent, minus the tangent, minus the cosecant, 
 and the secant, of an angle 90° less. 
 
 31. I. Functions of (90° — A) in terms of those of A. 
 By § 30, sin (90° - A) = cos (- A) = cos A (§ 28). 
 
 cos (90° - A) = - sin (- A) = sin A 
 tan (90° - A) = - cot (- A) = cot A. 
 cot (90° - A) = - tan (- A) = tan A. 
 sec (90° - A) = - csc (- A) = csc A. 
 csc (90° - A) = sec ( - A) = sec A 
 These formulae were proved for acute angles in § 4. 
 
 II. Functions of (180° — A) in terms of those of A. 
 
 By § 30, sin (180° - A) = cos (90°-^) = sin A (I) 
 cos (180° - A) = - sin (90° -4) = - cos A 
 tan (180° - A) = - cot (90°-4) = - tan A. 
 cot (180° - A) = - tan (90°-4) = - cot A 
 sec (180° - .4) = - csc (90° -A) = - sec A. 
 csc (180° - A) = sec(90°-4) = csc A. 
 
24 Plane Trigonometry. 
 
 III. Functions of (180° + A) in terms of those of A 
 
 sin (180° + A) = cos ( 90° + A)=- sin A. 
 cos (180° + A) = - sin ( 90° + A) = - cos A. 
 tan (180° + A) = - cot ( 90° +A)= tan A. 
 cot (180° + A) = - tan ( 90°+ A) = cot A 
 sec (180° + A) = - esc ( 90° + A) = - sec A 
 esc (180° + A) = sec ( 90° + .4) = - esc A. 
 
 IV. Functions of (270° — A) in terms of those of A 
 
 sin (270° - A) = cos (180° - A) = - cos A (by II). 
 
 cos (270° - A) = - sin (180° - A) = - sin .4. 
 
 tan (270° - A) = - cot (180° - A) = cot A 
 
 cot (270° - A) = - tan (180° - A) = tan A. 
 
 sec (270° - A) = - esc (180° - .4) = - esc A. 
 
 esc (270° - A) = sec (180° -.4)= -sec A 
 
 V. Functions of (270° + A) in terms of those of A. 
 
 sin (270° + A) = cos (180° + A)=- cos A (by III). 
 
 cos (270° + A) =- sin (180° + A)= sin A 
 
 tan (270° + A) = - cot (180° + .4) = - cot A. 
 
 cot (270° + A) = - tan(180° + .4)= - tan A 
 
 sec (270° +A)= -esc (180° + .4)= esc A 
 
 esc (270° +A)= sec (180° + A) = - sec A 
 
 VI. Functions of (360° - 4) in terms of those of A 
 
 sin (360°- 41) = cos (270° - A) = - sin A (by IV). 
 cos (360°- A )=- sin (270° - A) = cos A 
 tan(360°- A) = - cot (270° ~A)=- tan A 
 cot (360°-.4) = -tan (270° - A) = - cot A 
 sec (360° - A) = - esc (270°- A) = sec A 
 esc (360° -A)= sec (270° - A) = - esc A 
 
 32. From the formulae of the preceding articles the fol- 
 lowing rule is derived : 
 
 Any function of 0°, or an even multiple ~f 90°, plus or 
 minus A, is the same function of A; and any function of an 
 odd multiple of 90°, plus or minus A, is the complementary 
 function of A. 
 
Trigonometric Functions. 25 
 
 For the algebraic sign of the result, plot the given angle, 
 regarding A as acute; the algebraic sign of the given func- 
 tion of this angle as plotted is the algebraic sign of the result. 
 
 1. Find the value of sec (990° - A). 
 sec (990° -A) = sec (11 • 90° - A) = - esc A. 
 
 2. Find the value of tan (- 180° + A). 
 tan (- 180° +A)= tan (- 2 • 90° + A) = tan A - 
 
 3. Find the value of cos 520°. / 
 cos 520°=cos(5 • 90°+ 70°)= -sin 70°= -cos 20°. \L 
 
 Express each of the following as a function of 
 an acute angle less than 45° : 
 
 4. cot 155°. 7. sec(-160°). 10. sin 1180° 24'. 
 
 5. sin 600°. 8. tan (-220°). 11. cos (-310° 40'). 
 
 6. cos 287°. 9. esc (-310°). 12. tan (920° 35'). 
 
 Find the numerical value of the following : 
 
 13. cot 405°. 15. esc 600°. 17. cos (-420°). 
 
 14. sin 480°. 16. tan 690°. 18. sin (-225°). 
 
 Prove that ^ 
 
 19. sin 210° tan 225° + cos 300° cot 315" = - 1. 
 
 20. cos 570° • sin 510° - sin 330° cos 390° = 0. 
 
 21. sip ( 180 °-^) tan(90 °+A) +-r-^ = l-secJ. 
 
 sin(270°-^) V ' sin(270°-^L) 
 
 22 2s ™(~ A ) , cos(180°+>4) cot(-A) =Q 
 "' cos(90°+.<4) cos(-A) tan (270°+^) 
 
26 
 
 Plane Trigonometry. 
 
 III. GENERAL FORMULAE. 
 
 33. It follows directly from the definitions of § 17 that, 
 if x is any angle, 
 
 sin x = - 
 
 cos x = ■ 
 
 esc a; 
 1 
 
 tana; = 
 
 cot a; 
 
 1_ 
 sec x tan x 
 
 34. To prove the formula 
 
 sec a; ; 
 
 1 
 
 cot X = 
 
 CSC X = - 
 
 COS X 
 
 1 
 
 sin x 
 
 tana;: 
 
 sinai 
 coscc 
 
 (3) 
 
 (*) 
 
 P 
 
 M 
 
 Y' 
 
 Fig. 
 
 Km. 8. 
 
 There may be four cases : x in the first quadrant (Fig. 1), 
 x in the second quadrant (Fig. 2), x in the third quadrant 
 (Fig. 3), or x in the fourth quadrant (Fig. 4). 
 
 In each case, let the positive angle XOP represent the 
 angle x, and draw PM perpendicular to XX'. 
 
General Formulae. 27 
 
 Then in each figure, by the definitions of § 17, 
 
 ord. P 
 ord. P dist. P sin x 
 
 tan* = 
 
 abs. P abs. P cos >. 
 
 dist. P 
 
 35. To prove the formula 
 
 , cos X ,«» 
 
 cot x = (5) 
 
 sin x 
 
 By §33, cot x = -A_ = J_. (§ 34) = £2?.?. 
 tan a; sma ' sin a; 
 
 cos a; 
 
 36. To prove £fte formula 
 
 sin 2 a; + cos 2 a; = 1. (6) 
 
 Note. Sin 2 x signifies (sin x) 2 ; that is, the square of the sine of x. 
 
 There may be four cases : x in the first quadrant, x in the 
 second quadrant, x in the third quadrant, or x in the fourth 
 quadrant. 
 
 In each figure of § 34, we hare by Geometry, 
 
 MP* + OM 2 =OP 2 - 
 
 tv -a- u Ttp 2 MP\OM 2 1 
 Dividing by OP , ==J + -=j = 1. 
 
 OP OP 
 
 But in each figure, 
 
 ¥?- = (sin a;) 2 , and OIL = (cos «)*; 
 OP* ■ OP 2 ' 
 
 MP 
 for, whether — — is + or — , its square is +. 
 
 Whence, sin 2 x + cos 2 x = l. 
 
 Formula (6) may be written in the forms 
 
 sin 2 x = 1 — cos 2 x, and cos 2 x = 1 — sin 2 x. 
 
28 
 
 Plane Trigonometry. 
 
 37. To prove the formulce 
 
 sec 2 x = 1 + tan 2 x, 
 and esc 2 x = 1 + cot 2 a;. 
 
 By (6), 1 = cos 2 a; 4- sin 2 x. 
 
 V) 
 
 (8) 
 (A) 
 
 Dividing by cos 2 x, 
 
 = 1 + 
 
 sm'a; 
 
 cos a; 
 
 Whence by (3) and (4), sec 2 x = 1 + tan 2 x. 
 
 Again, dividing (A) by sin 2 a;, we have 
 
 1 _ i i cos 2 a; 
 sin 2 a; sin 2 x 
 
 Whence by (3) and (5), esc 2 x = 1 + cot 2 x. 
 
 38. To express sin (a; + y) and cos (a: + y) in terms of the 
 sines and cosines of x and y. 
 
 I. When x 
 
 and y are acute. 
 
 
 
 
 
 C 
 
 / 
 
 \o 
 
 
 / E 
 
 x\ 
 
 B 
 
 E 
 
 ~%\ 
 
 
 
 As 
 
 
 \vj 
 
 B 
 
 /^x, 
 
 
 
 
 
 \/x\ 
 
 
 J. 
 Fig. 1 
 
 I . 
 
 D 
 
 u 
 
 1 
 
 J 
 
 FlO. 2. 
 
 
 
 There may be two cases : x + y acute (Fig. 1), and x + y 
 obtuse (Fig. 2). 
 
 In each figure, let Z. DOB = x and ZBOC=y. 
 
 Then, ZZ>00=a; + 2/. 
 
 From any point in 00 draw lines CA and OB perpen- 
 dicular to OB and 0-B, respectively ; also, draw lines BD 
 and BE perpendicular to OD and AC, respectively. 
 
General Formulae. 29 
 
 Since EC is perpendicular to OD, and BC to OB, angles 
 BCE and DOB are equal ; that is, Z BCE = x. 
 In either figure, by § 17, 
 
 S ^ {x+) = AC = DB + EC = DB EC 
 K "' OC 00 00 oc 
 
 = DB OB EC BC 
 OB 00 BO 00' 
 
 Then, sin (a; + y) = sin x cos y + cos x sin y. (9) 
 
 Again, by § 17, in Fig. 1, cos (3 + 2,) = ^-, 
 
 and inFig. 2, cos(a; + w) = — =-^?. 
 
 s ' ^ T ^ 00 OC 
 
 Then in either figure, 
 
 OD-EB OD EB 
 
 cos (% + y)-- 
 
 00 OC 00 
 
 OD OB EB BC 
 OB OC BC OC 
 
 Then, cos (x + y) = cos x cos y — sin x sin y. (10) 
 
 39. Formulae (9) and (10) are very important, and it is 
 necessary to prove them for all values of ' x and y. 
 
 They have already been proved when x and y are any 
 two acute angles ; or, what is the same thing, when they 
 are any two angles in the first quadrant. 
 
 Xow let a and b be any two angles in the first quadrant. 
 
 By § 29, sin [90° + (a + 6)] = cos (a 4- 6), 
 and cos [90° + (a + 6)] = — sin (a + b). 
 
 Whence, by (9) and (10), 
 
 sin [90° + (a + 6)] = cos a cos 6 — sin a sin b, (A) 
 and cos [90° + (a + 6)] = — sin a cos b — cos a sin 6. (B) 
 
30 Plane Trigonometry. 
 
 By § 29, cos a = sin (90°+ a), and — sin a = cos (90°+ a). 
 
 Then, (A) and. (B) may be written in the forms 
 sin [(90° + a) + 6] = sin (90° + a) cos b + cos (90° + a) sin b, 
 cos [(90° + a) + 6] = cos (90° + a) cos 6 - sin (90° + a) sin 6 ; 
 
 which are in accordance with (9) and (10). 
 
 But 90° + a is an angle in the second quadrant. 
 
 Therefore, (9) and (10) hold when one of the angles is 
 in the second quadrant, and the other in the first. 
 
 In like manner, by supposing a to be any angle in the 
 first quadrant, and b any angle in the second, (9) and (10) 
 may be proved to hold when both angles are in the second 
 quadrant. 
 
 Again, by supposing a and b to be any two angles in the 
 second quadrant, (9) and (10) may be proved to hold when 
 one angle is in the second quadrant and the other in the 
 third; and so on. 
 
 Hence, (9) and (10) hold for any values of x and y what- 
 ever, positive or negative. 
 
 40. Putting, in (9) and (10), — y in place of y, 
 
 sin (a; — y) — sin x cos (— y)+ cos x sin (— y) 
 
 = sin x cos y + cos x (— sin y), by § 28, 
 = sin x cos y — cos x sin y. (ll) 
 
 cos (x — y) = cos a; cos (— y) — sin x sin (— y) 
 = cos x cos y — sin x (— sin y) 
 = cos x cos y + sin x sin y. (12) 
 
 41. By (4), 
 
 ^n(x + y)= sin( - x + y) 
 cos (a; + y) 
 
 sin x cos y + cos x sin tt , ,. s , .. . 
 
 = iL - J — -. r- s by (9) and (10). 
 
 cos x cos y — sin x s;n y 
 
General Formulae. 31 
 
 Dividing each term of the fraction by cos x cos & 
 
 sin x cos y . cos x sin y 
 
 , . cos x cos y cos x cos y 
 
 tan (x + y)= 2 = :— ^ 
 
 v ' cos x cos y sm x sin y 
 
 cos a; cos y cos a; cos y 
 
 tan a; + tan y 
 1 — tana; tan y 
 
 In like manner, we may prove 
 
 (13) 
 
 , , N tan x — tan y ,, .« 
 
 tan f.r — «) = *— (14) 
 
 v * y 1 + tanaitany v y 
 
 Again, by (5), cot (x + y)= c P s ^ + y ) 
 
 sin (a; + y) 
 
 _ cos .r cos y — sin x sin y 
 sin x cos y+ cos a; sin y 
 
 Dividing each term of the fraction by sin x sin y, 
 
 cos .r cos y sin x sin y 
 
 sin a; sin y sin a; sin « 
 
 cot(x + y) = - ^ ^-£ 
 
 v ' sm a; cos y cos x sin y 
 
 sin x sin y sin a; sin y 
 cot .r cot y — 1 
 
 (15) 
 
 cot y + cot X 
 In like manner, we may prove 
 
 .. . COt.VCOtw + 1 ,,-x 
 
 cot (x — y) = ^-i — (16) 
 
 v *' cot y — cot x y 
 
 42. From (9), (10), (11), and (12), we have 
 
 sin(a +6) = sinocos6 + cosasin6. (A) 
 
 sin (a — 6) = sin a cos b — cos a sin 6. (B) 
 
 cos (a + 6) = cos a cos 6 — sin a sin &. (C) 
 
 cos (a — 6) = cos a cos 6 + sin a sin 6. (D) 
 
32 Plane Trigonometry. 
 
 Adding and subtracting (A) and (B), and then (C) and (JD). 
 
 sin (a + 6) + sin (a — b) = 2 sin a cos 6. 
 sin(a + &) — sin(a — b) = 2cosasin&. 
 cos (a + 6) + cos (a — 6) = 2 cos a cos b. 
 cos (a + &) — cos (a — b) = — 2 sin a sin 6. 
 Let a + b = x, and a — b = y. 
 Adding, 2o = « + y, or a = %(x + y). 
 
 Subtracting, 2b = x—y, or b = £ (a; — y). 
 Substituting these values, we have 
 
 sino;+sin2/= 2 sin £ (a; -j-y) cos £ (a; — y). (17) 
 sina;— sin?/ = 2cos£(a; + i/) sin£(a; — y). (18) 
 cos a; + cos y = 2 cos \ (x + j/) cos J (a; — ?/). (19) 
 cos x — cos ?/ = — 2 sin £ (x + ?/) sin£(a; — y). (20) 
 
 43. By (17) and (18), we have 
 
 sin x + sin y __ 2 sin £ (x + y) cos £ (x — y) 
 sin x — siny~ 2 cos £ (a; + y) sin £ (a; — y) 
 
 = tan |- (x + y) cot J (a; — y) 
 
 tan + fr + y) 
 
 tan^(a; — ?/)' J v ' 
 
 44. Functions of 2 a;. 
 Putting y = x in (9), we have 
 
 sin 2 a; = sin x cos x + cos a; sin x 
 
 = 2 sin x cos a;. (22) 
 
 Putting y = x in (10), we obtain 
 
 cos 2 a; = cos x cos a? — sin x sin a; 
 
 = cos 2 as — sin 8 x. (23) 
 
General Formulae. 33 
 
 We also have by § 36, 
 
 cos 2 x = (1 — sin 2 re) — sin 2 x = 1 — 2 sin 2 x. (24) 
 
 cos 2 a; = cos 2 x — (1 — cos 2 a;) = 2 cos 2 x — 1. (25) 
 
 Putting y = a; in (13) and (15), 
 
 . tan a; + tan x 2 tan a; ,„_, 
 
 tan 2 a; = — ^- = — (26) 
 
 1 — tan x tan x 1 — tan 2 x 
 
 . o cot a; cot a; — 1 cot 2 a; — 1 ,„„-. 
 
 cot 2 a; = = — (27) 
 
 cot x + cot x 2 cot x 
 
 45. Functions of \ x. 
 
 From (24) and (25) we have, by transposition, 
 
 2 sin 2 x = 1 — cos 2 a;, and 2 cos 2 a; = 1 + cos 2 a;. 
 
 Putting £ a: in place of x, and therefore x in place of 2 a;, 
 
 we have 
 
 2 sin 2 £ a; = 1 — cos a;, (28) 
 
 2 cos 2 £ a; = 1 + cos as. (29) 
 
 Again, putting \ x in place of x in (22), 
 
 2 sin £ a; cos £ x — sin a;. (A) 
 
 Dividing (28) by (A), 
 
 2 sin 2 \ x _ 1 — cos x 
 
 2 sin £ a; cos £ a; sin a; 
 
 Whence, by (4), tan % x = 1 ~ cos x . (30) 
 
 sin iu 
 
 Dividing (29) by (A), 
 
 2 cos 2 \ x _ 1 + cos a; 
 
 2 sin^a; cos-|-a; sin a; 
 Whence, by (5), cot \ x = 1 + cos x . (31) 
 
 Sill X 
 
34 Plane Trigonometry. 
 
 EXERCISES. 
 
 46. 1. Prove the relation sec 2 x esc 2 x = sec 2 x + esc 2 x. 
 
 -r. /a\ o o 1 sin 2 x + cos 2 x , , CN 
 
 By (3), sec 2 a; esc 2 a: = — - — —- - = ^ , — , by (6) 
 
 cos 2 a; sin 2 a; cos 2 x sm 2 a; 
 
 sin 2 a; . cos 2 a; 
 
 cos 2 a; sin 2 a; cos 2 a; sin 2 a; 
 
 - + — = sec 2 x + esc 2 x. 
 
 cos 2 x sin 2 x 
 
 2. Prove the relation = tan 2 x. 
 
 cos 5 a; + cos a; 
 
 By (18) and (19), 
 
 sin 5x — sin a; _ 2 cos \{h x + x) sin|(5x — x) _ sin 2 x _ t 2 
 cos5x+cosx 2 cos £(5 a; + x) cos ^(5 a; — x) cos 2 a; 
 
 3. Prove the relation ^ -rV) ~ _ £ an y 
 
 1 + tan (x + y) tan x 
 
 B (14) tan(x + y)-tanx [( +y) _ x}= tan ,,. 
 
 J v v ' l + tan(x + ?/)tanx u ^ " J J y 
 
 4. Prove the relation sin 3 x = 3 sin a; — 4 sin 3 x. 
 
 By (18), sin 3 a; — sinx = 2 cos|(3x + x) sin |(3x — a;). 
 
 Then, sin 3 x = sin x 4 2 cos 2 x sin x 
 
 = sin x + 2 (1 — 2 sin 2 x) sin x, by (24) 
 
 = sin x + 2 sin x — 4 sin 8 x = 3 sin x — 4 sin 3 x. 
 
 The artifice used above is advantageous in finding the 
 sine or cosine of any odd multiple of x. 
 
 Prove the following relations : 
 
 5. tan x sin x + cos x = sec x. 
 o siny _ 1 — cos y 
 1 -f cos y sin y 
 
 „ 1 + sin x 1 — sin a; ., 
 
 7. — ■ — : : — = 4 tan x sec a;. 
 
 1 — sin a; 1 + sin a; 
 
 o sin x , 1 + cos x „ 
 
 8. ^ 1 '- = 2 esc x. 
 
 1 + cos » sin a; 
 
General Formulae. 3$ 
 
 9. (sin x + cos x) (tan x + cot a;) = sec x + esc x. 
 
 n esc A , esc A , . 
 
 0. — ~-\ — -=2secM. 
 
 esc A — 1 esc A + 1 
 
 1. sec 2 .4 + tan 2 5 = sec 2 B + tan 2 A 
 
 2.4 
 
 + sinj. 
 
 ■4 
 
 - = sec -4 — tan A. 
 
 o /esc 6 — cot 9 n , n 
 
 3- \ ; ; = esc 9 — cot 9. 
 
 'csc0 + cot0 
 
 4 sin {A + .B) sin (4 - B) = ^ ^ _ ^.^ 
 
 cos 2 ^4 cos 2 S 
 5. cos (x — y)+ sin (a; + ?/) = (sin x + cos a) (sin y + cos 2/). 
 
 g sin (x + y) _ tan a; + tan y 
 sin (a: — y) tan a; — tan y 
 
 m cos (a; + y) _ cot a; cot y — 1 _ 
 cos (a; — y) cot x cot 2/ + 1 
 
 8. cos x cos (a; — y) + sin a; sin (x — y) = cos 2/. 
 
 9. sin 5 x cos 2 x — cos 5 a; sin 2 as = sin 3 x. 
 
 20. sin (60° + A) cos (30° + .4) 
 
 - cos (60° + A) sin (30° + .4) = |. 
 
 21. sin (a; + y + z) = sin x cos 2/ cos z + cos x sin 2/ cos z 
 
 + cos a; cos y sin z — sin x sin 2/ sin z. 
 
 22. cos (a; + y + z) = cos x cos ?/ cos z — sin x sin y cos z 
 
 — sin a; cos y sin z — cos x sin t/ sin z. 
 no sin 3 a; cos 3 x _ g 
 sin a: cos a; 
 
 24. sin (a; + y) sin (x — y) = sin 2 # — sin 2 2/. 
 
 25. cos (» + 2/) cos (x — y) = cos 2 x — sin 2 1/. 
 
 26. tan (a; + 45°) + cot (x~ 45°) =0. 
 
 27. tan (45° - x) + cot (45° + x) = 2 sec 2 a; - 2 tan 2 a;. 
 
 28 tan(^ + ^) + tan(^-^) =tan2A 
 ' l-tan(.4 + J3)tan(.4-.B) 
 
$6 Plane Trigonometry. 
 
 on 1 + sin 2 x _ ft&a x + 1\ 
 
 30. 
 
 1 — sin 2 a; \tana; — 1, 
 cos 3 a; + sin 3 x 2 — sin 2 a; 
 
 cos x + sin x 2 
 
 oi • o 2tana; o r cot^l+tan^l A 
 
 31. sin2a; = 33. ! -=see2A 
 
 l + tan 2 a; cot A— tan A 
 
 32. cos2a, = 1 + taR i a . - 34. csc2 0-cot2 0=tan0. 
 
 35. 1 + tan 2 a; tan a; = see 2 a;. 
 
 <jr c,„„o„, csc 2 a: „- , „. 1 — cos a; 
 
 oo. secza; = — ; -• 37. tan -l-a; = 
 
 esc* a; — 2 ' 1 + cos x 
 
 38. tan (45° + x) - tan (45° - x) = 2 tan 2 a;. 
 
 39. Prove the relation of Ex. 4 by putting x = 2x and 
 ?/ = x in (9). 
 
 40. cos 3 x = 4 cos 3 a; — 3 cos a;. 
 
 41. tan3a; = 3tana; - tan3a; . 
 
 1—3 tan 2 x 
 
 42. sin 4 a; = 4 sin x cos a; — 8 sin 3 x cos a;. 
 
 43. cos 4 a; = 1 — 8 cos 2 x + 8 cos 4 a;. 
 
 44. 4 cos x cos (60° + x) cos (60° — a;) = cos 3 x. 
 
 .c- sin 3 x + sin x , 
 
 45. ■ — - — ! = tan 2 a;. 
 
 cos 3 x + cos a; 
 
 AC sin 3 aj — sin 5 x ,. 
 
 46. = — cot 4 x. 
 
 cos 3 x — cos 5 a; 
 
 4 - sin 50° - sin 10° ^ /^ 
 cos 50° - cos 10° ' 
 
 48. cos 80° + cos 40° = cos 2 0°. 
 
 49. (sin x + sin yf + (cos x + cos y) 2 = 4 cos 2 ^~ y \ 
 
 50. cos 2 (a; +2/)— sin 2 a: = cos (2 a; + ?/) cos ?/. 
 
 51. cos (2x + y)+2 sin a; sin (x + y) = cos y. 
 
 52. cos 20° + cos 100° + cos 140° = 0. 
 
 53. sin 18° + sin 42° + cos 168° =0. 
 
General Formulae. 37 
 
 54. By putting x = 45° and # = 30° in (11) and (12), 
 
 • iKO V6-V2 1KO V6 + V2 
 
 prove sin 15 = — — , cos 15 = — — ^ • 
 
 55. By putting a; = 30° in (30) and (31), prove 
 
 tanl5° = 2-V3, cot 15° = 2 + V3. 
 
 56. By putting a; = 45° in (28) and (29), prove 
 
 sin 221° = 1.V2-V2, cos 221° = 1V2+V2. 
 
 57. By putting x = 45° in (30) and (31), prove 
 
 tan22£°=V2-l, cot 22£° = V2 + 1. 
 
 58. sin a; (1 + tan a;) + cos x (1 + cot a;) = sec x + esc a;. 
 
 59. cot A — cot2A = esc 2 A. 
 
 60. 
 
 1 + sin 2 a; __ cos a; + sin x 
 cos 2 x cos a; — sin x 
 
 61. 2 sin fc - A\ cos (| + B\ = cos (^ - E)-sin(A +B). 
 
 co csc a; + cot x , 2 , 
 
 62. ! = cot 2 \ x. 
 
 csc a;— cot a; 
 
 63. cos 4 x — sin 4 x = cos 2 x. 
 
 64. sin (90° + A) + sin (210° + A) + sin (210° -A)=0. 
 
 65. sin a; + sin 3 x + sin 5 a; + sin 7 a; =4 sin 4 a; cos 2 a: cos x. 
 
 66. cos 5 x = 5 cos a; — 20 cos 3 a; + 16 cos 5 a;. 
 
 2 tan ( — — x ) 
 
 67. -^ J — = cos 2 a 
 
 l + tan 2 (^-aA 
 
 68. tan a; tan (60° + x) tan (120° + x) = — tan 3 x. 
 
 69. sin^t + l + sm^ =secM(cse ^ + 1)j 
 1 — sin A sin ^4 
 
 70. J 1 + sin2g =tan(45° + fl). 
 
 \1- S in20 v y 
 
 „, , . 4 tan a; — 4 tan 3 x 
 
 71. tan 4 a; 
 
 1 — 6 tan 2 x + tan 4 x 
 72. tan 2 as cot x — 1 = sec 2 a;. 
 
38 Plane Trigonometry. 
 
 IV. MISCELLANEOUS THEOREMS. 
 
 47. Circular Measure of an Angle. 
 
 An angle is measured by finding its ratio to another angle, 
 adopted arbitrarily as the unit of measure. 
 
 The usual unit of measure for angles is the degree, which 
 is an angle equal to the ninetieth part of a right angle. 
 
 Another method of measuring angles, and one of great 
 importance, is known as the Circular Method; in which the 
 unit of measure is the angle at the centre of a circle subtended 
 by an arc whose length in equal to the radius. 
 
 Note. The unit of circular measure is called a radian. 
 
 Thus,, let AOB be any central angle, and -400 the unit 
 of circular measure; that is, the angle at the centre sub- 
 tended by an arc whose length is equal to OA. 
 
 Then, circular measure AOB — ^ A01 i . 
 
 ZAOC 
 
 But by Geometry, ^OB = arcAB arc AB % 
 
 J J ' AAOC arc AC OA 
 
 Whence, circular measure AOB = aic . 
 
 OA 
 
 That is, the circular measure of an angle is the ratio of its 
 subtending arc to the radius of the circle. 
 
 48. By § 47, the circular measure of a right angle is the 
 ratio of one-fourth the circumference to the radius. 
 But if B denotes the radius, the circumference is 2 ttS. 
 
Miscellaneous Theorems. 39 
 
 Whence, circular measure 90° = T of 2 wR = -• 
 
 i? 2 
 
 It follows from the above that the circular measure of 
 
 180° is tt; of 60°, -; of 45°, -; etc. 
 ' '3' 4' 
 
 That is, an angle expressed in degrees may be reduced to 
 circular measure by finding its ratio to 180°, and multiplying 
 the result by ir. 
 
 23 
 Thus, since 115° is — of 180°, the circular measure of 
 on 36 
 
 115° is =££. 
 36 
 
 49. Conversely, an angle expressed in circular measure 
 may be reduced to degrees by multiplying by 180° and dividing 
 by ir ; or, more briefly, by substituting 180° for tt. 
 
 Thus, if = -L of 1S0° = S4°. 
 15 15 
 
 50. In the circular method, such expressions may occur 
 as "the angle f," "the angle 1," etc. 
 
 These refer to the unit of circular measure ; thus, the 
 angle f signifies an angle whose subtending arc is two-thirds 
 of the radius. 
 
 The angle 1 signifies the angle whose subtending arc is 
 equal to the radius, or the unit of circular measure. 
 
 The angle 1, reduced to degrees by the first rule of § 49, 
 
 gives 
 
 180^ _ — 180^ — = 57295S ° approximately. 
 *■ 3.14159". ' ™ J 
 
 Then the rule of § 49 may be modified as follows : 
 
 An angle expressed in circular measure may be reduced U 
 degrees by multiplying by 57.2958°. 
 
 Thus, the angle f 
 
 = I X 57.2958° = 38.1972° = 38° 11' 49.92". 
 
40 Plane Trigonometry. 
 
 EXAMPLES. 
 51. Express each of the following in circular measure : 
 
 1. 120°. 3. 67° 30'. 5. 86° 24'. 7. 163° 7' 30\ 
 
 2. 315°. 4. 146° 15'. 6. 53° 20'. 8. 88° 53' 20". 
 Express each of the following in degree measure : 
 
 „.«£. n.fr. 13. I- 15. |. 
 
 10 .^- 12.fr. 14. |- 16. |- 
 
 (In the following examples take w = - 2 y-)- 
 
 17. Eind in degrees the angle subtended at the centra 
 of a circle whose radius is 15 feet by an arc of 10 feet. 
 
 The angle = — radians 
 15 
 
 = ?X— = 38.2°. 
 
 3 ii- 
 lS. The radius of a circle is 42 feet. Eind the angle sub- 
 tended at the centre of the circle by an arc of 22 feet. 
 
 Ans. 30°. 
 
 19. The radius of a circle is 18 feet. Find the length of 
 the arc subtending an angle of 20" at the centre. 
 
 Ans. 6.3 feet. 
 
 20. How long does it take the minute hand of a clock to 
 turn through — 3| radians ? Ans. 35 minutes. 
 
 21. A fly wheel makes 7 revolutions a second. How long 
 will it take it to turn through 4 radians ? Ans. -JL- second. 
 
 22. The radius of a circle is 4 feet. Eind the length of 
 an arc subtending an angle of 126° at the centre. 
 
 Ans. 8.8 feet. 
 
 23. A train is traveling at the rate of 40 miles an hour on 
 a curve of two-mile radius. Through what angle does it 
 turn in 22 seconds ? Ans 7° 
 
Miscellaneous Theorems. 41 
 
 24. The radius of a graduated scale is 4 feet, and the 
 graduations are 5' apart. Find the distance between suc- 
 cessive graduations. Ans. ■££% inch. 
 
 52. Inverse Trigonometric Functions. 
 
 The relation between x and y in the above triangle may 
 be expressed by saying that y is the sine of x, or by saying 
 that a; is an angle whose sine is y. 
 
 The first statement is written in the form y = sin x. To 
 write the second statement a new symbol is now introduced, 
 namely sin -1 y. This symbol is read, " the angle whose sine 
 is y," " inverse sine y" or " anti-sine y." 
 
 The second statement of the relation between x and y in 
 the above figure can now be written x = sin -1 y. 
 
 In like manner to denote " the angle whose cosine is y," we 
 write cos -1 ?/ (also read " inverse cosine y," or " anti-cosine y "), 
 and so on for the other functions. 
 
 Note. The student must be careful not to confuse the above nota- 
 tion with the exponent — 1 ; the — 1st power of sin x is expressed as 
 (sin x)' 1 or esc x, and not sin -1 x. Sometimes sin -1 a, tan- 1 6, etc., are 
 written arc sin a, arc tan 6, etc. 
 
 When an angle is expressed in this notation, the angle is 
 called an inverse trigonometric function, or anti-trigonometrio 
 function, or inverse circular function. 
 
 53. From any formula involving direct functions a rela- 
 tion between inverse functions may be derived by aid of the 
 principles of § 52. 
 
4 2 
 
 Plane Trigonometry. 
 
 1. Find the value of tan (2 sin -1 •§). 
 
 Let sin" 1 § = A, then sin A = £ . 
 
 By (26), tan 2 .4: 
 
 2 tan J. 
 I -tan 2 ^' 
 
 By§6,tan^ = f. .-. tan 2 A = -^% = ~ ■ 
 
 2. Prove the relation 2 cos -1 a = cos -1 (2 a 2 — 1). 
 Let cos -1 a = A, then cos ^4 = a. 
 
 By (23), cos2^. = 2cos 2 Jl— 1. 
 
 Substituting, cos (2 cos -1 a) = 2 a 2 — 1, 
 -and 2 cos -1 a = cos- 1 (2 a 2 — 1). 
 
 3. Prove that one value of tan -1 7 + tan -1 4 is — -• 
 
 6 4 
 
 tan -1 7 = x, and tan -1 $ = y. 
 tan x = 7, and tan y = § . 
 
 tan(x + 2,)=l±i=-l. 
 1 — *f- 
 
 Let 
 Then, 
 
 By (13) 
 
 But tan(- ^=-1, 
 
 .-. tan -1 7 + tan-i f = - - • 
 
 4. Prove the relation cot -1 a — sec -1 5 = cos -1 "*" ~~ ■ 
 
 Let cot -1 a = x, and sec -1 b = y. 
 
 Then, cot x = a, and sec y = b. 
 
 By (12) , cos (x — !/) = cos x cos y + sin x sin y. (A) 
 
 To find the sines and cosines of x and y we use the method of § 6. 
 
 Substituting the values of cos x, cos y, sin x, and sin y obtained from 
 these triangles in (A), we have (A) 
 
 V6 2 -l 
 
 cos (x — y) = 
 
 a 1 1 
 
 Va 2 + 1 6 Va 2 + 1 
 . a + V 6 2 - 1 
 6 Va 2 + 1 
 
 & 
 
Miscellaneous Theorems. 43 
 
 Whence, 
 
 x — y or cot -1 a — sec -1 6 = cos -1 a + "^ . 
 
 6 Va 2 + 1 
 
 EXAMPLES. 
 Prove that 
 
 5. sin- 1 1| = cot- 1 -fa = cos- 1 ^. 
 
 6. tan -1 1^ = sec -1 |f = sin -1 ^. 
 Find the values of 
 
 7. sin (cos- 1 if). 10. tan (2 sin" 1 1). 
 
 8. cos (cot- 1 2). 11. cos (2 sin- 1 1). 
 
 9. tan (sec- 1 H). 12. sin (2 tan" 1 1). 
 
 13. Prove that one value of 
 
 sin" 1 — + 00^3 is-- 
 
 14. tan" 1 j + 2 tan" 1 ! is |- 
 
 15. sin -1 a; + cos -1 a; is ^ ) - 
 
 16. cos^-^ + csc^^iis^- 
 
 V82 4 4 
 
 Prove the following relations : 
 
 17. tan- 1 | + tan- 1 1 V = tan 1 |. 
 
 18. sin" 1 ^ + tan- 1 \ = cos" 1 -^= • 
 
 5 V2 
 
 19. tan- 1 ! + 00^4 = I cos" 1 !. 
 
 20. sin- 1 |f + cos- 1 if = sin" 1 f. 
 
 21 . cos -1 a + cos -1 6 = cos -1 (a& — Vl— a 3 VI — 
 
 22. tan- 1 a + cot- 1 6 = tan- 1 ^^^y 
 
44 Plane Trigonometry. 
 
 23. tan -1 a = sin -1 — ==== • 
 
 n/ _i _i Va 2 — 1 
 
 24. esc 1 a = cos * 
 
 a 
 
 ok • -i i -u i _, aft + VI — a 2 VT- 
 
 25. sm 'a + cos 1 6 = tan x 
 
 6Vl-a 2 -aVT^ r & 3 
 
 OC -1 -17 x Va 2 -1+V& 2 -1 
 
 26. sec * a — csc * o = cos -1 — ! • 
 
 ao 
 
 27. 2 sin" 1 a = tan" 1 ? a ^ 1 ~ a ' 
 
 1-2 a 2 
 
 28. tan" 1 — tan" 1 ^-+1 = tan" 1 -L. 
 
 a — 1 a 2 cr 
 
 29. sin (cos" 1 A) = tan f sin -1 -?- V 
 
 30. cos (tan -1 £■) = cot f cos -1 ) • 
 
 54. Trigonometric Equations. 
 
 An equation which involves one or more trigonometric 
 functions of an unknown angle is called a trigonometric 
 equation. 
 
 The solution of a trigonometric equation involves the 
 finding of all values of the unknown angle which will satisfy 
 the equation. 
 
 No fixed rules for solving trigonometric equations can be 
 given. Usually it is best to change all the trigonometric 
 functions into a single trigonometric function of one angle 
 and solve the resulting equation algebraically for this func- 
 tion. Suppose the function is the sine and the solution 
 
 gives sin x = -, thena; = 5, -ir -—■*, etc. If the solution 
 2 boo 
 
 7T 5 9 
 
 gives tano; = l, ® = -r, jt> t""; etc. It is evident that to 
 
Miscellaneous Theorems. 
 
 45 
 
 any given trigonometric function there corresponds an infinite 
 number of angles. It is possible, however, to find a gen- 
 eral value for the angles in each case, and we will now pro- 
 ceed to find them. 
 
 Case I. Angles having the same sine or the same 
 cosecant. 
 
 Given sin x = a. (1) 
 
 If a is positive, the terminal side of 
 the angle will lie in the first or 
 second quadrant, as OP lt or OP* 
 Let A represent the angle XOP^ 
 the smallest numerically of all the 
 angles satisfying Eq. (1). 
 
 Then the expression «7r + (— T) n A represents a series of 
 angles the terminal sides of which will for even values of n 
 coincide with OP x and for odd values 
 of n coincide with OP 2 . 
 
 (Let n in this and the two follow- 
 ing cases denote any integer positive 
 or negative including zero.) 
 
 If a in Eq. (1) is negative, the 
 terminal side of the angle will lie 
 in the third or fourth quadrant, as 
 OP 3 or OP t . Let A represent the negative angle XOP it 
 the smallest numerically of all the angles satisfying 
 Eq. (1). Then the expression mr + ( — 1)"A represents a 
 series of angles the terminal sides of which will for even 
 values of n coincide with 0P 4 and for odd values of n coin- 
 cide with 0P S . 
 
 Since rnz + (— 1)"A includes all the angles and only those 
 angles whose sines are a, it is the general value of x in 
 Eq. (1). 
 
 Since the sine and cosecant are reciprocal functions, it 
 follows that this same general value holds for all angles 
 which have the same cosecant. 
 
46 
 
 Plane Trigonometry. 
 
 Note. The special angles sin- 1 (1), sin- 1 (0), and sin" 1 (.— 1) 
 evidently have the general values 2 mr + - , nir, and 2 mr — — i 
 respectively. 
 
 Case II. Angles having the same cosine or the same 
 secant. 
 
 (2) 
 
 — X 
 
 ven 
 
 cosa; = 
 
 a. 
 
 
 
 
 
 r 
 
 
 
 
 r 
 
 
 ■K 
 
 
 
 
 
 
 ^^ 
 
 
 ^\ 
 
 
 -N 1 
 
 .' 
 
 ^U - 
 
 
 x' ■ 
 
 
 
 
 \~~ -* 
 
 
 
 
 
 
 Y 
 
 
 r> 
 
 
 F* 
 
 Fig. 
 
 3. 
 
 
 
 Fig. 4. 
 
 Using Figs. (3) and (4), and applying a line of reasoning 
 similar to that used in Case I, it can be easily shown that 
 2 utt± A is a general value for x in Eq. (2). 
 
 Since the cosine and secant are reciprocal functions, it 
 follows that this same general value holds for all angles 
 which have the same secant. 
 
 Case III. Angles having the same tangent or the same 
 cotangent. 
 
 Given tan x = a. (3) 
 
 X L 
 
 '*, 
 
 \Y\ 
 
 Y 
 Fig. 5. 
 
 Px 
 
 Y 
 Fig. & 
 
Miscellaneous Theorems. 4? 
 
 Using Figs. (5) and (6) and applying a line of reasoning 
 similar to that used in Case I, it can be easily shown that 
 n-K + A is a general value for x in Eq. (3). 
 
 Since the cotangent and tangent are reciprocal functions, 
 it follows that this same general value holds for all angles 
 which have the same cotangent. 
 
 In these three cases we have used A to represent the 
 smallest numerical value of the angle which satisfies the 
 given relation. This value is called the principal value of 
 the angle. 
 
 1. Find the general value of x when sin x= —\. 
 The principal value is — -• 
 
 Hence, Case I, the general value is me— (— 1)"( ?)• 
 
 2. Find the general value of 6 when cos = -■ 
 
 V2 
 The principal value is fir. 
 Hence, Case II, the general value is 2 n-n- ± f tt. 
 
 3. Find the general value of x, when cos x = — \, (a) 
 and tan x = — VS, (6). 
 
 From (a), x can lie in either the second or the third quad- 
 rant. From (b) x can lie in either the second or fourth 
 quadrant. To satisfy both equations x must lie in the 
 second quadrant. 
 
 The principal value is f w. 
 
 Hence, Case II, the general value is 2 wir + -f t. 
 
 EXAMPLES. 
 
 In the following two examples show that the same angles 
 are indicated by both given expressions : 
 
 4. mr+(-l)"l, and2n,r + £- 
 
 5. nir — —, and n-K + £ tt. 
 
 6 
 
48 Plane Trigonometry. 
 
 In each of the following examples give the principal and 
 the general values of the angles : 
 
 6. sin x = \. 10. cos x = 0. 
 
 7 csca;= %• 11 - tana;=-l. 
 
 V3 12. cota;=-V3. 
 
 8. cosa; = ^- 13. see 2 a; = 4. 
 
 2 
 
 /n 14. cos 2 x = — A. 
 
 9. seca;=— V2. 2 
 
 Find the general value of x when 
 
 15. sma; = -, and cos x= — • 
 
 16. tan x = — 1, and cos x = • 
 
 V2 
 
 55. The following examples will illustrate some of the 
 methods used in solving trigonometric equations. 
 
 1. Solve the equation 3 sin 2 x — 2 sin x — 1 = 0. 
 
 This is in the quadratic form ax 2 + bx + c = 0. 
 
 Hence by formula, sin x = — — — — — • 
 
 Then, sin x = 1, (1) 
 
 or sin x = — J. (2) 
 
 From (I), x = 2 mr + - . Note, Case I, § 54. 
 
 From (2), x =ror-(-l)»(19° 28.1'). Case I, § 54. 
 
 Note. The principal value of x in (2) is found from a table of 
 natural sines. 
 
 2. Solve the equation cos 2 x = 3 cos x + 1. 
 
 By (25) cos 2 a: = 2 cos 2 a; — 1. 
 
 Then the equation becomes 2 cos 2 a; — 3 cos x — 2 = 0. 
 Factoring, (2 cos a; + 1) (cos x — 2) = 0. 
 
Miscellaneous Theorems. 49 
 
 Hence, 
 
 cos x — 2 = 0, 
 
 (1) 
 
 2 cos x + 1 = 0. 
 
 (2) 
 
 COS £ = 2. 
 
 
 or 
 
 From (1) we have, 
 
 There is no angle which will satisfy this equation, since 1 is the 
 largest cosine possible. 
 
 From (2) we have, cos x = — \. 
 
 Therefore by Case II, § 54, x = 2 m ± § ir. 
 
 3. Solve the equation cos x — V3 sin x + 1 = 0. 
 
 Substituting sin x = Vl — cos' 2 x, we have 
 
 cos x + 1 = VSVl— cos 2 a;. 
 
 Squaring, cos 2 x + 2 cos x + 1 = 3 — 3 cos 2 x, 
 
 or, 2 cos 2 x + cos a; — 1 = 0. 
 
 Factoring, (2 cos x — 1) (cos a; + 1) = 0. 
 
 Then, cos x + 1 = 0, (1) 
 
 or 2 cos a; - 1 = 0. (2) 
 
 From (1) we have cos x =— 1. 
 
 By Case II, § 54, .-. x = 2 mr + tt. 
 
 From (2), cos a; = J. 
 
 By Case II., § 54, .-. % = 2mr ± £. 
 
 o 
 
 Upon testing these values in the original equation, we find that 
 
 2 nir — — will not satisfy it, and therefore the only roots are 2 nw + it 
 
 o 
 
 and 2 nir + v - ■ 
 3 
 
 4. Solve the equation tan 3 = cot 2 0. 
 
 By §4, 
 
 cot 2 8 = tan (- _ 2 s\- 
 
 Therefore, 
 
 tan30 = tan/'--26A- 
 
 By Case III, § 54, 
 
 30 = ra7r + - — 2 0. 
 
 Hence, 
 
 50= K5T+-, 
 
 2 
 
 and 
 
 6 10 
 
50 Plane Trigonometry. 
 
 5. Solve the equation sin 7 + sin $ = sin 4 0. 
 
 By (17), sin 7 + sin 9 = 2 sin 4 cos 3 0. 
 
 Then, 2 sin 4 9 cos 3 9 = sin 4 9, 
 
 or sin 4 0(2 cos 3 - 1) = 0. 
 
 Then, sin 4 = 0, (1) 
 
 or 2 eos 3 - 1 = 0. (2) 
 
 From (1) by Case I, § 54, 4 = ra and = ^ . 
 
 From (2) we have cos 3 = J. 
 
 By Case II, § 54, 3 = 2 nir ± | , 
 
 , „ 2 W . 7T 
 
 and e = -T ± V 
 
 Equations of the form a cos x±b sin a; = c may be solved 
 by the following method : 
 
 Dividing the equation by Va 2 + 6 2 , it becomes 
 a : cosz±. b 
 
 Va 2 + ft 2 Va 2 + 6 2 Va 2 + 6 2 
 
 By § 6 we see that a is the cosine and is the sine 
 
 Va 2 + ft 2 Va 2 + ft 2 
 
 of an angle which may be found exactly or approximately from a table. 
 The equation may then be written in the form 
 
 cos <f> cos x ± sin cj> sin x ■ 
 
 Va 2 + 6 2 
 Whence by (10) or (12), 
 
 c 
 
 cos (k =1- <p) = — — 
 
 Va 2 + 6 2 
 Therefore, a; = ± tf> + cos-* ( c \ . 
 
Miscellaneous Theorems. 
 
 6. Solve the equation cos x — V3 sin x = — 1. 
 Dividing by 2, we have 
 
 -cos a; sinx = — -. 
 
 2 2 2 
 
 By § 8, - is the cosine, and -^ the sine of - • 
 £t 2 3 
 
 Hence the equation becomes 
 
 5» 
 
 .(. 
 
 oo.(, +£) = -!. 
 
 By Case II, §54, x + % = 2n*±K, 
 
 and x = 2nw + - or 2 wtt — tt. 
 
 o 
 
 EXAMPLES. 
 Solve the following equations : 
 
 7. cos 2 a;— sin 2 a; = |. 10. cos 2 x — 2 sin a; = — \. 
 
 8. sin a; + esc x = 2. 11. tan 2 x + cot 2 x = 2. 
 
 9. sina; — V3 cos a;= 1. 12. sec 2 a; + tan x = 1. 
 
 13. esc 2 a; — cot x = 3. 
 
 14. tan 2 a;-(l+V3)tana;+ V3 = 0. 
 
 15. tan sec = — V2. 21 - 4 cos 2 &+ 3 cos a; = 1 
 
 16. tan0 + cot0 = 2. 22. cos 2 x - 4 sin a; = |. 
 
 . a; , a; /k 23. sin 2 = sin ft 
 
 17. sin - + cos - = V2. 
 
 2 2 24. cos 2 a; = sin x. 
 
 18. V3 sin x — cos a; = V2. 25. sin 3 x = cos a;. 
 
 19. 6 sin + cos = 2. 26. cosm0=sinr0. 
 
 20. 2sin0 + cos0 = 2. 27. sec esc = — 2. 
 
 28. tan 2 a; = tan a;. 
 
 29. sin 5 x — sin 3 a; + sin x = 0. 
 
 30. cos 7 — cos + sin 4 = 0. 
 
 31. cos + cos 2 + cos 3 = 0. 
 
 32. sin 4 x — sin 2 a; — cos 3 a; = 0. 
 
52 Plane Trigonometry. 
 
 56. Solution of equations involving inverse trigonometric 
 functions. 
 
 The method of solving equations involving inverse trigo 
 nometric functions is illustrated by the two following 
 examples. 
 
 1. Solve the equation sin -1 x = 2 cos -1 ^. 
 Let sin -1 x = A, and cos -1 x — B. 
 Then, sin A = x, and cos B = x. 
 Taking the sine of both members of the equation, 
 
 sin (sin -1 x) = sin (2 cos -1 x). 
 sin A = sin 2 B 
 By (22), = 2 sin B cos B. 
 
 Substituting, x = 2 xVl — x 2 . By § (6), 
 
 Whence, x = 0, or ± \ Vo. 
 
 2. Solve the equation tan -1 x + tan -1 (1 — a;) = tan -1 1. 
 Taking the tangent of both members of the equation, 
 
 By (13) x + (1-aQ _4 
 
 l-ic(l-x) 3' 
 Whence, 4 a; 2 — 4x + l = 0, 
 
 and x = - ■ 
 
 2 
 
 EXAMPLES. 
 
 Solve the following equations : 
 
 3. cos -1 x = sec -1 x. Ans. ± 1. 
 
 4. tan" 1 (as + 1) — tan" 1 (x — 1) = tan -1 ^. ^4ns. ± 6. 
 
 5. sin-^ + sin-^W. ^ MS . 13. 
 
 x \x J 2 
 
 6. tan- 1 (2 a;) + tan" 1 (3 as) = ~. Ans. {• or - 1. 
 
 7. sin- 1 (3 a; — 2) + cos" 1 (x) = cos" 1 VI — a* Ans. \ or 1. 
 
 8. tan -1 2 a; + tan -1 3 x = f tt. .4ns. — A or 1. 
 
 9. tan" 1 a; = 2 cos -1 j? • ^Ins. or V3. 
 
Miscellaneous Theorems. 
 
 53 
 
 10. cos-^ + cos- 1 -^ 
 
 1 t 
 
 Ans. 5. 
 
 57. The following table expresses the value of each of 
 the six principal functions in terms of the other five: 
 
 sin 
 cos 
 tan 
 cot 
 sec 
 
 CSC 
 
 ... 
 
 
 tan 
 
 1 
 
 
 CSC 
 
 Vsec 2 — 1 
 
 Vl — COS 2 
 
 Vl + tan 2 
 1 
 
 Vl + cot 2 
 cot 
 
 sec 
 sec 
 
 Vcsc 2 — 1 
 
 VI - sin 2 
 sin 
 
 Vl + tan 2 
 
 1 
 
 tan 
 
 Vl + cot 2 
 
 _1_ 
 cot 
 
 CSC 
 
 1 
 
 Vl — cos' 2 
 
 Vsec 2 — 1 
 1 
 
 Vl — sin' 2 
 
 cos 
 cos 
 
 Vcsc 2 — 1 
 
 Vl - sin 2 
 
 Vcsc 2 — 1 
 
 CSC 
 
 sin 
 1 
 
 Vl — COS 2 
 
 J_ 
 
 COS 
 
 1 
 
 Vsec 2 — 1 
 sec 
 
 Vl + cot 2 
 
 Vl + tan 2 
 
 Vl - sin 2 
 
 1 
 sin 
 
 cot 
 
 Vcsc 2 — 1 
 
 Vl + tan 2 
 
 Vl + cot 2 
 
 Vl — cos 2 
 
 tan 
 
 Vsec 2 — 1 
 
 The reciprocal forms were proved in § 33. 
 The others may be derived by aid of §§ 33, 34, 35, 36, 
 and 40, and are left as exercises for the pupil. 
 As an illustration, we will prove the formula 
 
 cos A-- 
 
 Vcsc 2 A — 1 
 
 By § 39, 
 
 csc^4 
 
 cos A = Vl — sin 2 A =•» II 
 
 VcscM-l 
 
 cse 2 ^. 
 
 cscJ. 
 
 They may also be conveniently proved by the method of 
 § 6 ; thus, let it be required to prove the formula for each 
 of the other functions in terms of the secant. 
 
 We have 
 
 sec A = 
 
 sec A 
 
54 
 
 Plane Trigonometry. 
 
 B 
 
 / 
 
 *f 
 
 >/ 
 
 y 
 
 A*— : 
 
 4 
 
 o 
 
 
 
 Since the secant is the ratio of the hypotenuse to th 
 adjacent side, we take AB = sec A, and AG =1. 
 
 Whence, BO = ^J AB* -AG 1 = Vsec* 4 -1. 
 Then by definition, 
 
 . . Vsec 2 A — l 
 sin A = — > 
 
 sec .4 
 
 cot -4 = • 
 
 COS^.: 
 
 sec .4 
 
 csc.4 = 
 
 Vsec 2 A — l 
 sec 4 
 
 Vsec 2 A — l 
 
 tan -4 = Vsec 2 A — l, 
 
 58. Line Values of the Functions. 
 
 Let XOB be any angle. 
 
 With as a centre, and a radius equal to 1, describe cir- 
 cumference AB, cutting OX at A, OB at B, and OF at C. 
 
Miscellaneous Theorems. 
 
 5S 
 
 Draw line BD perpendicular to XX ; also, lines AE and 
 CF perpendicular to OX and OY, respectively, meeting OB, 
 or OB produced, at E and F, respectively. 
 
 Then by § 17, the functions of Z XOB are : 
 
 Sin. 
 
 Fig. 1. 
 Fig. 2. 
 Fig. 3. 
 Fig. 4. 
 
 DB 
 OB 
 DB 
 OB 
 DB 
 OB 
 DB 
 OB 
 
 (-) 
 (-) 
 
 Cos. 
 
 OP 
 OB 
 OP 
 OB 
 
 (-) 
 
 0B K 
 
 OP 
 
 OB 
 
 Tan. 
 
 PB 
 
 OP 
 
 0D y 
 
 PB 
 
 OP 
 
 0P K ' 
 
 Cot. 
 
 OP 
 PB 
 
 OP 
 PB 
 
 OP 
 PB 
 
 op 
 
 DB 
 
 (-) 
 
 (-) 
 
 Sec. 
 
 OB 
 OD 
 OB 
 OD 
 OB 
 OD 
 OB 
 OD 
 
 (-) 
 (-) 
 
 Csc. 
 
 OB 
 DB 
 OB 
 DB 
 OB 
 DB 
 
 DB K 
 
 (-) 
 
 Now right triangles OBD, OF A, and OOF are similar, 
 since their sides are parallel each to each. 
 Then since OA = 0O= 1, we have 
 
 OD OA ' 
 
 DB OG ' 
 
 °B = °Z=OE, 
 
 OD OA ' 
 
 OB _ 0F_ nw 
 DB~OQ-° F - 
 
£6 Plane Trigonometry. 
 
 Whence, since OB = 1, the functions of Z XOB are : 
 
 
 Sin. 
 
 Cos. 
 
 Tan. 
 
 Cot. 
 
 Sec. 
 
 Csc. 
 
 Fig. 1. 
 Fig. 2. 
 Fig. 3. 
 Fig. 4. 
 
 DB{+) 
 DB(+) 
 DB(-) 
 DB(-) 
 
 0D{+) 
 OD(-) 
 OD(-) 
 0D(+) 
 
 AE(+) 
 AE(-) 
 AE(+) 
 AE(-) 
 
 CF(+) 
 OF(-) 
 CF(+) 
 CF(-) 
 
 0E( + ) 
 OE(-) 
 OE(-) 
 0E(+) 
 
 0F( + ) 
 0F(+) 
 0F(-) 
 OF(-) 
 
 That is, if the radius of the circle is 1. 
 
 The sine is the perpendicular drawn from XX' to the 
 intersection of the circumference with the terminal line. 
 
 The cosine is the line drawn from the centre to the foot of 
 the sine. 
 
 The tangent is that portion of the geometrical tangent to 
 the circle at the intersection of its circumference with OX 
 included between OX and the terminal line, produced if 
 necessary. 
 
 The cotangent is that portion of the geometrical tangent 
 to the circle at the intersection of its circumference with 
 Y included between Y and the terminal line, produced 
 if necessary. 
 
 The secant is that portion of the terminal line, or terminal 
 line produced, included between the centre and the tangent. 
 
 The cosecant is that portion of the terminal line, or termi- 
 nal line produced, included between the centre and the 
 cotangent. 
 
 And with regard to algebraic signs, 
 
 Sines and tangents measured above XX' are positive, and 
 below, negative; cosines and cotangents measured to the right 
 of YY' are positive, and to the left, negative; secants and 
 cosecants measured on the terminal line itself are positive, 
 and on the terminal line produced through 0, negative. 
 
 The above are called the line values of the trigonometric 
 functions. 
 
Miscellaneous Theorems. 
 
 57 
 
 They simply represent the values of the functions when 
 the radius is 1 ; that is, the numerical value of the sine of an 
 angle is the same as the number which expresses the length 
 of the perpendicular drawn to XX' from the intersection of 
 the circumference and terminal line. 
 
 59. To trace the changes in the sine, cosine, tangent, cotan- 
 gent, secant, and cosecant of an angle as the angle increases 
 from 0° to 360°. 
 
 Let ATt i be a circle whose radius is 1. 
 
 Let the terminal line start from the position OA, and re- 
 volve about point as a pivot towards the position 00. 
 
 Then since the sine of the angle commences with the 
 value 0, and assumes in succession the values DjB^ D 2 B a 
 OC, D 3 B 3 , A-B* et; c. (§ 58), it is evident that, as the angle 
 increases from 0° to 90°, the sine increases from to 1; 
 from 90° to 180°, it decreases from 1 to ; from 180° to 270°, 
 it decreases (algebraically) from to — 1 ; and from 270° to 
 360°, it increases from — 1 to 0. 
 
 Since the cosine commences with the value OA, and 
 assumes in succession the values OD v OD 2 , 0, OD 3 , OD 4 , etc., 
 from 0° to 90°, it decreases from 1 to ; from 90° to 180°, 
 it decreases from to — 1 ; from 180° to 270°, it increases 
 from —1 to 0; and from 270° to 360°, it increases from 
 to 1. 
 
58 Plane Trigonometry. 
 
 Since the tangent commences with the value 0, and as- 
 sumes in succession the values AE U AE 2 , 00 (see Note to 
 I 25), AE S , AE it etc., from 0° to 90°, it increases from 
 to qo ; from 90° to 180°, it increases from — 00 to ; from 
 180° to 270°, it increases from to 00 ; and from 270° to 
 360°, it increases from — 00 to 0. 
 
 Since the cotangent commences at 00 , and assumes in suc- 
 cession the values CF^ OF 2 , 0, CF^, OF 4 , etc., from 0° to 90°, 
 it decreases from 00 to 0; from 90° to 180°, it decreases 
 from to — 00 ; from 180° to 270°, it decreases from 00 to ; 
 and from 270° to 360°, it decreases from to — 00 . 
 
 Since the secant commences with the value OA, and as- 
 sumes in succession the values OE^ OE^ 00, OE& 013 v 
 etc., from 0° to 90°, it increases from 1 to 00 ; from 90° to 
 180°, it increases from — 00 to — 1 ; from 180° to 270°, it 
 decreases from — 1 to — 00 ; and from 270° to 360°, it de- 
 creases from oo to 1. 
 
 Since the cosecant commences at 00, and assumes in suc- 
 cession the values OF x , OF 2 , OC, OF s , OF t , etc., from 0° to 
 90°, it decreases from 00 to 1 ; from 90° to 180°, it increases 
 from 1 to 00 ; from 180° to 270°, it increases from — 00 to 
 — 1 ; and from 270° to 360°, it decreases from — 1 to — 00. 
 
 i-n t- •*• tt i r sin» , tan* 
 60. Limiting Values of and . 
 
 xx 
 
 SlU. X tSiIl T 
 
 To find the limiting values of the fractions and —. — - 
 
 when x is indefinitely decreased. 
 
 Note. We suppose x to be expressed in circular measure (§ 47). 
 
 P 
 
Miscellaneous Theorems. 59 
 
 Let OPXP be a sector of a circle ; Z POP being < 180°. 
 Draw lines PT and PT tangent to the arc at P and P', 
 respectively; also, lines OTand PP' intersecting at M. 
 
 By Geometry, PT=PT. 
 
 Then, OT bisects PP at right angles, and also bisects 
 arc PP' at X. 
 Let Z XOP = A XOP' = x. 
 
 By Geometry, arc PP' > chord PP, and < PTP. 
 
 Whence, arc PX>PM, and <PT. 
 
 m , . arc PX^PM n A ^PT 
 
 Therefore, -^- > — , and < — . 
 
 Or by § 47, circ. meas. x > sin x, and < tan x. 
 
 Eepresenting the circular measure of a; by a; simply, and 
 dividing through by sin x, we have 
 
 x _ .1 j ^. tanas 1 , c OA \ 
 
 > 1, and < - — or (§ 34). 
 
 sin x sin x cos x 
 
 „, sin a; * , 
 
 Whence, <1, and >cosa:. 
 
 ' x 
 
 But when x is indefinitely decreased, cos x approaches the 
 limit 1 (§ 22). 
 
 Hence, ^5-^ approaches the limit 1 when x is indefinitely 
 
 decreased. 
 
 tan* sin a; sin a; 1 
 
 & alI1 > a; a; cos a; a; cos a; 
 
 But Biax and approach the limit 1 when x is indefi- 
 
 x cos a; 
 
 nitely decreased 
 
 Hence, - 
 decreased. 
 
 Hence, — ^ approaches the limit 1 when x is indefinitely 
 
60 Plane Trigonometry. 
 
 V. LOGARITHMS. 
 
 61. Every positive number may be expressed, exactly or 
 approximately, as a power of 10. 
 
 Thus, 100 = 10 2 ; 13 = 10 UMI "; etc. 
 
 When thus expressed, the corresponding exponent is called 
 its Logarithm to the Base 10. 
 
 Thus, 2 is the logarithm of 100 to the base 10; a relation 
 which is written log 10 100 = 2, or simply log 100 = 2. 
 
 62. Logarithms of numbers to the base 10 are called 
 Common Logarithms, and, collectively, form the Common 
 System. 
 
 They are the only ones used for numerical computations. 
 
 Any positive number, except unity, may be taken as the 
 base of a system of logarithms ; thus, if a? = m, where a 
 and m are positive numbers, then x = log a m. 
 
 Note. A negative number is not considered as having a logarithm. 
 
 63. We have by Algebra, 
 
 10° = 1, 10-^ip.l, 
 
 10' = 10, io-» = A = .oi, 
 
 10 2 = 100, 10- s = -i- = .001, etc. 
 
 10 s 
 
 Whence by the definition of § 61, 
 
 log 1 = 0, log .1 = - 1 = 9 - 10, 
 
 log 10 = 1, log .01 = - 2 = 8 - 10, 
 
 log 100 = 2, log .001 = - 3 = 7 - 10, etc. 
 
 Note. The second form for log.l, log .01, etc., is preferable in 
 practice. If no base is expressed, the base 10 is understood. 
 
Logarithms. 61 
 
 64. It is evident from § 63 that the logarithm of a num- 
 ber greater than 1 is positive, and the logarithm of a num- 
 ber between and 1 negative. 
 
 65. If a number is not an exact power of 10, its common 
 logarithm can only be expressed approximately. 
 
 The integral part of the logarithm is called the character- 
 istic, and the decimal part the mantissa. 
 
 For example, log 13 = 1.1139. 
 
 Here, the characteristic is 1, and the mantissa .1139. 
 
 For reasons which will appear hereafter, only the man- 
 tissa of the logarithm is given in a table of logarithms of 
 numbers ; the characteristic must be found by aid of the 
 rules of §§ 66 and 67. 
 
 66. It is evident from § 63 that the logarithm of a num- 
 ber between 
 
 1 and 10 is + a decimal ; 
 
 10 and 100 is 1 + a decimal ; 
 
 100 and 1000 is 2 + a decimal ; etc. 
 
 Therefore, the characteristic of the logarithm of a number 
 with one place to the left of the decimal point, is ; with 
 two places to the left of the decimal point, is 1 ; with three 
 places to the left of the decimal point, is 2 ; etc. 
 
 Hence, the characteristic of the logarithm of a number 
 greater than 1 is 1 less than the number of places to the left oj 
 the decimal point. 
 
 For example, the characteristic of log 906328.5 is 5. 
 
 67. In like manner, the logarithm of a number between 
 
 1 and .1 is 9 + a decimal — 10 ; 
 .1 and .01 is 8 + a decimal — 10 ; 
 .01 and .001 is 7 + a decimal — 10 ; etc. 
 
62 Plane Trigonometry. 
 
 Therefore, the characteristic of the logarithm of a decimal 
 with no ciphers between the decimal point and first signifi- 
 cant figure, is 9, with — 10 after the mantissa ; of a decimal 
 with one cipher between the point and first significant figure 
 is 8, with — 10 after the mantissa ; of a decimal with two 
 ciphers between the point and first significant figure is 7, 
 with — 10 after the mantissa ; etc. 
 
 Hence, to find the characteristic of the logarithm of a num- 
 ber between and 1, subtract the number of ciphers between 
 the decimal point and first significant figure from 9, writing 
 — 10 after the mantissa. 
 
 For example, the characteristic of log .007023 is 7, with 
 — 10 written after the mantissa. 
 
 Note 1. It is customary in practice to omit the — 10 after the man- 
 tissa of a negative logarithm ; but it should be allowed for in the result. 
 
 Beginners should always write it. 
 
 Note 2. Some writers combine the two portions of the character- 
 istic, and write the result as a negative characteristic before the 
 mantissa. 
 
 Thus, instead of 7.6036 - 10, the student will frequently find 3.6036, 
 a minus sign being written over the characteristic to denote that it 
 alone is negative, the mantissa being always positive. 
 
 PROPERTIES OP LOGARITHMS. 
 
 68. In any system, the logarithm of 1 is 0. 
 
 For by Algebra, a = 1 ; whence by § 62, log„ 1 = 0. 
 
 69. In any system, the logarithm of the base is 1. 
 For a 1 = a; whence, log a a = 1. 
 
 70. In any system whose base is greater than 1, the loga- 
 rithm of is — oo. 
 
 For if a is greater than 1, a~" = — = — = 0. 
 
 o? oo 
 Whence by § 62, log„ = — co. 
 
 Note. No literal meaning can be attached to such a result aa 
 log a 0=— oo. 
 
Logarithms. 63 
 
 It must be interpreted as follows : 
 
 If, in any system whose base is greater than unity, a number ap- 
 proaches the limit 0, its logarithm is negative, and increases without 
 limit in absolute value. 
 
 71. In any system, the logarithm of a product is equal to 
 the sum of the logarithms of its factors. 
 
 Assume the equations 
 
 a * = m \., whence by §62, J» = J°&« 
 a". = n ) (y = log. n 
 
 Multiplying the assumed equations, 
 
 a" x a" = mn, or a x+y = mn. 
 Whence, log„ mn = x + y = log„ m + log n. 
 
 In like manner, the theorem may be proved for the prod- 
 uct of three or more factors. 
 
 72. By aid of § 71, the logarithm of a composite number 
 may be found when the logarithms of its factors are known. 
 
 1. Given log 2 = .3010 and log 3 = .4771; find log 72. 
 log 72 = log(2 x2x2x3x3) 
 
 = log2 + log2 + log2 + log3 + log3 (§71) 
 = 3 x log 2 + 2 x log 3 = .9030 + .9542 = 1.8572. 
 
 EXAMPLES. 
 
 Given log 2 = .3010, log 3 = .4771, log 5 = .6990, and 
 log 7 = .8451, find: 
 
 2. log 35. 6. log 147. 10. log 288. 14. log 2205. 
 
 3. log 30. 7. log 225. 11. log 686. 15. log 7875. 
 
 4. log 98. 8. log 175. 12. log 504. 16. log 5832. 
 
 5. log 84. 9. log 420. 13. log 375. 17. log 14112 
 
64 Plane Trigonometry. 
 
 73. In any system, the logarithm of a fraction is equal to 
 the logarithm of the numerator minus the logarithm of the 
 denominator. 
 
 Assume the equations 
 
 «* = ''H; whence, j 2 ^ ^ 
 a? = n ) (y = log a n. 
 
 Dividing the assumed equations, 
 
 a" m , „ m, 
 
 — = — , or a"'" = — 
 a" n n 
 
 Whence, log a — = x — y = log a m — log a n. 
 
 74. 1. Given log 2 = .3010; find log 5. 
 
 log 5 = log i5 = log 10 - log 2 (§ 73) = 1 - .3010 = .6990. 
 
 EXAMPLES. 
 Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find: 
 
 1 10 
 
 2- logy 
 
 5. log 33^. 
 
 q 1 162 
 8. log—- 
 
 11. log23f 
 
 3. log I. 
 
 c i 27 
 6. log-- 
 
 9. log 4f. 
 
 10 1 196 
 
 12. log—. 
 
 4. log 45. 7. log 105. 10. log 525. 13. log96f- 
 
 75. In any system, the logarithm of any power of a quantity 
 is equal to the logarithm of the quantity multiplied by the ex- 
 ponent of the power. 
 
 Assume the equation a x = m ; whence, x = log m. 
 
 Raising both members of the assumed equation to theptb 
 power, 
 
 a px = mP ; whence, log m p = px = p log„ m. 
 
Logarithms. 65 
 
 76. In any system, the logarithm of any root of a quantity 
 is equal to the logarithm of the quantity divided by the index 
 of the root. 
 
 For, logoV m = log a (m r ) = - log„ m (§ 75). 
 
 77. 1. Given log 2 = .3010 ; find log 2*. 
 
 log 2* = 5 x log 2 = 5 x .3010 = . 5017. 
 
 o o 
 
 Note. To multiply a logarithm by a fraction, multiply first by 
 the numerator, and divide the result by the denominator. 
 
 2. Given log 3 = .4771 ; find log -Vs. 
 
 io g ^3 = 1 -°^ = im =0596 _ 
 
 8 8 
 
 EXAMPLES. 
 Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find : 
 
 3. log2 8 . 6. log42 5 . 9. log^/3. 12. log ^28. 
 
 4. log 7*. 7. logl5i 10. log-\/7. 13. log a/324. 
 
 5. log5 f . 8. log48$. 11. log -\/B". 14. log a/735. 
 15. Find log (2* x 3^). 
 
 By § 71, log(2^ x 3^)= log2^ + log 3* = $log2 + f log3 
 = .1003 + .5964 =.6967. 
 
 Find the values of the following : 
 
 16. log^|. 18. logg- 20. logM 22. log(£)* 
 
 17. log5ty2. 19. log^- 21. log A- 23. log (2* x 21*). 
 
 5* VT 
 
66 Plane Trigonometry. 
 
 78. To prove the relation 
 
 log„m 
 log„6 
 
 Assume the equations 
 
 a x = m) , (a; = log m, 
 
 V ; whence, { 
 b" = m) iy = log 6 a». 
 
 From the assumed equations, 
 
 of = b v . 
 
 Taking the yth root of both members, 
 
 x 
 
 a* = b. 
 Therefore, log„ b — -, or 
 
 That is, log 6 m 
 
 y' "' log„6 
 log„m 
 
 log„6 
 
 79. To prove the relation 
 
 log,, a x log„ 6 = 1. 
 Putting m = a in the result of § 78, we have 
 
 Whence, log 6 a x log. 6 = 1. 
 
 80. in i^e common system, the mantissce of the logarithms 
 of numbers having the same sequence of figures are equal. 
 
 Suppose, for example, that log 3.053 = .4847. 
 
 Then, log 305.3 = log (100 x 3.053) = log 100 + log 3.053 
 = 2 + .4847 = 2.4847 ; 
 
 log .03053 = log (.01 x 3.053) = log .01 + log 3.053 
 = 8 - 10 + .4847 = 8.4847 - 10 ; etc. 
 
Logarithms. 67 
 
 It is evident from the above that, if a number be multi- 
 plied or divided by any integral power of 10, producing 
 another number with the same sequence of figures, the man- 
 tissa of their logarithms will be equal. 
 
 The reason will now be seen for the statement made in 
 § 65, that only the mantissse are given in a table of loga- 
 rithms of numbers. 
 
 For, to find the logarithm of any number, we have only 
 to take from the table the mantissa corresponding to its 
 sequence of figures, and the characteristic may then be 
 prefixed in accordance with the rules of §§ 66 or 67. 
 
 Thus, if log 3.053 = .4847, then 
 
 log 30.53 = 1.4847, log .3053 = 9.4847 - 10, 
 
 log 305.3 = 2.4847, log .03053 =8.4847-10, 
 
 log 3053. = 3.4847, log .003053 = 7.4847 - 10, etc. 
 
 This property is enjoyed only by the common system of 
 logarithms, and constitutes its superiority over others for 
 the purposes of numerical computation. 
 
 81. 1. Given log 2 = .3010, log 3 = .4771 ; find log .00432. 
 We have log 432 = log (2* x 3 8 ) = 4 log 2 + 3 log 3 = 2.6353. 
 Then by § 80, the mantissa of the result is .6353. 
 Whence by § 67, log .00432 = 7.6353 - 10. 
 
 EXAMPLES. 
 
 Given log 2 = .3010, log 3 = .4771, and log 7 = .8451, find: 
 
 2. log 2.4. 6. log .00135. 10. log .1029. 
 
 3. log 16.8. 7. log 5880. 11. log 201.6. 
 
 4. log .81. 8. log .0245. 12. log y/7J>. 
 
 5. log .0192. 9. log .000486. 13. log (12.6)*- 
 
68 Plane Trigonometry. 
 
 USE OF THE TABLE OF LOGARITHMS OF NUMBERS. 
 
 (For directions as to the use of the Table of Logarithms 
 of Numbers, see pages 1 to 4 of the Introduction to the 
 author's New Four Place Logarithmic Tables.) 
 
 EXAMPLES. 
 
 82. Find the logarithms of the following numbers ; 
 
 1. 80. 6. .03294. 11. .0007178. 
 
 2. 6.3. 7. .5205. 12. 5.1809. 
 
 3. .298. 8. 20.08. 13. 1036.5. 
 
 4. 772.3. 9. 92461. 14. .086676. 
 
 5. 1056. 10. .0040322. 15. .000011507. 
 
 Find the numbers corresponding to the following loga- 
 rithms : 
 
 16. 1.8055. 21. 8.1646-10. 26. 1.6482. 
 
 17. 9.4487-10. 22. 7.5209-10. 27. 6.0450-10. 
 
 18. 0.2165. 23. 2.0095. 28. 4.8016. 
 
 19. 3.9487. 24. 0.9774. 29. 8.1142-10. 
 
 20. 2.7371. 25. 9.3178-10. 30. 5.7015-10. 
 
 APPLICATIONS. 
 
 83. The approximate value of an arithmetical quantity, 
 in which the operations indicated involve only multiplica- 
 tion, division, involution, or evolution, may be conveniently 
 found by logarithms. 
 
 The utility of the process consists in the fact that addi- 
 tion takes the place of multiplication, subtraction of divi- 
 sion, multiplication of involution, and division of evolution. 
 
 Note. In computations with four-place logarithms, the results 
 cannot usually he depended upon to more than/ow significant figures. 
 
Logarithms. 69 
 
 84. 1. Find the value of .0631 x 7.208 x .51272. 
 By § 71, log (.0631 x 7.208 x .51272) 
 
 = log .0631 + log 7.208 + log .51272. 
 
 log .0631 = 8.8000-10 
 
 log 7.208 = 0.8578 
 
 log. 51272= 9.7099-10 
 Adding, log of result = 19.3677 - 20 
 
 = 9.3677 - 10. (See Note 1.) 
 Number corresponding to 9.3677 — 10 = .2332. 
 
 Note 1. If the sum is a negative logarithm, it should be written 
 in such a form that the negative portion of the characteristic may be 
 - 10. 
 
 Thus, 19.3677 - 20 is written in the form 9.3677 - 10. 
 
 2. Find the value of ^5- 
 7984 
 
 By § 73, log ||M = log 336.8 - log 7984. 
 
 log 336.8 = 12.5273 - 10 (See Note 2.) 
 log 7984 = 3.9022 
 
 Subtracting, log of result = 8.6251 — 10 
 
 Number corresponding = .04218. 
 
 Note 2. To subtract a greater logarithm from a less, or to subtract 
 a negative logarithm from a positive, increase the characteristic of the 
 minuend by 10, writing — 10 after the mantissa to compensate. 
 
 Thus, to subtract 3.9022 from 2.5273, write the minuend in the form 
 12.5273 - 10 ; subtracting 3.9022 from this, the result is 8.6251 - 10. 
 
 3. Find the value of (.07396) 5 . 
 
 By § 75, log (.07396)5 = 5 x log .07396. 
 
 log. 07396 = 8.8690 -10 
 5 
 
 44.3450 - 50 
 
 = 4.3450 - 10 (See Note 1.) 
 = log .000002213. 
 
70 Plane Trigonometry. 
 
 4. Find the value of V-035063. 
 
 By § 76, log V.035063 = § log .035063. 
 
 log .035063 = 8.5449 - 10 
 
 20. - 20 (See Note 3.) 
 
 3)28.54~49-30 
 
 9.6150 - 10 = log .3274. 
 
 Note 3. To divide a negative logarithm, write it in such a form 
 that the negative portion of the characteristic may be exactly divisible 
 by the divisor, with — 10 as the quotient. 
 
 Thus, to divide 8.5449 - 10 by 3, we write the logarithm in the 
 form 28.5449 — 30 ; dividing this by 3, the quotient is 9.5150 - 10. 
 
 85. Arithmetical Complement. 
 
 The Arithmetical Complement of the logarithm of a num- 
 ber, or, briefly, the Cologarithm of the number, is the loga- 
 rithm of the reciprocal of that number. 
 
 Thus, colog 409 = log -i- = log 1- log 409. 
 
 log 1 = 10. - 10 (Note 2, § 84.) 
 log 409= 2.6117 
 .-. colog409= 7.3883-10. 
 
 Again, colog .067 = log —— = log 1 — log .067. 
 
 logl= 10. -10 
 log .067 = 8.8261 - 10 
 .-. colog .067 = 1.1739. 
 
 It follows from the above that the cologarithm of a number 
 may be found by subtracting its logarithm from 10 — 10. 
 
 Note. The cologarithm may be obtained by subtracting the last 
 significant figure of the logarithm from 10, and each of the others 
 from 9, — 10 being written after the result in the case of a positive 
 logarithm. 
 
Logarithms. 71 
 
 8& Example. Find the value of - 51384 
 
 8.709 x .0946 
 
 log - 51384 = log ( .51384 x -L. x _J_\ 
 5 8.709 x .0946 8 V 8.709 .0946 j 
 
 = log .51384 + log — — + log - 1 
 
 .709 .0946 
 
 = log .51384 + colog 8.709 + colog .0946. 
 
 log. 51384 = 9.7109 -10 
 
 colog 8. 709 = 9.0601 - 10 
 colog .0946 = 1.0241 
 
 9.7951 - 10 = log .6239. 
 
 It is evident from the above example that the logarithm 
 of a fraction either of whose terms is the product of factors, 
 may be found by the following rule : 
 
 Add together the logarithms of the factors of the numerator, 
 and the cologarithms of the factors of the denominator. 
 
 EXAMPLES. 
 
 Note. A negative number has no common logarithm (§ 62, Note). 
 
 If such numbers occur in computation, they should be treated as if 
 they were positive, and the sign of the result determined irrespective 
 of the logarithmic work. 
 
 Thus, in Ex. 3, § 87, the value of 439.2 x (- 7.1367) is obtained by 
 finding the value of 439.2 x 7.1367, and putting a negative sign before 
 the result. See also Ex. 33. 
 
 87. Find by logarithms the values of the following : 
 
 1. 3.145 x .6839. 4. (- 9.0654) x (- .010785) 
 
 2. 847.6 x .02287. 5. .36552 x .025208. 
 
 3. 439.2 x (- 7.1367). 6. - .0019036 x 57.143. 
 
 - 486.7 9 - .2709 .. 8062.4 
 
 76.52' ' .08683 ' ' 9.5073* 
 
 a 1.0548 10 6.802 lg .0001798 
 
 34,96 .0051264 - .033166 
 
72 Plane Trigonometry. 
 
 -„ 38.961 x .695 15 (- .87028) x 37 , 
 
 4994 x .0045' ' (- .0659) x (- 42.32) 
 
 14 715 x (- .02416) , 6 .08214 x (- 73.4) 
 
 (-.516) x 142.07' ' .84x2808.7 
 
 17. (7.795)*. 22 . (.095129)*. 27. VIOO. 
 
 18. (.8328/. 23. (.00010594)1 28. VI995. 
 
 19. (-25.144) 3 . 24. VS. 29. V.072563. 
 
 s 
 
 20. (.01)?. 25. V2. 30. V.0026139. 
 
 21. (-964.8)1 26. V^6. 31. V-. 00095174. 
 
 32. Find the value of ^^- 
 
 3* 
 By § 86, 
 
 log 1^ = log 2 + log y/E + colog 3$ 
 3* 
 
 = log2 + $log5 + f colog3. 
 
 log 2 = 0.3010 
 log 5 = 0.6990 ; divide by 3 = 0.2330 
 
 colog 3 = 9.5229 - 10 ; multiply by | = 9.6024 - 10 
 
 0.1364 = log 1.369. 
 
 33. Find the value of - 3 ' - - 03296 
 
 7.962 
 
 log^ 
 
 03_296 = , j .03296 = , (J . 0g296 _ j ? g62) 
 962 T 5 7.962 TV b / 
 
 log .03296 = 8.5180 -10 
 log 7.962 =0.9010 
 
 3)27.6170 -^30 
 
 9.2057 -10 = log. 1606. 
 
 Result, -.1606. 
 
Logarithms. 
 
 Find the values of the following : 
 
 73 
 
 34. 4$x7i 
 3* 
 
 39. 
 
 4400\f 44. ^3x^5xa/7. 
 69287 
 
 35 
 
 36. 
 
 37. 
 
 38, 
 
 40. J™ 
 
 \ 94( 
 
 io79 
 
 46 
 
 (•QQi) f 
 
 ^7 
 
 V^8 
 
 (-10) 1 
 
 41. 
 
 42. 
 
 43. 
 
 940 
 
 5i 
 
 45. 
 
 46. 
 
 / 76-lx.Oi 
 V 1.307 
 
 .0593\l i 
 
 V=3 
 
 75.44 
 
 -V1000 47. 
 
 (-.6)* 
 
 «/3 s/7 
 
 48. 
 
 31.4 x. 415 
 
 ^.0009657 | 
 ^.0049784 
 
 -(.25693)^ 
 (-.8346)* 
 
 49. (25.467) 10 x (- .052) 12 . 
 
 50. ^5106.5 x .00003109. 
 
 51. (837.5 x .0094325)1 
 
 52. (4.8672)^ x (.17544)* 
 V31J29 x^65~48 
 
 54. 
 
 55. 
 
 V- .7664x1.2809 
 (.00259)* 
 
 ■y/Mm 
 
 .374 x V.0078359 
 
 53. 
 
 V721.33 
 57. .083184 x (.2682) 8 x (56.1)*. 
 .0005616 x a/424.65 
 
 56 -^04142 x (- .947*). 
 38.014 
 
 58. 
 
 59. 
 
 60. 
 
 61. 
 
 (6.73) 4 x (.03194)* 
 485.7 x (.07301) 7 x -\/MI 
 (9.1273) 6 x (.7095)* 
 
 X(- .95048)" x (8473)^ 1 
 \|_ (- 2080.9) x -v^0572 J 
 
 ^-.003012x1.955 
 C- .843) 8 x a/17959 x (- 560.6)* 
 
74 Plane Trigonometry. 
 
 EXAMPLES IN THE USE OF TRIGONOMETRIC 
 TABLES. 
 
 (For directions, see pages 4 to 8 of the Introduction to the 
 author's New Four Place Logarithmic Tables.) 
 
 88. Tables of Logarithmic Sines, Cosines, etc. 
 
 Find the values of the following : 
 
 1. log tan 35° 39'. 6. log sin 30° 37.2'. 
 
 2. log sin 61° 58'. 7. log cos 55° 21' 48". 
 
 3. log cot 12° 34'. 8. log cot 48° 3' 43". 
 
 4. log cos 26° 56'. 9. log sec 80° 7'. 
 
 5. log tan 82° 3'. 10. log esc 65° 12'. 
 
 Find the angles corresponding in the following : 
 
 11. log tan = 0.9164. 16. log cot = 0.2154. 
 
 12. log cos = 9.9221 - 10. 17. log sin = 9.1891 - 10. 
 
 13. log sin = 9.8619 - 10. 18. log tan = 8.9668 - 10. 
 
 14. log cot = 9.4700 - 10. 19. log esc = 0.1888. 
 
 15. log cos = 9.2204 - 10. 20. log sec = 0.4032. 
 
 Tables of Natural Sines, Cosines, etc. 
 
 •Find the values of the following : 
 
 21. sin 17° 13'. 23. tan 35° V. 
 
 22. cos 75° 38'. 24. cot 68° 46'. 
 
 Find the angles corresponding in the following: 
 
 25. sin = .7385. 27. tan = 1.1897. 
 
 26. cos = .9280. 28. cot = 1.8207. 
 
Solution of Right Triangles. 
 
 75 
 
 VI. SOLUTION OF RIGHT TRIANGLES. 
 
 89. The elements of a triangle are its three sides and its 
 three angles. 
 
 We know by Geometry that a triangle is, in general, com- 
 pletely determined when three of its elements are known, 
 provided one of them is a side. 
 
 The solution of a triangle is the process of computing the 
 unknown from the given elements. 
 
 90. To solve a right triangle, two elements must be given 
 in addition to the right angle, one of which must be a side. 
 
 The various cases which can occur may all be solved by 
 aid of the following formulae : 
 
 • a a 
 sin A = — 
 c 
 
 sin B = 
 
 cos 5 = 
 
 tan A = - 
 
 tan B = 
 
 c a 
 
 When the given elements are a side and an 
 
 91. Case I, 
 angle. 
 
 The formula for computing either of the remaining sides 
 may be found by the following rule : 
 
 Take that function of the angle which involves the given side 
 and the required side. 
 
 1. Given c = 23, B = 21° 33'. Find a and 6. 
 In this case, the formulas to be used are 
 
 Whence, 
 
 cos B =-, and sin B = — 
 c' c 
 
 a = c cos B, and 6 = c sin B. 
 
 (A) 
 
76 Plane Trigonometry. 
 
 Solution by Natural Functions. 
 a = 23 x cos 21° 33' = 23 x .9301 = 21.39. 
 6 = 23 x sin 21° 33' = 23 x .3673 = 8.448. 
 
 Solution by Logarithms. 
 
 Taking the logarithms of both members, in formulae (A), 
 
 log a = log c + log cos B, and log b = log c + log sin B. 
 
 log c = 1.3617 logc = 1.3617 
 
 log cos B = 9.9685 - 10 log sin .8 = 9. 5651 - 10 
 
 loga = 1.3302 log& = 0.9268 
 
 a = 21.39. 6 = 8.448. 
 
 Note. In examples under Case I. in which the given sides are 
 numbers of not more than two significant figures, and the operations 
 indicated involve only multiplication, it is usually shorter to employ 
 Natural Functions. 
 
 In such a case, the results cannot be depended upon to more than 
 four significant figures. 
 
 2. Given a = .2359, A = 67° 18'. Find b and c. 
 
 In this case, tan A = -, and sin A = - . 
 
 b c 
 
 Whence, 6= — - — , and c = — - — • 
 
 tan A sin A 
 
 By logarithms, 
 
 log 6 = log a — log tan A, and log c = log a — log sin A. 
 
 logo = 9.3727 -10 logo = 9.3727 -10 
 
 log tan A = 0.3785 log sin^4 = 9.9650 - 10 
 
 logft = 8.9942 -10 logc = 9.4077 -10 
 
 6 = .09868. c = .2557. 
 
 92. Case II. When both given elements are sides. 
 
 First calculate one of the angles by aid of either formula 
 involving the given elements, and then compute the remain- 
 ing side by the rule of Case I. 
 
Solution of Right Triangles. 77 
 
 Ex. Given b = .1512, c = .3081. Find A and a. 
 
 We first find A by the formula cos A = -, and then a by the 
 „ c 
 
 formula sin A = -, or a = c sin A 
 c 
 
 By logarithms, 
 
 log cos A = log 6 — log c, and log a = log c + log sin A. 
 log 6 = 9.1796 - 10 log c = 9.4887 - 10 
 
 logc= 9.4887 -10 log sin ^4 = 9.9401 - 10 
 
 log cos A = 9. 6909 - 10 log a = 9. 4288 - 10 
 
 ^ = 60° 36.4'. a = .2684. 
 
 93. In the Trigonometric solution of an example under 
 Case II., it is necessary to find first one of the angles, and 
 the remaining side may then be calculated. 
 
 But it is possible to compute the third side directly, with- 
 out first finding the angle, by Geometry. 
 
 Thus, in the example of § 92, we have 
 
 a? + W = <?. 
 
 Whence, a = -Vc? — b 2 = V(c + b) (c — V). 
 By logarithms, log a = | [log (c + 6) + log (c — 6)]. 
 
 c + b = .4593 ; log = 9.6621 - 10 
 c - b = .1569 ; log = 9.1956 - 10 
 2 )18.8577 -"20 
 loga = 9.4289 -10 
 a= .2685. 
 
 If the given sides are a and b, the expression for c is 
 Va 2 + b 2 , which is not adapted to logarithmic computation. 
 In such a case, it is usually shorter to proceed as in § 92. 
 
 EXAMPLES. 
 
 94. Solve the following right triangles : 
 
 1. Given A = 15°, c = 7. 3. Given B = 50°, b = 20. 
 
 2. Given B = 68°, a = 5. 4. Given a = .35, c = .62. 
 
78 
 
 Plane Trigonometry. 
 
 5. Given a = 27, 6 = 42. 8. Given b = 586, c= 763, 
 
 6. Given A = 38°, a = 8.09. 9. Given ^4 = 9°, b = 937. 
 
 7. Given B = 65°, c = .014. 10. Given a = 3.41, b = 2.87. 
 
 11. Given A = 31° 50', 
 
 12. Given A = 46° 15', 
 
 13. Given b = .0469, 
 
 14. Given B = 79° 28', 
 
 15. Given B = 67° 47', 
 
 16. Given A = 43° 30', 
 
 17. Given a = 3402, 
 
 18. Given B = 82° 6', 
 
 19. Given 6 = 578.9, 
 
 20. Given A = 26° 12', 
 
 21. Given B = 14° 53', 
 
 22. Given B = 43° 24', 
 
 23. Given a = 99.46, 
 
 24. Given .4 = 62° 44', 
 
 25. Given A = 74° 17', 
 
 26. Given B = 29° 56', 
 
 27. Given a = 63827, 
 
 28. Given A = 58° 39', 
 
 a = 48.04. 
 c = 5280. 
 c = .0515. 
 b = 842. 
 c = .00954. 
 6 = 26185. 
 6 = 2317. 
 a = .08937. 
 c = 2492. 
 c = .4694. 
 6 = 1353. 
 a = .89658. 
 c = 156.8. 
 6 = 4.2492. 
 a = .000020386. 
 c = .00078144. 
 c = 92275. 
 c = 35.733. 
 
 29. Given B = 35° 8', 6 = 17269. 
 
 30. Given a =.0067239, 6 = .0038453. 
 
 Solve the following isosceles triangles, in which A and B 
 are the equal angles, and a, 6, and c the sides opposite 
 angles A, B, and C, respectively : 
 
 31. Given A = 71°, 6 = 39. 
 
 32. Given B = 36° 40', c = .4688. 
 
Solution of Right Triangles. 79 
 
 33. Given 0=83° 52', 6 = 710.6. 
 
 34. Given a = 6875, c = 11318. 
 
 35. Given B = 29° T, a = 2.569. 
 
 36. Given A = 54° 39', c = 1.7255. 
 
 37. Given = 135° 26', c = .06377. 
 
 MISCELLANEOUS PROBLEMS. 
 
 95. If AG is the diagonal of rectangle ABGD, and the 
 side AB is horizontal and BG vertical, 
 Z BAG is called the angle of elevation 
 of point O from point A, and Z. AGD 
 the angle of depression of point A from 
 point G. 
 
 96. 1. From the top of a lighthouse, 150 feet above the 
 sea, the angles of depression of two boats, in line with the 
 lighthouse, are observed to be 12° and 30°, respectively. 
 Find the distance between the boats. 
 
 Let A be the position of the first boat, A' of the second, B the top 
 of the lighthouse, and C its foot. 
 
 Then, 
 AA' = AC -A'O-BO cot A- BC cot BAG 
 = 150 (cot 12° -cot 30°) 
 = 150(4.7046-1.7321) 
 = 150 x 2.9725 = 445.9 ft. 
 
 2. From the top of a lighthouse, 250 feet above the sea, 
 the angle of depression of a buoy is observed to be 31°. 
 Find the horizontal distance of the buoy. 
 
 3. If the radius of a circle is 834, what is the length of a 
 chord which subtends an arc of 46° ? 
 
 4. A regular hexagon is circumscribed about a circle 
 whose diameter is 59. Find the length of its side. 
 
80 Plane Trigonometry. 
 
 5. Find the angle of elevation of a road which rises a 
 distance of 349 feet in a horizontal distance of five-eighths 
 of a mile. 
 
 6. A regular polygon of nine sides is inscribed in a circle 
 whose diameter is 68. Find the length of its side. 
 
 7. How far from the foot of a tower 121 feet in height 
 must an observer stand so that the angle of elevation of its 
 top may be 23° ? 
 
 8. A regular polygon, whose side is 7.6 and angle 144°, is 
 circumscribed about a circle. Find the radius of the circle. 
 
 9. Find the angle of elevation of the sun when a monu- 
 ment whose height is 214.8 feet casts a shadow 167.4 feet 
 in length. 
 
 10. Find the length of the diagonal of a regular pentagon 
 whose side is 9.437. 
 
 11. At a distance of 41.6 feet from the base of a tower, 
 the angle of elevation of its top is observed to be 59° 36'. 
 Find its height. 
 
 12. The middle point of a chord of a circle, 24 units in 
 length, is distant 7 units from the middle point of its sub- 
 tended arc. How many degrees and minutes are there in 
 the arc ? 
 
 13. If the diameter of a circle is 6374, find the angle at 
 the centre subtended by an arc whose chord is 2138. 
 
 14. If the radius of a circle is 9.54, and the distance from 
 the middle point of a chord to the middle point of its sub- 
 tended arc is 3.87, how many degrees and minutes are there 
 in the arc ? 
 
 15. From the top of a tower, the angle of depression of 
 the extremity of a horizontal base line, 236.1 feet in length 
 measured from the foot of the tower, is observed to be 
 29° 48'. Find the height of the tower. 
 
Solution of Right Triangles. 81 
 
 16. A chord of a circle, whose length is 14.95, subtends 
 an arc of 135° 52'. What is the distance from the middle 
 point of the chord to the middle point of the arc ? 
 
 17. If a vertical pole casts a shadow which is three-fourths 
 its own length, what is the angle of elevation of the sun ? 
 
 18. The radius of the inscribed circle of an equilateral 
 triangle is .307. Find its perimeter, and the diameter of 
 the circumscribed circle. 
 
 19. The side of a regular octagon is 23.68. Find the 
 radii of its inscribed and circumscribed circles. 
 
 20. A chord of a circle subtends an arc of 70° 24'. If the 
 length of the chord is 853.4, find the radius of the circle. 
 
 21. At a point 250 feet from the foot of a cliff surmounted 
 by a lighthouse, the angle of elevation of the top of the 
 lighthouse is 50°, and of its foot 30°. Find the height of 
 the cliff, and of the lighthouse. 
 
 22. From the top of a cliff 378 feet above the sea, the 
 angles of depression of two boats, in line with the observer, 
 are observed to be 11° 50' and 29° 20', respectively. Find 
 the distance between the boats. 
 
 23. At a distance of 169 feet from the foot of a tower 
 surmounted by a pole, the tower subtends an angle of 35°, 
 and the pole an angle of 12°. Find the length of the pole. 
 
 24. How many degrees and minutes are there in the arc 
 included between two parallel chords, on the same side of 
 the centre of a circle, whose distances from the centre are 5 
 and 7, respectively, the radius of the circle being 11 ? 
 
 25. From the top of a tower, the angle of depression of a 
 stake is 31° 29'. What will be the angle of depression of 
 the stake from a point half way to the top ? 
 
 26. The diagonal of a regular pentagon is 43.92. Find 
 the radius of its inscribed circle. 
 
82 Plane Trigonometry. 
 
 27. A railway runs from A to B, a horizontal distance oi 
 1250 feet, at an angle of elevation of 8° 12', and then from 
 B to C, a horizontal distance of 375 feet, at an angle of eleva- 
 tion of 7° 26'. How many feet is above the plane oi A? 
 
 28. From a point 200 feet from the foot of a tower sur- 
 mounted by a pole, the angle of elevation of the top of the 
 pole is 38°; from a point 150 feet further, the angle of 
 elevation of the foot of the pole is 22°. Find the height 
 of the pole. 
 
 29. If the radius of the earth is 3956 miles, find the 
 radius in miles of the arctic circle, latitude 66° 32' N. 
 
 30. If the diameter of the earth is 7912 miles, what is 
 the distance of the remotest point of the surface visible 
 from the top of a mountain, 1\ miles above the sea ? 
 
 31. A flagpole 23 feet long surmounts a tower whose 
 height is 98 feet. What angle does the flagpole subtend at 
 a point on the ground 315 feet from the base of the tower ? 
 
 32. An observer notes that a spire bears due north from 
 him, the angle of elevation of its top being 22° 17'- On 
 going due east 550 feet, the spire bears 49° west of north. 
 What is the height of the spire ? 
 
 33. A regular pyramid stands on a square base, whose 
 side is 50 feet. Each side of the base makes an angle of 
 69° with the lateral edge. Find the altitude of the pyramid. 
 
 34. A vessel is sailing due north at a uniform rate of 
 speed. At 7.30 a.m., a lighthouse is observed to bear 70° 
 west of north; at 8 a.m. it is due west, at a distance of 12 
 miles. Find the distance and bearing of the lighthouse 
 at 9.30 a.m. 
 
 FORMULA FOR THE AREA OF A RIGHT TRIANGLE. 
 97. Case I. Given the hypotenuse and an acute angle. 
 
Solution of Right Triangles. 8j 
 
 b 
 
 Denoting the area by K, we have by Geometry, 
 
 2 K= ab. 
 But by § 5, a = c sin A, and 6 = c cos A. 
 Whence, 2 K = c 2 sin A cos A = % c 2 sin 2 ^4, by (22). 
 Then, 4 .fiT = c 2 sin 2 A (32) 
 
 In like manner, 4 iT= c 2 sin 2 B. (33) 
 
 C asb II. Given an angle and its opposite side. 
 
 By § 2, b = a cot A. 
 
 Whence, 2K=axacotA = a 2 cotA. (34) 
 
 In like manner, 2 iT = 5 2 cot JB. (35) 
 
 Case III. Given an angle and its adjacent side. 
 
 By § 2, b = a tan B. 
 
 Whence, 2 K=a x a tan B = a 2 tan B. (36) 
 
 In like manner, 2 K=b 2 tan A. (37) 
 
 Case IV. Given the hypotenuse and another side. 
 Since a 2 + & 2 = c 2 ? we have 
 
 2K=ab = a Vc 2 — a 2 = aV(c + a) (c — a). (38) 
 
 In like manner, 2 £"= 5V(c + 6)(c — 6). (39) 
 
 Case V. Given the two sides about the right angle. 
 
 In this case, 2K= ab. (40) 
 
84 Plane Trigonometry. 
 
 EXAMPLES. 
 
 9E 1. Given c = 10.36, B = 75° ; find the area, 
 By (33), 4 K = c 2 sin 2 5. 
 
 Whence, log (4 jBT) = 2 log c + log sin 2 .B. 
 
 logc = 1.0153 ; multiply by 2 = 2.0306 
 2 B = 150° ; log sin = 9.6990 - 10 
 
 log (4 K)= 1.7296 
 
 .-. 4JT= 53.65, and £"=13.41. 
 
 Note. To find log sin 150°, take either log 00s 60° or log sin 30°. 
 (See page 7 of the Introduction to the author's New Pour Place 
 Logarithmic Tables.) 
 
 Find the areas of the following right triangles : 
 
 2. Given A = 46°, a = 2.717. 
 
 3. Given B = 35° 16', a = .557. 
 
 4. Given a = 283.17, 6 = 94.93. 
 
 5. Given b = 4.564, c = 7.176. 
 
 6. Given A = 53° 9', c = 13.84. 
 
 7. Given A = 20° 57', b = .05027. 
 
 8. Given a = .0861, c = .4806. 
 
 9. Given B = 67° 48', c = 67.409. 
 
 10. Given B = 75° 34', b = .0032056. 
 
 11. Given A = 81° 23', c = 195.84. 
 
General Properties of Triangles. 
 
 85 
 
 VII. GENERAL PROPERTIES OP 
 TRIANGLES. 
 
 99. In any triangle, the sides are proportional to the sines 
 of their opposite angles. 
 
 I. To prove a : b = sin A : sin B. (41) 
 
 C 
 
 There will be two cases, according as angles A and B are 
 both acute (Fig. 1), or one of them obtuse (Fig. 2). 
 In each case, draw line CD perpendicular to AB. 
 
 Then in each figure, 
 Also in Fig. 1, 
 And in Fig. 2, 
 
 CD = b sin A (§ 5). 
 CD = a sin B. 
 CD = a sin CBD 
 
 = a sin (180° — B) = a sin B 
 
 (§ 32). 
 Then in either case, b sin A = a sin B. 
 
 Whence by the theory of proportion, 
 
 a:b = sin A : sin B. 
 
 In like manner, b :c = sin B : sin C, (42) 
 
 and c:a= sin C : sin A. (43) 
 
 100. In any triangle, the sum of any two sides is to their 
 difference as the tangent of half the sum of the opposite angles 
 is -to the tangent of half their difference. 
 
 By (41). a : b = sin A : sin B. 
 
86 Plane Trigonometry. 
 
 Whence by composition and division, 
 
 Or, 
 
 a + b : a — b = sin _4+sin_B: sin A— sin B. 
 
 a + b _ sin A + sin B 
 a — b sin A — sin B 
 
 B sin A + sin B = tan 1 (4 + B) , , . 
 
 ' sin A- sin £ tan %(A-E)' J K ' 
 
 Whence, 
 
 a + b = tan.\ (A + B) 
 a — b ta,n l(A — B)' 
 
 In like manner, ' = 2_5> — — — L 
 
 b-c tan \ (B - G) 
 
 and 
 
 c + q = tan j(C + A) 
 c — a tani(C — A) 
 
 (44) 
 (45) 
 (46) 
 
 101. In any triangle, the square of any side is equal to the 
 sum of the squares of the other two sides, minus twice their 
 product into the cosine of their included angle. 
 
 I. To prove a 2 = ft 2 -+- c 2 — 2 be cos A. 
 Case I. When the included angle A is acute. 
 
 (47) 
 
 There will be two cases, according as angle B is acute 
 (Fig. 1), or obtuse (Fig. 2). 
 In each case, draw line CD perpendicular to AB. 
 In Fig. 1, BD = c — AD, and in Fig. 2, BD = AD — c, 
 Squaring, we have in either case, 
 
 BD' = AD 1 + c 2 -2cxAD. 
 
General Properties of Triangles. 
 
 Adding Uff to both members, 
 
 B& + CD 2 = AD 2 + Off + <? - 2c x AD. 
 But, W + CI? = a 2 , and AD 2 + CD 2 = 6 2 . 
 Also, by § 5, AD = b cos A. 
 
 Whence, a 2 = 6 2 + c 2 — 2 6c cos A 
 
 Case II. When the included angle A is obtuse. 
 
 
 
 87 
 
 Draw line CD perpendicular to AB. 
 
 We have BD = AD + c. 
 
 Squaring, and adding CD to both members, 
 
 Bff+CD 2 = AD 2 + C~ff + c 2 + 2cxAD. 
 
 But, BD 2 +CD 2 = a 2 , and AD* + CD 1 = ft 2 . 
 
 And by § 5, 
 AD = & cos CL4Z> = & cos (180° - A) = - b eos ^ (§ 32). 
 
 Whence, a 2 = 6 2 + c 2 — 2 6c cos A. 
 
 In like manner, 6 2 = c 2 + a 2 — 2 ca cos -B, (48) 
 
 and c 2 = a 2 + b 2 - 2 a6 cos 0. (49) 
 
 102. To express the cosines of the angles of a triangle in 
 terms of the sides of the triangle. 
 
 By (47), a 2 = 6 2 + c 2 - 2 6c cos A. 
 
 Transposing, 2 6c cos A = 6 2 + c 2 — a 2 . 
 
88 Plane Trigonometry. 
 
 Whence, cos A = W + f ~ a? - (50) 
 
 2 6c 
 
 In like manner, cos B = — 31- > (51) 
 
 and C osO= a2 + 5 V~ c2 - (52) 
 
 103. To express the sines, cosines, and tangents of the half 
 angles of a triangle in terms of the sides of the triangle. 
 
 By (50), 
 
 1 - COS ^ = 1 -£— = -pr 
 
 2 6c 2 6c 
 
 Whence by (28), 
 
 o -2i a a 2 — (6 — c) 2 
 2 sin'' \A = ■£- <-• 
 
 1 2 be 
 
 r* ■ 21 a (a — b + c)(a + b — c) 
 
 Or, sitf 1 A = i L — -^ — ! '• 
 
 4 6c 
 
 Denoting the sum of the sides, a + b + c, by 2 s, we have 
 
 a — b + c = (a + b + c) — 2b = 2 s — 2b = 2(s — b), 
 
 and a + b — c = (a + 6 + c) — 2 c = 2 s — 2 c = 2 (s — c). 
 
 Whence, sin 2 A -4 = ' ~~, ;^ ~ ' • 
 
 2 4 6c 
 
 Or, sin | A = ^ (— b )(—°\ (sa) 
 
 In like manner, sin l.B=\ K 8 ~ c )( 8 ~ a ) , (54) 
 
 * ca 
 
 and siniG = A |^^3. (55 ) 
 
General Properties of Triangles. 89 
 
 Again, by (50), 
 
 2 6c 2 6c 
 
 Whence by (29), 
 
 2 cos 2 £ A 
 
 2 6c 
 
 Or, cos 2 1 A = {b + c + a)(b + c-a\ 
 
 ' 2 46c 
 
 But, 6 + c + a = 2s, 
 and b + c— a= (b + c + a) — 2 a = 2 (s — a). 
 
 Whence, cos 2 14 = 4a ('-«) . 
 
 46c 
 
 Or. C os|4=Alfel^. (56) 
 
 ™ OG 
 
 In like manner, cos \ B =a) ^ ~ h (57) 
 
 and cosl(7=J^3. 
 
 * db 
 
 Dividing (S3) by (56), we have, 
 
 (58) 
 
 sm 
 cos \A 
 
 \A_ l (s-b)(s^6) JST' 
 \A V 6c y>s(s-a) 
 
 Whence by (4), tan 1 A = J ( S & )( s ~ c ) . ( 59 ) 
 
 ' » s(s — a) ' 
 
 In like manner, tanJJB=J 5 /K* ffl ) , (60) 
 
 * s (s — 6) ' 
 
 and t an|<7=J2^H^. (61) 
 
 * s(s — c) 
 
9 o 
 
 Plane Trigonometry. 
 
 Note. Since each angle of a triangle is less than 180°, its half is 
 less than 90° ; henee, the positive sign must be taken before the radi- 
 cal in each formula of § 103. 
 
 FORMULAE FOR THE AREA OF AN OBLIQUE TRIANGLE. 
 104. Case I. Given two sides and their included angle. 
 I. When the given parts are b, c, and A. 
 
 by ' 
 
 There will be two cases, according as A is acute (Fig. 1), 
 or obtuse (Fig. 2). 
 In each case, draw line CD perpendicular to AB. 
 Then denoting the area by K, we have by Geometry, 
 
 2 iT= c x CD. 
 
 But in Fig. 1, CD = b sin A (§ 5). 
 
 And in Fig. 2, CD = b sin CAD 
 
 = 6 sin (180° - A) = & sin A (§ 32). 
 Then in either case, 
 
 2Z= be sin A. 
 
 In like manner, 2K=ca sin B, 
 
 and 2K=ab sin C. 
 
 Case II. Given a side and all the angles. 
 I. When the given parts are a, A, B, and C 
 By (64), 2K=abBinC. 
 
 (62) 
 (63) 
 (64) 
 
General Properties of Triangles. 91 
 
 t>„4. u„ /.-.s 6 sin B , a sin 2? 
 
 But by (41), - = - — -, or 6 = — — —• 
 
 a sin .4 sm^ 
 
 Whence, 2 .Sr= a x asinE x sin C 
 
 sin .4 
 
 . a 2 sin i? sin 
 sin^L 
 
 (65) 
 
 In like manner, 2 K = b * sin C sin A , (66) 
 
 and 2g= c'ski8ii 1 .B | (67) 
 
 sm G 
 
 Case III. Given the three sides. 
 
 By (62), 2 5"= 6c sin A = 2 6c sin \ A cos | ^4, by (22). 
 
 Dividing by 2, and substituting the values of sin \ A and 
 cos \ A from (53) and (56), we have, 
 
 K = 5e J('-ft)('-c) >(»-«) 
 » 6c * 6c 
 
 = V*(s— a)(s— 6)(s-c). (68) 
 
Plane Trigonometry. 
 
 VIII. SOLUTION OF OBLIQUE TRIANGLES. 
 
 In the solution of oblique triangles, we may distinguish 
 four cases. 
 
 105. Case I. Given a side and any two angles. 
 
 The third angle may be found by Geometry, and then by 
 aid of § 99 the remaining sides may be calculated. 
 
 The triangle is possible for any values of the given ele- 
 ments, provided the sum of the given angles is < 180°. 
 
 1. Given 6 = 20, A = 104°, B = 19° ; find O, a, and c. 
 
 We have C = 180° - (A + B)= 180° - 123° = 57°. 
 
 By §99, a = sin^ i and c = sin£ 
 
 6 sin B b sin B 
 
 Then, a = & sin A cse B, and c = &sinCcsc.B. 
 
 Whence, log a = log & + log sin A + log esc B, 
 
 and log c = log b + log sin C + log esc B. 
 
 log 6 = 1.3010 log 6 = 1.3010 
 
 log sin A = 9.9869 - 10 log sin C - 9.9236 - 10 
 
 log esc B = 0.4874 log esc B = 0.4874 
 
 log a = 1.7753 log c = 1.7120 
 
 a = 59.61. c = 51.52. 
 
 Note. To find the log cosecant of an angle, subtract the log sine 
 from 10 — 10. To find log sin 104°, take either log cos 14° or log sin 
 76°. (See page 7 of the author's New Four Place Logarithmic 
 Tables.) 
 
 EXAMPLES. 
 
 Solve the following triangles : 
 
 2. Given a = 180, A = 38°, B =75° 20'. 
 
 3. Given 6 = 8.19, B = 52°, Q = 109°. 
 
 4. Given c = .0246, .4 = 83° 30', B = 38° 50'. 
 
Solution of Oblique Triangles. 93 
 
 5. Given b = 67.13, ^ = 26° 18', = 44° 35'. 
 
 6. Given c= .45924, ^L = 74° 43', C =61° 29'. 
 
 7. Given a = 3024, B = 133° 34', C = 22° 57'- 
 (For additional examples under Case I, see § 112.) 
 
 106. Case II. Given two sides and their included angle. 
 
 Since one angle is known, the sum of the remaining angles 
 may be found, and then their difference may be calculated 
 by aid of § 100. 
 
 Knowing the sum and difference of the angles, the angles 
 themselves may be found, and then the remaining side may 
 be computed as in Case I. 
 
 The triangle is possible for any values of the data. 
 
 1. Given a = 82, c = 167, B = 98° ; find A, C, and 6. 
 By Geometry, C + A = 180° - B = 82°. 
 
 By § 100, c -±^ = ta "UC + ^). 
 
 JS ' c-a tan \ (C - A) 
 
 Or, tani(C-4) = ^^tanJ(C + ^). 
 
 Then, 
 log tan £ ( G - A) = log (c - a) + colog (c + a) + log tan \ ( C + A). 
 
 c-a = 85. log = 1.9294 
 
 c + a = 249. colog = 7.6038 - 10 
 
 J (C + A) = 41°. log tan = 9.9392 - 10 
 
 log tan £ (C - .4) = 9.4724 - 10 
 
 £(C--4)=16°31.7'. 
 
 Then, C=KC+i)+}(C-4)=H 31.7 l ) 
 
 and i = KC + i)-KC-4)=24°28.S', 
 
 To find the remaining side, we have by § 99, 
 
 6 = osin5_ ({ginBoBc4g 
 sin A 
 
94 Plane Trigonometry. 
 
 Whence, log 6 = log a + log sin B + log esc A. 
 
 log a = 1.9138 
 log sin B = 9.9958 -10 
 log esc A = 0.3828 
 
 log 6 = 2.2924 
 
 b = 196.1. 
 
 EXAMPLES. 
 Solve the following triangles : 
 
 2. Given a = 67, c = 33, B = 36°. 
 
 3. Given a = 986, b = 544, C = 134°. 
 
 4. Given & = .149, c = .427, A = 71°. 
 
 5. Given a = 3.95, & = 6.64, C = 68° 30'. 
 
 6. Given a = 2937, c = 6185, B = 55° 46'. 
 
 7. Given b = .01292, c=. 00286, ^ = 26° 32'. 
 (For additional examples under Case II., see § 112.) 
 
 107. Case III. Given the three sides. 
 
 The angles might be calculated by the formulae of § 102 ; 
 but as these are not adapted to logarithmic computation, it 
 is usually more convenient to use the formulae of § 103. 
 
 Each angle should be computed trigonometrically ; for we 
 then have a check on the work, since their sum should be 
 180°. 
 
 If all the angles are to be computed, the tangent formulae 
 are the most convenient, since only four different numbers 
 occur in the second members. 
 
 If but one angle is required, the cosine formula involves 
 the least work. 
 
 The triangle is possible for any values of the data, pro 
 vided no side is greater than the sum of the other two. 
 
Solution of Oblique Triangles. 95 
 
 If all the angles are required, and the tangent formulae 
 are used, it is convenient to modify them as follows ; by (59), 
 
 t3n '.l^j , - o X'- 6 X*- < = 1 j (s-a)(s-b-)(s-c) 
 % ' s{s—a) 3 s—a* s 
 
 Denoting ^Q-«)(s-&)(s-<0 by r> we ^^ 
 
 r 
 taniJ.=- 
 
 s — a 
 
 In like manner, tan \B = , and tan A C = ■ 
 
 s—b s—c 
 
 1. Given a = 2.5, 6 = 2.S, c = 2.2 ; find A, B, and G 
 Here, 2s = a4-6 + c = 7.5, and s = 3.75. 
 Then, s — a = 1.25, s — b = .95, s — c = 1.55. 
 By logarithms, 
 
 log r = £ [log(s - a)+ log(s - 6) + log (s - c)+ cologs]. 
 Also, logtan^ji = logr — log(s — a), 
 
 log tan JB = log r — log(s — 6), 
 
 log tan J O = log r — log (s — c> 
 
 log (s- a) =0.0969 logr = 9.8455- 10 
 
 log (s- 6) =9.9777 -10 log (g - 6)= 9.9777 - 10 
 
 log (s-g)= 0.1903 IogtanJJS = 9.8678 -10 
 
 cologs = 9.4260 -10 £B = 36°24.6'. 
 
 2)19.6909 - 20 B _ 7 oo 49 g/. 
 
 log r = 9.8455 -10 
 
 log (s- a) =0.0969 
 
 log tan 1 4 = 9.7456 — 10 
 
 }A = 29° 16.3'. 
 .4 = 58° 32.6'. 
 
 Check, 4 + B+C= 180° 1.2'. 
 
 log r = 9.8455 — 10 
 log (s-e) = 0.1903 
 log tan J C= 9.6552 - 1« 
 
 J C= 24° 19.7'. 
 C = 48°39.4'. 
 
96 Plane Trigonometry. 
 
 2. Given a = 7, 6 = 11, c = 9.6 ; find B. 
 
 By §103, cosiB= ^K^.. 
 
 Or, log cos J JS = J[log s + log (s — 6) + oolog c + colog a] . 
 Here, 2s = 27.6 ; whence, s = 13.8, s - 6 = 2.8. 
 logs =1.1399 
 log (s- 6) = 0.4472 
 colog c = 9.0177 - 10 
 colog a = 9.1549 -10 
 2 )19.7597 -20 
 log cos J .B = 9.8799 -10 
 
 J B = 40° 40.9', and B = 81° 21.8'. 
 
 EXAMPLES. 
 Solve the following triangles : 
 
 3. Given a = 5, 6 = 7, c = 6. 
 
 4. Given a = 10, 6 = 9, c = 8. 
 
 5. Given a = .56, 6 = .43, c = .89. 
 
 6. Given a = 70.5, 6 = 56.2, c = 63.9 ; find A. 
 
 7. Given a = .0292, 6 = .0185, c = .0357; find B. 
 
 8. Given a = 302, 6 = 427, c = 674 ; find O. 
 (For additional examples under Case III., see § 112.) 
 
 108. Case IV. Given two sides, and the angle opposite to 
 one of them. 
 
 It was stated in § 89 that a triangle is in general com- 
 pletely determined when three of its elements are known, 
 provided one of them is a side. The only exceptions occur 
 in Case IV. 
 
Solution of Oblique Triangles. 97 
 
 To illustrate, let us consider the following example : 
 Given a = 52.1, 6 = 61.2, ^ = 31° 26'; find B, O, and c. 
 
 By§99, sinB = 6 ) Qr sinB = b S \nA. 
 
 sin A a a 
 
 Whence, log sin B = log 6 + colog a + log sin A. 
 
 log 6 = 1.7868 
 colog a = 8.2832 -10 
 log sin ^ = 9.7173 -10 
 log sin 5 = 9.7873 -10 
 
 B = 37° 47.5', from the table. 
 
 But in finding the angle corresponding, attention must be paid to 
 the fact that an angle and its supplement have the same sine (§ 32). 
 
 Therefore another value of B will be 180° - 37° 47.5', or 142° 12.5' ; 
 and calling these values B\ and _B 2 , we have 
 
 B, = 37° 47.5', and B 2 = 142° 12.5'. 
 
 The reason for the ambiguity is at once apparent when we attempt 
 to construct the triangle from the data. 
 
 as* 
 
 «! D 
 
 We first lay off angle DAF= 31° 26', and on AF take AC = 61.2. 
 With O as a centre, and a radius equal to 52.1, describe an arc cutting 
 AD at Bi and _Z? 2 . Then either of the triangles ABiG or AB 2 
 satisfies the given conditions. 
 
 The two values of B which were obtained are the values of angles 
 ABiC and AB 2 C, respectively; and it is evident geometrically that 
 these angles are supplementary. 
 
 To complete the solution, denote angles ACBi and ACB 2 by C\ and 
 d, and sides ABi and AB- 2 by Ci and c., respectively. 
 
 Then, d = 180° - {A + B{) = 180° - 69° 13.5' = 110° 46. 5', 
 and C 2 = 180 o -(^-rB 2 )=180 o -173 o 38.6'= 6° 21.5'. 
 
98 Plane Trigonometry. 
 
 Again, by §99, «i = si "^, 
 a &mA 
 
 and 
 
 c 2 _ sin Oi 
 a sin A 
 
 
 Whence, ci = a sin C\ esc A, 
 
 and 
 
 C2 = a sin 2 esc A. 
 
 
 logo = 1.7168 
 
 
 loga = 1.7168 
 
 
 logsinCi = 9.9708 -10 
 
 
 log sin C 2 = 9.0443- 
 
 -10 
 
 log esc A = 0.2827 
 
 
 log esc .4 = 0.2827 
 log c 2 = 1.0438 
 
 
 log d = 1.9703 
 
 
 d = 93.40. 
 
 
 c 2 = 11.06. 
 
 
 109. Whenever an angle of an oblique triangle is deter- 
 mined from its sine, both the acute and obtuse values must 
 be retained, unless one or both can be shown to be inadmis- 
 sible ; hence there may sometimes be two solutions, some- 
 times one, and sometimes none, in an example under Case IV. 
 
 1. Let the data be a, b, and A, and suppose b < a. 
 
 By Geometry, B must be < A; hence, only the acute value 
 of B can be taken ; in this case there is but one solution. 
 
 2. Let the data be a, b, and A, and suppose b > a. 
 Since B must be > A, the triangle is impossible unless A 
 
 is acute. 
 
 Again, since s l n - = -, and b is > a, sin 23 is > sinA 
 sin .4 a 
 
 Hence, both the acute and obtuse values of B are > A, 
 and there are two solutions, except in the following cases : 
 
 If log sin 23 = 0, then sin 23 = 1 (§ 68), and B = 90°, and 
 the triangle is a right triangle ; if log sin B is positive, then 
 sin B is > 1, and the triangle is impossible. 
 
 The above results may be stated as follows : 
 
 If, of the given sides, that adjacent to the given angle is 
 the less, there is but one solution, which corresponds to the 
 acute value of the opposite angle. 
 
 If the side adjacent to the given angle is the greater, there 
 are two solutions, unless the log sine of the opposite angle is 
 or positive ; in which cases there are one solution (a right 
 triangle), and no solution, respectively. 
 
Solution of Oblique Triangles. 99 
 
 110. "We will illustrate these points by examples : 
 
 1. Given a = 7.42, 6 = 3.39, 4 = 105° 13'; find B. 
 
 Since 6 is < a, there is but one solution,, corresponding to the acute 
 ralue of B. 
 
 By §99, B in.B = 5iillA 
 
 a 
 
 log 6= 0.5302 
 cologa = 9.1296-10 
 logsin.^ 9.9845 -10 
 log sin B = 9.6443 — 10 
 B = 26° 9.6'. 
 
 2. Given 6 = 3, c = 2, C=100°; find S. 
 
 Since 5 is > c, and C is obtuse, the triangle is impossible. 
 
 3. Given a = 22.764, c = 50, A = 27° 4.8'; find C 
 
 We have;, sin G = °®2A. 
 
 a 
 
 logc = 1.6990 
 cologa = 8.6428 -10 
 log sin ^ = 9.6582 -10 
 log sin C = 0.0000 
 Therefore, sin C = 1, and O = 90°. 
 
 Here there is but one solution ; a right triangle. 
 
 4. Given a =.83, 6 =.715, 5 = 61° 47'; find^. 
 
 We have, s in4 = ^l2?. 
 
 o 
 
 log a = 9.9191 -10 
 Colog& = 0.1457 
 log sin £ = 9.9451 -10 
 log sin ^1 = 0.0099 
 Since log sin A is positive, the triangle Is impossible. 
 
ioo Plane Trigonometry. 
 
 EXAMPLES. 
 
 111, Solve the following triangles : 
 
 1. Given a = 7.3, 6 = 6.6, .4 = 56°. 
 
 2. Given 6 = 86, c = 159, (7=115°. 
 
 3. Given 6 = 60.93, c = 76.09, B = 133° 41'. 
 
 4. Given 6 = 38, c = 48, 5 = 34°. 
 
 5. Given a = .279, c = .227, 0= 65° 45'. 
 
 6. Given a = 3215, c = 6754, A = 28° 26'. 
 , 7. Given a = .06358, c = .08604, O = 19° 14'. 
 
 8. Given a = 186.7, 6 = 394.2, B = 114° 28'. 
 
 9. Given a = .462, c = .647, A = 31° V. 
 (For additional examples under Case IV., see § 112.) 
 
 MISCELLANEOUS EXAMPLES. 
 
 112. Solve the following triangles : 
 
 1. Given a = 934, 6 = 756, C = 73° 16'. 
 
 2. Given c = 8.706, B = 38° 45', C=31°59'. 
 8. Given a = 61, 6 = 85, c = 48. 
 
 4 Given a = .425, c=.454, C=37°9'. 
 
 5. Given 6 = .0479, c = .0144, A = 121° 28'. 
 
 6. Given a = 7824, c = 3202, A = 140° 53'. 
 
 7. Given 6 = .0005639, .4 = 44° 24', 5=116° 9'. 
 
 8. Given a = 1.5, 6 = 1.3, c = 1.9. ' 
 
 9. Given a = 576, 6 = 813, .4 = 23° 25'. 
 
 10. Given 6 = 2615, c = 6086, ^1 = 115° 10'. 
 
 11. Given 6 = 9.874, c = 7.486, B = 81° 47'. 
 
 12. Given a = 71387, B = 42° 56', C = 76° 7'. 
 
Solution of Oblique Triangles. 
 
 101 
 
 13. Given 6 = 51.434, 
 
 14. Given a = .008727, 
 
 15. Given a = .031, 
 
 16. Given a = .19597, 
 
 17. Given a = 3.5374, 
 
 18. Given a = .40932, 
 
 19. Given a = 31.06, 
 
 20. Given a = .019186, 
 
 21. Given a = 353.85, 
 
 22. Given 6 = 24883, 
 
 c = 47.955. 0=72° 54'. 
 c = .007065, B = 84° 56'. 
 6 = .024, c = .028. 
 6 = .13927, B = 45° 17'. 
 6 = 9.6036, A = 97° 46'. 
 .4 = 53° 13', 0=67° 32'. 
 6 = 51.49, 0=47° 43'. 
 6 = .033728, 5 = 125° 33'. 
 c = 579.42, B =19° 37'. 
 c = 20609, = 48° 6'. 
 
 AREA OF AN OBLIQUE TRIANGLE. 
 
 113. 1. Given a = 18.063, A = 96° 30', B = 35° ; find ^ 
 
 By § 104, 2_g-= a2sin ^ sinC =a2sin.BsiiiC l csoA 
 
 sin .4. 
 Whence, 
 
 log (2 K) = 2 log a + log sin I? + log sin C + log esc A. 
 Here, C = 180° - (A + 5) = 48° 30'. 
 
 log c = 1.2668; multiply by 2 = 2.5136 
 
 log sin B = 9.7586 -10 
 log sin 0= 9.8745 -10 
 log csc .4 = 0.0028 
 log (2 E) = 2. 1495 
 
 2X= 141.1, and K= 70.55. 
 
 EXAMPLES. 
 Find the areas of the following triangles : 
 
 2. Given a = 26.4, c = 47.9, 5 = 67°. 
 
 3. Given a = 8.05, B = 65° 30', C=81°40'. 
 
 4. Given a = 7, 6 = 9, c = 6. 
 
io2 Plane Trigonometry. 
 
 5. Given c = .518, A = 67° 45', 5 = 37° 19'. 
 
 6. Given 6 = 15.32, c = 36.78, A = 105° 43'. 
 
 7. Given & = 210.6, B = 32° 21', = 108° 56'. 
 
 8. Given c = .004096, A = 17° 45', C = 46° 8'. 
 
 9. Given a = .73, 6 = .55, c = .63. 
 
 10. Given a = .0006854, b = .0009743, C = 61° 44'. 
 
 11. Given a = 7.219, A = 23° 33', B = 124° 12'. 
 
 12. Given a = 5.321, c = 8.467, B = 152° 51'. 
 
 13. Given a = 39.5, 6 = 47.3, c = 50.8. 
 
 14. Given b = 250.8, ^4 = 77° 53', C = 55°29'. 
 
 15. Given b = .19146, c = .42829, A = 59° 7'. 
 
 16. Given a = .078, 6 = .091, c = .084. 
 
 17. Given b = 109.41, A = 77° 46', 5 = 43° 32'. 
 
 18. Given a = 5.7434, 6 = 8.6326, O = 129° 17'. 
 
 19. Given a = 307.4, b = 351.9, c = 335.7. 
 
 20. Given a = .0083214, A = 34° 44', = 105° 23'. 
 
 21. Given a = .064325, c = .033777, B = 141° 38'. 
 
 MISCELLANEOUS PROBLEMS. 
 
 114. 1. To find the distance of an inaccessible object A 
 from a position B, I measure a base-line BC 675 feet in 
 length, and observe the angles ABG and ACB to be 101° 17' 
 and 36° 55', respectively. Find the distance AB. 
 
 2. In a field ABCD, the sides AB, BC, CD, and DA 
 
 are 16, 23, 18, and 29 rods, respectively, and the diagonal 
 AC is 34 rods. Find the area of the field. 
 
 3. From the top of a cliff the angles of depression of 
 two stakes in the plain below, in line with the observer, 
 and 725 feet apart, are found to be 35° 10' and 19° 40', respec- 
 tively. Find the height of the cliff above the plain. 
 
Solution of Oblique Triangles. iqj 
 
 4. The area of a triangle is 437, and two of its sides are 
 36 and 43. Find the angle between them. 
 
 5. From a point in the same horizontal plane with the 
 base of a tower, the angle of elevation of its top is 42°, and 
 from a point 200 feet farther away, it is 26°. Find the 
 height of the tower, and the distance of its base from each 
 point of observation. 
 
 6. Two vessels start at the same point, at the rates of 
 9.7 and 5.5 miles an hour, respectively, the first due east, and 
 the second due southwest. Find the distance between them 
 at the end of an hour and a half, and the bearing of each 
 from the other. 
 
 7. Two sides of a triangle are .85 and .74, and the differ- 
 ence between their opposite angles is 18° 27'. Solve the 
 triangle. 
 
 8. The area of a triangle ABO is 980, its angle A is 
 56° 20', and its side b is 44. Find B, c, and a. 
 
 9. The sides of a triangle are 5, 7, and 9, respectively. 
 Find the radius of the inscribed circle. 
 
 (By Geometry, the area of a triangle is equal to one-half its perim- 
 eter multiplied by the radius of the inscribed circle.) 
 
 10. Two sides of a parallelogram are 8 and 5, and include 
 an angle of 61°. Find the diagonals. 
 
 11. The diagonals of a field ABCD intersect at E at an 
 angle of 78°. If AH, BE, CE, and DE are 27, 31, 59, and 
 
 64 feet, respectively, find the area of the field. 
 
 12. The bases of a trapezoid are 49 and 95, and the angles 
 at the extremities of tLe latter are 64° and 71°. Find the 
 non-parallel sides. 
 
 13. Two vessels, A and B, are sailing due northeast. At 
 a certain time, B lies 8 miles due south of A, and at the 
 expiration of an hour 75° east of south. If the rate of A 
 is 6 miles an hour, find the rate of B. 
 
104 Plane Trigonometry. 
 
 14. From a point in the same horizontal plane with tht 
 base of a tower, the angle of elevation of its top is 39° ; and 
 from a point 150 feet vertically above the first, the angle of 
 depression of the top is 43°. Find the height of the tower, 
 and its distance from the first point of observation. 
 
 15. From two points on either side of, and in line with, a 
 tower, 300 feet apart, the angles of elevation of its top ar« 
 observed to be 31° and 27°, respectively. Find the height 
 of the tower. 
 
 16. From a point in the same horizontal plane with the 
 base of a tower, the angle of elevation of its top is 22°, and 
 its bearing 31° west of north. From another point 400 feet 
 west of the first, the bearing is 26° east of north. Find the 
 height of the tower. 
 
 17. One of the non-parallel sides of a trapezoid is 15, the 
 angle between it and the longer base is 78°, the angle at the 
 other extremity of the longer base is 62°, and the shorter 
 base is 9. Find the other two sides. 
 
 18. Two sides of a parallelogram are 103 and 54, and one 
 of the diagonals is 137. Find the angles of the parallelo- 
 gram, and the other diagonal. 
 
 19. From a ship, two lighthouses bear due northwest. 
 After sailing 18 miles in a direction 35° west of south, the 
 lighthouses bear 6° west of north and 9° east of north, re- 
 spectively. Find the distance between the lighthouses. 
 
 20. The sides AB and BC, of quadrilateral ABCD, are 9 
 and 5, respectively, and the angles, A, B, and C are 84°, 
 109°, and 96°, respectively. Find the sides AD and CD. 
 
 21. From a position at the foot of a hill surmounted by a 
 tower, the angle of elevation of the top of the tower is 56°. 
 After walking 1260 feet toward the foot of the tower, up a 
 slope whose angle with a horizontal plane is 29°, the tower 
 subtends an angle of 65°. How far is the top of the tower 
 above the horizontal plane of the foot of the hill ? 
 
Solution of Oblique Triangles. 105 
 
 22. The sides AB, BC, and CD, of quadrilateral ABCD, 
 are 23, 41, and 36, respectively, and the angles B and C are 
 116° and 131°, respectively. Find the side AD and the 
 angles A and D. 
 
 23. The diagonals of a parallelogram are 58 and 92, 
 respectively, and intersect at an angle of 55°. Find the 
 sides and angles of the parallelogram. 
 
 24. From two points on the slope of a hill, in the same 
 vertical plane with the summit, the angles of elevation of 
 the top are 11° and 18°, respectively. The points are 300 feet 
 apart, and the second 40 feet above the horizontal plane of 
 the first. How far is the top of the hill above the hori- 
 zontal plane of the first point ? 
 
 25. To find the distance between two inaccessible buoys, 
 A and B, a line CD, 150 feet in length, is measured on the 
 shore. At C the angles ACD and BCD are observed to be 
 83° and 69°, respectively, and at D the angles ADC and 
 BDC are observed to be 74° and 97°, respectively. Find 
 the distance AB. 
 
 26. A bluff, with a lighthouse on its edge, is observed 
 from a boat, the angle of elevation of the top of the light- 
 house being 25°. After rowing 1000 feet directly toward the 
 lighthouse, the angles of elevation of its top and bottom are 
 found to be 53° and 39°, respectively. Find the height of 
 the bluff, and of the lighthouse. 
 
APPENDIX TO PLANE TRIGONOMETRY. 
 
 PARALLEL SAILING. 
 
 In Parallel Sailing, a vessel sails along a parallel of latitude from 
 one position to another in the same latitude. 
 
 Let 0' be the centre of the earth ; P, the north .pole ; PA' and PB', 
 meridians ; A'B' t the equator ; AB, a parallel, intersecting PA' at A 
 and PB' at B, and having its centre at ; draw 
 lines OA, OB, O'A', O'B', O'A, and O'P. p r 
 
 The number of geographical miles* in AB is 
 called the departure between A and B. 
 
 By geometry, 
 
 arc A'B' . 
 arc AB 
 
 O'A' , 
 OA 
 
 O'A . 
 
 OA ' 
 
 sec O'AO. 
 
 Whence, arc A'B'= arc AB x sec A' O'A. 
 
 That is, difference of longitude between A and 
 B = departure between A and B x sec lat. A. 
 
 Ex. A vessel whose position is lat. 25° 20' N., Ion. 36° 10' W., sails 
 due west 140 geographical miles. Find the longitude of the position 
 reached. 
 
 Here, diff. long. = 140 sec 25° 20'. 
 
 log 140 = 2. 1461 
 log sec 25° 20' = 0.0439 
 
 Then, 
 
 2. 1900 = log 154.9. 
 longitude = 36° 10' + 154.9' = 38° 44.0' W. 
 
 If a vessel sails from a position whose latitude and longitude are 
 
 known, due west or due east, until it is in a certain longitude, and the 
 
 distance in geographical miles is required, it follows from the above 
 
 that 
 
 departure = diff. long.t x cos latitude. 
 
 * A geographical mile is a minute of arc on the equator, 
 t In geographical miliw. In all cases hereafter in which the word 
 "mile" is used, it will be understood to mean a geographical mile. 
 
 106 
 
Appendix. 
 
 107 
 
 EXAMPLES. 
 
 1. A vessel in lat. 36° 48' N., Ion. 56° 15' W., sails due east 226 
 miles. Find the longitude of the position reached. 
 
 2. A vessel in lat. 48° 54' N., Ion. 10° 55' W., sails due west until it 
 is in Ion. 15° 12' W. Find the number of miles sailed. 
 
 MIDDLE LATITUDE SAILING. 
 
 If a vessel sails from A to B in such a man- 
 ner that its path makes the same angle with 
 every meridian that it crosses, it traverses a 
 curve called a rhumb line. 
 
 Let BC be a parallel drawn through B, 
 meeting the meridian PA at C; then, CA is 
 the difference in latitude between A and B. 
 
 If the distance traversed is small, we may 
 take ABC as a plane triangle, right-angled at 
 C, having its angle A equal to the course. 
 
 CB is the departure and AB the distance ; 
 ther. 
 
 dep. CB = dist. AB x sin A, 
 diff. lat. AC = dist. AB x cos A. 
 
 If AB is too long to neglect the curvature of 
 the earth, it may be divided into parts, AB, 
 DE, etc., each being of such length that the Fig. 3. 
 
 curvature of the earth may be neglected in it. 
 
 Draw parallels DO, EH, etc., meeting meridians PA, PD, etc, 
 at G, H, etc., respectively. 
 
 Then, ADG, BES, etc., may be regarded as plane triangles, right- 
 angled at G, H, etc., having ZBAG = ZEDS etc. = course. 
 
 Then, GD + EE + — = AD sin A + DEsin A + ■■■ 
 
 = (AD + DE + • ••) sin A, 
 
 Or, total departure = total distance x sin A. 
 
 Also, AG+ DH+ — = AD cos A + DE cos A + — 
 
 = (AD + DE+ —)cosA. 
 
 Or, total diff. lat. = total distance x cos A. 
 
io8 
 
 Plane Trigonometry. 
 
 Fig. 4. 
 
 The above relations may be represented by triangle ABC, right- 
 angled at C, having its sides AB, BC, and CA 
 equal to the distance, departure, and difference of 
 latitude, respectively, and its ZA equal to the course. 
 
 In Middle Latitude Sailing, we take the 
 total departure as measured on the parallel MN 
 (Fig. 2), midway between the parallels of A and 
 B ; it is evident from the figure that MN is ap- 
 proximately equal to the sum of QD, HE, KF, 
 and LB. 
 
 (This is sufficiently accurate if the run is not of 
 great length, nor too far away from the equator. ) 
 
 By the principles of parallel sailing, we have 
 
 dep. MN = diff. Ion. AB x cos lat. M. 
 
 The latitude of M is one-half the sum of the latitudes of A and B j 
 this is called the middle latitude. 
 
 Then, dep. = diff. Ion. x cos middle lat. 
 
 This may be represented graphically by an- 
 nexing to Fig. 4 the right triangle BCD, having 
 its hypotenuse BD equal to the difference of 
 longitude, and its angle GBD equal to the mid- 
 dle latitude. 
 
 Any problem in middle latitude sailing may 
 be solved by constructing a figure, as above, and 
 noting what is given and required ; the letter A 
 should be at the starting-point of the vessel. 
 
 Ex. A ship, in lat. 42° 30' N. , Ion. 68° 51' "W. , 
 sails S. 33° 45' E. 300 miles. Find the latitude and longitude of the 
 position reached. 
 
 Here, A = 33° 45', AB = 300. 
 Then, diff. lat. AC = 500 cos 33° 45'. 
 log 300 = 2.4771 
 log cos 33° 45' = 9.9198 - 10 
 
 2.3969 = log 249.4. 
 
 Then, lat. = 42° 30' - 249.4' * = 38° 20. 6' N. 
 
 mid. lat. = \ (42° 30' + 38° 20.6') = 40° 25.3'. 
 diff. Ion. BD = BC x sec mid. lat. 
 
 = 300 sin 33° 45' sec 40° 25.3'. 
 
 Fig. S. 
 
 Fig. 6. 
 
 * A geographical mile is equal to a minute of latitude. 
 
Appendix. 109 
 
 log 300 = 2.4771 
 log sin 33° 45' = 9. 7448 - 10 
 log sec 40° 25.3'= 0.1185 
 
 2.3404 = log 219. 
 Then, Ion. = 58° 51' - 219' = 55° 12' W. 
 
 EXAMPLES. 
 
 1. A ship in lat. 26° 15' >'., Ion. 61° 43' W., sails N.W. 253 miles. 
 Find the latitude and longitude of the position reached. 
 
 2. A ship sails from a, position whose lat. is 49° 56' N., Ion. 15° 16' 
 W., to another whose lat. is 47'18'X., Ion. 20° 10' W. Find the 
 course and distance. 
 
 (The difference in latitude and difference in longitude (in miles) 
 are known, and the middle latitude.) 
 
 3. A vessel in lat. 37° >"., Ion. 32° 16' "W. , sails N. 36° 56' W., and is in 
 lat. 41° N. Find the distance and the longitude of the position reached. 
 
 4. A ship in lat. 42° 30' N., Ion. 58° 51' W., sails, in a direction 
 between south and east, until her departure is 163 miles and her lati- 
 tude 38° 22' N. Find her course and distance and the longitude of the 
 position reached. 
 
 5. A vessel in lat. 47°44'N., Ion. 32° 44' W., sails 171 miles, in a 
 direction between north and east, until her latitude is 50° 2' N. Find 
 her course and the longitude of the position reached. 
 
 6. What is the course and distance in sailing from Pemambuco 
 (lat. 3° 27' S., Ion. 34° 50' W.) to the Cape of Good Hope (lat. 34° 23 S., 
 Ion. 18°29'E.)? 
 
 7. A ship in lat. 47° 15' N., Ion. 20° 48' W., sails, in a direction be- 
 tween south and west, 208 miles, until the departure is 162 miles. 
 Find the course and the latitude and longitude of the position reached. 
 
 8. A vessel in lat. 51° 16' S., Ion. 34° 13' E., sails B.N.E. until the 
 departure is 156 miles. Find the distance sailed and the latitude and 
 longitude of the position reached. 
 
 TRAVERSE SAILING. 
 
 In Traverse Sailing, a vessel sails from one position to another on 
 two or more different courses. 
 
 The path which it follows is called a traverse. 
 
 Each portion of the traverse is worked out independently and the 
 results combined. 
 
no Plane Trigonometry. 
 
 Ex. A vessel sails from a position in lat. 23° 20' N., Ion. 64° 30' W., 
 E.N.E. 135 miles, and then S. by E. 148 miles. Find the latitude and 
 longitude of the position reached and its bearing and distance from 
 the starting-point. 
 
 Proceeding in the same manner as in the illustrative example in 
 middle latitude sailing, we find that, on the first course, the position 
 reached is in lat. 26° 11.66' N., Ion. 62° 11.5' W. 
 
 In the same manner we find that, on the second course, the position 
 reached is in lat. 23° 46.66' N., Ion. 61° 39.65' W. 
 
 Then, as in Ex. 3 under middle latitude sailing, we find that, if 
 sailing from lat. 25° 20' N., Ion. 64° 30' W., to lat. 23° 46.56' N., 
 Ion. 61°39.65' W., the course would be S. 58° 54.3' E., and the 
 distance 180.9 miles. 
 
 (The traverse would not be worked out by the principles of middle 
 latitude sailing unless its parts were of such length that the curvature 
 of the earth could not be neglected.) 
 
 EXAMPLES. 
 
 1. A vessel in lat. 21° 45' S., Ion. 73° 10' E., sails S.W. by S. 158 
 miles, then N.N.W. 172 miles. Find the latitude and longitude of the 
 position reached and its bearing and distance from the starting-point. 
 
 2. A ship in lat. 39° N., Ion. 42° 30' W., sails N. 15° 23' E. 161 
 miles, then S. 72° 14' E. 186 miles, then S. 48° 42' W. 195 miles. 
 Find the latitude and longitude of the position reached and its bearing 
 and distance from the starting-point. 
 
 ANSWERS. 
 
 Parallel Sailing. — 1. 51° 32.7' W. 2. 168.9. 
 
 Middle Latitude Sailing. — 1. Lat. 29° 13.9' N.; Ion. 65° 5.1' W. 
 2. Course, S. 50° 53.1' W. ; distance, 250.5 miles. 3. Distance, 300.3 
 miles ; Ion. 36° 8.2' W. 4. Course, S. 33° 18.9' E. ; distance, 296.7 miles ; 
 Ion. 55° 16.8' W. 5. Course, N. 36° 11.1' E. ; lou. 30° 10.4' W. 
 
 6. Course, S. 58° 28.6' E.; distance, 3550 miles. 7. Course, S. 51° 9' W.; 
 lat. 45° 4.5' N. ; Ion. 24° 41.9' W. 8. Distance, 168.8 miles ; lat. 50° 11.4' S. ; 
 Ion. 38° 19.4' E. 
 
 Traverse Sailing. — 1. Lat. 21° 17.5' S.; Ion. 70° 23.52' E.; distance, 
 157.3 m. ; course, N. 79° 55.8' "W. 2. Lat. 38° 29.84' N. ; Ion. 40° 48.91' W.j 
 distance, 84.42 m. ; course, S. 69° 3.9' E. 
 
SPHERICAL TRIGONOMETRY. 
 
 oXKo 
 
 IX. GEOMETRICAL PRINCIPLES. 
 
 115. If a triedral angle be formed, with its vertex at the 
 centre of a sphere, it intercepts on the surface a spherical 
 triangle. 
 
 The triangle is bounded by three arcs of great circles, 
 called its sides, which measure the face angles of the tri- 
 edral angle. 
 
 The angles of the spherical triangle are the spherical 
 angles formed by the adjacent sides ; and each is equal to 
 the angle between two straight lines drawn, one in the plane 
 of each of its sides, perpendicular to the intersection of these 
 planes at the same point. 
 
 The sides of a spherical triangle are usually expressed in 
 degrees. 
 
 116. A spherical triangle is called right when it has a 
 right angle ; quadrantal when it has one side a quadrant. 
 
 117. Spherical Trigonometry treats of the trigonometric 
 relations between the sides and angles of a spherical tri- 
 angle. 
 
 The face and diedral angles of the triedral angle are not 
 altered by varying the radius of the sphere ; and hence the 
 relations between the sides and angles of a spherical triangle 
 are independent of the length of the radius. 
 
 Ill 
 
1 1 1 Spherical Trigonometry. 
 
 118. We shall limit ourselves in the present work to such 
 triangles as are considered in Geometry, where each angle 
 is less than two right angles, and each side less than the 
 semi-circumference of a great circle; that is, where each 
 element is less than 180°. 
 
 119. The proofs of the following properties of spherical 
 triangles may be found in any treatise on Solid Geometry : 
 
 1. Any side of a spherical triangle is less than the sum 
 of the other two sides. 
 
 2. The sum of the sides of a spherical triangle is less 
 than 360°. 
 
 3. The sum of the angles of a spherical triangle is greater 
 than 180°, and less than 540°. 
 
 4. If A'JB'C is the polar triangle of spherical triangle 
 ABC, that is, if A, B, and C are poles of sides B'C, CA', 
 and A'B', respectively, then ABC is the polar triangle of 
 spherical triangle A B'C. 
 
 5. In two polar triangles, each angle of one is measured 
 by the supplement of that side of the other of which it is 
 the pole ; that is, 
 
 a' = 180° -A. b' = 180° - B. c' = 180° - O. 
 
 A' = 180° - a. B' = 180° -b. C = 180° - c. 
 
 6. If two angles of a spherical triangle are unequal, the 
 sides opposite are unequal, and the greater side lies opposite 
 the greater angle ; conversely, if two sides of a spherical 
 triangle art unequal, the angles opposite are unequal, and 
 the greater angle lies opposite the greater side. 
 
Geometrical Principles. 
 
 ii 
 
 120. A spherical triangle is called tri-rectangular when it 
 has three right angles ; each side is a quadrant, and each 
 vertex is the pole of the opposite side. 
 
 121. I. Let C be the right angle of right spherical tri- 
 angle ABC, and suppose a < 90° and b < 90°. 
 
 Complete the tri-rectangular triangle A'B'C; also, since 
 B' is the pole of AC, and A 1 of BC, construct the tri- 
 rectangular triangles AB'D and A'BE. 
 
 Then since B lies within triangle AB'D, AB or c is < 90°. 
 
 Since BC is < B'C, ZAis<Z BAD, or < 90°. 
 Since AC is < AC, ZBis<Z ABE, or < 90°. 
 
 IL Suppose a < 90° and 6 > 90°. 
 
 B 
 
 Complete the lune ABAC. 
 Then in right triangle ABC, AC =180" - b. 
 That is, sides a and AC of triangle ABC are each < 90°j 
 and by I., A'B and angles A' and ABC are each < 90°. 
 
 But, c = 180° - AB, A = A, and B = 180° - ABC. 
 
 Whence, c is > 90°, A < 90°, and B > 90°. 
 
114. Spherical Trigonometry. 
 
 Similarly, if a is > 90° and b < 90°, then c is > 90° k 
 A > 90°, and B < 90°. 
 
 III. Suppose a > 90' and b > 90°. 
 
 ,3 B 
 
 Complete the lune AGBC. 
 Then in right triangle ABC, 
 
 AC = 180° - b, and BC = 180° - a. 
 
 That is, sides AC and BC of triangle ABC are each 
 <90°; and by L, AB and angles BAG' and ABC are 
 each < 90°. 
 
 But, A = 180° - 5.4(7, and S = 180° - ABC. 
 
 Whence, c is < 90°, A > 90°, and B > 90°. 
 
 Hence, in any right spherical triangle : 
 
 1. If the sides about the right angle are in the same quad- 
 rant, the hypotenuse is < 90° ; if they are in different quad- 
 rants, the hypotenuse is > 90°. 
 
 2. An angle is in the same quadrant as its opposite side. 
 
 122. In the figure of § 119, we have, by § 119, 1, 
 a' < V + c'. 
 
 Putting for a', b', and c' the values given in § 119, 5, we 
 have 
 
 180° - A < 180° - B + 180° - C, or B + C-A<180°. 
 
 Again, by § 118, B + C+ 180° > A. 
 
 Whence, B+G-A is >- 180°. 
 
 Therefore, B + C - A is between 180° and — 180°. 
 
 Similarly, G+A — B and A + B— G are between 180* 
 and - 180°. 
 
Right Spherical Triangles. 
 
 "5 
 
 X. RIGHT SPHERICAL TRIANGLES. 
 
 123. Let Cbe the right angle of right spherical triangle 
 ABC, and the centre of the sphere. 
 
 Draw radii OA, OB, and 00. 
 
 At any point A' of OA, draw lines A'B' and A'C in 
 planes OAB and OAC, respectively, perpendicular to OA, 
 meeting OB and 00 at B' and C, respectively ; also, draw 
 line B'C. 
 
 Then, OA is perpendicular to plane A' B'C. 
 
 Whence, each of the planes A'B'C and OBC is perpen- 
 dicular to plane OAC, and hence B'C is perpendicular 
 to OAC. 
 
 Therefore, B'C is perpendicular to A'C and OC. 
 
 By § 115, sides a, b, and c measure angles BOC, COA, 
 and AOB, respectively, and the angle A of the spherical 
 triangle is equal to angle B'A'C. 
 
 In right triangle OA'B', we have 
 
 ,, nD , OA' OC „ OA' 
 
 But in right triangles 05' C and 00'^', 
 
 0(7' 0-4' 
 
 _ = cosa, and _ = cos&. 
 
 Whence, 
 
 cos c = cos a cos 6. 
 
 (69) 
 
n6 Spherical Trigonometry. 
 
 B'C 
 
 A ■ ■ a ™ A,n, B ' G ' 0B ' sina /** 
 
 Again, sin A=smB'A'0' = -rr^. = -7™ = -= (70) 
 
 A'B' A'B' sin c 
 
 OB' 
 
 A'C 
 
 A J A WAini A ' G ' 0A ' taUb /.r.\ 
 
 And, cos^l = cos5'^.'C"=- jT ^ r , = -7T™= r (71) 
 
 ' J.'.B' ^'.B' tanc v ' 
 
 0A< 
 
 In like manner, sin B = — — > (72) 
 
 sine 
 
 and cos B = (73) 
 
 tan c 
 
 124. From (70) and (71), we obtain 
 
 ■ a _ sin A _ sin a tan c _ sina 
 
 cos A sin c tan b cos c tan 6 
 
 Whence by (69), tan A = ^BJ? = ^5_« ( 74 ) 
 
 cos a cos b tan 6 sin 6 
 
 In like manner, tan 5 = ^A (75) 
 
 sin a 
 
 125. By (4), sin a = cos a tan a ; then (70) may be written 
 
 tana 
 
 . cos a tan a tan c 
 
 sin A = = 
 
 cos c tan c cos c 
 
 cos a 
 
 Whence by (69) and (73), 
 
 a cos B , . 
 
 sin ^4 = -• (76) 
 
 cos v ' 
 
 In like manner, sin.B= — — . / 77 , 
 
 cos a ' 
 
Right Spherical Triangles. 117 
 
 126. From (69), (76), and (77), we have 
 
 , COS A COS B , A . -r, /„„\ 
 
 cos c = cos a cos = x = cot^i cot B. (78) 
 
 sin B sin A 
 
 127. The proofs of § 123 cannot be regarded as general, 
 for in the construction of the figure we have assumed a and 
 6, and therefore c and A (§ 121), to be less than 90°. 
 
 To prove formulae (69) to (73) universally, we must con- 
 sider two additional cases : 
 
 Case I. When one of the sides a and b is < 90°, and the 
 other > 90°. 
 
 c B 
 
 « V 
 
 In right spherical triangle ABC, let o be < 90° and b > 90°. 
 Complete the lune ABA'C; then, in spherical triangle 
 A'BC, 
 
 A'B=im°-c, A'C=l%Q°-b, A'=A, and A'BC=180°-B. 
 
 But by § 121, c is > 90°, A < 90°, and B > 90°. 
 Hence, each element, except the right angle, of right 
 spherical triangle A'BC is <90°; and we have by § 123, 
 
 cos A'B = cos o cos A'C, 
 
 . ., sin a • A ,- Dn _ B\a.A'C 
 
 sin A' = -— , Sin A. BL = - — — — , 
 
 sin A'B sm A'B 
 
 ., tanJ.'C o^„ Aivn— t&lla 
 
 COS A' = rr^-, COS A. BL = — • 
 
 tan A'B tan. A'B 
 
 Putting for A'B, A'C, A' and A'BC their values, we have 
 cos (180° - c) = cos a cos (180° - 6), 
 
n8 Spherical Trigonometry. 
 
 sin A = ^ , sin(180°-i?)= sin ( 180 °- 6 > , 
 
 sin(180°-c)' ^ ; sin (180° - c)' 
 
 cos A= t a n(180"-b) cos(18()0 _ jB) = . 
 
 tana 
 
 tan (180° -c) v ' tan (180° -c) 
 
 Whence, by § 32, — cos c = cos a(— cos 6), 
 
 a sin a • t, sin 6 
 
 Sin A = - — , Bin -B = - — , 
 
 sin c sin c 
 
 , — tan 6 „„„ t> tan a 
 cos A = , — cos B = ; 
 
 — tan c — tan c 
 
 and we obtain formulae (69) to (73) as before. 
 
 In like manner, the formulae may be proved to hold when 
 a is > 90° and b < 90°. 
 
 Case II. When both a and b are > 90°. 
 
 In right spherical triangle ABC, let a and b be > 90°. 
 Complete the lune ACBC. 
 By § 121, c is < 90°, A > 90°, and B > 90°. 
 Hence, each element, except the right angle, of right 
 spherical triangle ABC is <90°; and we have by § 123, 
 
 cos c = cos AC cos BC, 
 
 sin BAC = ™*?°L, sin ABC = sinA0 ', 
 
 sin c sin c 
 
 CO s 2L1C = *J?LA°L } C os iBC = **L5£. 
 
 tan c tan c 
 
Right Spherical Triangles. 119 
 
 Putting for AC, BC, BAC and ABC their values, 
 cos c = cos (180° - a) cos (180° — b), 
 
 sin (180° -A) = s^ (180° -a) 
 sine 
 
 cos(180°-^) = tan ( 180 °- 6 ), 
 v ' tanc ' 
 
 sin(180' - B) = si" (180° -fr) 
 sine 
 
 cos (180° - E) = tan ( 180 °- ffi >. 
 tanc 
 
 Whence, by § 32, cos e = (— cos a) ( — cos 6), 
 
 ■_ A sin a . -r, sin 6 
 
 sm A = — , sin B = , 
 
 sm c sin c 
 
 „ A — tan 6 „ — tan a 
 
 — <zosA= , — cosjB = ; 
 
 tan c tan c 
 
 and we obtain formulae (69) to (73), as before. 
 
 128. The formulae of §§ 123 to 126 are collected below 
 for convenience of reference : 
 
 cos c = cos o cos b. 
 
 sin .4 
 
 sin a 
 sine 
 
 . t, sin 6 
 
 sin B = — — • 
 smc 
 
 cos A- 
 
 tan b 
 tan c 
 
 cos 2* = *^ 
 tanc 
 
 tan A 
 
 tana 
 sin b 
 
 tanB- tan6 . 
 sin a 
 
 sin A 
 
 cos-B 
 cost 
 
 sinB _cosJ 
 cos a 
 
 
 cosc = 
 
 = cot A cot B. 
 
120 Spherical Trigonometry. 
 
 The student should compare the formulae for the sineSj 
 cosines, and tangents of A and B with the corresponding 
 formulas in §§ 2 and 4. 
 
 129. Napier's Rules of Circular Farts. 
 
 These are two rules which include all the formulae of 
 §128. 
 
 co. B 
 
 co. c 
 
 co. A 
 
 In any right spherical triangle, the elements a and b, and 
 the complements of elements A, B, and c (written in abbre- 
 viated form, co. A, co. B, and co. c), are called the circular 
 parts. 
 
 If we suppose them arranged in the order in which the 
 letters occur in the triangle, any one of the five may be 
 taken and called the middle part; the two immediately 
 adjacent are called the adjacent parts, and the remaining 
 two the opposite parts. 
 
 Then Napier's rules are : 
 
 I. The sine of the middle part is equal to the product of 
 the tangents of the adjacent parts. 
 
 II. The sine of the middle part is equal to the product of 
 the cosines of the opposite parts. 
 
 130. Napier's rules may be proved by taking each cir- 
 cular part in succession as the middle part, and showing 
 that the results agree with the formulae of § 128. 
 
 1. If a be taken as the middle part, b and co. B are the 
 adjacent parts, and co. c and co. A the opposite parts. 
 Then the rules give 
 
 sin a = tan b tan (co. B), and sin a = cos (co. c) cos (co. A). 
 
Right Spherical Triangles. 121 
 
 Or by § 31, sin a = tan b cot B, and sin a = sin c sin A ; 
 which are equivalent to (75) and (70). 
 
 2. If b be taken as the middle part, a and co. A are the 
 adjacent parts, and co. c and co. B the opposite parts. 
 
 Then, sin b = tan a tan (co. A) = tan a cot A, 
 
 and sin 5 = cos (co. c) cos (co. B) = sine sin .B ; 
 
 which are equivalent to (74) and (72). 
 
 3. If co. c be taken as the middle part, co. A and co. B 
 are the adjacent parts, and a and 6 the opposite parts. 
 
 Then, 
 
 ein (co. c) = tan (co. A) tan (co. _B), and sin (co. c) = cos a cos 6. 
 
 Or, cos c = cot .4 cot B, and cos c = cos a cos 6 ; 
 which agree with (78) and (69). 
 
 4. If co. A be taken as the middle part, b and co. c are 
 the adjacent parts, and a and co. B the opposite parts. 
 
 Then, 
 sin (co. A) =tan b tan (co. c), and sin (co. A) =cos a cos (co. i?). 
 
 Or, cos .4 = tan 6 cot c, and cos A = cos a sin B ; 
 which are equivalent to (71) and (77). 
 
 5. If co. B be taken as the middle part, a and co. c are 
 the adjacent parts, and b and co. A the opposite parts. 
 
 Then, 
 sin (co. B)= tan a tan (co. c), and sin (co. _B)=cos b cos (co. A). 
 
 Or, cos B = tan a cot c, and cos 5 = cos bsinA; 
 which are equivalent to (73) and (76). 
 
 Writers on Trigonometry differ as to the practical value of 
 Napier's rules ; but in the opinion of the highest authorities, 
 it seems to be regarded as preferable to attempt to remem- 
 ber the formulse by comparing them with the analogous 
 formulae for plane right triangles, as stated in § 128. 
 
122 Spherical Trigonometry. 
 
 SOLUTION OF RIGHT SPHERICAL TRIANGLES 
 
 131. To solve a right spherical triangle, two elements 
 must be given in addition to the right angle. 
 
 There may be six cases : 
 
 1. Given the hypotenuse and an adjacent angle. 
 
 2. Given an angle and its opposite side. 
 
 3. Given an angle and its adjacent side. 
 
 4. Given the hypotenuse and another side. 
 
 5. Given the two sides a and b. 
 
 6. Given the two angles A and B. 
 
 132. Each of these cases may be solved by the aid of §128. 
 If any two elements are given, the formula for computing 
 
 each remaining element may be found as follows : 
 
 Take the formula which involves the given parts and the 
 required part. 
 
 If all the remaining elements are required, the following 
 rule will be found convenient : 
 
 Take the three formula?, which involve the given parts. 
 
 133. It is convenient to have a check on the logarithmic 
 work, which may always be done without the necessity of 
 looking out any new logarithms. 
 
 Examples of this will be found in § 136. 
 The check formula for any particular case may be selected 
 from the set in § 128 by the following rule : 
 
 Take the formula which involves the three required parts. 
 
 Note. If Napier's rules are used, the following rule will indicate 
 which of the circular parts corresponding to the given elements and 
 any required element is to be regarded as the middle part. 
 
Right Spherical Triangles. 123 
 
 If these three circular parts are adjacent, take the middle one as the 
 middle part, and the others are then adjacent parts. 
 
 If they are not adjacent, take the part which is not adjacent to 
 either of the others as the middle part, and the others are then oppo- 
 site parts. 
 
 For the check formula, proceed as above with the circular parts 
 corresponding to the three required elements. 
 
 Thus, if c and A are the given elements, 
 
 1. To find a, consider the circular parts a, co. c, and co. A; of 
 these, a is the middle part, and co. c and co. A are opposite parts. 
 Then, by Napier's rules, 
 
 sin a = cos (co. c) cos (co. A) = sin c sin A. 
 
 2. To find 6, the circular parts are 6, co. c, and co. A ; in this case 
 co. A is the middle part, and 6 and co. c are adjacent parts. Then, 
 
 sin (co. A) = tan 6 tan (co. c) , or cos A = tan 6 cot c. 
 
 3. To find B, the circular parts are co. B, co. c, and co. A ; co. c is 
 the middle part, and co. A and co. B are adjacent parts. Then, 
 
 sin (co. c) = tan (co. A) tan (co. B) , or cos c = cot A cot B. 
 
 4. For the check formula, the circular parts are a, b, and co. B ; 
 a is the middle part, and 6 and co. B are adjacent parts. Then, 
 
 sin a = tan 6 tan (co. B) = tan 6 cot B. 
 
 134. In solving spherical triangles, careful attention 
 must be paid to the algebraic signs of the functions; the 
 cosines, tangents, and cotangents of angles between 90° and 
 180° being taken negative (§ 21). 
 
 It is convenient to place the sign of each function just 
 above or below it, as shown in the examples of § 136 ; the 
 sign of the function in the first member being then de- 
 termined in accordance with the principle that, in multipli- 
 cation or division, like signs produce +, and unlike signs 
 produce — . 
 
 Note. In the examples after the first of § 136, the signs are 
 omitted in every case where both functions in the second member 
 are positive. 
 
124 Spherical Trigonometry. 
 
 135. In finding angles corresponding, if the function is a 
 cosine, tangent, or cotangent, its sign determines whether 
 the angle is acute or obtuse ; that is, if it is +, the angle is 
 acute; and if it is — , the angle is obtuse, and the supple- 
 ment of the acute angle obtained from the tables must be 
 taken (§ 32). 
 
 If the function is a sine, since the sine of an angle is 
 equal to the sine of its supplement (§ 32), both the acute 
 angle obtained from the tables and its supplement must 
 be retained as solutions, unless the ambiguity can be 
 removed by the principles of § 121. 
 
 EXAMPLES. 
 
 136. 1. Given B = 33° 50', a = 108° ; find A, b, and c. 
 
 By the rule of § 132, the formulae from § 128 are, 
 
 =,•„ t> cos A ,.__ t, tan 6 „ D tana 
 
 sin B = 1 tan_B = 1 cos B = 
 
 cos a sin a tan c 
 
 .— — + + ++ — fori y. 
 
 That is, cos A = cos a sin B, tan 6 = sin a tan B, tan c = i^HLH.. 
 
 cos B 
 
 Hence, log cos A = log cos a 4- log sin B. 
 
 log tan 6 = log sin a + log tan B. 
 log tan c = log tan a — log cos B. 
 
 Since cos A and tan c are negative, the supplements of the acute 
 angles obtained from the tables must be taken (§ 135). 
 
 Note 1. When the supplement of the angle obtained from the 
 tables is to be taken, it is convenient to write 180° minus the element 
 in the first member, as shown below in the cases of A and c. 
 
 By the rule of § 133, the check formula for this case is 
 
 cos A — — — , or log cos A = log tan b — log tan c. 
 tan c 
 
 The values of log tan 6 and log tan c may be taken from the first 
 part of the work, and their difference should be equal to the result 
 previously found for log cos A. 
 
Right Spherical Trian^ies. ii-c 
 
 XV ^ * = ».*W0 — 10 1 c tan * = ; -t^?S 
 
 lor sir. .5 = 9. * +: L ~ — 10 ;.-c .-.-s 5 = s . - i -4 — 10 
 
 lai iOS_i = ft^C-T — 10 kg HD <■ = 0. v.>? 
 
 -^. v —J = £0- 0.-V. isv — c = "V >. ; - 5 . 
 
 J = SK^ o4. 3 . <■ = :>;-.v -?.2 . 
 
 1-c sir. « = ; . - >:■ - 10 Otot*. 
 
 1^ta;iB=: a.f;-?'."- — 10 In; r.ir. h — :-.>C4-> — 10 
 
 teetan > = 1 >.•*•.- — 10 In; tan <•=.". SfS5 
 
 • = S^ SI. 1 . V^ . * .4 = S>.iScT — 10 
 
 i G>e:i <•=:•' 30*. -4 = 100-; fc.A tl . fc, ar.i a 
 In :i:> «s?. the :irsf f .■>— r.ils aw. 
 
 scr. c an e 
 
 Thai is. sir. « = sir. .• sir. J. Tan r- = jaa - .-•.-ts _4. eet B = ;v\s r tan J. 
 
 E;rv. the siir .-, is determined £t.xe: ;:s sir.; ; 1ml the .\rrrUrrrr :s 
 rf —.--f .1 by jie rrrr:irl;S of J 111 : for * and A r..r.si **? in ti# sszie 
 qnadnui?. Tfcexeftie « is .'::..«;. and the sv.rrltir.-r.; of the angle 
 ^fetaitted invn the table rrr.s; l>f iiiker. 
 
 By j 138, :i; ;i«ei famuli is 
 
 .i = tan l> eet B. 
 
 s r. * 
 
 Hots 2. The vii'.-i forsrals sr.:"Vi sl^vsys be esrrvss?:; in :crrr.s 
 of the :"r.r.:::;r.s r.sc-.i in i; := rrrir..r; the i^rirtxi P-^*; "--*- h* the 
 oist- abOTC, the eheek i>T=r;i» i* :rir:sf :rrr.-i s.- at i.c inv-cvrg est s 
 ris^.iof tanS. 
 
 1.; s. . c = 9.$**;" — 10 l.-e .-.->s <■ = 1 ii?o — 10 
 
 :.-g sir .4 = 9.9SS4 — 10 l.ytan J= 0.T-5gr 
 
 >V sir. « = >"" - 10 Ire ecc B = ©i~i 
 
 ■ s ; — • — is- 10. IS. ~ — _? = !*"- ^O.t?\ 
 j = 111' »'. 3 = 1 5i- i.i'. 
 
 r.\-3.-..- =o.43«» 
 
 i-.-c --■* a = 9.as:-:> <*** 
 
 l.V ixa r = 9.S8WS - 10 l.-c r^r > = 9.«»£»5 - 10 
 
 :>."-"-* = ??--".?. lcgeo«JB = Oji"i 
 
 1- = 15S-" 1 i . loc srr a = i-. ,>?T5 — 10 
 
126 Spherical Trigonometry. 
 
 Note 3. We observe here a difference of .0001 in the two values 
 of log sin a. This does not necessarily indicate an error in the work, 
 for such a small difference might easily be due to the fact that the 
 logarithms are only approximately correct to the fourth decimal place. 
 
 3. Given a = 132° 6', b = 77° 51' ; find A, B, and c. 
 In this case, the three formulae are, 
 
 . "~ . tan a . t» tan b ~ ~~ "^ x 
 
 tan A = i tan B = i cos c = cos a cos 6. 
 
 sin 6 sin a 
 
 + 
 
 The check formula is 
 
 cos c = cot A cot B, or cos c tan A tan B = 1. 
 
 That is, log cos c + log tan A + log tan B = log 1=0. 
 
 log tan a = 0.0440 log cos a = 9.8263 - 10 
 
 log sin b = 9.9901 - 10 log cos 6 = 9.3232 - 10 
 
 log tan A = 0.0539 log cos u = 9. 1495 - 10 
 
 180° - A = 48° 32.8'. 180° - c = 81° 53.4'. 
 
 A = 131° 27.2' c = 98° 6.6'. 
 
 log tan 6 = 0.6670 Check. 
 
 log sin a = 9.8704 - 10 log cos c = 9.1495 - 10 
 
 log tan B = 0. 7 966 log tan .4 = 0. 0539 
 
 B = 80° 55.4'. log tan B = 0. 7966 
 
 log 1 = 0.0000 
 
 4. Given A = 105° 59', a = 128° 33'; find b, B, and c. 
 The formulae are, 
 
 .+ , tan a ■ + -r, cos A 
 sin o = 1 sin B — 
 
 tan .4 cos a sin .4 
 
 The check formula is sin B = 5^- 
 
 sin c 
 
 In this example, each required part is determined from its sine ; 
 and as the ambiguity cannot be removed by § 121, both the acute 
 angle obtained from the tables and its supplement must be retained in 
 each case. 
 
Right Spherical Triangles. 127 
 
 log tan a = 0.0986 log sin a = 9.8932 - 10 
 
 log tan A = 0.5430 log sin A = 9. 9828 - 10 
 
 log sin 6 = 9.5556 - 10 log sin c = 9.9104 - 10 
 6 = 21° 3.9', c= 54° 26.7', 
 
 or 158° 56.1'. or 125° 33.3'. 
 
 log cos A = 9.4399 - 10 Oh k 
 
 log cos a = 9.7946 — 10 , . , „,.,. ln 
 
 . log sin 6 = 9.5556 — 10 
 
 logsin_B = 9.6453 - 10 , . n ni „. ln 
 
 6 log sin c = 9.9104 — 10 
 
 B = 26° 13.5', log sin 5 = 9.6452 -10 
 
 or 153° 46.5'. 
 
 It does not follow, however, that these values can be combined pro- 
 miscuously ; for by § 121, since a is > 90°, with the value of 6 less 
 than 90° must be taken the value of c greater than 90°, and the value 
 of B less than 90° ; while with the value of 6 greater than 90° must be 
 taken the value of c less than 90°, and the value of B greater than 90°. 
 
 Thus the only solutions are : 
 
 1. 6 =21° 3.9', c = 125° 33.3', .8 = 26° 13.5'. 
 
 2. 6 = 158° 56.1', c = 54° 26.7', B = 153° 46.5'. 
 
 Note 4. The figure shows geometrically why there are two solu- 
 tions in this case. 
 
 B_ 
 
 For if AB and AC be produced to A', forming lune ABA'C, tri- 
 angle A 1 BO has side a and angle A' equal, respectively, to side a and 
 angle A of triangle ABC, and both triangles are right-angled at G. 
 
 It is evident that sides A'B and A'O and angle A'BC are the sup- 
 plements of sides c and 6 and angle ABO, respectively. 
 
 Solve the following right spherical triangles : 
 
 5. Given a = 31°, c = 60°. 
 
 6. Given A = 21°, B = 73°. 
 
I2 8 Spherical Trigonometry. 
 
 7. Given a = 8°, 6 = 22°. 
 
 8. Given B = 25°, c = 34°. 
 
 9. Given A = 40°, a = 26°. 
 
 10. Given B = 127° 20', a = 82°. 
 
 11. Given 6=18°, c = 112° 10'. 
 
 12. Given ^4 = 120°, c = 161°50'. 
 
 13. Given J. = 159° 40', 6 = 135°. 
 
 14. Given B = 110° 50', b = 118° 30'. 
 
 15. Given a =49° 10', & = 100°. 
 
 16. Given A = 170° 50', 6 = 55°. 
 
 17. Given A = 28° 20', c = 108° 40'. 
 
 18. Given A = 104° 50', B = 156° 30'. 
 
 19. Given a = 164° 10', c = 133° 50'. 
 
 20. Given B = 99° 40', c = 50°30'. 
 
 21. Given b = 130° 40', c = 70° 10'. 
 
 22. Given a = 129° 30', & = 166°50'. 
 
 23. Given A = 24° 31', 6 = 19° 9'. 
 
 24. Given ^ = 83° 15', a = 76° 46'. 
 
 25. Given 5 = 115° 22', a = 145° 39'. 
 
 26. Given & = 43° 57', c = 62° 5'. 
 
 27. Given A = 81° 29', 5=131° 51'. 
 
 28. Given a =147° 35', c = 52°13'. 
 
 29. Given 4 = 139° 4', c = 63°47'. 
 
 30. Given 5 = 39° 43', a = 54° 26'. 
 
 31. Given b = 153° 18', c = 121° 54'. 
 
 32. Given A = 37° 56', & = 157° 12'. 
 
 33. Given B = 114° 38', c = 168°23'. 
 
Right Spherical Triangles. 129 
 
 34. Given a =66° 6', c = 109°44'. 
 
 35. Given A = 30° 48', c = 13° 27'. 
 
 36. Given B = 69° 16', a = 160° 55'. 
 
 37. Given a = 142° 42', b = 78° &. 
 
 38. Given A = 126° 53', B = 47° 34'. 
 
 39. Given B = 16° 24', c = 140° 37'. 
 
 40. Given 5 = 98° 17', 6 = 143° 8'. 
 
 137. Quadrantal Triangles. 
 
 By § 119, 5, the polar triangle of a quadrantal triangle is 
 a right spherical triangle. 
 
 Hence, to solve a quadrantal triangle, we have only to 
 solve its polar triangle, and take the supplements of the 
 results. 
 
 1. Given c = 90°, a =67° 38', & = 4S°50'; find A, B, 
 and C. 
 
 Denoting the polar triangle by A'S'C, we have by § 119, 5, . 
 
 a = 90°, A' = 112° 22', B> = 131° 10' ; to find a', b', and c . 
 By § 132, the formulae for the solution are 
 
 cos a' = cos A \ cosS'= cos£ ' , and cose* = cot A' cot B'. 
 sin B' sin A' 
 
 + + 
 
 The check formula is cose' = cos a' cos 6'. 
 
 log cos A' - 9. 5S04 - 10 log cot A' = 9.6143 - 10 
 
 log sin B' = 9.S767 - 10 log cot B> = 9.9417 - 10 
 
 log cos c* = 9.5560 — 10 
 c' = 6S :, oi.S'. 
 
 Check. 
 log cos a' = 9.7037 - 10 
 log cos 6' = 9.8524 -10 
 
 log cos c = 9.5561 — 10 
 
 log cos a' 
 
 = 9.7037 - 
 
 10 
 
 180° - a' 
 
 = 59° 38. 2'. 
 
 
 log cos B 1 
 
 = 9.8184 - 
 
 10 
 
 log sin A 
 
 = 9.9660 - 
 
 10 
 
 log cos b' 
 
 = 9.8524 - 
 
 10 
 
 180° - b< 
 
 = 44°36.7', 
 
 
ijo Spherical Trigonometry. 
 
 Then in the given quadrantal triangle, we have 
 ^ = 180°-a'= 59° 38.2', 
 £ = 180° -6'= 44° 36.7', 
 C=180°-c' = lll° 5.2'. 
 
 EXAMPLES. 
 Solve the following quadrantal triangles : 
 
 2. Given A = 157°, 0=121°. 
 
 3. Given a = 117°, 6=142°50'. 
 
 4. Given ^ = 43°, 5=106°. 
 
 5. Given b =162° 20', C=64°40'. 
 
 6. Given .4 = 30° 10 ', a =72° 30'. 
 
 7. Given .4 = 118° 16', 6 =137° 57'. 
 
 8. Given a = 51° 34', C=25°49'„ 
 
 9. Given B = 141° 13', C = 49°35'. 
 
 10. Given a =17° 41', B = 38° 24'. 
 
 11. Given B = 159° 2', & = 136° 28'. 
 
 138. Isosceles Spherical Triangles. 
 
 We know, by Geometry, that if an arc of a great circle be 
 drawn from the vertex of an isosceles spherical triangle to 
 the middle point of the base, it is perpendicular to the base, 
 bisects the vertical angle, and divides the triangle into two 
 symmetrical right spherical triangles. 
 
 By solving one of these, we can find the required parts 
 of the given triangle. 
 
 1. Given a = 115°, 6 = 115°, 0=71° 40'; find A, B, 
 and c. 
 
 Denoting the elements of one of the right triangles by A , £', C, 
 a', &', and c', where C" is the right angle, we have 
 
 C' = a = 116°, and A' = J C = 35° 50'. 
 
Right Spherical Triangles. 131 
 
 We have then to find the parts a and B in this triangle. 
 
 By § 128, sin A' = ^-^-, and cos c' = cot A' cot B>. 
 sin c* 
 
 - - + 
 
 Or, sin a' = sin c* sin A', and cot IP = cos c' tan A'. 
 
 log sine' =9.9573-10 log cose =9.6259-10 
 
 log sin A 1 ~ 9.7675 - 10 log tan A' = 9.8586 - 10 
 
 log sin a' = 9. 7248 - 10 log cot B' = 9.4845 - 10 
 
 a' = 32° 3. 0". 180° - B> = 73° 1.8 '. 
 
 Bf = 106° 58.2'. 
 Then in the given isosceles triangle, 
 
 A = B= B 1 = 106° 58.2', and c = 2 a> = 64° 6.0'. 
 
 EXAMPLES, 
 Solve the following isosceles spherical triangles . 
 
 2. Given A = 27°, S = 27°, c = 135°. 
 
 3. Given o =152°, b =152°, (7= 68". 
 
 4. Given a = 112° 36', & =112° 36', c=123°50' 
 
 5. Given A = 159° 14', B = 159° 14', a = 137° 47'. 
 
132 
 
 Spherical Trigonometry. 
 
 XI. OBLIQUE SPHERICAL TRIANGLES. 
 
 GENERAL PROPERTIES OF SPHERICAL TRIANGLES. 
 
 139. In any spherical triangle, the sines of the sides are 
 proportional to the sines of their opposite angles. 
 
 Let ABC be any spherical triangle, and draw arc CD per- 
 pendicular to AB. 
 
 There will be two cases according as CD falls upon AB 
 (Fig. 1), or AB produced (Fig. 2). 
 
 In right spherical triangle ACD, in either figure, we have 
 by (70), 
 
 sin CD 
 
 Also, in Fig. 1, 
 And in Fig. 2, 
 
 sin .4 : 
 
 sin 5 = 
 
 sin b 
 sin CD 
 
 sin a 
 
 sin B = sin (180° - CBD) 
 = sin CBD (§ 32) 
 
 sin CD 
 
 sma 
 
 Dividing these equations, we have in either case 
 
 sin CD 
 sin A sin 6 sin a 
 
 sin B sin CD sin 6 
 
 (79) 
 
 sma 
 
In like manner, 
 
 sin B 
 sin C 
 
 sin b 
 sine 
 
 id 
 
 sin .4 
 
 sin a 
 
 
 sinC 
 
 sine 
 
 Oblique Spherical Triangles. 133 
 
 (80) 
 (81) 
 
 140. In any spherical triangle, the cosine of any side is 
 equal to the product of the cosines of the other two sides, plus 
 the continued product of tJieir sines and tlie cosine of their 
 included angle. 
 
 In right spherical triangle BCD, in Mg. 1, § 139, we have, 
 by (69), 
 
 cos a = cos BD cos CD = cos (c — AD) cos CD. 
 
 And in Fig. 2, 
 
 cos a = cos BD cos CD = cos (AD — c) cos CD. 
 Then in either case, by (12), 
 
 cos a = cos c cos AD cos CD + sin c sin AD cos CD. 
 But in right spherical triangle ACD, by (69), 
 cos AD cos CD = cos b. 
 
 COS & 
 
 Also, sin AD cos CD = sin AD j-p; = cos tan .41). 
 
 ' cos AD 
 
 sin& sin 6 
 
 But, since tan = r, cos — - — r- 
 
 ' cos 6 tan 
 
 tan 4.Z) 
 Then, sin AD cos CD = sin 6 — - — j- = sin b cos .4, by (71). 
 
 "Whence, cos a = cos b cos c + sin 6 sin c cos A. (82) 
 In like manner, 
 
 cos b = cos c cos a + sin c sin a cos 5, (83) 
 
 and cos c = cos a cos b + sin a sin b cos C (84) 
 
134 Spherical Trigonometry. 
 
 141. Let ABC and A'B'O be a pair of polar triangles. 
 
 Applying formula (82) to triangle A'B'C, we obtain 
 
 cos a' = cos b' cos c' + sin 6' sin c' cos A'. 
 Putting for a', V, c', and A! the values given in § 119, 5, 
 cos (180° - A) = cos (180° - B) cos (180° - G) 
 
 + sin (180° - B) sin (180° - C) cos (180° - a). 
 Whence, by § 32, 
 
 — cos A = (— cos B) (— cos C) + sin B sin C(— cos a). 
 That is, 
 
 cos A = — cos -B cos O + sin JB sin C cos a. (83) 
 Similarly, 
 
 cos B = — cos (7 cos ^4 + sin C sin .4 cos 6, (86) 
 
 and cos C = — cos J. cos B + sin .4 sin B cos c. (87) 
 
 The above proof illustrates a very important application 
 of the theory of polar triangles in Spherical Trigonometry. 
 
 If any relation has been found between the elements of a 
 spherical triangle, an analogous relation may be derived 
 from it, in which each side or angle is replaced by the 
 opposite angle or side, with suitable modifications in the 
 algebraic signs. 
 
 142. To express the sines, cosines, and tangents of the half- 
 angles of a spherical triangle in terms of the sides of the 
 triangle. 
 
Oblique Spherical Triangles. 133 
 
 From (82), § 140, 
 
 sin 6 sin c cos A = cos a — cos b cos c. 
 
 Whence, cos A = C0S a ~ C0S 6 cos c - (A) 
 
 sin 6 sin c 
 
 Subtracting both members from 1, we have 
 
 -I „„ /< 1 cos a — c °s & cos c 
 1 — cos .4 = 1 
 
 sin b sin c 
 
 _ cos b cos c + sin b sin c — cos a 
 sin 6 sin c 
 Whence, by (28), 
 
 2sinH^ = COS(6 ~ C) ~ C ° Sa - 
 sin b sin c 
 
 But by (20), 
 
 cos y — cos x = 2 sin \ (x + y) sin \ (x — y). (B) 
 
 Whence, 
 
 2 sin 2 1 A- 2 Sln * E" + (& ~ C ^ Sin i C a ~ ( 6 ~ c )], 
 T sin 6 sin c 
 
 or sinH^ = Sini(a + & ~1 Sini(CT ~ 6 + C) - 
 J sm sm c 
 
 Denoting the sum of the sides, a + b + c, by 2 s, wo have 
 
 a + &-c = (a + & + c)-2c = 2s-2c = 2(s-c), 
 
 anda-6 + c = (a + 6 + c)-26 = 2s-26 = 2(s-6). 
 
 ^^ . „ , . sin (s — 6) sin (s — c) 
 
 Whence, sm 2 \A = v ■ I . -• 
 
 ' 2 sin 6 sine 
 
 • , a sm (s — b) sin (s — c) . 
 
 Or, smii=\ • I ■ -• (88) 
 
 " > ^ \ sin & sin c v ' 
 
 . , _ /sin (s — c) sin (s — a) , on . 
 In like manner, sm \ B =yj \J 0flh \ ^ ( 89 ) 
 
 • , ^ /sin (s — a) sm (s — 6) 
 
 and sm i C =\/ ^-= — ' ■ \ -• (90) 
 
 a,mj ' 2 \ sin a sin 6 v ' 
 
136 Spherical Trigonometry. 
 
 Again, adding both members of (A) to 1, we have 
 
 -, , „ 4 t , cos a — cos b cos c 
 
 1 + cos A = 1 -f ; : 
 
 sin sin c 
 
 _ cos a — (cos & cos c — sin & sin c) 
 sin 6 sin c 
 Whence, by (29), 
 
 „ , , . cos a — cos (b + c) 
 
 2 cos 2 A -4 = -. — = — r^ — ■ — '- 
 
 * sin sin c 
 
 2 sin A (6 -f c + a) sin A (6 + c — a) , ._. 
 
 = — -. — t^—. — — -, by (B). 
 
 sin b sin c ' j \ / 
 
 Putting a + b + c = 2s, whence b + c — a = 2 (s — a), 
 
 cosH-^ = Si " SSill(s ~ CT) - 
 
 ' sin 6 sin c 
 
 Or, c 0S A^=V sin8si ^ S - a >. 
 
 " » sin h sin p. 
 
 r ,., , „ /sin s sin (s — b) 
 In like manner, cos A 5 = A/ : t -, 
 
 ' » sin (i sin n 
 
 sin c sin a 
 
 , . ~ _ /sin s sin (s — c) 
 
 and tos A = \/ -. i — ; — '-. 
 
 i » sm a sm b 
 
 Dividing (88) by (91), we have 
 
 (91) 
 (92) 
 (93) 
 
 c 
 
 sin \A _ / sin (g — &) sin (s — e) / sin 6 sin 1 
 cos A, ^4 — \ sin & sine *sins sin (s — a) 
 
 Whence, tan I A =J^~ &) ^ ( *~ C) - (94) 
 
 8 ' sm« sin (s — «) v ' 
 
 In like manner, tan A £ =J^ ( s ~ c > sin (s ~i£> (95^ 
 
 v sin s sin (s — b) v ' 
 
 and tanAC=JgM s - a ) sin ( s - 6 l , 96 x 
 
 » sin a sin (s — c) ' 
 
Oblique Spherical Triangles. 137 
 
 143. To express the sine?, cosines, and tangents of the half- 
 mltt of a spherical triangle in terms of the angles of the 
 triangle. 
 
 Frcm <85\ § 141, sin B sin C 00s o = cos A 4- cos B cos C. 
 
 Whence, cos a = eos A + ws B TOS C . m 
 
 sinBsiuC v * ' 
 
 Then, 1 - cos a = 1 - ws^ + wwBcosO 
 
 sin B sin C 
 
 Or, *> nn a ' a - ~ (eos B eos r ~ sinB sin °) ~ cos A 
 
 sin £ sin C 
 
 _ cos (B + C) + cos A 
 
 sin £ sin 
 Then by (19\ 
 
 o ,:.., , _ _ - cos 1 r (-» + C + ,-n cos j (J + C-A) 
 
 sin 2} sin (7 
 
 Denoting the suni of the angles, A 4- B 4- C, by 2 £, we 
 have B + C — A = l\,5 - A\ 
 
 Whence, sin 2 i a = - ^-^os^ -.4\ 
 
 sin £ sin C 
 
 /-w_ ■ 1 ! COS ■> COS ( o — *± ) • ^ 
 
 Or, sin&a=-\— V -, • (97) 
 
 \ sin /is 11 (7 V ' 
 
 cos .S cos (^ — A) 
 
 T T-L 11.' COS ^ COS (■> — _B) ,,__. 
 
 In like manner, sin i 6 = \i ; — —\ — - — -, (98) 
 
 » sin V sin A 
 
 , • , / cos -S cos ( S — C) ,„-.. 
 
 and sin i e = V = — r^ — ^ — - .(") 
 
 V sinxlsuuB v ' 
 
 Again, adding both members of (A) to 1, we have 
 
 1 , ., , cosvl 4- cos B cos C 
 
 1 + cos a = H . 
 
 sm B sm V 
 
 __ cos ^-1 4- cos B co s C 4- sin B sin C 
 sin B sin C 
 
138 Spherical Trigonometry. 
 
 rri. n 21 COS A + COS (B — G) 
 
 Then, 2 cos 2 A a = . ^ > „ L 
 
 2 sin -B sin G 
 
 2 cos ^ [^t + B- 0] cos I- [J. - (-B- O)] 
 sin B sin C 
 
 Or, cosHa = C °^ ( ^ + ^-^ COS ^~^ +C) - 
 
 ' i sm 1? sin G 
 
 But A+ B - C =2(S - C), and A- B + C =2(S - B). 
 
 „„ , cos (S — B) cos (S — C) 
 
 Whence, cos" 1 ia= => : — ±— — *- <-• 
 
 ' 2 sm jB sin G 
 
 „ , /COS (# — -B) COS (# — G) /inrkS 
 
 Or, cos 4 a = \ ^ — : — ' . \, l - (100) 
 
 2 \ smjBsinC. v ' 
 
 In like manner, 
 
 cosib= J^H^ G)^(f- A \ (161) 
 2 V sinCsini 
 
 j 1 /cos (# — .4) cos (S — B) /,no\ 
 
 and cos4c=\ * — : — j—. — ^- <-• (102) 
 
 Dividing (97) by (100), we have 
 
 , , I cosScos(S-A) /,«-\ 
 
 tan ^ = V- C os(^-^cos (< S-(7) - (1 ° 3) 
 In like manner, 
 
 tan ib =J- cos S cos V-*) , (104) 
 
 r \ cos OS- C) cos (S- -4)' ^ ; 
 
 j , , / cos $ cos (S — (7) /,„., 
 
 and tan4c=\ — — -^ — - 1 • (105) 
 
 2 V cos (£ — A) cos (£ — B) v J 
 
 NAPIER'S ANALOGIES. 
 144. Dividing (94) by (95), we have 
 
 \A_ /sin (s — 6) sin (s — c) f 
 iB \ sin .s sin (\s — a^ \ i 
 
 tan \A_ /sin (s — 6) sin (s — c) / sin s sin (s — 6) 
 
 tan \ B » sin s sin (s — a) * sin (s — c) sin (s — a) 
 
 sin £ J. cos 1.B _ /sin 2 (s — 6) _ sin (s — 6) 
 ' cos 4 A sin 1 5 — « sin 2 (s — a) sin (s — a) 
 
Oblique Spherical Triangles. 139 
 
 Whence by composition and division, 
 
 sin \ A cos \ B + cos \ A sin \ B _ sin (s — b) + sin (s — a) 
 sin !^ cos ^.B — cos £ -4 sin £ iJ — sin(s — 6) — sin(s — a) 
 
 Then by (9), (11), and (21), 
 
 sinQ-^4 + 'i-_B)_ tan£[s — b + s — a] 
 sin (| J. — I-B) - tan£[s — 6 — (s — a)]' 
 
 But s — b + s — a = 2s — a — b = c. 
 
 Whence, "i'fff + ff - , ^ ° » ■ (106) 
 
 sin -J- (J. — 5) tan-i- (a — 6) v ; 
 
 145. Multiplying (94) by (95), we have 
 
 tan i A tan \ B= / sin (s-b) sin (a-c) / sin (s-c) sin (a- a) 
 * sin s sin (s — a) » sin s sin (s — 6) 
 
 n sin \ A sin £ J5 _ /sin 2 (s — c) _ sin (s — c) 
 
 ' cos ^ j4 cos % B ~ » sin 2 s sin s 
 
 Whence by composition and division, 
 
 cos ^ A cos -J- 1? — sin \ A sin |- B _ sin s — sin (s — c) 
 cos \ A cos \ B + sin £ A sin £ £ — sin s + sin (s — c) 
 
 Or, by (21), 
 
 cos (\ A + \ B) _ tan -J- [s — (s — c)] 
 cos (^ .4 — ^ 5) — tan £ [s + s — c] " 
 
 But s + s — c = 2s — c = a + b. 
 
 Whence cos K-* -j- -B) _ tan jo ( . 
 
 wnence, C0S i (i _ B) - ta i (( , + 6) ^ U/ J 
 
 146. Applying formula (106) to triangle A'B'C, in the 
 figure of § 141, we obtain 
 
 sai\(A' + B r ) tan^c' 
 
 sin \ (A' -B') tan $(a'-b') 
 
140 Spherical Trigonometry. 
 
 But, 
 $(A' + B>)= i(180° - a + 180° - b)= 180° -%( a + b); 
 
 %(A'-B')=%(180°-a-18G° + b) = -l(a-b); 
 i c< = |(180 o -C)= 90° -1(7; 
 and I (a' - 6') = £ (180° - J. - 180° + B) = -\(A-B). 
 Whence, 
 
 sin[180 o -|(a + &)] _ tan (90° - £ C) 
 sin[— £(a — 6)] _ tan [—±(A — B)1 
 
 Therefore, by §§ 28, 31, and 32, 
 
 sin £(« + &) cot^C 
 
 — sinj(« — 6) — tan ^ ( J. — B) 
 
 sin^(q + 5) _ cotjg , . 
 
 ' sin^(a-6) tsm$(A-B)' K ' 
 
 In like manner, from (107), we obtain 
 
 cos%(A' + B') _ tan^c' 
 cos \ (A - B') ~ tan £ (a' + 6')" 
 But, 
 
 l(o' + &0=K 18O ° - ^ + 180° - 5)= 180° -\(A + B). 
 
 Whence, 
 
 cos [180° -■!■(« + &)] _ tan (90° — \G) 
 cos[-£(a-6)] - tan[180°-^(^4 + B)]' 
 
 Therefore, by §§ 28, 31, and 32, 
 
 — cos£(a + &)_ cot^C 
 
 cos I (a — 6) — — tan ^(A + B)' 
 
 cos j (of 6) _ cotjC 
 ^ cos^(a-&)~tan^(^l + B) ^ - 1 
 
Oblique Spherical Triangles. 14* 
 
 147. The formulae exemplified in §§ 144, 145, and 146 
 are known as Xcqrie) J s Analogies. In each case there may 
 be other forms according as other elements are used. 
 
 SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 
 
 148. In the solution of oblique spherical triangles, we 
 may distinguish six cases : 
 
 1. Given a side and the adjacent angles. 
 
 2. Given two sides and their included angle. 
 
 3. Given the three sides. 
 
 4. Given the three angles. 
 
 5. Given two sides atid the angle opposite to one of them. 
 
 6. Given two angles and the side opposite to one of them. 
 
 By application of the principles of § 119, 5, the solution 
 of an example unaer Case 2, 4, or 6, may be made to depend 
 upon the solution of an example under Case 1, 3, or 5, 
 respectively ; and rice versa. 
 
 Hence, it is not essential to consider more than three cases 
 in the solution of oblique spherical triangles. 
 
 The student must carefully bear in mind the remarks 
 ,made in §§ 134 and 135. 
 
 149. Case I. Given a side and the adjacent angles. 
 
 1. Given A = 70°, £ = 131° 20', c = 116°; find a, b, and C. 
 
 By Napier's Analogies (§§ 144, 145), we have 
 
 sin i (B + A) _ tan^c , cos j (B + A) _ tan^c 
 
 iini(.B- A) ~ tan ^(6 -a)' Sa cos J (B - A) ~ tan £ (6 + a) 
 
 Whence, tan J (6 - a) = sin | (B - A) esc \ (B + A) tan \c, 
 
 - + - + 
 
 and tan £(& + o) = cos J (B — A) sec J (B + A) tan \ c. 
 
142 Spherical Trigonometry. 
 
 From the data, \ (B - A) = 30° 40', %(B + A) = 100° 40', J c = 58°. 
 
 \ogsni %(B - A)= Q.mQ - \0 log cos \ (B - A)= 9.9346 - 10 
 
 log esc i (B + A) = 0. 0076 log sec \ (B + A) = 0. 7326 
 
 log tan I c = 0.2042 log tan \ c = 0.2042 
 
 log tan i (6 - a) = 9.9194 - 10 log tan \(fi + a) =0. 8714 
 
 \ (b - a) = 39° 42.8'. 180° ~l(b + a)= 82° 20.5'. 
 
 K& + a)=97°39.5'. 
 Then, a = J(& + a)- K& - «)= 57° 56.7', 
 
 and & = -K& + a)+K&-a)=137°22.3'. 
 
 To find C, we have hy §146, 
 
 . , _ sin 1(6 + oV , ._ .. 
 
 cot 1 C = . * ,. ( tan 4 (B - A) 
 
 2 sin J (6 -a) * v ' 
 
 = sin \ (6 + a) esc \ (& — a) tan J(B — .4). 
 
 log sin \ (6 + a) = 9.9961 - 10 
 log esc |(6- a) = 0.1946 
 logtani(.B--4)= 9.7730 - 10 
 log cot£C = 9.9637 -10 
 
 JC = 47°23.6', and C = 94° 47.2'. 
 Note 1. The value of C may also be determined by the formula 
 
 cot^C = COS f^ + " ) tanK-g + ^) (§146). 
 2 cos J (6 — o) z ^ ' vs ' 
 
 Note 2. The triangle is always possible for any values of the 
 given elements. 
 
 EXAMPLES. 
 Solve the following spherical triangles : 
 
 2. Given A = 87°, B = 61°, c = 112°. 
 
 3. Given B = 41°, C=122°, a = 37°. 
 
 4. Given .4 = 135°, 0=51°, & = 69°- 
 
 5. Given A = 147° 30', £ = 163° 10', c = 76°20'. 
 (For additional examples under Case I., see § 155.) 
 
Oblique Spherical Triangles. 143 
 
 150. Case II. Given two sides and their included angle: 
 
 1. Given 6=137° 20', c=116°, .4=70°; find B, C, and a. 
 
 By Napier's Analogies (§ 146), we have 
 
 sinj(6 + c)_ cot^A , cos £(& + <;)_ cot ^4 
 
 sin I (6 - c) _ tan \{B - C)' an cos£(& -c) _ tan £(£ + O)' 
 
 Whence, tan J (B — O) = sin J (6 — c) esc J (6 + c) cot £ J, 
 
 - + - + 
 
 and tan \(_B + C) = cos J (6 — c) sec £ (& + c) cot £ .4. 
 
 From the data, £ (6 - c) = 10° 40', J (6 + c) = 126° 40', J 4 = 35°. 
 
 log sin J(& - c) = 9.2674 - 10 log cos £(& - c) = 9.9924 - 10 
 
 log esc J (6 + c) = 0.0958 log sec j(& + c) = 0.2239 
 
 log cot J .4 = 0.1548 log cot 1 4 = 0.1548 
 
 log tan $(.B- C)=9.5180-10 logtan£(-B+ C)= 0.3711 
 
 \{B- C)=18°14.5'. 180°- \{B+ = 66° 57.1' 
 
 K-B+ C)=113°2.9'. 
 Then, B = K-B+G)+K-B-C , ) = 1 31°17.4', 
 
 and C = K-B + C) - £(-B - C) = 94° 48.4'. 
 
 To find a, we have by § 144, 
 
 tan j a = ** K* + g) tan J(B _ c) . 
 
 log sin J (B + C) = 9. 9639 - 10 
 logcsc£(-B- C) = 0.5044 
 log tan J(6 - c) = 9.2750 - 10 
 logtan£a = 9.7433-10 
 
 Ja = 28°58.3', and a = 57° 56.6'. 
 Note. The triangle is possible for any values of the given elements 
 
 EXAMPLES. 
 
 Solve the following spherical triangles : 
 
 2. Given a = 64°, b = 34°, C = 48°. 
 
 3. Given 6 = 42°, c = 96°, ^ = 110° 
 
t44 Spherical Trigonometry. 
 
 4. Given a = 146°, c = 69°, B = 125°. 
 
 5. Given a = 90° 50', 6 = 117° 50', C = 120°. 
 (For additional examples under Case II., see § 165.) 
 
 151. Case III. Given the three sides. 
 
 The angles may be calculated by the formulae of § 142. 
 
 If all the angles are to be computed, the tangent formulas 
 are the most convenient, since only four different angles 
 occur in the second members. 
 
 If but one angle is required, the cosine formula involves 
 the least work. 
 
 The triangle is possible for any values of the data, pro- 
 vided that no side is greater than the sum of the other two, 
 and that the sum of the sides is less than 360° (§ 119, 1 
 and 2). 
 
 If all the angles are required, and the tangent formulae 
 are used, it is convenient to modify them as follows. 
 
 By (9i), 
 
 tan }. A —Jj ^~( s ~ a "> sin (* ~ & ) sin ( s ~^> 
 ^ sin s sin 2 (s — a^ 
 
 sin s sin 2 (s — a) 
 
 1 IS 
 
 sin (s — a)* 
 
 sin (s — ft) sin (s — b) sin (s — c ) 
 sins 
 
 Denoting J sin ( s - ft) sin ( S - fc) sin ( S - c) fc we h 
 
 V SID * 
 
 tan \A = 
 
 sms 
 
 sin (s — a) 
 
 Similarly, tan I B = -^—. — , and tan A 0= -^— ; -. 
 
 * sin (s — 6)' 2 sin (s — c) 
 
 1. Given « = 57°, b = 137°, c = 116° ; find A, B, and O. 
 
 Here, 2s=a + 6-|-c = 310°. 
 
 Whence, g = 165°, s - a = 98°, s - 6 = 18°, s -c = 39°. 
 
Oblique Spherical Triangles. 145 
 
 log sin(s - a) = 9.9958 - 10 log k = 9.8294 - 10 
 
 log sin(s - 6) = 9.4900 - 10 log sin(s - 6) = 9.4900 - 10 
 
 log sin(s - c) = 9. 7989 - 10 log tan \ B = 0.3394 
 
 log esc s = 0.3741 J B = 65° 24.2'. 
 
 2 )19.6588 - 20 B = 130° 48.4'. 
 
 logfc = 9.8294- 10 log A = 9.8294 -10 
 
 og sin(s - a) = 9.9958 - 10 log sin(s - c) = 9.7989 - 10 
 
 log tan \ A = 9. 8336 - 10 log tan \ C = 0.0305 
 
 1^ = 34° 17.0'. \C = 47° 0.8'. 
 
 A = 68° 34.0'. = 94° 1.6'. 
 
 EXAMPLES. 
 
 Solve the following spherical triangles : 
 
 2. Given o = 69°, 6 = 74°, c = 63°. 
 
 3. Given a = 103°, 6 = 53°, c = 61°. 
 
 4. Given a = 91°, 6 = 118°, c = 132°. 
 
 5. Given a = 58°, 6 = 138°, c = 116°; find A. 
 (For additional examples under Case III., see § 155.) 
 
 152. Case IV. Given the three angles. 
 
 The sides may be calculated by the formulae of § 143. 
 
 If all the sides are to be computed, the tangent formulae 
 are the most convenient, since only four different angles 
 occur in the second members. 
 
 If but one angle is required, the sine formula involves the 
 least work. 
 
 The triangle is possible for any values of the data, pro- 
 vided that the sum of the angles is between 180° and 540° 
 (§ 119, 3), and that each of the quantities B+G—A, 
 C + A-B, and A + B-C is between 180° and -180° 
 (§ 122). 
 
 For such values of the angles, S is between 90° and 270°, 
 and each of the quantities S — A, S —B, and S — G between 
 90° and - 90°. 
 
146 Spherical Trigonometry. 
 
 Then, cos S is — , while the cosines of S — A, 8 — B, and 
 S - C are + (§ 21). 
 
 Hence, the expressions under the radical signs in the 
 formulae are essentially positive, and no attention need be 
 paid to the algebraic signs. 
 
 If all the sides are required, and the tangent formulae are 
 used, it is convenient to modify them as follows : 
 
 By (103), 
 
 taaia= JIZZ oosS^(S-A) ' 
 
 * V cos (S - A) cos (S-B) cos (S-C) 
 
 = aos (S-A)yj- 
 
 cos 8 
 
 cos (JS -A)cos (8 - B) cos (S - C) 
 
 Den ° ting ^- M(8-A)<J(j3-B)caaiS-0) hj * 
 
 tan % a = if cos (S — A). 
 In like manner, 
 tan \ b = if cos (S — B), and tan \c = Kcos(S — G). 
 
 1. Given ^4 = 150°, 5 = 131°, 0=115°; find a, b, and e 
 
 Here, 28=A+B+C= 396°. 
 
 Whence, £=198°, £-^ = 48°, ,S-J? = 67 , #-0=83°. 
 
 log 00s S- 9. 9782 - 10 log K = 0. 7375 
 
 log sec (S-A)= 0.1745 xog cos ( 8 - B) = 9. 5919 - 10 
 
 log sec (S-B) = 0.4081 log tan J 6 = 0.3294 
 
 log sec QS -C)= 0.9141 £6 = 64° 54. 2'. 
 
 2 )1.4749 & = 129° 48.4'. 
 
 log #=0.7375 log #=0.7375 
 
 log cos Off -.4)= 9.8255 -10 log cos (S - Q = 9.0859 - 10 
 
 log tan \ a = 0. 5630 log tan \ c = 9. 8234 - 10 
 
 $ a = 74° 42.2'. . J c = 33° 39.6'. 
 
 as = 149° 24.4'. c = 67° 19.2'. 
 
Oblique Spherical Triangles. 147 
 
 3y § 34, cos 198° = - sin 108° = - cos 18° ; wl 
 ilgebraic sign, log cos 198° = log cos 18°. 
 
 2. Given A = 123°, £ = 45°, C=58°; find a. 
 
 Note 1. By § 34, cos 198° = - sin 108° = - cos 18° ; whence, with- 
 out regard to algebraic sign, log cos 198° = log cos 18°. 
 
 By (97) sin } a =-v|- ™sSco S (S-A). 
 
 ' sin B sin O 
 
 Here, 
 
 28=A+B+ C=226°; whence, S= 113°, and £-^ = -10°, 
 
 log cos S= 9.5919-10 
 log cos (S-A) = 9.9934-10 
 log esc B= 0.1505 
 log esc C= 0.0716 
 
 2 )19.8074-20 
 log sin J a = 9.9037 - 10 
 
 $ a = 53° 14.4', and a = 106° 28.8'. 
 Note 2. By § 28, cos (- 10°) = cos 10°. 
 
 EXAMPLES. 
 
 Solve the following spherical triangles : 
 
 3. Given A = 52°, B = 59°, C = 83°. 
 
 4. Given A = 143°, B = 28°, C = 32°. 
 
 5. Given A = 142°, B = 159°, C = 133°. 
 
 6. Given A = 70°, B = 122°, C = 95° ; find h. 
 (For additional examples under Case IV., see § 155.) 
 
 153. Case V. Given two sides and the angle opposite to 
 one of them. 
 
 1. Given a = 58°, o = 138°, B = 134°. 50' ; find A, C, and c, 
 
 t. ,-«s sin A sin a . . . , . _ 
 
 By 79), -: — ^ = — — =-! ° r sini = sinacsc&sm.B. 
 * v " sin B smb 
 
 log sin a = 9.9284 -10 
 
 log esc 6 = 0.1745 
 log sin B = 9.8507 - 10 
 log sin A = 9.9636 -10 
 
 A = 63° 58.6' or 116° 1.4' (§ 136). 
 
148 Spherical Trigonometry. 
 
 To find C and c, we have by §§ 144 and 146, 
 
 cot J C = sin J(6 + a) esc J(6 — a) tan $(i? — A), 
 and tan J c = sin J(jB + A) esc J(B — A) tan J(6 — a). 
 
 Using the first value of A, 
 
 i(B + A) = 99° 24.3', and J(B - A) = 35° 25.7'. 
 
 Also, J( & + a) = 98°, and }(& - a) = 40°. 
 
 log sin $(& + a) = 9.9958 - 10 log sin J(JB + A) = 9.9941 - 10 
 
 log esc i(b - a) = 0.1919 log esc %{B - A) = 0.2368 
 
 log tan £(.B- 4) =9.8521 -10 logtanJ(6 - a) = 9.9238 - 10 
 
 log cot $ O = 0.0398 log tan J c = 0. 1547 
 
 J C= 42° 22.7'. }c = 54°59.6'. 
 
 O = 84° 45.4'. c = 109° 69.2'. 
 
 Using the second value of A, 
 
 J(B + ^)=125°25.7', and i(B - A) = 9°24.3'. 
 
 log sin }(& + a) = 9.9958 - 10 logsin J(B + A) = 9.9111 - 10 
 
 logoscj(6- a) =0.1919 log esc \{B -A) = 0.7867 
 
 log tan J(.B - A) = 9.2192 - 10 log tan J(6 - o) = 9.9238 - 10 
 
 log cot J C = 9.4069 - 10 log tan J c = 0.6216 
 
 J O = 75° 40.9'. £ c = 76° 33.6'. 
 
 0=151° 21.8'. c = 153° 7.2'. 
 
 Thus the two solutions are : 
 
 1. A = 63° 58.6', G = 84° 45.4', c = 109° 59.2'. 
 
 2. ^ = 116° 1.4', 0=151° 21. 8', c = 153° 7.2'. 
 
 As in the corresponding case in the solution of plane 
 oblique triangles (compare §§ 108 and 109), there may 
 sometimes be two solutions, sometimes only one, and some- 
 times none, in an example under Case V. 
 
 After the two values of A have been obtained, the num. 
 ber of solutions may be determined by inspection ; for, by 
 § 119, 6, if a is < b, A must be < B ; and if a is > 6, A must 
 be>5. 
 
Oblique Spherical Triangles. 149 
 
 Hence, only those values of A can be retained which are 
 greater or less than B according as a is greater or less than b. 
 
 Thus, in Ex. 1, a is given < b ; and since both values of 
 A are < B, we have two solutions. 
 
 Again, if the data are such as to make log sin A positive, 
 there will be no solution corresponding. 
 
 2. Given a = 58°, c = 116°, C=94°50'; find A. 
 
 In this case, sln = 5S*, or sin A = sin o esc c sin C. 
 
 sin C sin c 
 
 log sin a =9.9284-10 
 
 log esc c =0.0463 
 
 log sin C = 9.9985 -10 
 
 log sin ^4 = 9.9732 -10 
 
 A = 70° 5.0' or 109° 55.0'. 
 
 Since a is given < c, only values of A which are < C can be re- 
 tained ; then the only solution is A = 70° 5.0'. 
 
 3. Given b = 126°, c = 70°, J3 = 57°; find G 
 
 R7T1 i^ Q1T1 P 
 
 In this case, — = , or sin C = sin c esc 6 sin B. 
 
 sin B sin b 
 
 log sine =9.9730- 10 
 
 log esc 6 =0.0920 
 
 log sin .8 = 9.9236 -10 
 
 log sin O = 9.9886 - 10 
 
 C= 76° 56.7' or 103° 3.3'. 
 
 Since both values of C are > B, while c is given < 6, there is no 
 solution. 
 
 EXAMPLES. 
 
 Solve the following spherical triangles : 
 
 4. Given a = 29°, 6 = 14°, ^4 = 49°. 
 
 5. Given a = 98°, c = 36°, = 163°. 
 
150 Spherical Trigonometry. 
 
 6. Given 6 =132°, e = 56°, B = 116° 18'. 
 
 7. Given a = 104° 50', c = 153° 20', .4 = 70°. 
 
 8. Given a = 111° 20', 6 = 41° 40', S = 25°. 
 (For additional examples under Case V., see § 155.) 
 
 154. Case VI. Given two angles and the side opposite to 
 one of them. 
 
 1. Given 4 = 110°, £=131° 20', b = 137° 20'; find a, c, 
 and C. 
 
 In this case, 2£5 = EIL^ or sin a = sin ^1 esc B sin 6. 
 
 sin 6 sin B 
 
 log sin ^ = 9.9730 -10 
 log esc 5 = 0.1244 
 log sin 6 =9.8311 -10 
 log sin a =9.9285- 10 
 
 a = 58° 1.2' or 121° 58.8'. 
 To find c and C, we have by §§ 144 and 146, 
 
 tan J c = sin J (B + A) esc \ (B — A) tan \ (6 — a), 
 and cot £ C = sin J. (6 + a) esc J (6 — a) tan J (B — .4) . 
 
 Using the first value of a, 
 
 i(b + a)= 97° 40.6', and J (6 - o) = 39° 39.4'. 
 Also, $(B + A)= 120° 40', and J (B - A) = 10° 40'. 
 
 log sin \(B + A) = 9.9346 - 10 log sin \ (6 + a) = 9.9961 - 10 
 
 log esc i (B - A) = 0. 7326 log cso J (6 - a) = 0. 1951 
 
 log tan J (6 - a) = 9.9185 - 10 log tan \ (B - A) = 9.2750 - 10 
 
 log tan J c = 0. 5857 log cot £ C = 9. 4662 - 10 
 
 £c = 75°26.9'. \ C = 73° 41.5'. 
 
 c = 150° 53.8'. C = 147° 23.0'. 
 
 Using the second value of a, 
 
 i(b + o) = 129° 39.4', and J(6 - a) = 7° 40.6'. 
 
Oblique Spherical Triangles. 151 
 
 log sin \ (B + A) = 9.9346 - 10 log sin J (& + a) = 9.8865 - 10 
 
 log esc 1 (B - A) = 0.7326 log esc \ (6 - a) = 0.8742 
 
 log tan \ (6 - a) = 9. 1297 - 10 log tan \ {B - A) = 9.2750 - 10 
 log tan J c = 9.7969 - 10 log cot J C = 0.0357 
 
 £c = 32°3.9'. $C = 42°38.8'. 
 
 c = 64° 7.8'. C = 85° 17.6'. 
 
 Thus the two solutions are : 
 
 1. a = 58° 1.2', c = 150° 53.8', C= 147° 23.0'. 
 
 2. a = 121° 58.8', o = 64° 7.8', 0=85° 17.6'. 
 
 In examples in Case VI., as in Case V., there may some- 
 times be two solutions, sometimes only one, and sometimes 
 none. 
 
 As in Case V., only those values of a can be retained which 
 are greater or less than b according as A is greater or less 
 than B. 
 
 Also, if log sin a is positive, the triangle is impossible. 
 
 EXAMPLES. 
 
 Solve the following spherical triangles : 
 
 2. Given A = 84°, B = 19°, a = 28°. 
 
 3. Given B = 159°, C = 36°, b = 9°. 
 
 4. Given A = 25° 20', C = 153° 27', a = 73° 10'. 
 
 5. Given A = 142° 40', C=71°10', c = 39°30'. 
 
 6. Given A = 110°, B = 123° 20', b = 127°. 
 (For additional examples under Case VI., see § 155.) 
 
 MISCELLANEOUS EXAMPLES. 
 155. Solve the following spherical triangles : 
 
 1. Given a = 38°, 6 = 51°, c = 42°. 
 
 2. Given 5 = 116°, C=80°, c = 83°. 
 
152 Spherical Trigonometry. 
 
 3. Given A = 78°, B = 41°, c = 108°. 
 
 4. Given b = 99° 40', c = 64°20', j3 = 96°10'. 
 
 5. Given A = 76°, B = 81°, = 61°. 
 
 6. Given A = 62°, C = 102°, a = 64° 30'. 
 
 7. Given a = 72°, 6 = 47°, 0=33". 
 
 8. Given .4 = 133° 50', 5=66° 30', a = 81° 10'. 
 
 9. Given a = 101°, 6 = 49°, c = 60°. 
 
 10. Given B = 135°, C = 50°, a = 70° 20'. 
 
 11. Given a = 162° 20', & = 15°40', 5 = 125". 
 
 12. Given .4 = 138° 20', B = 31° 10', C = 35° 50'. 
 
 13. Given a = 109° 20', c = 82°, .4 = 107° 40'. 
 
 14. Given A = 132°, B = 140°, 6 = 127°. 
 
 15. Given a = 60°, c = 98°, 5=110°. 
 
 16. Given a = 55°, c = 138°10', xl = 42°30'. 
 
 17. Given A = 61° 40', C = 140° 20', c = 150° 20'. 
 
 18. Given a = 61°, 6 = 39°, c = 92°. 
 
 19. Given a = 40°, b = 118° 20', .4 = 29° 20'. 
 
 20. Given A = 110°, 5 = 131°, G = 147°. 
 
 21. Given a = 115° 20', c = 146°20', C=141°10'. 
 
 22. Given B = 73°, C = 81° 20', b = 122° 40'. 
 
 23. Given A = 31° 40', C = 122° 20', 6 = 40° 40'. 
 
 24. Given 6 = 108° 30', c = 40°50', C =39° 50'. 
 
 25. Given & = 120°20', c = 70°40', A = 50°. 
 
 26. Given B = 22° 20', C=146°40', c = 138°20'. 
 
Applications. 1 53 
 
 XII. APPLICATIONS. 
 
 156. Shortest Distance between Two Points on the Sur- 
 face of the Earth. 
 
 In problems concerning navigation, the earth may be 
 regarded as a sphere. 
 
 The shortest distance between any two points on the sur- 
 face is the arc of a great circle which joins them ; the angles 
 between this arc and the meridians of the points determine 
 the hearings of the points from each other. 
 
 P 
 
 - ; <? 
 
 Thus, if Q and R are the points, and PQ and PR their 
 meridians, Z PQR determines the bearing of R from Q, and 
 Z PRQ the bearing of Q from R. 
 
 If the latitudes and longitudes of Q and R are known, 
 the arc QR and angles PQR and PRQ may be determined 
 by the solution of a spherical triangle. 
 
 For if EE' is the equator, and PG the meridian of 
 Greenwich, 
 
 Z QPR = Z RPG - Z QPG = longitude R - longitude Q. 
 
 Also, PQ = PE —QE = 90° — latitude Q, 
 
 and PR = PE' + RE' = 90° + latitude R. 
 
 Thus, in spherical triangle PQR, two sides and the in- 
 cluded angle are known, and the remaining elements may 
 be computed. 
 
154 Spherical Trigonometry. 
 
 When QR has been found in degrees, its length in miles 
 may be calculated by finding its ratio to 360°, and multiply- 
 ing the result by the length of the circumference of a great 
 circle ; in the following problems, the radius of the earth is 
 taken as 3956 miles. 
 
 EXAMPLES. 
 
 157. In each of the following examples, find the shortest 
 distance in miles between the places named, and the bearing 
 of each from the other : 
 
 1. Havana (lat. 23° 9' 1ST., Ion. 82° 23' W.), and Gibraltar 
 (lat. 36° 9' K, Ion. 5° 21' W.). 
 
 2. Batavia (lat. 6° 8' S., Ion. 106° 52' E.), and San Fran- 
 cisco (lat. 37° 48' 1ST., Ion. 122° 24' W.). 
 
 3. Vera Cruz (lat. 19° 12' K., Ion. 96° 9' W.), and Cape 
 of Good Hope (lat. 34° 22' S., Ion. 18° 29' E.). 
 
 4. Auckland (lat. 36° 51' S., Ion. 174° 50' E.), and Callao 
 (lat. 12° 4' S., Ion. 77° 13' W.). 
 
 5. Boston lies in lat. 42° 20' K, Ion. 71° 4' W., and Glas- 
 gow in lat. 55° 52' K, Ion. 4° 16' W. In what latitude does 
 a great circle course from Boston to Glasgow cross the meri- 
 dian of 40° W. ? 
 
 6. Yokohama lies in lat. 35° 27' K, Ion. 139° 41' E., and 
 Cape Horn in lat. 55° 59' S., Ion. 67° 16' W. In what lati- 
 tude does a great circle course from Yokohama to Cape 
 Horn cross the meridian of 160° W. ? 
 
 158. The Astronomical Triangle. 
 
 Let O be the position of an observer on the surface of 
 the earth ; P the celestial north-pole ; Z the zenith. 
 
 The great circle JEE' having P for its pole, is called the 
 celestial equator; and the great circle HH', having Z for its 
 pole, is called the horizon. 
 
Applications. 
 
 »55 
 
 Let S be the position of a star ; PSM a meridian through 
 S ; ZSN a quadrant of a great circle through Z and S. 
 
 The arc SM is called the declination of the star ; it is 
 called declination north or south, according as the star is 
 north or south of the celestial equator. 
 
 The angle SPZ is called the hour-angle of the star ; the 
 arc SN, its altitude; the angle PZS, its bearing or azimuth. 
 
 The arc EZ is the latitude of the observer. 
 
 The spherical triangle SPZ is called the Astronomical 
 Triangle. 
 
 Its sides have the following values : 
 
 SP=PM— SM = 90° — the declination of the star; 
 SZ = ZN — SN = 90° — the altitude of the star ; 
 PZ= EP — EZ = 90° — the latitude of the observer. 
 
 Its angle SPZ is the hour-angle of the star, and its angle 
 SZP the azimuth. 
 
 If any three of these five elements are known, the solu- 
 tion of a spherical triangle serves to determine the other 
 two. 
 
 159. Determination of Longitude and Time. 
 
 If the altitude and declination of the sun are known, and 
 the latitude of the observer, the three sides of triangle SPZ 
 are known, and the hour-angle SPZ may be computed. 
 
156 Spherical Trigonometry. 
 
 If 24 hours be multiplied by the ratio of this angle to 
 360°, we obtain the time required for the sun to move from 
 S to the meridian EP. 
 
 If this time be subtracted from 12 o'clock, if the observa- 
 tion is made in the morning, or added, if made in the after- 
 noon, we obtain the hour of the day at the time and place of 
 observation. 
 
 If the Greenwich time of the observation be noted on a 
 chronometer, the difference between this and the local time 
 as calculated above serves to determine the longitude of the 
 place of observation. 
 
 In reducing time to longitude, it should be remembered 
 that 24 hours of time correspond to 360° of longitude ; that 
 is, one hour of time corresponds to 15° of longitude, one 
 minute to 15', and one second to 15". 
 
 EXAMPLES. 
 
 160. 1. At a certain place in latitude 40° N., the altitude 
 of the sun was found to be 41°. If its declination at the 
 time of the observation was 20° N., and the observation was 
 made in the morning, how long did it take the sun to reach 
 the meridian ? 
 
 2. A mariner observes the altitude of the sun to be 60°, 
 its declination at the hour of observation being 6° N. If 
 the latitude of the vessel is 12° S., and the observation is 
 made in the morning, find the hour of the day. If the 
 observation is taken at 11.40 a.m., Greenwich time, find the 
 longitude of the vessel. 
 
 (In this case, the side PZ of the astronomical triangle is 90° plus 
 12°.) 
 
 3. At what hour will the sun set in Montreal (lat. 45° 
 30' N.), if its declination at sunset is 18° N. ? 
 
 (At sunset, the sun's altitude is 0°, so that the side SZ of the 
 astronomical triangle becomes 90°.) 
 
Applications. 157 
 
 4. A mariner observes the altitude of the sun to be 35° 
 23', its declination being 10° 48' S. If the latitude of the 
 vessel is 26° 13' N., and the observation is made in the 
 afternoon, find the hour of the day. If the observation is 
 taken at 7.13 p.m., Greenwich time, find the longitude of 
 the vessel. 
 
 5. At what hour will the sun rise in Panama (lat. 8° 57' 
 N.), if its declination at sunrise is 23° 2' S. ? 
 
 6. What will be the bearing of the sun at 4 p.m. in Bio 
 Janeiro (lat. 22° 54' S.), if its declination at that time is 
 3°S.? 
 
 7. What will be the bearing of the sun at sunrise in 
 Boston (lat. 42°21'N.), if its declination at that time is 
 13° 24' N. ? 
 
 8. What will be the altitude of the sun at 9 a.m. in 
 Mexico (lat. 19°25'N.), if its declination at that time is 
 8° 23' N. ? 
 
Plane Trigonometry. 
 
 FORMULA. 
 PLANE TRIGONOMETRY. 
 
 §28. sin (— A) = — sin A. cos (—^4)= cos A 
 
 tan(— A) = — tanA. cot (— A) — — cot A. 
 
 sec (—.4) = sec A csc(— A) = — esc A. . 
 §29. 
 
 sin (90° + A) = cos A. cos (90° + A) = - sin A. 
 
 tan(90° + A) = - cot A. cot (90° + A) = - tan A 
 
 sec (90° + A) =- csc A esc (90° + A) = sec A 
 §33. 
 
 1,1 1 
 
 sin as = tan x = — ; — sec x = ■ 
 
 esc x cot x cos x 
 
 1,1 1 
 
 cos a; = cot a; = esc a; = 
 
 sec a; tana; sin a; 
 
 (1] 
 
 (2) 
 
 (3) 
 
 §34. tan * = ^5JL (4) 
 
 cos a; v ' 
 
 §35. cota; = ^. (5) 
 
 sin a; ' 
 
 § 36. sin 2 x + cos 2 x = 1. (6) 
 
 § 37. sec 2 a; = 1+ tan 2 a;. (7) 
 
 esc 2 x = l + cot 2 a. (8) 
 
 § 38. sin (x + y) = sin a; cos y + cos a; sin y. (9) 
 
 cos (a; + y) = cos a; cos y — sin a: sin y. (io) 
 
 §40. sin (a; — y) = sin a; cos y — cos a; sin y. (n) 
 
 cos (x — y) = cos a; cos y + sin a; sin y. (12) 
 
 . ... , , , . tan x + tan y . . 
 
 §41. ^(- + y)- 1 _ t J x J y - (13) 
 
 . , . tana; — tan y , . 
 
 tan ^- y ) = l + tan a tanV < 14 > 
 
Formulae. 159 
 
 , . . cot .r cot v — 1 f-, -x 
 
 COt (.1! + ll) = * . (15) 
 
 v coty + cotje 
 
 ,, cot .r cot y + 1 /n _ x 
 
 cot (x — y) = — — • (16) 
 
 v "' cot;/ — cot-e 
 
 § 42. sin x + sin y = 2 sin i (x + y) cos ^ (« — y). (17) 
 
 sinx — siny = 2 cos ^- (.»+(/> sin J- (.c — y). (18) 
 
 cos.i- + cosy= 2cos-i-(.r + j/)eosl(.e — y\ (19) 
 
 cos x — cos y = — 2 sin 1 (,r + y) sin i i^b — y). (20) 
 
 „ sin x + sin y _ tan ^ (x + y) ^ 
 sin .r — sin y tan ^ (a; — y) 
 §44. 
 
 sin 2 . i- = 2 sin x cos x. (22) cos 2 x = cos' .r — sin* .r. (23) 
 
 cos2x = l — 2sin a .e. (24) cos 2 x = 2 cos- x — 1. (25) 
 
 tan2j , = 2tan.v . (26) cot2a; = co^-l > (2?) 
 
 1 — tan 3 * v J 2cota; v ' 
 §45. 
 
 2sinU.e=l — cos a-. (28) 2cos 3 £* = l + cos«. (29) 
 
 tanl* = i^58£. (30) cot i , = l±™i£. (31) 
 sin a; " sinj; 
 §97. 
 
 4 K=<~ sin 2 J. (32) 4 2^=0* sin 2 5. (33) 
 
 21T= a- cot J. (34) 2iT=& 2 cot.B. (35) 
 
 2K=a-tanB. (36) 2iT=& a tanA (37) 
 2 -fiT= oV(c+a)(c-A (38) 2 K= b V(c + 6) (c - 6). (39) 
 22T=a&. (40) 
 
 § 99. a : 6 = sinJ : sin.B. (41) 
 
 6 : c = sin JB : sin C. (42) 
 
 c: a = sin C : s'mA. (43) 
 
i6o 
 
 Plane Trigonometry. 
 
 § 100. 
 
 a + b tan.i(A + B) 
 a — b~ tan $(A — B) 
 
 
 b + c tan 1(5+0) 
 b — c _ tan^(J3— C) 
 
 
 c + a ta,ni(C + A) 
 c—a tan£(0— A) 
 
 §101. 
 
 a 2 = 6 2 + c 2 -26ccosA 
 
 
 b 2 = <? + a 2 — 2 ca cos B. 
 
 
 c 2 = a 2 + b 2 - 2 a& cos 0. 
 
 (44) 
 (45) 
 
 (46) 
 
 (47) 
 (48) 
 (49) 
 102. cos A = V+#-a\ (5Q) 
 
 — ' DC 
 
 cos 5 = — -t- (51) 
 
 2ca 
 
 COS G = t— • (52) 
 
 2ab 
 
 §103. sin^=V (S ~ 5) 6 |r C) 
 
 ca 
 
 . ig=> /(«- a )(«-!a. 
 
 6c 
 
 (53) 
 
 eini5 = V (S - C) i S " a) - ^ 
 
 8131 * C = \ a& (55) 
 
 , . Is(s — a) 
 
 coslB^J^i (57) 
 
 C0S * C7 =V^^- ( 58 ) 
 
Formulae. 161 
 
 -^"V 5 ^ 5 » 
 
 i -r. /(s — c)(s — a) ,„. 
 
 ^^ V ,(.-6) ' W 
 
 * * s(s — c) x ' 
 
 §104. 
 
 2iT=& C sinA (62) i^ rfri ^f° . (65) 
 
 2K^casmB. (63) 2*=^^1 (66) 
 
 2K=absmC. (64) 2^= ^-~ •• (67) 
 
 sin_B 
 
 in .4 sii 
 sinC 
 
 K=-\/s (s -a)(s- b) (s - c). (68) 
 
 SPHERICAL TRIGONOMETRY. 
 
 § 123. cos c = cos a cos 6. (69) 
 
 sin ^ = !iM. (70 ) sin B = Si°*. (72) 
 
 sine smc 
 
 . tan 6 /_ s „ 7? tana ,--■. 
 
 cos A = • (71) cos is = • (73) 
 
 tanc tanc 
 
 §124. 
 
 tan ^ = tana. (?4) tan 2? = *^. (75) 
 
 sin 6 sum 
 
 §125. 
 
 8in^ = 52S|. (76) sin5 = c -2L4. (77) 
 
 cos 6 cos a 
 
 § 126. cos c = cot A cot .B. (78) 
 
 §139. s_in| = sina (?9) 
 
 sini3 smo 
 
 sin_B_ sin 6 
 sin C sin c 
 
 (80) 
 
1 62 Spherical Trigonometry. 
 
 sin C sin c 
 
 sin 
 
 sin c sin a 
 
 sin 
 
 cos ., 
 
 sin a sin ft 
 
 (81) 
 
 sin A sin a 
 
 § 140 cos a = cos & cos c + sin & sin o cos A. (82) 
 
 cos 6 = cos c cos a + sin c sin a cos B. (83) 
 
 cos c = cos a cos 6 + sin a sin 6 cos C. (84) 
 
 § 141. cos A = — cos 5 cos C + sin 5 sin C cos a. (85) 
 
 cos B = — cos CcosA + smC sin ^4 cos ft. (86) 
 
 cos C = — cos J. cos 5 + sin J. sin B cos c. (87) 
 
 § 142. sin 4. .4 = J sin( S -5)sin (^. (88) 
 
 » sm o sin c 
 
 P = J sin(s-c)sin(s-a) . (89) 
 
 M sin c sin a 
 
 1 C -= \/ si " (S ~ a) Si " (S ~ &) (90) 
 
 * V sinasinft v ' 
 
 , A sin s sm (s — a) ,„,,. 
 
 cos 4 ^4 = \ . , \ u (91) 
 
 2 « sinftsinc / 
 
 , -r, /sin s sin (s — 6) ,.„v 
 
 cos 4 B = -*/ ; ^ £• (92) 
 
 » sin c sin a 
 
 ig / sin S sin( S -cX (g3) 
 
 V sin a sin ft 
 
 H i = Jg L-^"(«-') . (94) 
 * sm s sin (s — a) 
 
 tml j = ^E -c)™ (»-") . (95) 
 
 2 * sin s sin (s- 6) v ' 
 
 tan4C= X fg^ - CT ) S | n ( S - & ) . (96) 
 
 » sin s sm (s — c) 
 
Formulae. 163 
 
 §143. s inia=J- c -^ Scos ( S - A ). (97) 
 
 * V sin 73 sin <7 K } 
 
 „;~ 1 j, / cos S cos (S — B) ,_„. 
 
 » sin f/sin ,4 v ' 
 
 Sin 
 
 c== / cos 5 cos (5-0). ^ 
 
 \ sin yl sin B v ' 
 
 co Icosi^-^cos^-Ol. 
 
 * \ sin 5 sin O K ' 
 
 b= lcOs(5-O)C0s(5-^. ( 
 
 \ sinOsin^l v ; 
 
 COS $</=•*, . ' . 
 
 » sin O sin A 
 
 M &ESO. (102) 
 
 * * sm^LsinS v 
 
 ten 4 a /_ cos 5 cos (5-^1) , ( 
 
 T >f 008(5-5)008(5-0) k ; 
 
 tan 1 6 =J- cos 5 cos (5 -5) . ( } 
 
 T M cos (5-0) cos (5 -.4) ^ ; 
 
 tanlc-tf - cos5cos(5-0) . , 
 
 s 144 sini(^ + .B) _ tan 4c . (1Q6) 
 
 sm%(A-B) tan 4 (a -6) v ' 
 
 U5 cos4(-^ + -B) = tan 4 c , } 
 
 3 cos%(A-B) tan 4 (a + 6) V J 
 
 8146 sin4(a+6) _ cotlg . (108) 
 
 8 sin 4 (a -6) tan 4(^-5) v ; 
 
 cos 4 (a + 6) _ cot^-0 
 cos £ (a — 6) tan 4, (A + £) 
 
 (109) 
 
ANSWERS. 
 
 
 
 
 §72, 
 
 page 63. 
 
 
 
 2. 
 
 1.5441. 
 
 6. 
 
 2.1673. 
 
 10. 2.4592. 
 
 14. 
 
 3.3434. 
 
 3. 
 
 1.4771. 
 
 7. 
 
 2.3522. 
 
 11. 2.8363. 
 
 15. 
 
 3.8963. 
 
 4. 
 
 1.9912. 
 
 8. 
 
 2.2431. 
 
 12. 2.702a 
 
 16. 
 
 3.7656. 
 
 5. 
 
 1.9242. 
 
 9. 
 
 2.6232. 
 §74; 
 
 13. 2.5741. 
 page 64. 
 
 17. 
 
 4.1494 
 
 2. 
 
 .1549. 
 
 5. 
 
 1.5229. 
 
 8. .5192. 
 
 11. 
 
 1.3734. 
 
 3. 
 
 .2431. 
 
 6. 
 
 .2273. 
 
 9. .6478. 
 
 12. 
 
 .8942. 
 
 4. 
 
 1.6532. 
 
 7. 
 
 2.0212. 
 §77; 
 
 10. 2.7202. 
 page 65. 
 
 13. 
 
 1.9842. 
 
 3. 
 
 2.4080. 
 
 8. 
 
 2.2415. 
 
 13. .2510. 
 
 19. 
 
 .9132. 
 
 4. 
 
 .6036. 
 
 9. 
 
 .0954. 
 
 14. .4095. 
 
 20. 
 
 .1643. 
 
 5. 
 
 1.0485. 
 
 10. 
 
 .1409. 
 
 16. .0409. 
 
 21. 
 
 .3726. 
 
 6. 
 
 8.1160. 
 
 11. 
 
 .0777. 
 
 17. .7264. 
 
 22. 
 
 .1118. 
 
 7. 
 
 .4704. 
 
 12. 
 
 .3618. 
 
 18. .1511. 
 
 23. 
 
 .8618. 
 
 § 81 ; page 67. 
 
 2. 0.3801. 5. 8.2831 - 10. 8. 8.3892 - 10. 11. 2.3043. 
 
 3. 1.2252. 6. 7.1303 - 10. 9. 6.6865 - 10. 12. 0.1469. 
 
 4. 9.9084 - 10. 7. 3.7693. 10. 9.0124 - 10. 13. 1.6505. 
 
Answers. 
 
 § 82 ; page 68. 
 
 4. 
 
 2.8878. 
 
 10. 
 
 7.6055 - 10. 
 
 18. 
 
 1.646. 
 
 24. 
 
 9.493. 
 
 5. 
 
 3.0237. 
 
 11. 
 
 6.8560 - 10. 
 
 19. 
 
 8886. 
 
 25. 
 
 .2079. 
 
 6. 
 
 8.5177 - 10. 
 
 12. 
 
 0.7144. 
 
 20. 
 
 545.9. 
 
 26. 
 
 44.48. 
 
 7. 
 
 9.7164-10. 
 
 13. 
 
 3.0155. 
 
 21. 
 
 .01461. 
 
 27. 
 
 .0001109 
 
 8. 
 
 1.3028. 
 
 14. 
 
 8.9379 - 10. 
 
 22. 
 
 .003318. 
 
 28. 
 
 63330. 
 
 9. 
 
 4.9659. 
 
 ■•5. 
 
 5.0610 - 10. 
 
 23. 
 
 102.2. 
 
 29. 
 
 .01301. 
 
 
 
 
 30. .00005029. 
 
 
 
 
 
 
 § 87; pages 
 
 71 to 73. 
 
 
 
 1. 
 
 2.151. 
 
 16. 
 
 - .002555. 
 
 31. 
 
 - .3702. 
 
 48. 
 
 .2985. 
 
 2. 
 
 19.38. 
 
 17. 
 
 3692. 
 
 34. 
 
 13.83. 
 
 49. 
 
 .04477. 
 
 3. 
 
 - 3135. 
 
 18. 
 
 .2777. 
 
 35. 
 
 2.487. 
 
 50. 
 
 .7945. 
 
 4. 
 
 .09778. 
 
 19. 
 
 - 15890. 
 
 36. 
 
 1.056. 
 
 51. 
 
 1.805. 
 
 5. 
 
 .009213. 
 
 20. 
 
 .03162. 
 
 37. 
 
 .00002143. 
 
 52. 
 
 179.5. 
 
 6. 
 
 - .1088. 
 
 21. 
 
 244.1. 
 
 38. 
 
 .007105. 
 
 53. 
 
 1.883. 
 
 7. 
 
 6.359. 
 
 22. 
 
 .002791. 
 
 39. 
 
 .6955. 
 
 54. 
 
 - 8894. 
 
 8. 
 
 .03017. 
 
 23. 
 
 .0000002373. 
 
 40. 
 
 .5428. 
 
 55. 
 
 1.344. 
 
 9. 
 
 - 3.119. 
 
 24. 
 
 2.236. 
 
 41. 
 
 - 36.03. 
 
 56. 
 
 - .01335. 
 
 10. 
 
 1327. 
 
 25. 
 
 1.149. 
 
 42. 
 
 - 11.11. 
 
 57. 
 
 37.82. 
 
 11. 
 
 847.8. 
 
 26. 
 
 - 1.220. 
 
 43. 
 
 .9432. 
 
 58. 
 
 .00001146 
 
 12. 
 
 - .005421. 
 
 27. 
 
 1.778. 
 
 44. 
 
 2.627. 59. 
 
 .00000000001782. 
 
 13. 
 
 1.205. 
 
 28. 
 
 .6683. 
 
 45. 
 
 2.534. 
 
 60. 
 
 4.698. 
 
 14. 
 
 .2357. 
 
 29. 
 
 .6458. 
 
 46. 
 
 - 1.795. 
 
 61. 
 
 - .03402. 
 
 15. 
 
 - 11.54. 
 
 30. 
 
 .1378. 
 
 47. 
 
 1.032. 
 
 
 
 § 88 ; page 74. 
 
 1. 
 
 9.8556 - 
 
 10. 
 
 8. 
 
 9.9535 - 10. 
 
 15. 
 
 80° 26.3'. 
 
 22. 
 
 .2482. 
 
 2. 
 
 9.9458 - 
 
 10. 
 
 9. 
 
 0.7654. 
 
 16. 
 
 31° 20.4'. 
 
 23. 
 
 .7033. 
 
 3. 
 
 0.6518. 
 
 
 10. 
 
 0.0420. 
 
 17. 
 
 8° 53.5'. 
 
 24. 
 
 .3886. 
 
 4. 
 
 9.9501 - 
 
 10. 
 
 11. 
 
 83° 5.2'. 
 
 18. 
 
 5° 17.6'. 
 
 25. 
 
 47° 36.3'. 
 
 5. 
 
 0.8550. 
 
 
 12. 
 
 33° 17.8'. 
 
 19. 
 
 40° 20.7'. 
 
 26. 
 
 21° 52.7'. 
 
 6. 
 
 9.7070 - 
 
 10. 
 
 13. 
 
 46° 40.9'. 
 
 20. 
 
 66° 43.3'. 
 
 27. 
 
 49° 57.0'. 
 
 7. 
 
 9.7547 - 
 
 10. 
 
 14. 
 
 73° 33.4'. 
 
 21. 
 
 .2960. 
 
 28. 
 
 28° 46.7'. 
 

 § 94 ; pages 77 to 79. 
 
 3 
 
 1. 
 
 a = 1.812, 6 = 6.761. 
 
 17. 
 
 .4 = 55° 44.8', c = 4116. 
 
 2. 
 
 6 = 12.38, c = 13.35. 
 
 18. 
 
 6 = .6441, 
 
 c = .6503. 
 
 3. 
 
 a = 16.78, c = 26.11. 
 
 19. 
 
 A = 76° 34.0', o = 2423. 
 
 4. 
 
 A = 34° 22.2', 6 = .5118. 
 
 20. 
 
 a = .2072, 
 
 6 = .4212. 
 
 5. 
 
 .4 = 32° 44.4', c = 49.92. 
 
 21. 
 
 a = 5091, 
 
 c = 5268. 
 
 6. 
 
 6 = 10.35, c = 13.14. 
 
 22. 
 
 6 = .8478, 
 
 c = 1.234. 
 
 7. 
 
 a = .005916, 6 = .01269. 
 
 23. 
 
 A = 39° 22.0', 6 = 121.2. 
 
 8. 
 
 .4 = 39° 49.1', a = 488.7. 
 
 24. 
 
 a = 8.243, 
 
 c = 9.275. 
 
 9. 
 
 a = 148.4, c = 948.6. 
 
 25. 
 
 6 = .000005736, c = .00002118 
 
 10. 
 
 A = 49° 55.0', c = 4.457. 
 
 26. 
 
 a = .0006772, 6 = .0003899. 
 
 11. 
 
 6 = 77.38, c = 91.08. 
 
 27. 
 
 4 = 43° 45.7', 6 = 66650. 
 
 12. 
 
 a = 3814, 6 = 3651. 
 
 28. 
 
 a = 30.51, 
 
 6 = 18.59. 
 
 13. 
 
 ^ = 24° 23.3', a = .02126. 
 
 29. 
 
 a = 24540, 
 
 c = 30010. 
 
 14. 
 
 a = 156.6, c = 856.4. 
 
 30. 
 
 A = 60° 14.1', c = . 007745. 
 
 15. 
 
 a = .003607, 6 = .008830. 
 
 31. 
 
 c = 25.40. 
 
 
 16. 
 
 a = 24840, c = 36090. 
 
 32. 
 
 a = .2923. 
 
 
 33. 
 
 c = 949.8. 34. A = 
 
 34°: 
 
 J6.7'. 
 
 35. c = 4.488. 
 
 36. 
 
 a = 1.491. 37. a = 
 
 .03446. 
 
 
 
 § 96 ; pages 79 to 82. 
 
 
 2. 
 
 416.1 ft. 6. 23.26. 
 
 10. 
 
 15.27. 
 
 14. 107° 3.6'. 
 
 3. 
 
 651.8. 7. 285.1 ft. 
 
 11. 
 
 70.91 ft. 
 
 15. 135.2 ft. 
 
 4. 
 
 34.07. 8. 11.695. 
 
 12. 
 
 121° 0.8'. 
 
 16. 5.036. 
 
 5. 
 
 6° 2.3'. 9. 52° 4.2'. 
 
 13. 
 
 39° 12.0'. 
 
 17. 53° 8.1'. 
 
 18. 
 
 Perimeter, 3.1908; diameter circumscribed circle, 1.2278. 
 
 19. 
 
 Radius inscribed circle, 28.58 
 
 ; circumscribed, 
 
 30.94. 
 
 20. 
 
 740.2. 21. Height of cliff, 144.4 ft.; of 
 
 lighthouse, 153.6 ft 
 
 22. 
 
 1131.3 ft. 25. 17° 1.6'. 
 
 28. 
 
 14.9 ft. 
 
 31. 3° 43.9'. 
 
 23. 
 
 62.9 ft. 26. 18.68. 
 
 29. 
 
 1575 mi. 
 
 32. 195.9 ft. 
 
 24. 
 
 12° 28.9'. 27. 229.02. 
 
 30. 
 
 109.0 mi. 
 
 33. 60.14 ft. 
 
 34. 
 
 Bearing, S. 42° 28.8' W.; distance. 
 
 , 17.77 mi. 
 
 
 
 §98; 
 
 page 
 
 > 84. 
 
 
 2. 
 
 3.564. 4. 13440. 
 
 6. 
 
 46.0. 
 
 8. .02036 
 
 3. 
 
 .1098. 5. 12.64. 
 
 7. 
 
 .0004838. 
 
 9. 795. 
 
 11. 2840. 
 
Answers. 
 
 § 105 ; pages 92, 93. 
 
 2. 6 = 282.9, c = 268.5. 
 
 3. a = 3.384, c = 9.828. 
 
 4. a = .02893, 6 = .01825. 
 
 5. a = 31.49, c = 49.88. 
 
 6. a = .5042, 6 = .3618. 
 
 7. 6 = 5499, c = 2959. 
 
 2. A: 
 
 §106; 
 118° 18.0', 6 = 44.73. 
 
 3. .4 = 29° 59.5', c = 1419. 
 
 4. = 88° 34.8, a = .4038. 
 
 page 94. 
 
 5. B = 76° 12.9', c = 6.362. 
 
 6. 0=96° 3.3', 6 = 5141. 
 
 7. 5= 146° 26.3', a = .01044 
 
 § 107 ; page 96. 
 
 3. ^ = 44°25.0', .B = 78°28.0', = 57°7.4'. 
 
 4. A = 71° 47.4', B = 58° 45.6', O = 49° 27.6'. 
 
 5. A = 29° 55.2', 5 = 22° 31 .4', = 127° 34.4'. 
 
 6. 71° 33.4'. 7. 31° 6.6'. 8. 134° 29.4'. 
 
 § 1U ; page 100. 
 
 5939. 
 = .1435. 
 
 = 278.4. 
 
 8728; 
 
 or, 0= 133° 38.3', 6 = .2351. 
 
 6. O = 90°, 
 
 7. ^ = 14° 5.4', 
 
 8. A = 25° 31.9', 
 
 9. C = 46° 21.7', 
 
 1. 5 = 48° 32.7', = 8.522. 
 
 2. 5 = 29° 21.4', a =102.2. 
 
 3. Impossible. 
 
 4. = 44° 56.2', a = 66.68; 
 or, C- 135° 3.8', a = 12.89. 
 
 5. Impossible. 
 
 §112; 
 
 1. A = 61° 25.7', c = 1018. 
 
 3. .4 = 44°37.8', 5 = 101° 60.0', C = 
 
 4. J. = 34" 25.0', 6 = .7135. 6. 
 
 5. .8 = 46° 2.1', a = .05676. 7. 
 
 8. A = 51° 52.6', B = 42° 59.0', C = 85°8.8'. 
 
 9. .8 = 34° 7.8', c = 1223; 11. C = 48° 37.3', a 
 or, B= 145° 52.2', c = 269.3, 
 10. 0=46° 37.8', a =7577. 
 
 14. ^4 = 54° 5.4', 6 = .01073. 
 
 15. A = 72° 43.8', £ = 47° 40.6', 0=59° 36.6'. 
 
 16. .4 = 90°, c = .1379. 20. A = 27° 34.2-, c 
 
 17. Impossible. 21. 0=134° 36.9', 6 
 
 18. 6 = .4392, c = .4723. 22. 5 = 63° 58.6', a 
 
 19. B =96° 22.2'. c = 38.25. or, B = 116° 1.4', a 
 
 pages 100, 101. 
 
 2. a = 15.52, 6 = 10 
 : 33° 33.4'. 
 0=14° 57.7', 6 = 
 a = .0004395, c = 
 
 29. 
 
 = 5074. 
 : .0002092 
 
 12. 6 = 55610, 
 
 13. Impossible. 
 
 c = 
 
 7.597. 
 79270. 
 
 = .01875. 
 = 273.3. 
 = 25660 j 
 = 7573. 
 

 
 
 Answers. 
 
 
 1 
 
 
 
 
 § 113 : pages 101, 102. 
 
 
 
 2. 
 
 582. 
 
 7. 
 
 24530. 
 
 12. 10.28. 
 
 17. 
 
 7255. 
 
 3. 
 
 53.8. 
 
 8. 
 
 .000003186. 
 
 13. 883.2. 
 
 18. 
 
 19.19. 
 
 4. 
 
 20.98. 
 
 9. 
 
 .1682. 
 
 14. 34840. 
 
 19. 
 
 47210. 
 
 5. 
 
 .0780. 
 
 10. 
 
 .0000002941. 
 
 15. .03519. 
 
 20. 
 
 .00003759. 
 
 6. 
 
 271.3. 
 
 11. 
 
 28.77. 
 
 16. .003042. 
 
 21. 
 
 .000675. 
 
 § 114; pages 102 to 105. 
 1.608.4 ft. 2. 420.0 sq. rd. 3.525.8 ft. 4. 34° 22.2' or 145° 37.8'. 
 
 5. Height of tower, 212.8 ft.; distances, 236.4 ft., 436.4 ft. 
 
 6. Distance, 21.20 mi.; bearing of first from second, N. 74° 1.7' E. 
 
 7. Opposite angles, 76° 9.2', 57° 42.2'; remaining Side, .6313. 
 
 8. 5 = 51° 30.5', c= 53.51, a = 46.80. 9. 1.658. 
 
 10. 7.087, 11.30. 11. 3995 sq. ft. 12. 61.51, 58.48. 
 
 13. 14.922 miles an hour. 14. Height, 69.71 ft.; distance, 86.08 ft 
 
 15. 82.70 ft. 16. 173.2 ft. 17. 19.92, 16.62. 
 
 18. One angle, 118° 6.6'; diagonal, 91.02. 19. 9.012 mi. 
 
 20. AD = 9.282, CD = 10.65. 21. 1538 ft. 
 
 22. AD = 74.98, A = 68° 58.3'. 
 
 23. One angle, 59° 44.8' ; sides, 66.99, 37.77. 24. 84.28 ft. 
 
 25. 272.4 ft. 
 
 26. Bluff, 438.7 ft.; lighthouse, 280.1 ft 
 
 § 136; pages 127 to 129. 
 
 A = 36° 29.4', B = 69° 42.1', 6 = 54° 18.9'. 
 
 a = 21° 18.0', 6 = 49= 54.7', c = 53°8.2'. 
 
 A = 20° 33.8', B = 70° 59.5', c = 23° 18.3'. 
 
 8. A = 68° 51.6', a = 31° 26.4', 6 = 13° 40.2'. 
 
 9. 5 = 58° 27.5', 6 = 35° 32.4', c = 42° 59.3'; 
 or, B = 121° 32.5', 6 = 144° 27. 6', c = 137° 0.7'. 
 
 10. ^1 = 83° 38.8', 6 = 127° 36.2', c = 94° 52.3'. 
 
 11. .4 = 97° 36.4', a = 113° 22.4', B =19° 29.4'. 
 
 12. a = 164° 20', 5 = 31° 16.9', 6 = 9° 19.1'. 
 
 13. a = 165° 18.8', B = 104° 13.4', c = 46° 50.4'. 
 
 14. A = 48° 10.9', a = 44° 29.2', c = 109° 52.5' ; 
 or, A= 131° 49.1', a = 135° 30.8', c = 70°7.5'. 
 
 5. 
 6. 
 7. 
 
6 Answers. 
 
 15. A = 49° 35.8', B = 97° 36.0', c = 96° 31.2'. 
 
 16. a = 172°28.1', 5 = 84° 45.4', c = 124° 39.3', 
 
 17. a = 26° 42.8', B = 99° 47.4 , 6 = 110° 59.7'. 
 
 18. o = 129° 56. 7', 6 = 161° 32. 5', c = 52° 28.2'. 
 
 19. ^ = 157° 46.8', B = 74° 12.0, 6 = 43° 56.9'. 
 
 20. A = 165° 0.6', a = 168° 29.2', 6 = 130° 28.2'. 
 
 21. A = 114° 49.3', a = 121° 22.9', 5 = 126° 14.4'. 
 
 22. A = 100° 38.0', 5 = 163° 8.0', c = 51° 44.4'. 
 
 23. a = 8° 30. 5, 5 = 66° 55.5', c = 20° 53.7'. 
 
 24. B = 30° 53.3', 6 = 30° 12.9', c = 78°35.0'; 
 or, B = 149° 6.7', 6 = 149° 47.1', c = 101° 25.0'. 
 
 25. 4 = 138° 15.5', 6 = 130° 2.3', c = 57° 55.4'. 
 
 26. .4 = 59° 17.1', a = 49° 26.0', £ = 51° 46.0'. 
 
 27. a = 78° 32.1', 6 = 132° 25.0', c = 97° 42.6'. 
 
 28. .4 = 137° 17.7', B =119° 29.5', 6 = 136° 31.7. 
 
 29. a = 144° 0.6', B = 110° 57.9', b = 123° 6.1'. 
 
 30. .4 = 68° 10.6', 6 = 34° 3.0', c = 61° 11.3'. 
 
 31. 4 = 71° 45. 5', a = 53° 44.7', B = 148° 2. 5'. 
 
 32. a = 16° 48.5', B= 124° 31.6', c = 151° 56.7'. 
 
 33. A = 25° 4.8'. a = 4° 53.9', 6 = 169° 27.2'. 
 
 34. A = 76° 16.7', J3 = 144°l.l', & = 146° 26.2'. 
 36. a = 6° 50. 4', S = 59° 54. 0', 6 = 11° 36.6'. 
 
 36. A = 152° 7.1', b = 40° 48. 8', c = 135° 39.6'. 
 
 37. ^ = 142° 5.8', B = 82° 43.4', c = 99° 26.4'. 
 
 38. a = 144° 24.4', b = 32° 28.8', c = 133° 19.3'. 
 
 39. A = 102° 48.8', a = 141° 46.9', 6 = 10° 19.1'. 
 
 40. ^ = 10°22.5', a = 6° 16.1', c = 142° 41.2'; 
 or, ^ = 169°37.5', a = 173° 43.9', c = 37° 18.8'. 
 
 § 137 ; page 130. 
 
 2. a = 152° 52.8', 6 = 104° 46.7', JS = 124° 1.1'. 
 
 3. A = 138° 42.7', B = 153° 25.0', C = 132° 14.3'. 
 
 4. a = 44° 8.1 , 6 = 101° 3.9', O = 78° 22.3'. 
 
 5. a = 82° 14.4', A = 63° 34.8', B = 164° 4.8'. 
 
 6. 6 = 20° 21.2', B = 10° 33.7', C = 148° 11.9' ; 
 or 6 = 159° 38.8', B= 169° 26.3', C = 31° 48.1'. 
 
Answers. j 
 
 1. a = 108° 29.5', B = 141° 31.9', C= 111° 45.7'. 
 
 8. A = 19° 56.3', 6 = 138° 36.4', B = 163° 15.7'. 
 
 9. a = 46° 49.3', A = 33° 43.3', 6 = 124° 37.8'. 
 
 10. A = 11° 12.1', 6 = 76° 13.8', C = 140° 14.8'. 
 
 11. a = 50° 56.0', A = 23° 47.2', C= 31° 17.6'; 
 or, a = 129° 4.0', A = 1 56° 12.8', C = 148° 42.4'. 
 
 § 138 ; page 131. 
 2. C= 159° 59. 8', a = 69° 44.6'. 4. C = 145° 45.0', A = 141° 17.G". 
 3 A = 120° 46.6', c = 30° 26.6'. 5. C = 148° 37.6', c = 80° 36.8'. 
 
 § 149 ; page 142. 
 
 2. a = 98° 19.9', 6 = 60° 3.9', C= 110° 38.6'. 
 
 3. c = 72° 14.1', 6 = 47° 27.3', 4 = 32° 24.2'. 
 
 4. a =120° 34.6', c = 71° 7.6', B = 50° 3.6'. 
 
 5. 6 = 153° 48.4', a =125° 0.8', C= 140° 24.6'. 
 
 §150; pages 143, 144. 
 
 2. 4 = 110° 46.6', B = 35° 34.2', c = 45° 36.0'. 
 
 3. C= 78° 55.5', 5 = 41° 19.7', a = 107° 47.6'. 
 
 4. .4 = 145° 11.8', C = 107° 39.8', 6 = 126° 37.4'. 
 
 5. B = 121° 43.2', 4 = 105° 52.8', c= 115° 48.6'. 
 
 § 151; page 145. 
 
 2. 4 = 74° 12.4', 5 = 82° 12.0', C = 66° 41.4'. 
 
 3. A = 78° 32.0', 5 = 87° 7.0', C = 93° 29.6'. 
 
 4. A = 120° 21.4', .B = 130° 21.8', C = 140° 7.0'. 
 
 5. 70° 8.8' 
 
 § 152 ; page 147. 
 
 3. a = 37° 7.2', 6 = 41° 1.6', c = 49° 28.2'. 
 
 4. a = 101° 34.4', 6 = 49° 50.4', c = 59° 37.2'. 
 
 5. a = 125° 16.2', & = 151° 37.4', c = 75° 55.4'. 
 
 6. 126° 43.4'. 
 
 § 153 ; pages 149, 150. 
 
 4. B = 22° 7.4', C= 112° 17.0', c = 36° 28.0'. 
 
 5. Impossible. 6 C=90°, A = 138° 31.6', a = 146° 41.9 
 
 7. C=154°7.7', £ = 60° 45. 8', 6 =63° 52.2'. 
 
 8. 4 = 36° 18.2', C = 160° 52.8', c = 148° 58.4'; 
 or 4 = 143° 41.8', C = 38° 23.0', c = 77° 39.2'. 
 
or, 
 
 or, 
 
 or, 
 
 Answers. 
 
 § 154 ; page 151. 
 
 2. & = 8°50.4', c = 27° 37.0', C=79°9.2'. 
 
 3. Impossible. 4. c = 90°, B = 8° 14.0', & = 18 c 41.2'. 
 
 5. a = 155° 56.8', 6 = 138° 33.6', _B = 100°0'. 
 
 6. a = 63° 55.7', c = 156° 5.8', 0= 154° 55.0'; 
 a = 116° 4. 3' c = 72° 43. 8', C = 87° 24. 0'. 
 
 § 155 ; pages 151, 152. 
 
 1. 
 
 A = 51° 58.8', 
 
 B = 
 
 83° 55.2', 
 
 C = 
 
 58° 53.8'. 
 
 2. 
 
 6 = 115° 3.3', 
 
 a = 
 
 85° 16.0', 
 
 A = 
 
 81° 24.0'. 
 
 3. 
 
 a = 95° 37. 8', 
 
 6 = 
 
 41° 52.2', 
 
 C = 
 
 110° 48.2'. 
 
 4. 
 
 C = 65°23.3', 
 
 A = 
 
 96° 8.0', 
 
 a = 
 
 99° 40.0'. 
 
 5. 
 
 a = 68° 25.2', 
 
 6 = 
 
 71° 10.8', 
 
 c = 
 
 56° 56.8'. 
 
 6. 
 
 c = 90°, 
 
 B = 
 
 63° 43.4', 
 
 b = 
 
 66° 26.8'. 
 
 7. 
 
 A = 121° 32.8', 
 
 B = 
 
 40° 56.8', 
 
 c = 
 
 37° 25.8'. 
 
 8. 
 
 Impossible. 
 
 
 
 
 
 9. 
 
 A = 142° 32.8', 
 
 B = 
 
 27° 52.6', 
 
 o = 
 
 32° 27.2'. 
 
 10. 
 
 6 = 120° 16.6', 
 
 c = 
 
 69° 19.6', 
 
 A = 
 
 50° 26.2'. 
 
 11. 
 
 Impossible. 
 
 
 
 
 
 12. 
 
 a = 100° 8.4', 
 
 6 = 
 
 50° 1.8', 
 
 c = 
 
 60° 6.0'. 
 
 13. 
 
 C = 90°, 
 
 B = 
 
 : 113° 36. 6', 
 
 6 = 
 
 114° 51.9'. 
 
 14. 
 
 o = 67° 24.0', 
 
 C = 
 
 : 164° 6.4', 
 
 c = 
 
 : 160° 6.4' ; 
 
 
 a = 112° 36.0', 
 
 c= 
 
 : 128° 20. 6', 
 
 c = 
 
 : 103° 2.4'. 
 
 15. 
 
 0=86° 59.7', 
 
 A = 
 
 60° 50.9', 
 
 b = 
 
 111°17.0\ 
 
 16. 
 
 = 146° 37.9', 
 
 B = 
 
 55° 1.2', 
 
 b = 
 
 96° 34.4'. 
 
 17. 
 
 o = 43° 3.1', 
 
 6 = 
 
 : 129° 8.4', 
 
 B = 
 
 89° 23.8'; 
 
 
 a = 136° 56.9', 
 
 6 = 
 
 = 20° 35.8', 
 
 B = 
 
 : 26° 58.6'. 
 
 18. 
 
 .4 = 35° 31.0', 
 
 B = 
 
 : 24° 42.6', 
 
 C = 
 
 = 138° 24.8'. 
 
 19. 
 
 .8 = 42° 7.9', 
 
 C = 
 
 : 160° 12.8', 
 
 c = 
 
 :153°37.8'3 
 
 
 B = 137° 52.1', 
 
 C = 
 
 = 49° 38.8', 
 
 c = 
 
 : 89° 51.2'. 
 
 20. 
 
 a = 69° 34.4', 
 
 6 = 
 
 : 136° 10.6', 
 
 c = 
 
 : 150° 1.6'. 
 
 21. 
 
 Impossible. 
 
 
 22 
 
 . Impossible. 
 
 23. 
 
 c = 64° 19.4', 
 
 a = 
 
 : 34° 3.0', 
 
 J5 = 
 
 = 37° 39.6'. 
 
 24. 
 
 B = 68° 18.0', 
 
 A = 
 
 : 132° 33. 8', 
 
 a = 
 
 : 131° 15.8' ; 
 
 
 B= 111° 42.0', 
 
 A = 
 
 : 77° 4.6', 
 
 a = 
 
 : 95° 50.0'. 
 
 25. 
 
 B = 134° 57.3', 
 
 , C = 
 
 :50°41.1', 
 
 a = 
 
 = 69° 8.8'. 
 
 26 
 
 b = 27° 22.1', 
 
 a = 
 
 : 117° 9.2', 
 
 A = 
 
 : 47° 21.2'. 
 
Answers. 9 
 
 § 157 ; page 154. 
 
 1. Bearing of Havana from Gibraltar, N. 77° 40.3' W. ; of Gibraltar 
 from Havana, N. 59" 5.1' E. ; distance, 4593 mi. 
 
 2. Bearing of Batavia from San Francisco, N. 67° 25.5' W. ; of San 
 Francisco from Batavia, N. 47° 12.3' E. ; distance, 8650 mi. 
 
 3. Bearing of Cape of Good Hope from Vera Cruz, S. 60° 45.6' E. ; 
 of Vera Cruz from Cape of Good Hope, N. 86° 42.4' W. ; distance, 
 8330 mi. 
 
 4. Bearing of Callao from Auckland, S. 69° 29.9' E. ; of Auckland 
 from Callao, S. 50° 2.5' W. ; distance, 6671 mi. 
 
 5. 54° 35.9' N. 6. 51° 48.0' S. 
 
 § 160; pages 156, 157. 
 
 1. 3 h. 30 m. 40 s. 
 
 2. Hour of the day, 10 h. 23 m. 33.6 s., a.m. ; longitude, 19° 6.6' W. 
 
 3. 7 h. 17 m. 14.4 s., p.m. 
 
 4. Hour of the day, 2 h. 45 m. 4 s., p.m. ; longitude, 66° 59' W. 
 
 5. 6 h. 15 m. 21.6 s., a.m. 6. N. 80° 24.6' W. 7. N. 71° 43.7 E. 
 8. 45° 5'. 
 
FOUR PLACE 
 
 LOGARITHMIC TABLES 
 
 TOGETHEK WITH A 
 
 TABLE OF NATTJEAL SIXES, COSINES, 
 TAXGEXTS, AND COTAXGEXTS 
 
 PREPARED BT 
 
 WEBSTER WELLS, S.B. 
 
 PBOFTSSOR OF MATHEMATICS IX THE MASSACHUSETTS 
 CtSTU'UlE OF TECHNOLOGY 
 
 D. C. HEATH & CO., PUBLISHERS 
 BOSTON" SEW YORK CHICAGO 
 
COPTMGHT, 1887 AND 1900, 
 
 Bv WEBSTER WELLS 
 
 1q6 
 
Use of the Tables. i 
 
 USE OP THE TABLES. 
 
 I. USE OF THE TABLE OP LOGARITHMS OF NUMBERS. 
 
 This table (pages 12 and 13) gives the mantissae of the 
 logarithms of all integers from 100 to 1000, calculated to 
 four places of decimals. 
 
 To find the Logarithm op a Number of Three 
 Figures. 
 
 Look in the column headed " No." for the first two signifi. 
 cant figures of the given number. 
 
 Then the required mantissa will be found in the corre- 
 sponding horizontal line, in the vertical column headed by 
 the third figure of the number. 
 
 Finally, prefix the characteristic in accordance with the 
 rules of §§ 66 or 67. 
 
 For example, log 168 =2.2253; 
 
 log .0344 = 8.5366 - 10 ; etc. 
 
 For a number consisting of one or two significant figures, 
 the column headed may be used. 
 
 Thus, let it be required to find log 83 and log 9. 
 
 By § 80, log 83 has the same mantissa as log 830, and log 
 9 the same mantissa as log 900. 
 
 Hence, log 83 = 1.9191, and log 9 = 0.9542. 
 
 To find the Logarithm of a Number of more than 
 Three Figures. 
 
 Ex. 1. Required the logarithm of 327.6. 
 
 From the table, log 327 = 2.5145, 
 
 and log 328 = 2.5159. 
 
2 Use of the Tables. 
 
 That is, an increase of one unit in the number produces 
 an increase of .0014 in the logarithm. 
 
 Therefore, an increase of .6 of a unit in the number will 
 produce an increase of .6 x .0014 in the logarithm, or .0008 
 to the nearest fourth decimal place. 
 
 Hence, log 327.6 = 2.5145 + .0008 = 2.5153. 
 
 Note I. The above method is based on the assumption that the 
 differences of logarithms are proportional to the differences of their 
 corresponding numbers ; which, though not strictly accurate, is suffi- 
 ciently exact for practical purposes. 
 
 Note II. The difference between any mantissa in the table and the 
 mantissa of the next higher number of three figures, is called the tabu- 
 lar difference. The subtraction may be performed mentally. 
 
 The following rule is derived from the above : 
 
 Find from the table the mantissa of the first three significant 
 figures, and the tabular difference. 
 
 Multiply the latter by the remaining figures of the number 
 with a decimal point before them. (See Note III.) 
 
 Add the result to the mantissa of the first three significant 
 figures, and prefix the proper characteristic. 
 
 Note III. In finding the correction to the nearest unit's figure, the 
 decimal portion should be omitted provided that, if it is .5 or more 
 than .5, the unit's figure is increased by 1. Thus, 13.26 would be taken 
 as 13 ; 30.5 as 31 ; 22.803 as 23. 
 
 Ex. 2. Find the logarithm of .021508. 
 
 Mantissa 215 = 3324 Tabular difference = 21 
 
 2 _M 
 
 3326 Correction = 1.68 = 2, nearly. 
 
 Eesult, 8.3326 - 10. 
 
 To FIND THE NUMBER CORRESPONDING TO A LOGARITHM, 
 
 Ex. 1. Required the number whose logarithm is 1.6571. 
 Find in the table the mantissa 6571. 
 
Use of the Tables. 3 
 
 In the corresponding line, in the column headed " No.," 
 we find 45, the first two figures of the required number, and 
 at the head of the column we find 4, the third figure. 
 
 Since the characteristic is 1, there must be two places to 
 the left of the decimal point (§ 66). 
 
 Hence, the number corresponding to 1j6571 is 45.4. 
 
 Ex. 2. Eequired the number whose logarithm is 2.3934. 
 
 We find in the table the mantissse 3927 and 3945, whose 
 corresponding numbers are 247 and 248, respectively. 
 
 That is, an increase of 18 in the mantissa produces an 
 increase of one unit in the number corresponding. 
 
 Therefore, an increase of 7 in the mantissa will produce an 
 increase of -fa of a unit in the number, or A, nearly. 
 
 Hence, the number corresponding is 247 + .4, or 247.4. 
 
 The following rule is derived from the above : 
 
 Find from the table the next less mantissa, the three figures 
 corresponding, and the tabular difference. 
 
 Subtract the next less from the given mantissa, and divide 
 the remainder by the tabular difference. (See Note V.) 
 
 Annex the quotient to the first three figures of the number, 
 and point off the result. (See Note IV.) 
 
 Note IV. The rules for pointing oft are the reverse of the rules 
 for characteristic given in §§ 66 and 67. 
 
 1. If —10 is not written after the mantissa, add 1 to the character- 
 istic, giving the number of places to the left of the decimal point. 
 
 2. If — 10 is written after the mantissa, subtract the positive part of 
 the characteristic from 9, giving the number of ciphers to be placed 
 between the decimal point and first significant figure. 
 
 Ex. 3. Find the number whose logarithm is 8.5265 — 10 
 
 5265 
 Next less mantissa = 5263 ; three figures corresponding, 336. 
 Tabular difference = 13)2.00(.15 = .2, nearly. 
 13 
 70 
 
4 Use of the Tables. 
 
 By the rule of Note IV., there will be one cipher between the decimal 
 point and first significant figure. 
 
 Hence, the number corresponding = .03362. 
 
 Note V. The correction can usually be depended upon to one 
 decimal place ; the division should be carried out to two decimal places 
 in order to determine the last figure accurately. (See Note III.) 
 
 II. USE OF THE TABLE OF LOGARITHMIC SINE 
 COSINES, ETC. 
 
 This table (pages 14 to 19) gives the logarithms of the 
 sines, cosines, tangents, and cotangents of all angles at inter- 
 vals of 10 minutes from 0° to 90°. 
 
 For angles between 0° and 45°, the degrees and minutes 
 will be found in the left-hand column, and the functions in 
 the columns designated by the names at the top; that is, 
 sines in the first column, cosines in the second, tangents in 
 the third, and cotangents in the fourth. 
 
 For angles between 45° and 90°, the degrees and minutes 
 will be found in the right-hand column, and the functions in 
 the columns designated by the names at the foot; that is, 
 cosines in the first column, sines in the second, cotangents 
 in the third, and tangents in the fourth. 
 
 If only the mantissa of the logarithm is found, the charac- 
 teristic may be determined from the nearest logarithm in the 
 same column in which the characteristic is given. 
 
 Since the sines and cosines of all acute angles, the tan- 
 gents of angles between 0° and 45°, and the cotangents of 
 angles between 45° and 90°, are less than unity, the charac- 
 teristics of their logarithms have been increased by 10, and 
 — 10 must be written after the mantissa; in all other cases. 
 the true value of the characteristic is given in the table. 
 
 Thus, log sin 38° 30' = 9.7941 - 10 ; 
 
 log tan 65° 20' = 0.3380; 
 log cot 79° 10' = 9.2819 - 10; 
 log cos 89° 40' = 7.7648 - 10 ; etc. 
 
Use of the Tables. 5 
 
 To fixd the Logarithmic Slxe, Co>r\~E. Taxgkst, or 
 
 OOTAXGEXT. OF 1XT AlTTE AxGLE EXPRESSED 
 
 ix Degrees asd Mixvtes. 
 
 JVni f:\--M the table the logarithmic sine, cosine, tangent, or 
 fc<MUK it of the nejet less multiple of ten minutes, and the 
 ■-iij?- vn«.v for V corresponding. i^See Note VI.) 
 
 -V'rffy-Jv th '■< diference by the number of minutes remaining. 
 
 If si'tz or taii'i*.?. add ) . . 
 
 '. " . Wins correction. 
 
 1/ cosine or cotauoe-nt, su'tra-:! \ 
 
 Note VI. The columns immediately to rhe right of those headed 
 '•?:-■..'* "C\»s.." and "Tan.," contain the respective differences fori'; 
 the richt-h.uid column o: di&rtuces is also tobe used with the column 
 hcsdt.i -Cot." 
 
 It will he ohserred that the differences do i.h stand in the same 
 h^rijwntal line with tie logarithms, but opposite the intervals between 
 eo:iSvNr.;:ive I.srarichais. For ancles Ht>«- 4-3-. the difference next be- 
 Iok sh; f.'.I be taken ; for .-.uj'.es abor* 4-3- , the difference next abort. 
 
 Note VTX Theru'e lissnoies that the ditto rtnees otthe logarithmic 
 fcno:: .■>:•.* are proportional to the differences of their corresponding 
 ancles, rti ■">. unless the ancle is near to OP or 90*, is in general suffi- 
 ciently exact for practical purposes. ^See page 9.) 
 
 Note VIXJL I: the ancle is expressed in decrees, minutes, and 
 k:x-..1s. the stN.vn.is shrald be reduced to the decimal part of a minute 
 beiore applying the rule. 
 
 Ex. 1. Find loc tan 17" U». 
 
 Ice taa 17- 10 = i\45;>> — 10 
 
 15 
 
 Result, 9.4916 - 10 
 
 kg cos 35" 30* = 9.71S1 - 10 
 
 Besult, 9.;i74 - 10 
 
 IX 1= 4.5 
 
 
 4 
 Corr. = 1S.0 
 
 
 dl r= sj 
 
 
 S.5 
 105 
 
 
 68 
 
 7.36 = ; 
 
 ■.nearly. 
 
6 Use of the Tables. 
 
 To find the Acute Angle corresponding to a gives 
 Logarithmic Sine, Cosine, Tangent, or Cotangent. 
 
 Take from the table, if sine or tangent the next less, if cosine 
 or cotangent the next greater, logarithmic function, the angle 
 corresponding, and the difference for V. (See Note IX.) 
 
 Find the difference between the given logarithm and that 
 taken from the table, and divide it by the difference for V, 
 giving the correction in minutes. 
 
 Add the result to the angle corresponding to the next less, or 
 next greater, function. 
 
 Note IX. In searching for the next less (or greater) logarithm, 
 attention must be paid to the fact that the functions are found in 
 different columns according as the angle is below or above 45°. 
 
 If, for example, the next less logarithmic sine is found in the column 
 with "Sin." at the top, the angle corresponding must be taken from 
 the left-hand column ; but if it is found in the column with '• Sin." at 
 the foot, the angle corresponding must be taken from the right-hand 
 column. Similar considerations hold with respect to the other three 
 functions. 
 
 Ex. 1. Find the angle whose log sin = 9.9594 — 10. 
 
 9.9594 - 10 
 Next less log sin = 9.9590 — 10 ; angle corresponding = 65° 30'. 
 
 D. V = .6)4.0(6.66 = 6.7, nearly. 
 Adding the correction, the result is 65° 36.7'. 
 
 Ex. 2. Find the angle whose log cot = 0.1696. 
 
 Next greater log cot = 0.1710 ; angle corresponding = 34° 0' 
 0.1696 
 
 D. 
 
 V 
 
 = 2.7)14.0(5.18 = 
 13 5 
 
 = 5.2 
 
 , nearly. 
 
 
 
 
 
 60 
 
 27 
 230 
 
 
 Result, 
 
 34° 
 
 5.3', 
 
Use of the Tables. 7 
 
 To find the Logarithmic Secant or Cosecant of 
 any Acute Angle. 
 
 1 1 
 
 Since sec x = and esc x = , we have by § 85, 
 
 cos a; sin a; 
 
 log see x = colog cos x, and log esc x = colog sin x. 
 
 Hence, to find the logarithmic secant, subtract the logarith- 
 mic cosine from 10 — 10; and to find the logarithmic cosecant, 
 subtract the logarithmic sine from 10 — 10. 
 
 Ex. Find log sec 22° 38'. 
 
 From the table, log cos 22° 38' = 9.9652 - 10 
 
 Subtracting from 10 — 10, we have 
 
 log sec 22° 38' = 0.0348. 
 
 Note X. The logarithmic cotangent of an angle may be obtained 
 by subtracting the logarithmic tangent from 10 — 10. 
 
 To find the Logarithmic Functions of an Angle not 
 
 LYING BETWEEN THE LlMITS 0° AND 90°. 
 
 By § 34, any function of any angle may be expressed as a 
 function of a certain acute angle ; and hence the table of the 
 functions of acute angles serves to determine the functions 
 of angles of any magnitude whatever, positive or negative. 
 
 Ex. Find log sin 152° 16'. 
 
 By § 34, sin 152° 16' = cos 62° 16'. 
 
 Then, log sin 152° 16' = log cos 62° 16' = 9.6678 - 10. 
 
 Another method would be to find the logarithmic sine of 27° 44', 
 the supplement of 152° 16' (§ 32). 
 
 Note XI. If the natural function is negative, as for example in 
 the case of the cosine of an angle between 90° and 180°, there is no 
 logarithmic function, strictly speaking. (See Note before § 87.) 
 
 In solving examples involving such functions, we proceed as if tha 
 functions were positive, and determine the algebraic sign of the result 
 irrespective of the logarithmic work. Illustrations of this will be found 
 
 Chapters X. and XI. 
 
8 Use of the Tables. 
 
 m. USE OF THE TABLE OP NATURAL SINES AND 
 COSINES. 
 
 This table (pages 20 and 21) gives the natural values of 
 the sines and cosines of all angles at intervals of 10 minutes 
 from 0° to 90°, calculated to four places of decimals. 
 
 Its use is similar to that of the table of logarithmic func- 
 tions, except that the tabular differences for V are not given, 
 but are to be calculated from the table when required. 
 
 Ex. 1. Find sin 48° 52'. 
 
 The difference between sin 48° 50' and sin 49° 0' is .0019, one-tenth 
 Of which is .00019. 
 
 sin 48° 50' = .7528 D. V = 1.9 
 
 4 2 
 
 .7532, Am. Corr. = 3.8 = 4, nearly. 
 
 Ex. 2. Find the angle whose cos = .5506. 
 
 The difference between the next greater and next less functions, 
 .5519 and .5495, is .0024 ; one-tenth of which is .00024. 
 
 Next greater cos = .5519 ; angle corresponding = 56° {JO*. 
 .5506 
 D. V =2.4)13.0(5.41 = 5.4, nearly. 
 12 
 100 
 96 
 
 40 Result, 66° 35.4'. 
 
 IV. USE OP THE TABLE OF NATURAL TANGENTS 
 AND COTANGENTS. 
 
 This table (pages 22 and 23) gives the tangents and 
 cotangents of all angles at intervals of 10 minutes from J 
 to 90° ; its use is similar to that of the table of natural sinea 
 and cosines. 
 
Use of the Tables. $ 
 
 V. MORE ACCURATE METHOD FOR FTSDnvG THE 
 LOGARITHMIC FTTXCTIOXS OF ANGLES JJEAR 
 TO 0° OR iXl°. 
 
 Ir was stared in Xote TIL. page 5. that in general the 
 differences of the logarithmic functions are approximately 
 proportional to the differences of their corresponding augles. 
 It •will be seen from the table that this is not the case with 
 the logarithmic sines, tangents, and cotangents of angles 
 near to 0°. nor with The logarithmic cosines, tangents, and 
 cotangents of angles near to iX>*. 
 
 Tims, the difference for 1' in the case of the logarithmic 
 sine or tangent of an angle between 40' and o0' is 96.9, 
 while for an angle between o0' and 1° ir is 79... 
 
 A very accurate method for finding The logarithmic sine or 
 tangent of an angle near to s , or the logarithmic cosine or 
 cotangent of an angle near to 90^. is to first calculate the 
 natural function by aid of the Table of natural sines and 
 cosines, or of natural tangents and cotangents, and then 
 find the logarithm of the result. 
 
 To find the angle corresponding in similar cases, find the 
 number corresponding to the logarithmic function, and then. 
 by aid of the tables of natural functions, calculate the angle 
 corresponding to the result- 
 Ex. 1. Find log sin 3 «hV. 
 
 From the table of natural sines and cosines, we obtain 
 natural sin (P -xV = .01tX!??. 
 
 VThenee, log an CP 56' = S.2119 - 10. 
 
 Tbts iw-r.lt is correct to the last place of decimals ; by the ordinary 
 method we siouid have obtained 5.2102 — 10. 
 
 Ex. 2. Find the angle whose log tan = $.0302 — 10. 
 
 The number c orresponding to this logarithm is .01072. 
 from the table of natural tangents a;-.d cotangents, the angle ■siiass 
 natural tangent is .01072 is .iV.So'. 
 
IO Use of the Tables. 
 
 This is correct to the last place of decimals ; the ordinary method 
 would have given 37.15'. 
 
 Note XII. To find with accuracy the log cotangent of an angle 
 near to 0°, find the log tangent of the angle by the above method, and 
 then subtract the result from 10 — 10. (See Note X., page 7.) 
 
 To find the angle corresponding to a log cotangent in a similar case, 
 find the log tangent of the angle (Note X.), and then find the angle 
 corresponding as above. 
 
 Note XIII. To find the log tangent of an angle near to 90°, find 
 the log tangent of its complement, and subtract the result from 10 — 10. 
 (See Note XII.) 
 
 To find the angle corresponding in a similar case, find the angle 
 corresponding to its cologarithm, and take the complement of the 
 result. 
 
 Note XIV. The more accurate method should be employed in 
 finding the log sines, tangents, or cotangents of angles between 0° and 
 5°, or the log cosines, tangents, or cotangents of angles between 85" 
 and 90°, and the angles corresponding in similar cases. For angles 
 between 5° and 85° the ordinary method is sufficiently exact. 
 
FOUR PLACE LOGARITHMIC TABLES 
 
12 
 
 LOGARITHMS OF NUMBERS. 
 
 No. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 0000 
 
 0043 
 
 0086 
 
 0128 
 
 0170 
 
 0212 
 
 0253 
 
 0294 
 
 0334 
 
 0374 
 
 ii 
 
 0414 
 
 0453 
 
 0492 
 
 0531 
 
 0569 
 
 0607 
 
 0645 
 
 0682 
 
 0719 
 
 0755 
 
 12 
 
 0792 
 
 0828 
 
 0864 
 
 0899 
 
 0934 
 
 0969 
 
 1004 
 
 1038 
 
 1072 
 
 1 106 
 
 13 
 
 1139 
 
 "73 
 
 1206 
 
 1239 
 
 1271 
 
 1303 
 
 1335 
 
 1367 
 
 1399 
 
 1430 
 
 14 
 
 1461 
 
 1492 
 
 1523 
 
 '553 
 
 1584 
 
 1614 
 
 1644 
 
 1673 
 
 1703 
 
 1732 
 
 IS 
 
 1 761 
 
 1790 
 
 1818 
 
 1847 
 
 1875 
 
 1903 
 
 1931 
 
 1959 
 
 1987 
 
 2014 
 
 16 
 
 2041 
 
 2068 
 
 2095 
 
 2122 
 
 2148 
 
 2175 
 
 2201 
 
 2227 
 
 2253 
 
 2279 
 
 17 
 
 2304 
 
 2330 
 
 2355 
 
 2380 
 
 2405 
 
 2430 
 
 2455 
 
 2480 
 
 2504 
 
 2529 
 
 18 
 
 2553 
 
 2577 
 
 2601 
 
 2625 
 
 2648 
 
 2672 
 
 2695 
 
 2718 
 
 2742 
 
 2765 
 
 19 
 
 2788 
 
 2810 
 
 2833 
 
 2856 
 
 2878 
 
 2900 
 
 2923 
 
 2945 
 
 2967 
 
 2989 
 
 20 
 
 3010 
 
 3032 
 
 3054 
 
 3075 
 
 3096 
 
 3118 
 
 3i39 
 
 3160 
 
 3181 
 
 3201 
 
 21 
 
 3222 
 
 3243 
 
 3263 
 
 3284 
 
 3304 
 
 3324 
 
 3345 
 
 3365 
 
 3385 
 
 3404 
 
 22 
 
 3424 
 
 3444 
 
 3464 
 
 3483 
 
 3502 
 
 3522 
 
 3541 
 
 35 6 ° 
 
 3579 
 
 3598 
 
 23 
 
 3617 
 
 3636 
 
 3655 
 
 3674 
 
 3692 
 
 37 11 
 
 3729 
 
 3747 
 
 3766 
 
 3784 
 
 24 
 
 3802 
 
 3820 
 
 3838 
 
 3856 
 
 3874 
 
 3892 
 
 3909 
 
 3927 
 
 3945 
 
 3962 
 
 25 
 
 3979 
 
 3997 
 
 4014 
 
 4031 
 
 4048 
 
 4065 
 
 4082 
 
 4099 
 
 41 16 
 
 4133 
 
 26 
 
 4150 
 
 4166 
 
 4183 
 
 4200 
 
 4216 
 
 4232 
 
 4249 
 
 4265 
 
 4281 
 
 4298 
 
 27 
 
 43H 
 
 433° 
 
 4346 
 
 4362 
 
 4378 
 
 4393 
 
 4409 
 
 4425 
 
 4440 
 
 445° 
 
 28 
 
 4472 
 
 4487 
 
 45° 2 
 
 4518 
 
 4533 
 
 4548 
 
 4564 
 
 4579 
 
 4594 
 
 4609 
 
 29 
 
 4624 
 
 4639 
 
 4654 
 
 4669 
 
 4683 
 
 4698 
 
 4713 
 
 4728 
 
 4742 
 
 4757 
 
 30 
 
 4771 
 
 4786 
 
 4800 
 
 4814 
 
 4829 
 
 4843 
 
 4857 
 
 4871 
 
 4886 
 
 4900 
 
 31 
 
 4914 
 
 4928 
 
 4942 
 
 4955 
 
 4969 
 
 4983 
 
 4997 
 
 501 1 
 
 5024 
 
 5038 
 
 32 
 
 5051 
 
 5065 
 
 5079 
 
 5092 
 
 5i°5 
 
 5"9 
 
 5"32 
 
 5145 
 
 5159 
 
 5172 
 
 33 
 
 5i85 
 
 5198 
 
 5211 
 
 5224 
 
 5237 
 
 5250 
 
 5263 
 
 5276 
 
 5289 
 
 5302 
 
 34 
 
 S3IS 
 
 5328 
 
 534° 
 
 5353 
 
 5366 
 
 5378 
 
 5391 
 
 5403 
 
 54'6 
 
 5428 
 
 35 
 
 5441 
 
 S453 
 
 5465 
 
 5478 
 
 5490 
 
 55<52 
 
 5514 
 
 5527 
 
 5539 
 
 5551 
 
 36 
 
 5563 
 
 5575 
 
 5587 
 
 5599 
 
 5611 
 
 5623 
 
 5635 
 
 5647 
 
 5658 
 
 5670 
 
 37 
 
 5682 
 
 5 6 94 
 
 5705 
 
 57 J 7 
 
 5729 
 
 5740 
 
 575 2 
 
 5763 
 
 5775 
 
 5786 
 
 38 
 
 5798 
 
 5809 
 
 5821 
 
 5832 
 
 5843 
 
 5855 
 
 5866 
 
 5877 
 
 5888 
 
 5899 
 
 39 
 
 59" 
 
 5922 
 
 5933 
 
 5944 
 
 5955 
 
 5966 
 
 5977 
 
 5988 
 
 5999 
 
 6010 
 
 40 
 
 6021 
 
 6031 
 
 6042 
 
 6053 
 
 6064 
 
 6075 
 
 6085 
 
 6096 
 
 6107 
 
 6117 
 
 41 
 
 6128 
 
 6138 
 
 6149 
 
 6160 
 
 6170 
 
 6180 
 
 6191 
 
 6201 
 
 6212 
 
 6222 
 
 42 
 
 6232 
 
 6243 
 
 6253 
 
 6263 
 
 6274 
 
 6284 
 
 6294 
 
 6304 
 
 6314 
 
 6325 
 
 43 
 
 6335 
 
 6345 
 
 6355 
 
 6365 
 
 6375 
 
 6385 
 
 6395 
 
 6405 
 
 6415 
 
 6425 
 
 44 
 
 6435 
 
 6444 
 
 6454 
 
 6464 
 
 6474 
 
 6484 
 
 6493 
 
 6503 
 
 6513 
 
 6522 
 
 45 
 
 6532 
 
 6542 
 
 6551 
 
 6561 
 
 6571 
 
 6580 
 
 6590 
 
 6599 
 
 6609 
 
 6618 
 
 46 
 
 6628 
 
 6637 
 
 6646 
 
 6656 
 
 6665 
 
 6675 
 
 6684 
 
 6693 
 
 6702 
 
 6712 
 
 47 
 
 6721 
 
 6730 
 
 6739 
 
 6749 
 
 &7 5 8 
 
 6767 
 
 6776 
 
 6785 
 
 6794 
 
 6803 
 
 48 
 
 6812 
 
 6821 
 
 6830 
 
 6839 
 
 6848 
 
 6857 
 
 6866 
 
 6875 
 
 6884 
 
 6893 
 
 49 
 
 6902 
 
 691 1 
 
 6920 
 
 6928 
 
 6937 
 
 6946 
 
 6955 
 
 6964 
 
 6972 
 
 6981 
 
 5° 
 
 6990 
 
 6998 
 
 7007 
 
 7016 
 
 7024 
 
 7°33 
 
 7042 
 
 7050 
 
 7059 
 
 7067 
 
 51 
 
 7076 
 
 7084 
 
 7°93 
 
 7101 
 
 7110 
 
 7118 
 
 7126 
 
 7135 
 
 7H3 
 
 7152 
 
 52 
 
 7160 
 
 7168 
 
 7177 
 
 7185 
 
 7193 
 
 7202 
 
 7210 
 
 7218 
 
 7226 
 
 7235 
 
 53 
 
 7243 
 
 7251 
 
 7259 
 
 7267 
 
 7275 
 
 7284 
 
 7292 
 
 7300 
 
 7308 
 
 73i6 
 
 54 
 
 7324 
 
 7332 
 
 7340 
 
 7348 
 
 735 6 
 
 7364 
 
 7372 
 
 738o 
 
 7388 
 
 7396 
 
 No. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
LOGARITHMS OF NUMBERS. 
 
 13 
 
 So. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 55 
 
 7404 
 
 7412 
 
 7419 
 
 7427 
 
 7435 
 
 7443 
 
 745i 
 
 7459 
 
 7466 
 
 7474 
 
 56 
 
 7482 
 
 7490 
 
 7497 
 
 7505 
 
 7513 
 
 7520 
 
 7528 
 
 7536 
 
 7543 
 
 755' 
 
 57 
 
 7559 
 
 7566 
 
 7574 
 
 7582 
 
 7589 
 
 7597 
 
 7604 
 
 7612 
 
 7619 
 
 7627 
 
 58 
 
 7634 
 
 7642 
 
 7649 
 
 7657 
 
 7664 
 
 7672 
 
 7679 
 
 7686 
 
 7694 
 
 7701 
 
 59 
 
 7709 
 
 7716 
 
 77^3 
 
 773i 
 
 7738 
 
 7745 
 
 775 2 
 
 7760 
 
 7767 
 
 7774 
 
 60 
 
 7782 
 
 7789 
 
 7796 
 
 7803 
 
 7810 
 
 7818 
 
 7825 
 
 7832 
 
 7839 
 
 7846 
 
 61 
 
 7853 
 
 7860 
 
 78bS 
 
 7875 
 
 7SS2 
 
 7SS9 
 
 7896 
 
 79°3 
 
 7910 
 
 7917 
 
 62 
 
 79^4 
 
 793i 
 
 7938 
 
 7945 
 
 7952 
 
 7959 
 
 7966 
 
 7973 
 
 7980 
 
 7987 
 
 63 
 
 7993 
 
 8000 
 
 S007 
 
 8014 
 
 802 1 
 
 8028 
 
 8035 
 
 8041 
 
 8048 
 
 8055 
 
 64 
 
 S062 
 
 8069 
 
 8°75 
 
 S0S2 
 
 8089 
 
 8096 
 
 8102 
 
 8109 
 
 8116 
 
 8122 
 
 65 
 
 8129 
 
 S136 
 
 8142 
 
 8149 
 
 8156 
 
 8162 
 
 8169 
 
 8176 
 
 8182 
 
 8189 
 
 66 
 
 8i95 
 
 S202 
 
 8209 
 
 S215 
 
 8222 
 
 S22S 
 
 8235 
 
 8241 
 
 8248 
 
 8254 
 
 67 
 
 8261 
 
 8267 
 
 S274 
 
 S280 
 
 S287 
 
 8293 
 
 8299 
 
 8306 
 
 8312 
 
 8319 
 
 68 
 
 8325 
 
 8331 
 
 8338 
 
 S344 
 
 835' 
 
 8357 
 
 8363 
 
 837° 
 
 8376 
 
 8382 
 
 69 
 
 8388 
 
 8395 
 
 S401 
 
 S407 
 
 8414 
 
 8420 
 
 8426 
 
 8432 
 
 8439 
 
 8445 
 
 70 
 
 8451 
 
 8457 
 
 S463 
 
 8470 
 
 8476 
 
 8482 
 
 8488 
 
 8494 
 
 8500 
 
 8506 
 
 71 
 
 8513 
 
 8519 
 
 8525 
 
 8531 
 
 8537 
 
 8543 
 
 8549 
 
 8555 
 
 8561 
 
 8567 
 
 72 
 
 8573 
 
 8579 
 
 8585 
 
 8591 
 
 8597 
 
 8603 
 
 8609 
 
 8615 
 
 S621 
 
 8627 
 
 73 
 
 8633 
 
 8639 
 
 S645 
 
 S651 
 
 S657 
 
 8663 
 
 8669 
 
 8675 
 
 8681 
 
 8686 
 
 74 
 
 8692 
 
 S698 
 
 S704 
 
 8710 
 
 8716 
 
 8722 
 
 8727 
 
 8733 
 
 8739 
 
 8745 
 
 75 
 
 8751 
 
 8756 
 
 8762 
 
 8768 
 
 8774 
 
 8779 
 
 8785 
 
 8791 
 
 8797 
 
 8802 
 
 76 
 
 880S 
 
 8814 
 
 8S20 
 
 S825 
 
 8831 
 
 8837 
 
 S842 
 
 8848 
 
 8854 
 
 8859 
 
 77 
 
 8865 
 
 8871 
 
 8876 
 
 8882 
 
 8S87 
 
 8893 
 
 8899 
 
 8904 
 
 8910 
 
 8915 
 
 78 
 
 8921 
 
 8927 
 
 S932 
 
 8938 
 
 8943 
 
 8949 
 
 8954 
 
 8960 
 
 8965 
 
 8971 
 
 79 
 
 8976 
 
 8982 
 
 8987 
 
 8993 
 
 8998 
 
 9004 
 
 9009 
 
 9015 
 
 9020 
 
 9025 
 
 80 
 
 9031 
 
 9036 
 
 9042 
 
 9047 
 
 9053 
 
 9058 
 
 9063 
 
 9069 
 
 9074 
 
 9079 
 
 81 
 
 9085 
 
 9090 
 
 9096 
 
 9101 
 
 9106 
 
 9112 
 
 9117 
 
 9122 
 
 9128 
 
 9133 
 
 82 
 
 913S 
 
 9143 
 
 9149 
 
 9'54 
 
 9159 
 
 9165 
 
 9170 
 
 9175 
 
 9180 
 
 9186 
 
 83 
 
 9191 
 
 9196 
 
 9201 
 
 9206 
 
 9212 
 
 9217 
 
 9222 
 
 9227 
 
 9232 
 
 9238 
 
 84 
 
 9243 
 
 9248 
 
 9253 
 
 9258 
 
 9263 
 
 9269 
 
 9274 
 
 9279 
 
 9284 
 
 9289 
 
 85 
 
 9294 
 
 9299 
 
 93°4 
 
 93°9 
 
 9315 
 
 9320 
 
 9325 
 
 933° 
 
 9335 
 
 934° 
 
 86 
 
 9345 
 
 935° 
 
 9355 
 
 9360 
 
 9365 
 
 937° 
 
 9375 
 
 9380 
 
 9385 
 
 939° 
 
 87 
 
 9395 
 
 9400 
 
 9405 
 
 9410 
 
 9415 
 
 9420 
 
 9425 
 
 943° 
 
 9435 
 
 9440 
 
 88 
 
 9445 
 
 9450 
 
 9455 
 
 9460 
 
 9465 
 
 9469 
 
 9474 
 
 9479 
 
 9484 
 
 9489 
 
 89 
 
 9494 
 
 9499 
 
 95°4 
 
 9509 
 
 95 J 3 
 
 9518 
 
 9523 
 
 9528 
 
 9533 
 
 9538 
 
 go 
 
 9542 
 
 9547 
 
 9552 
 
 9557 
 
 9562 
 
 9566 
 
 9571 
 
 9576 
 
 9581 
 
 9586 
 
 9 1 
 
 959° 
 
 9595 
 
 9600 
 
 9605 
 
 9609 
 
 9614 
 
 9619 
 
 9624 
 
 9628 
 
 9633 
 
 92 
 
 9638 
 
 9643 
 
 9647 
 
 9652 
 
 9657 
 
 9661 
 
 9666 
 
 9671 
 
 9675 
 
 9680 
 
 93 
 
 9685 
 
 96S9 
 
 9694 
 
 9699 
 
 97°3 
 
 9708 
 
 97 r 3 
 
 97'7 
 
 9722 
 
 9727 
 
 94 
 
 9731 
 
 973 6 
 
 9741 
 
 9745 
 
 9750 
 
 9754 
 
 9759 
 
 9763 
 
 9768 
 
 9773 
 
 95 
 
 9777 
 
 9782 
 
 9786 
 
 9791 
 
 9795 
 
 9800 
 
 9805 
 
 9809 
 
 9814 
 
 9818 
 
 96 
 
 9823 
 
 9827 
 
 9832 
 
 9836 
 
 9841 
 
 9845 
 
 9850 
 
 9854 
 
 9859 
 
 9863 
 
 97 
 
 9868 
 
 9872 
 
 9877 
 
 9881 
 
 9886 
 
 9890 
 
 9894 
 
 9899 
 
 9903 
 
 9908 
 
 98 
 
 9912 
 
 9917 
 
 9921 
 
 9926 
 
 993° 
 
 9934 
 
 9939 
 
 9943 
 
 9948 
 
 9952 
 
 99 
 
 So. 
 
 9956 J 9961 
 
 9965 
 
 9969 
 
 9974 
 
 9978 
 
 9983 
 
 9987 
 
 9991 
 
 9996 
 9 
 
 
 
 1 
 
 2 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
14 
 
 LOGARITHMIC SINES, COSINES, 
 
 Angle. 
 
 0° O' 
 
 o° 10' 
 
 o° 2d 
 
 o° 30' 
 
 o° 40' 
 
 o° 5c/ 
 
 1° 0' 
 
 i u 10' 
 
 1° 2C/ 
 
 1° 30' 
 
 1° 40' 
 
 1° 50' 
 
 2° 0' 
 
 2° IO' 
 
 2° 20' 
 
 2° 30' 
 
 2° 40' 
 
 2° 50' 
 
 3° 0' 
 
 3° 10' 
 3 20' 
 3° 30' 
 3° 40' 
 
 3° 5°' 
 40 ; 
 
 4 10' 
 
 4 20 / 
 
 4° 30' 
 
 4° 40' 
 
 4° SO' 
 
 5° 0' 
 
 5° io 7 
 
 5° 2°' 
 
 5° 3o' 
 
 5° 4o' 
 
 5° 50' 
 
 6° 0' 
 
 6° 10' 
 
 6° 2cy 
 
 6° 30' 
 6° 40' 
 6° so 7 
 7° 0' 
 7 10' 
 7 20' 
 7° 3o' 
 
 Sin. 
 
 74637 
 .7648 
 .9408 
 
 8.0658 
 .1627 
 
 8.2419 
 
 .3088 
 .3668 
 ■4179 
 •4637 
 JOJO 
 
 8.5428 
 
 •5776 
 .6097 
 
 ■6397 
 .6677 
 .6940 
 
 i.7188 
 
 •7423 
 ■7645 
 •7857 
 .8059 
 .8251 
 i^436 
 
 .8613 
 .8783 
 .8946 
 .9104 
 ■9256 
 8.9403 
 
 •9545 
 .9682 
 .9816 
 
 •9945 
 9.0070 
 
 9.0192 
 
 •03 1 1 
 .0426 
 
 ■0539 
 .0648 
 
 •0755 
 
 9-0859 
 .0961 
 .1060 
 ■"57 
 
 Cos. 
 
 D.l'. 
 
 301. 1 
 I76.O 
 I25.O 
 96.9 
 79.2 
 66.9 
 58.O 
 511 
 
 45-8 
 4i-3 
 37-8 
 34-8 
 32.1 
 30.0 
 28.0 
 .26.3 
 24.8 
 
 23-5 
 22.2 
 21.2 
 20.2 
 19.2 
 18.5 
 17.7 
 17.0 
 16.3 
 15.8 
 15.2 
 14-7 
 14.2 
 
 13-7 
 134 
 12.9 
 
 12-5 
 
 12.2 
 11.9 
 11. 5 
 
 "•3 
 10.9 
 10.7 
 10.4 
 10.2 
 
 9-9 
 9-7 
 
 D.l'. 
 
 Cos. 
 
 .OOOO 
 .OOOO 
 .OOOO 
 .OOOO 
 .OOOO 
 
 9^9999 
 •9999 
 •9999 
 •9999 
 .9998 
 
 ■9998 
 9.9997 
 
 •9997 
 .9996 
 .9996 
 •9995 
 ■9995 
 
 9.9994 
 
 ■9993 
 •9993 
 .9992 
 .9991 
 .9990 
 
 9.9989 
 ■9989 
 
 •9987 
 
 ■9985 
 9-9983 
 
 .9982 
 .9981 
 .9980 
 •9979 
 •9977 
 
 9-9976 
 
 •9975 
 •9973 
 •9972 
 .9971 
 
 ■9969 
 9.9968 
 .9966 
 .9964 
 ■9963 
 
 Sin. 
 
 D. 1'. 
 
 D. 1'. 
 
 Xan. 
 
 74637 
 .7648 
 .9409 
 
 8.0658 
 .1627 
 
 8.2419 
 
 ■3089 
 •3669 
 .4181 
 .4638 
 ■5053 
 
 8-5431 
 
 •5779 
 .6101 
 .6401 
 .6682 
 •6945 
 
 8.7194 
 
 •7429 
 •7652 
 •7865 
 .8067 
 .8261 
 
 .8624 
 
 •8795 
 .8960 
 .9118 
 
 .9272 
 8.9420 
 
 •9563 
 .9701 
 
 •9836 
 
 .9966 
 
 9-OQ93 
 
 9.0216 
 
 ~^336 
 ■0453 
 ■0567 
 .0678 
 .0786 
 
 9.0891 
 
 ■0995 
 .1096 
 
 •"94 
 
 Cot. 
 
 D. 1'. 
 
 301. 1 
 1 76. 1 
 124.9 
 96.9 
 79.2 
 67.O 
 58.O 
 51.2 
 
 45-7 
 41.5 
 
 37-8 
 34-8 
 32.2 
 30.0 
 28.1 
 26.3 
 24.9 
 
 23-5 
 22.3 
 21.3 
 20.2 
 19.4 
 18.5 
 17.8 
 17. 1 
 .6.5 
 15.8 
 15.4 
 14.8 
 
 14-3 
 13-8 
 
 '3-5 
 13.0 
 12.7 
 12.3 
 12.0 
 11. 7 
 11.4 
 11. 1 
 10.8 
 10.5 
 10.4 
 10.1 
 9-8 
 
 D.l'. 
 
 Cot. 
 
 2.5363 
 .2352 
 .0591 
 
 I.9342 
 ■8373 
 
 I-758I 
 
 .6911 
 
 •6331 
 .5819 
 •5362 
 4947 
 1.4569 
 
 .4221 
 •3899 
 •3599 
 •33i8 
 
 •305S 
 
 1.2806 
 
 •257' 
 .2348 
 
 •2135 
 ■1933 
 •1739 
 
 '■ I 554 
 
 •1376 
 .1205 
 .1040 
 .0882 
 .0728 
 
 1.0580 
 
 •0437 
 .0299 
 .0164 
 .0034 
 0.9907 
 
 0-9784 
 .9664 
 •9547 
 •9433 
 .9322 
 
 ■9214 
 0.9109 
 
 .9005 
 .8904 
 
 Tan. 
 
 _4 
 
 90° 0' 
 
 89 50' 
 89 40' 
 89 30' 
 89 20' 
 89 ic/ 
 89° 0' 
 88° so 7 
 88° 40' 
 88° 30 7 
 88° 20' 
 88° 10' 
 88° 0' 
 87 50' 
 87 40' 
 87°3o / 
 87 2& 
 87 10' 
 87° O' 
 86° 50' 
 86° 40' 
 86° 30' 
 86° 20' 
 86° 10' 
 86° 0' 
 85°5o / 
 85 40' 
 85° 30' 
 85 20' 
 85 10' 
 85° 0' 
 84° 50' 
 84°40 / 
 84 30' 
 84 20' 
 84 10' 
 8*°0' 
 
 8 3 °5o' 
 S3 40' 
 83 30' 
 83 2C/ 
 83 10' 
 83° 0' 
 82 50' 
 82 40' 
 82 30' 
 
 Angle. 
 
TAXGEXTS AXD COTAXGENTS. 
 
 15 
 
 Angle. 
 
 Sin. 
 
 B.l'. 
 
 Cos. 
 
 B.l'. 
 
 Tan. 
 
 B.l'. 
 
 Cot. 
 
 
 7° 30 1 
 
 9"57 
 
 9-5 
 
 9-9963 
 
 
 9- "94 
 
 
 O.8806 
 
 82° 30' 
 
 7° 4o' 
 
 .1252 
 
 .9961 
 
 "t 
 
 .1291 
 
 9-7 
 
 .8709 
 
 82° 20* 
 
 7° So' 
 
 8° <y 
 
 •1345 
 
 9-3 
 9-1 
 8-9 
 
 8-7 
 8-5 
 8.4 
 8.2 
 8.0 
 
 7-9 
 7.8 
 7-6 
 
 ■9959 
 
 .1 
 2 
 
 -1385 
 
 9-4 
 9-3 
 9.1 
 
 8-9 
 8-7 
 8.6 
 84 
 8.2 
 8.1 
 80 
 
 .8615 
 
 82° IO' 
 
 82° C 
 
 9.1436 
 
 9.9958 
 
 9.1478 
 
 0.8522 
 
 8° 10' 
 
 ■1525 
 
 .9956 
 
 .2 
 
 .1569 
 
 .8431 
 
 8i° sd 
 
 8° 2C/ 
 
 .1612 
 
 -9954 
 
 .1658 
 
 -8342 
 
 8i° 40' 
 
 8° 3°' 
 
 .1697 
 
 •9952 
 
 
 -1745 
 
 -8255 
 
 81° 30- 
 
 8° 40- 
 
 .1781 
 
 -995° 
 
 
 .1831 
 
 .8169 
 
 gl°2o' 
 
 8° 50' 
 9° 0' 
 9° 10' 
 
 .1863 
 
 .9948 
 
 .2 
 
 .2 
 
 -1915 
 
 .8085 
 
 8i° io' 
 81 c 0' 
 
 80= 50' 
 
 9-1943 
 
 9.9946 
 
 9.1997 
 
 0.8003 
 
 .2022 
 
 -9944 
 
 .2078 
 
 •7922 
 
 9° 20' 
 
 .2100 
 
 •9942 
 
 
 .2158 
 
 7-8 
 
 7-7 
 7.6 
 
 74 
 7-3 
 7-3 
 7-i 
 7.0 
 
 6.9 
 6.8 
 
 6.6 
 
 6-7 
 6-5 
 6.4 
 6-3 
 6-3 
 6.1 
 6 1 
 
 -7842 
 
 8o 4 o' 
 
 9° SO* 
 
 .2176 
 
 .9940 
 
 
 .2236 
 
 -7764 
 
 80° 30* 
 
 9 ° 4 o' 
 
 .2251 
 
 7-5 
 
 •9938 
 
 
 •2313 
 
 •7687 
 
 8o°2o' 
 
 9° 50' 
 
 io° <y 
 
 io° 10' 
 
 -2324 
 
 7-3 
 7-3 
 7-1 
 7.0 
 6.8 
 6.8 
 66 
 
 •9936 
 
 .2 
 •3 
 
 .2389 
 
 .7611 
 
 80° io* 
 80 c O' 
 79° 5C 
 
 9-2397 
 .2468 
 
 9-9934 
 
 9-2463 
 
 0-7537 
 
 •9931 
 
 -2536 
 
 -7464 
 
 IO C 20 7 
 
 •2538 
 
 ■9929 
 
 
 .2609 
 
 ■739« 
 
 79° 40' 
 
 io° 30' 
 
 .2606 
 
 -9927 
 
 ■3 
 2 
 
 .2680 
 
 ■7320 
 
 79 3°' 
 
 IO° 40* 
 
 .2674 
 
 •9924 
 
 .2750 
 
 •7250 
 
 79° 20* 
 
 io° 50' 
 
 11° O' 
 
 n° io- 
 
 .2740 
 
 6.6 
 64 
 6-4 
 6-3 
 6.1 
 6.1 
 6.0 
 
 5-9 
 5-8 
 
 .9922 
 
 •3 
 .2 
 
 •3 
 
 .2819 
 
 .7181 
 
 79° id 
 79 c 0' 
 
 78° 50" 
 
 9.2806 
 
 9-99'9 
 .9917 
 
 9.2887 
 ~~*953 
 
 0.71 13 
 
 .2870 
 
 •7047 
 
 11° 20' 
 
 •2934 
 
 .9914 
 
 .3020 
 
 .6980 
 
 78° 40' 
 
 n° 30* 
 
 •2997 
 
 .9912 
 
 •3 
 
 -3085 
 
 .6915 
 
 78° 30' 
 
 H° 40» 
 
 .3058 
 
 .9909 
 
 •3149 
 
 .6851 
 
 78° 20* 
 
 11° 50- 
 12° C 
 
 12° id 
 
 •3"9 
 
 9907 
 9.9904 
 
 •3 
 -3 
 
 ■3212 
 9-3275 
 
 .6788 
 
 78° 10' 
 78° 0' 
 77050- 
 
 9-3I79 
 
 0.6725 
 
 •3238 
 
 .9901 
 
 •3336 
 
 .6664 
 
 12° 2d 
 
 .3296 
 
 ■9899 
 
 •3 
 •3 
 •3 
 -3 
 ' -3 
 ■3 
 
 ■3397 
 
 6 1 
 
 ■6603 
 
 77040- 
 
 12° 3d 
 
 •3353 
 
 5-7 
 
 .9896 
 
 -3458 
 
 5-9 
 5-9 
 5.8 
 
 5-7 
 5-7 
 5-6 
 
 .6542 
 
 77° 30' 
 
 12° 40' 
 
 .3410 
 
 5-7 
 5-6 
 5-5 
 5-4 
 
 ■9893 
 
 ■3517 
 
 .6483 
 
 77 20- 
 
 12° SO' 
 
 13° O' 
 
 13° 10' 
 
 .3466 
 
 •9890 
 
 •3576 
 
 .6424 
 
 77 io- 
 77- C 
 76° 50- 
 
 9-3521 
 -3575 
 
 9.9887 
 
 93634 
 
 0.6366 
 
 .9884 
 
 •3691 
 
 ■6309 
 
 13° 2C/ 
 
 .3629 
 
 5-4 
 
 .9881 
 
 -3748 
 
 .6252 
 
 76040- 
 
 13 30* 
 
 .3682 
 
 5-3 
 
 .9878 
 
 -0 
 -3 
 •3 
 •3 
 -3 
 •3 
 •4 
 -3 
 ■3 
 •4 
 
 .3804 
 
 .6196 
 
 76° 30- 
 
 •3° 4°* 
 
 •3734 
 
 S- 2 
 
 -987S 
 
 -3859 
 
 5-5 
 
 .6141 
 
 76° 20- 
 
 I3 C SO' 
 14° O' 
 
 14° 10' 
 
 .3786 
 
 5.2 
 5- 1 
 5.0 
 
 .9872 
 
 -39H 
 
 DO 
 
 5-4 
 
 5-3 
 5-3 
 5-3 
 5- 1 
 5-2 
 5- 1 
 
 .6086 
 
 76° 10' 
 
 76=0- 
 
 75° 50' 
 
 9-3837 
 
 9.9869 
 .9866 
 
 9-3968 
 
 0.6032 
 
 •3887 
 
 4021 
 
 •5979 
 
 14° 2d 
 14° 3d 
 
 -3937 
 .3986 
 
 5° 
 4-9 
 
 .9863 
 ■98S9 
 
 .4074 
 .4127 
 
 .5926 
 •5S/3 
 
 75° 40- 
 75 D 3o' 
 
 14° 40* 
 
 -4035 
 
 4-9 
 4.8 
 
 « 
 
 .9856 
 
 •4178 
 
 .5822 
 
 75 °2o' 
 
 14° 5<y 
 
 15° 0' 
 
 •4083 
 
 -9853 
 
 -4230 
 
 .5770 
 
 75° 10' 
 75° O' 
 
 9-4i3 
 
 9.9849 
 
 9.4281 
 
 0.5719 
 
 
 Cos. 
 
 B.l'. 
 
 Sin. 
 
 B. 1'. 
 
 Cot. 
 
 B.l'. 
 
 Tan. 
 
 Angle. 
 
16 
 
 LOGARITHMIC SINES, COSINES, 
 
 Angle. 
 
 Sin. 
 
 D.l'. 
 
 Cos. 
 
 D.l'. 
 
 Tan. 
 
 D. 1'. 
 
 Cot. 
 
 
 15° 0' 
 
 15° io' 
 
 94130 
 
 4-7 
 4.6 
 4.6 
 
 9.9849 
 
 •3 
 
 9.4281 
 
 5.0 
 
 5-o 
 4.9 
 
 4-9 
 4.8 
 4.8 
 
 4-7 
 4-7 
 
 0.5719 
 
 75° 0' 
 74° 50- 
 
 ■4177 
 
 .9846 
 
 4331 
 
 .5669 
 
 15° 20' 
 
 .4223 
 
 ■9843 
 
 •3 
 
 .4381 
 
 ■5 6l 9 
 
 74° 4°' 
 
 15° 30' 
 
 .4269 
 
 •9839 
 
 4 
 •3 
 4 
 4 
 •3 
 
 ■4430 
 
 •557o 
 
 74° 30' 
 
 1 5° 40' 
 
 •4314 
 
 4-5 
 4-5 
 44 
 4.4 
 
 .9836 
 
 •4479 
 
 •5521 
 
 74 20' 
 
 15° 5°' 
 16° 0' 
 
 1 6° 10' 
 
 ■4359 
 
 9-44°3 
 
 •4447 
 
 .9832 
 9.9828 
 
 45 2 7 
 94575 
 
 •5473 
 
 74° 10' 
 
 74° 0' 
 73° 5° r 
 
 0.5425 
 
 ■9825 
 
 .4622 
 
 -5378 
 
 1 6° 20' 
 
 .4491 
 
 44 
 
 .9821 
 
 4 
 
 .4669 
 
 •5331 
 
 73° 40' 
 
 1 6° 30' 
 
 •4533 
 
 4.2 
 4-3 
 
 .9817 
 
 4 
 •3 
 
 4 
 4 
 4 
 4 
 
 .4716 
 
 4-7 
 
 4.6 
 4.6 
 
 4-5 
 
 4-5 
 4-5 
 44 
 44 
 
 .5284 
 
 73° 30* 
 
 1 6° 40' 
 
 ■457° 
 
 .9814 
 
 .4762 
 
 .5238 
 
 Tf id 
 
 1 6° 50' 
 17° 0' 
 17° 10' 
 
 .4618 
 
 4.2 
 4.1 
 
 4-i 
 4.1 
 
 .9810 
 
 .4808 
 
 •5!92 
 
 73° IO ' 
 73° 0' 
 
 72° 50' 
 
 94659 
 .4700 
 
 9.9806 
 .9802 
 
 94853 
 
 °-5H7 
 .5102 
 
 .4898 
 
 17° 2C/ 
 
 .4741 
 
 .9798 
 
 4943 
 
 •5057 
 
 72° 40' , 
 
 17° 30' 
 
 .4781 
 
 4.0 
 4.0 
 
 •9794 
 
 4 
 4 
 
 4987 
 
 •5013 
 
 72° 3°' 
 
 17° 40' 
 
 .4821 
 
 •9790 
 
 •5°3i 
 
 .4969 
 
 72° 2d 
 
 17° 50' 
 18° 0' 
 1 8° 10' 
 
 .4861 
 
 4.0 
 3-9 
 3-9 
 3-8 
 3-8 
 3-7 
 3-8 
 3-6 
 
 3-7 
 3-6 
 3-6 
 3-5 
 3-6 
 3-5 
 34 
 34 
 34 
 34 
 3-3 
 3-3 
 3-3 
 3-3 
 
 3-2 
 3-2 
 
 3-t 
 
 3-2 
 
 3-i 
 3-i 
 3-o 
 
 ■9786 
 
 9.9782 
 
 •9778 
 
 •4 
 
 4 
 ■4 
 4 
 4 
 •5 
 
 •5075 
 
 4-4 
 4-3 
 4-3 
 4.2 
 4.2 
 4.2 
 4.2 
 4-1 
 4.1 
 4.0 
 4.0 
 4.0 
 4.0 
 4.0 
 
 3-9 
 
 3-9 
 3-8 
 3-9 
 3-8 
 3-8 
 
 3-7 
 3-8 
 
 .4925 
 
 72° \d 
 72° 0' 
 71° 50- 
 
 9.4900 
 
 9.5118 
 
 0.4882 
 4839 
 
 4939 
 
 .5161 
 
 1 8° 2d 
 
 ■4977 
 
 •9774 
 
 •5203 
 
 4797 
 
 71° 40' 
 
 18° 30' 
 
 •5 OI 5 
 
 .9770 
 
 .5245 
 
 4755 
 
 71° 30' 
 
 18° 40' 
 
 •5052 
 
 .9765 
 
 •5287 
 
 4713 
 
 71° 20' 
 
 18° 5CV 
 
 .5090 
 
 .9761 
 
 4 
 •4 
 •5 
 4 
 ■5 
 4 
 ■5 
 4 
 
 •5 
 4 
 •5 
 •5 
 •5 
 ■4 
 
 •5 
 •5 
 •5 
 ■5 
 ■5 
 ■5 
 
 •5 
 6 
 
 ■5329 
 
 .4671 
 
 71 10' 
 
 19° 0' 
 
 19° i<y 
 
 9.5126 
 •5163 
 
 9-9757 
 •9752 
 
 9-537° 
 
 0.4630 
 
 71° 0' 
 70° 50' 
 
 •541 1 
 
 .4589 
 
 19° 20' 
 
 •5 r 99 
 
 .9748 
 
 •5451 
 
 4549 
 
 70° 4c/ 
 
 1 9 ° 30' 
 
 •5235 
 
 ■9743 
 
 ■5491 
 
 .4509 
 
 70 30' 
 
 19° 40' 
 
 .5270 
 
 •9739 
 
 •553i 
 
 •4469 
 
 70° 20' 
 
 19° 50' 
 20° 0' 
 
 ■53o6 
 9-5341 
 
 •9734 
 
 ■557i 
 
 ■4429 
 0.4389 
 
 70° id 
 70° 0' 
 
 9-9730 
 
 9.5611 
 
 20° 10' 
 
 •5375 
 
 •9725 
 
 •5650 
 
 435° 
 
 69° 50' 
 
 20° 2C/ 
 
 •5409 
 
 .9721 
 
 .5689 
 
 431 1 
 
 69° 40' 
 
 20° 30' 
 
 •5443 
 
 .9716 
 
 •5727 
 
 4273 
 
 69° 30' 
 
 20° 40' 
 
 •5477 
 
 •97" 
 
 .5766 
 
 4234 
 
 69° 20' 
 
 20° 5c/ 
 
 21° O 
 21° 10' 
 
 •55'Q 
 
 9-5543 
 
 •557° 
 
 .9706 
 
 ■5804 
 9.5842 
 
 •5879 
 
 ■4196 
 0.4158 
 
 69° 10' 
 69° 0' 
 
 68° 50' 
 
 9-97°2 
 .9697 
 
 4121 
 
 21° 20' 
 
 .5609 
 
 .9692 
 
 ■5917 
 
 .4083 
 
 68° 40' 
 
 21° 3c/ 
 
 .5641 
 
 .9687 
 
 •5954 
 
 3-7 
 
 .4046 
 
 68° 30' 
 
 21° 40' 
 
 •5°73 
 
 .9682 
 
 •5991 
 
 3-7 
 
 .4009 
 
 68° 2d 
 
 21° yd 
 22° 0' 
 
 22° IO' 
 
 ■5704 
 
 ■9677 
 
 9.9672 
 
 .9667 
 
 .6028 
 
 3-7 
 3-6 
 
 3-6 
 3-6 
 3-6 
 
 •3972 
 
 68° 10' 
 68° 0' 
 
 67° 50' 
 
 9-5736 
 •5767 
 
 9.6064 
 
 o-3936 
 
 .6100 
 
 .3900 
 
 22° 2o' 
 
 •5798 
 
 .9661 
 
 •5 
 
 .6136 
 
 .3864 
 
 67° 40' 
 
 22° 30' 
 
 .5828 
 
 .9656 
 
 .6172 
 
 .3828 
 
 67° 30' 
 
 
 Cos. 
 
 D.l'. 
 
 Sin. 
 
 D. 1'. 
 
 Cot. 
 
 D.l'. 
 
 Tan. 
 
 Angle. 
 
TANGENTS AND COTANGENTS. 
 
 17 
 
 Angle. 
 
 Sin. 
 
 D.l'. 
 
 Cos. D. 
 
 1'. 
 
 Tan. 
 
 D. 1'. 
 
 Cot. 
 
 
 22° 30' 
 
 9.5828 
 
 3-i 
 
 9.9656 
 
 5 
 5 
 6 
 
 5 
 g 
 
 9.6172 
 
 3-6 
 
 O.3828 
 
 67° 30' 
 
 22° 40' 
 
 •5859 
 
 •9°5 x 
 
 .6208 
 
 •3792 
 
 67° 20' 
 
 22° 50' 
 
 23° 0' 
 
 23° 10' 
 
 .5889 
 
 3° 
 3-o 
 2.9 
 3-o 
 
 .9646 
 
 .6243 
 
 3-5 
 3-° 
 3-5 
 
 ■3757 
 
 67° id 
 67° 0' 
 
 66° 50' 
 
 9-5919 
 •5948 
 
 9.964O 
 •9°35 
 
 9.6279 
 
 0.3721 
 
 .6314 
 
 .3686 
 
 23° 2C/ 
 
 •5978 
 
 .9629 
 
 
 .6348 
 
 3-4 
 
 •3652 
 
 66° 40' 
 
 23° 30' 
 
 .6007 
 
 2.9 
 
 .9624 
 
 5 
 6 
 
 ■6383 
 
 3-5 
 
 •3°i7 
 
 66° 3d 
 
 23° 40' 
 
 .6036 
 
 2.9 
 
 .9618 
 
 .6417 
 
 3-4 
 
 •3583 
 
 66° 20' 
 
 23° 50/ 
 
 24° O' 
 24° 10' 
 
 .6065 
 
 2.9 
 2.8 
 2.8 
 2.8 
 2 8 
 
 ■9613 
 9.9607 
 
 5 
 6 
 
 5 
 6 
 g 
 
 .6452 
 
 3-5 
 3-4 
 3-4 
 
 3-3 
 3-4 
 3-3 
 
 •3548 
 
 66° 10' 
 66° 0' 
 
 65° 50' 
 
 9.6093 
 
 9.6486 
 .6520 
 
 Q-35'4 
 .3480 
 
 .6121 
 
 .9602 
 
 24° 20' 
 
 .6149 
 
 .9596 
 
 ■°553 
 
 •3447 
 
 65° 40' 
 
 24° 3°' 
 
 .6177 
 
 2.8 
 
 .9590 
 
 g 
 
 .6587 
 
 •3413 
 
 65° 30' 
 
 24° 40' 
 
 .6205 
 
 .9584 
 
 
 .6620 
 
 •338o 
 
 65° 2d 
 
 24° 50' 
 25° C 
 25° 10' 
 
 .6232 
 
 2.7 
 2.7 
 
 2-7 
 
 •9579 
 
 5 
 6 
 
 6 
 
 g 
 
 .6654 
 
 3-4 
 3-3 
 3-3 
 3-2 
 3-3 
 3-2 
 3-3 
 3-2 
 
 3-2 
 3-2 
 
 3-1 
 3-2 
 3-i 
 
 3-2 
 
 31 
 31 
 
 •3346 
 
 65° 10' 
 65° 0' 
 
 64° 50' 
 
 9.6259 
 
 9-9573 
 
 9.6687 
 
 03313 
 .3280 
 
 .6286 
 
 .9567 
 
 .6720 
 
 25° 2d 
 
 ■°3I3 
 
 2-7 
 
 .9561 
 
 6 
 
 .6752 
 
 ■3248 
 
 64° 40' 
 
 25° 30' 
 
 .6340 
 
 2.7 
 
 2 6 
 
 •9555 
 
 g 
 
 .6785 
 
 •3215 
 
 64° 30' 
 
 25° 40' 
 
 .6366 
 
 2 6 
 
 •9549 
 
 g 
 
 .6817 
 
 ■3183 
 
 64° 20' 
 
 25° so' 
 26° O' 
 
 ■6392 
 9.6418 
 
 2.6 
 
 26 
 
 •9543 
 
 6 
 
 7 
 g 
 
 .6850 
 
 •3150 
 
 64° 10' 
 64° 0' 
 
 9-9537 
 
 9.6882 
 
 0.31 18 
 
 26° 10' 
 
 .6444 
 
 2.6 
 2-5 
 
 2 6 
 
 •953° 
 
 .6914 
 
 .3086 
 
 63° 50- 
 
 26° 2C/ 
 
 .6470 
 
 •9524 
 
 g 
 
 .6946 
 
 •3054 
 
 63° 40' 
 
 26° 30' 
 
 .6495 
 
 .9518 
 
 g 
 
 .6977 
 
 •3023 
 
 63° 30' 
 
 26° 40' 
 
 .6521 
 
 
 .9512 
 
 
 .7009 
 
 .2991 
 
 63° 20' 
 
 26° 5C/ 
 
 27° 0' 
 27° id 
 
 .6546 
 
 2 -5 
 2.4 
 
 2-5 
 
 ■9505 
 9.9499 
 
 7 
 6 
 
 7 
 g 
 
 .7040 
 9-7072 
 
 .2960 
 
 63° io' 
 63° 0' 
 
 62° 50' 
 
 9-657° 
 •6595 
 
 0.2928 
 
 .9492 
 
 ■7 io 3 
 
 .2897 
 
 27° 2d 
 
 .6620 
 
 2-5 
 
 .9486 
 
 
 •7'34 
 
 .2866 
 
 62° 40' 
 
 27° 30' 
 
 .6644 
 
 2.4 
 
 •9479 
 
 7 
 g 
 
 .7165 
 
 3 1 
 3-i 
 3-o 
 3-i 
 3° 
 3-o 
 3-1 
 3-o 
 3-o 
 3° 
 2.9 
 
 3-o 
 2.9 
 3-o 
 2.9 
 2.9 
 
 •2835 
 
 62° 30' 
 
 27° 40' 
 
 .6668 
 
 2.4 
 
 ■9473 
 
 
 .7196 
 
 .2804 
 
 62° 20' 
 
 27° 50' 
 28° 0' 
 28° 10' 
 
 .6692 
 9.6716 
 T6740 
 
 2.4 
 
 2.4 
 2.4 
 
 .9466 
 
 7 
 7 
 6 
 
 .7226 
 
 •2774 
 
 62° id 
 62° 0' 
 61° $d 
 
 9-9459 
 
 97257 
 
 0.2743 
 
 •9453 
 
 •7287 
 
 •2713 
 
 28° 20' 
 
 .6763 
 
 2 -3 
 
 .9446 
 
 7 
 
 ■7317 
 
 .2683 
 
 61° 40' 
 
 28° 30' 
 
 .6787 
 
 2.4 
 
 •9439 
 
 7 
 7 
 7 
 7 
 7 
 7 
 7 
 7 
 7 
 8 
 
 ■7348 
 
 .2652 
 
 6i° 30' 
 
 28° 40' 
 
 .6810 
 
 2.3 
 
 •9432 
 
 •7378 
 
 .2622 
 
 61° 2d 
 
 28° 5c/ 
 29° O' 
 
 29° 10' 
 
 29° 20 7 
 
 .6833 
 
 2.3 
 
 2-3 
 
 2.2 
 
 2-3 
 
 ■9425 
 9.9418 
 
 .7408 
 
 .2592 
 
 6i° 10' 
 
 61 °0' 
 
 60° 50' 
 60° 4d 
 
 9.6856 
 .6878 
 .6901 
 
 97438 
 
 0.2562 
 
 •94" 
 .9404 
 
 .7467 
 •7497 
 
 •2533 
 .2503 
 
 29° 3°* 
 
 .6923 
 
 ' 
 
 ■9397 
 
 .7526 
 
 •2474 
 
 60° 30' 
 
 29° 40' 
 
 .6946 
 
 2-3 
 
 ■9390 
 
 •755 6 
 
 .2444 
 
 60° 2d 
 
 29° 50' 
 30° 0' 
 
 .6968 
 
 2.2 
 
 ■9383 
 
 ■7585 
 9.7614 
 
 .2415 
 
 60° 10' 
 60° 0' 
 
 9.6990 
 
 9-9375 
 
 0.2386 
 
 Cos. 
 
 D.l'. 
 
 Sin. D 
 
 l'. 
 
 Cot. 
 
 D. 1'. 
 
 Tan. 
 
 Angle. 
 
18 
 
 LOGARITHMIC SINES, COSINES, 
 
 Angle. 
 
 Sin. 
 
 D.l'. 
 
 Cos. 
 
 D.l'. 
 
 Tan. 
 
 D. 1'. 
 
 Cot. 
 
 ^— — — 
 
 30° 0' 
 
 9.6990 
 
 
 9-9375 
 
 •7 
 
 ■7 
 .8 
 
 9.7614 
 
 3-° 
 2.9 
 2.8 
 
 0.2386 
 
 60° 0' 
 
 30° 10' 
 
 .7012 
 
 
 .9368 
 
 .7644 
 
 .2356 
 
 59° 5°; 
 
 30° 20' 
 
 ■7°33 
 
 
 .9361 
 
 •7673 
 
 .2327 
 
 59° 40' 
 
 3°° 3°' 
 
 •7055 
 
 
 •9353 
 
 •7 
 ,8 
 
 .7701 
 
 2.9 
 2.9 
 2.9 
 2.8 
 2.9 
 2.8 
 2.9 
 2.8 
 
 .2299 
 
 59° 3^ 
 
 30° 40' 
 
 7076 
 
 
 •9346 
 
 ■7730 
 
 .2270 
 
 59° 20' 
 
 30° 50' 
 
 .7097 
 
 2.1 
 
 ■9338 
 
 ■7 
 .8 
 g 
 
 •7759 
 
 .2241 
 
 59° 10' 
 
 31° 0' 
 
 31° 10' 
 
 9-7" 8 
 7 r 39 
 
 2.1 
 
 9-9331 
 
 9.7788 
 
 0.2212 
 
 59° 0' 
 
 58° 50' 
 
 ■93 2 3 
 
 .7816 
 
 .2184 
 
 31° 20' 
 
 .7160 
 
 
 •93J5 
 
 •7 
 8 
 
 •7845 
 
 •2155 
 
 58° 40' 
 
 31° 30' 
 
 .7181 
 
 2.0 
 
 .9308 
 
 •7873 
 
 .2127 
 
 58030- 
 
 3'° 40' 
 
 .7201 
 
 2.1 
 
 .9300 
 
 .8 
 
 .7902 
 
 .2098 
 
 58° 20' 
 
 31° 50' 
 32° 0' 
 32° 10' 
 
 .7222 
 
 2.0 
 2.0 
 
 .9292 
 
 .8 
 .8 
 g 
 
 •793° 
 9-7958 
 
 2.8 
 2.8 
 2 8 
 
 .2070 
 
 58° 10' 
 58° 0' 
 
 57° 5°' 
 
 9.7242 
 
 9.9284 
 
 0.2042 
 
 .7262 
 
 .9276 
 
 .7986 
 
 .2014 
 
 32° 20' 
 
 .7282 
 
 
 .9268 
 
 8 
 
 .8014 
 
 2 8 
 
 .1986 
 
 57° 40' 
 
 32° 30' 
 
 .7302 
 
 
 .9260 
 
 8 
 
 .8042 
 
 28 
 
 •1958 
 
 57° 30' 
 
 32° 40' 
 
 .7322 
 
 
 .9252 
 
 8 
 
 .8070 
 
 2.7 
 2.8 
 2.8 
 
 .1930 
 
 57° 20- 
 
 32° 50' 
 33° 0' 
 
 ■7342 
 
 i-9 
 1-9 
 
 .9244 
 
 .8 
 .8 
 
 •8097 
 
 .1903 
 
 57° 10' 
 57° O' 
 
 9-736i 
 
 9.9236 
 
 9.8125 
 
 0.1875 
 
 33° 1°' 
 
 •738o 
 
 .9228 
 
 •9 
 
 8 
 
 ■8i53 
 
 2.7 
 2 8 
 
 .1847 
 
 56° st" 
 
 33° 2°' 
 
 .7400 
 
 i-9 
 i-9 
 1.9 
 
 i-9 
 1.8 
 
 i-9 
 1 8 
 
 .9219 
 
 .8180 
 
 .1820 
 
 56° 40' 
 
 33° 30' 
 
 .7419 
 
 .9211 
 
 8 
 
 .8208 
 
 2.7 
 2 8 
 
 .1792 
 
 56° 30' 
 
 33° 4°' 
 
 ■7438 
 
 .9203 
 
 ■9 
 
 .8 
 
 •9 
 g 
 
 ■8235 
 
 •'765 
 
 56° 20' 
 
 33° 5^ 
 34° 0' 
 
 34° iC 
 
 ■7457 
 
 9-7476 
 
 ■7494 
 
 ■9194 
 9.9186 
 
 .8263 
 
 2-7 
 
 2.7 
 
 •1737 
 
 56° 10' 
 56° 0' 
 
 55° 50' 
 
 9.8290 
 
 0.1710 
 
 .9177 
 
 ■8317 
 
 .1683 
 
 34° 20! 
 
 •75 '3 
 
 .9169 
 
 •9 
 •9 
 ■9 
 .8 
 
 ■9 
 •9 
 •9 
 •9 
 
 •8344 
 
 2.7 
 2.7 
 
 .1656 
 
 55° 40' 
 
 34° 3°' 
 
 •753i 
 
 1-9 
 1 8 
 
 .9160 
 
 •8371 
 
 .1629 
 
 55° 3°' 
 
 34° 40' 
 
 ■7550 
 
 .9151 
 
 •8398 
 
 2.7 
 2.7 
 2.7 
 2.7 
 2.7 
 
 2-7 
 2 6 
 
 .1602 
 
 55° 20' 
 
 34° 5°' 
 35° 0' 
 
 •7568 
 
 1.8 
 1.8 
 
 .9142 
 
 •8425 
 9.8452 
 
 •1575 
 
 55° ior 
 55° 0' 
 
 9.7586 
 
 9-9134 
 
 0.1548 
 
 35° IO' 
 
 .7604 
 
 1 8 
 
 ■9125 
 
 .8479 
 
 .1521 
 
 54° 50* 
 
 35° 2c/ 
 
 .7622 
 
 1 8 
 
 .9116 
 
 .8506 
 
 .1494 
 
 54° 40' 
 
 35° 3°' 
 
 .7640 
 
 i-7 
 1 8 
 
 .9107 
 
 •8533 
 
 .1467 
 
 54° 30' 
 
 35° 40' 
 
 •7657 
 
 .9098 
 
 •8559 
 
 
 .1441 
 
 54° 20- 
 
 35° 5^ 
 36° 0' 
 
 36° 10' 
 
 •7675 
 
 i-7 
 1.8 
 
 i-7 
 i-7 
 i-7 
 
 .9089 
 
 •9 
 
 ■9 
 
 1.0 
 
 •9 
 
 .8586 
 9.8613 
 
 2.7 
 
 2.7 
 
 2.6 
 
 .1414 
 
 54° 10' 
 54° 0' 
 
 53° 5°' 
 
 9.7692 
 
 9.9080 
 
 0.1387 
 
 .7710 
 
 .9070 
 
 .8639 
 
 .1361 
 
 36° 20' 
 
 .7727 
 
 .9061 
 
 .8666 
 
 2.7 
 
 2.6 
 2.6 
 
 •1334 
 
 53° 40' 
 
 36° 30- 
 
 •7744 
 
 .9052 
 
 •9 
 
 .8692 
 
 .1308 
 
 53° 3°' 
 
 36° 40' 
 
 .7761 
 
 .9042 
 
 
 .8718 
 
 .1282 
 
 53° 20- 
 
 36° 5c/ 
 
 37° 0' 
 
 37° iC 
 
 .7778 
 
 '•7 
 i-7 
 1.6 
 
 i-7 
 1 6 
 
 •9033 
 
 ■9 
 1.0 
 
 •9 
 
 .8745 
 
 2-7 
 2.6 
 
 2.6 
 
 •1255 
 
 53° 10' 
 53° 
 
 52° 50- 
 
 9-7795 
 .7811 
 
 9.9023 
 
 9.8771 
 
 0.1229 
 
 .9014 
 
 .8797 
 
 .1203 
 
 37° 20' 
 
 .7828 
 
 .9004 
 
 
 .8824 
 
 2.7 
 
 2.6 
 
 .1176 
 
 52° 40' 
 
 37° 3°' 
 
 ■7844 
 
 
 •8995 
 
 •9 
 
 .8850 
 
 .1150 
 
 52° 3C 
 
 
 Cob. 
 
 D.l'. 
 
 Sin. 
 
 D. 1'. 
 
 Cot. 
 
 D. 1'. 
 
 Tan. 
 
 Angle. 
 
TANGENTS AND COTANGENTS. 
 
 19 
 
 Angle. 
 
 37° 30* 
 37° 4o' 
 37° 5°' 
 38° 0' 
 38 10' 
 38° id 
 
 38° 30' 
 38 40' 
 38 5c* 
 39° O 
 
 39° 10' 
 39 20* 
 
 39° 3°' 
 39° 40' 
 39° 5°' 
 
 40 <y 
 
 40° 10' 
 
 40° 2d 
 40 30' 
 40 40' 
 40 50' 
 41° O' 
 41 10' 
 
 41° 2C/ 
 4 I° 30' 
 41° 40' 
 4 I° 5 d 
 
 42° 0' 
 42 id 
 42 2d 
 42 30' 
 42 40' 
 42 50' 
 43° O' 
 43° 10' 
 43° 2d 
 43° 3°' 
 43° 4°> 
 43° S^ 
 44° O' 
 
 44° 10' 
 44 20' 
 44° 30 7 
 44° 40' 
 44° 50' 
 45° O' 
 
 Sin. 
 
 9.7844 
 .7861 
 •7877 
 
 9-7893 
 
 .79IO 
 .7926 
 •7941 
 
 •7957 
 ■7973 
 
 9.7989 
 
 D.l'. 
 
 .8004 
 .8020 
 .8035 
 .8050 
 .8066 
 
 9.8081 
 
 .8096 
 .8111 
 .8125 
 .8140 
 •8155 
 
 9.8169 
 
 .8184 
 .8198 
 .8213 
 .8227 
 .8241 
 
 9.8255 
 
 .8269 
 .8283 
 .8297 
 .8311 
 •8324 
 
 ^338 
 
 •8351 
 .8365 
 .8378 
 •8391 
 ■8405 
 
 •8431 
 .8444 
 
 ■8457 
 .8469 
 .8482 
 
 9.8495 
 
 Cos. 
 
 i-7 
 1.6 
 1.6 
 
 i-7 
 1.6 
 
 i-5 
 1.6 
 1.6 
 1.6 
 
 '■5 
 i.6 
 1 -5 
 1-5 
 1.6 
 
 i-5 
 i-5 
 
 '•5 
 
 1.4 
 
 '•5 
 '•5 
 i-4 
 
 i-5 
 i-4 
 "•5 
 
 1-4 
 1.4 
 1.4 
 
 1-4 
 
 1-4 
 
 1.4 
 
 1-4 
 i-3 
 
 1-4 
 
 i-3 
 
 1.4 
 i-3 
 i-3 
 
 1.4 
 
 i-3 
 1-3 
 i-3 
 i-3 
 1.2 
 
 i-3 
 i-3 
 
 Cos. 
 
 D.l'. 
 
 9.8995 
 .8985 
 •8975 
 
 9.8965 
 
 ■8955 
 .8945 
 
 ■8935 
 .8925 
 
 .8915 
 
 9.8905 
 
 .8895 
 
 .8874 
 .8864 
 .8853 
 
 9.8843 
 
 .8832 
 .8821 
 .8810 
 .8800 
 ■8789 
 
 D.l'. 
 
 9.8778 
 
 .8767 
 .8756 
 ■8745 
 •8733 
 .8722 
 
 9I7T1 
 
 .8699 
 
 .8676 
 .8665 
 .8653 
 
 9.8641 
 
 9.8495 
 
 Sin. 
 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.O 
 I.I 
 I.O 
 
 I.O 
 
 I.I 
 
 I.O 
 
 I.I 
 I.I 
 I.I 
 
 I.O 
 
 I.I 
 I.I 
 I.I 
 I.I 
 I.I 
 
 1.2 
 
 I.I 
 I.I 
 
 1.2 
 I.I 
 1.2 
 I.I 
 1.2 
 1.2 
 1.2 
 I.I 
 1.2 
 1.2 
 1.2 
 1-3 
 1.2 
 1.2 
 
 1-3 
 1.2 
 
 '■3 
 
 1.2 
 
 Tan. D. 1'. 
 
 9.8850 
 .8876 
 .8902 
 
 9.8928 
 
 .8954 
 .8980 
 .9006 
 .9032 
 .9058 
 
 9.9084 
 
 .9110 
 
 ■9135 
 .9161 
 .9187 
 .9212 
 
 9.9238 
 
 .9264 
 .9289 
 
 •931S 
 ■9341 
 .9366 
 
 9-9392 
 
 •9417 
 •9443 
 .9468 
 
 •9494 
 ■95'9 
 
 9-9544 
 
 .9570 
 
 •9595 
 .9621 
 .9646 
 .9671 
 
 9.9697 
 
 .9722 
 
 •9747 
 .9772 
 .9798 
 •9823 
 9.9848 
 
 •9874 
 .9899 
 .9924 
 •9949 
 ■9975 
 
 0.0000 
 
 D. V. Cot. 
 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 2.6 
 
 2-5 
 2.6 
 2.6 
 
 2-5 
 2.6 
 
 2.6 
 
 2-5 
 2.6 
 2.6 
 
 2.5 
 
 2.6 
 
 2-5 
 
 2.6 
 
 2-5 
 2.6 
 
 2-5 
 
 2.5 
 
 2.6 
 
 2-5 
 2.6 
 
 2-5 
 2-5 
 2.6 
 
 2-5 
 2-5 
 
 2.5 
 
 2.6 
 2-5 
 2-5 
 2.6 
 
 2-5 
 2-5 
 2-5 
 2.6 
 
 2-5 
 
 Cot. 
 
 0.1150 
 
 .1124 
 .1098 
 
 0.0916 
 
 .0890 
 .0865 
 .0839 
 .0813 
 .0788 
 
 0.0762 
 
 .0583 
 
 •°557 
 
 .0532 
 
 .0506 
 .0481 
 
 0.0456 
 
 .0430 
 .0405 
 •0379 
 •0354 
 .0329 
 
 0.0303 
 
 .0278 
 .0253 
 
 .0228 
 .0202 
 
 .0177 
 
 0.0000 
 
 52 30' 
 5 2 20' 
 52 id 
 52° 0' 
 Si^o* 
 5i°4o' 
 51° 30- 
 
 5I°20' 
 
 51° 10' 
 
 51° O' 
 
 50 50' 
 50 40' 
 5o°3o' 
 
 50°2O / 
 
 50 10' 
 50° 0' 
 
 D. 1'. Tan. Angle 
 
 49° 5°' 
 49° 40' 
 49° 30' 
 49 2d 
 49 id 
 49° 0' 
 48° 50' 
 48 40' 
 48 30' 
 48 20' 
 48 10' 
 48° 0' 
 
 47° 5°; 
 
 47° 40 7 
 47° 3o' 
 47 20' 
 47° io' 
 47° 0' 
 4 6°5o' 
 46 40' 
 4 6°3o' 
 46°20 / 
 46 10' 
 46° 0' 
 
 K 5< i 
 
 45° 40' 
 45° 3o' 
 45° 2d 
 45° 10' 
 45° 0' 
 
20 
 
 NATURAL SINES AND COSINES. 
 
 A. 
 
 Sin. 
 
 Cos. 
 
 
 A. 
 
 Sin. 
 
 Cos. 
 
 
 A. 
 
 Sin. 
 
 Cos. 
 
 
 0° 
 
 10' 
 
 20' 
 30' 
 40' 
 50' 
 1° 
 10' 
 20' 
 30' 
 
 4 d 
 5°' 
 
 2° 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 3° 
 
 10' 
 20' 
 30' 
 40' 
 So' 
 4° 
 10' 
 20' 
 30' 
 40' 
 5°' 
 5° 
 id 
 20' 
 30' 
 40' 
 5°' 
 6° 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 7° 
 10' 
 20' 
 30' 
 
 .000000 
 
 1. 0000 
 
 90° 
 
 5°; 
 40' 
 
 30' 
 20' 
 10' 
 
 89° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 88° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 87° 
 
 50; 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 86° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 
 85° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 84° 
 50; 
 40' 
 30' 
 20' 
 10' 
 83° 
 
 50' 
 40' 
 30' 
 
 30' 
 40' 
 SO' 
 
 8° 
 10' 
 20' 
 30' 
 40' 
 50' 
 9° 
 10' 
 20' 
 30' 
 40' 
 
 50' 
 10° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 11° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 12° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 13° 
 
 10' 
 20' 
 30' 
 
 50' 
 
 14° 
 
 10' 
 2d 
 30' 
 40' 
 50'' 
 15° 
 
 •I305 
 ■1334 
 •1363 
 
 •9914 
 .9911 
 .9907 
 
 30' 
 20' 
 10' 
 82° 
 
 5°; 
 40' 
 30; 
 20' 
 10' 
 
 81° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 80° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 
 79° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 78° 
 
 5o' 
 40' 
 30' 
 20' 
 10' 
 77° 
 
 s °; 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 76° 
 
 5°; 
 40' 
 
 30' 
 20' 
 10' 
 
 75° 
 
 15° 
 
 IO' 
 
 2d 
 30' 
 
 40' 
 & 
 
 16° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 17° 
 
 10' 
 
 20' 
 
 id 
 
 ^d 
 50' 
 
 18° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 19° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 5°' 
 
 20° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 5°' 
 
 21° 
 
 10' 
 
 20' 
 
 yd 
 
 40' 
 
 50' 
 
 22° 
 id 
 20' 
 30' 
 
 .2588 
 
 ■9659 
 
 75° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 74° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 73° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 72° 
 50' 
 40' 
 30' 
 20' 
 10' 
 71° 
 
 4 d 
 30' 
 20' 
 10' 
 
 70° 
 
 40' 
 30' 
 20' 
 10' 
 
 69° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 
 68° 
 
 50; 
 40' 
 30' 
 
 .002909 
 .005818 
 .008727 
 .011635 
 .014544 
 
 1. 0000 
 
 1 .0000 
 
 I. OOOO 
 
 •9999 
 
 ■9999 
 
 .2616 
 .2644 
 .2672 
 .2700 
 .2728 
 
 •9652 
 .9644 
 .9636 
 .9628 
 .9621 
 
 •1392 
 
 •9903 
 
 .1421 
 .1449 
 .1478 
 .1507 
 •1536 
 
 .9899 
 .9894 
 .9890 
 .9886 
 .9881 
 
 .017452 
 
 .9998 
 
 .2756 
 
 .9613 
 
 .02036 
 .02327 
 .02618 
 .02908 
 .03199 
 
 .9998 
 ■9997 
 •9997 
 .9996 
 
 •9995 
 
 .2784 
 .2812 
 .2840 
 .2868 
 .2896 
 
 ■2924 
 
 .9605 
 .9596 
 .9588 
 .9580 
 ■9572 
 •9563 
 
 .1564 
 
 .9877 
 
 ■1593 
 .1622 
 .1650 
 .1679 
 .1708 
 
 .9872 
 .9868 
 .9863 
 .9858 
 •9853 
 
 .03490 
 
 ■9994 
 
 .03781 
 .04071 
 .04362 
 .04653 
 •04943 
 
 •9993 
 .9992 
 
 •9990 
 .9989 
 .9988 
 
 .2952 
 
 ■2979 
 .3007 
 
 ■3035 
 .3062 
 
 •9555 
 .9546 
 
 •9537 
 •9528 
 .9520 
 
 •1736 
 
 .9848 
 
 •1765 
 .1794 
 .1822 
 .1851 
 .1880 
 
 •9843 
 .9838 
 
 •9833 
 .9827 
 .9822 
 
 .05234 
 
 .9986 
 
 .3090 
 
 •95" 
 
 •055 2 4 
 .05814 
 .06105 
 
 •0639S 
 .06685 
 
 .9985 
 
 •9983 
 .9981 
 .9980 
 .9978 
 
 ■3"8 
 ■3145 
 •3173 
 .3201 
 .3228 
 
 •9502 
 ■9492 
 ■9483 
 •9474 
 •9465 
 
 .1908 
 
 .9816 
 
 •1937 
 .1965 
 .1994 
 .2022 
 .2051 
 
 .9811 
 •9805 
 ■9799 
 ■9793 
 •9787 
 
 .06976 
 
 .9976 
 
 •3256 
 
 •9455 
 
 .07266 
 •07S56 
 .07846 
 .08136 
 .08426 
 
 •9974 
 .9971 
 .9969 
 .9967 
 .9964 
 
 •3283 
 •331 1 
 •3338 
 •3365 
 3393 
 
 .9446 
 ■9436 
 .9426 
 
 •9417 
 .9407 
 
 .2079 
 
 .9781 
 
 .2108 
 .2136 
 .2164 
 .2193 
 .2221 
 
 •9775 
 •9769 
 ■9763 
 •9757 
 •975° 
 
 .08716 
 
 .9962 
 
 3420 
 
 •9397 
 
 .09005 
 
 .09295 
 
 .09585 
 
 .09874. 
 
 .10164 
 
 •9959 
 •9957 
 •9954 
 •9951 
 .9948 
 
 3448 
 
 3475 
 3502 
 
 3529 
 
 3557 
 
 •9387 
 •9377 
 •9367 
 ■9356 
 •9346 
 
 .2250 
 
 ■9744 
 
 .2278 
 .2306 
 
 •2334 
 ■2363 
 
 •2391 
 
 •9737 
 •9730 
 .9724 
 .9717 
 .9710 
 
 ■10453 
 
 ■9945 
 
 3584 
 
 •9336 
 
 .10742 
 .11031 
 .11320 
 .11609 
 .11898 ' 
 
 .9942 
 ■9939 
 ■9936 
 •993 2 
 .9929 
 
 361 1 
 
 363.8 
 3665 
 3692 
 37 '9 
 
 •9325 
 ■9315 
 ■9304 
 •9293 
 9283 
 
 .2419 
 
 •9703 
 
 •2447 
 .2476 
 .2504 
 .2532 
 .2560 
 
 .9696 
 .9689 
 .9681 
 .9674 
 .9667 
 
 .12187 
 
 •9925 
 
 3746 
 
 .9272 
 
 .12476 
 .12764 
 •13053 
 
 .9922 
 .9918 
 .9914 
 
 3773 
 •3800 
 •3827 
 
 .9261 
 •9250 
 •9239 
 
 .2588 
 
 ■9659 
 
 
 Cos. 
 
 gin. 
 
 A. 
 
 
 Cos. 
 
 Sin. 
 
 A. 
 
 
 Cos. 
 
 Sin. 
 
 A. 
 
NATURAL SINES AND COSINES. 
 
 21 
 
 Sin. 
 
 Cos. 
 
 A. Sin. Cos. 
 
 A. Sin. Cos 
 
 30' 
 40' 
 50' 
 
 23° 
 
 .3827 
 
 •3854 
 .3881 
 
 ■3907 
 
 30' 
 40' 
 5°' 
 24° 
 10' 
 20' 
 ■50' 
 4C 
 50' 
 
 25° 
 a' 
 20' 
 30' 
 40' 
 50' 
 26° 
 
 30' 
 
 40' 
 
 5°' 
 
 27° 
 
 10' 
 
 20' 
 
 •jo' 
 
 40' 
 
 5°' 
 
 28° 
 
 id 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 29° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 5°' 
 
 30° 
 
 •3934 
 .3961 
 
 •3987 
 .4014 
 .4041 
 
 ■4067 
 
 .4094 
 .4120 
 ■4H7 
 ■4173 
 .4200 
 
 .4226 
 
 •4253 
 .4279 
 
 •43°5 
 •4331 
 •4358 
 
 •4384 
 
 .4410 
 
 ■4436 
 .4462 
 .4488 
 
 •45 1 4 
 
 ■454° 
 .4566 
 .4592 
 .4617 
 
 ■4643 
 .4669 
 
 •4695 
 
 .4720 
 .4746 
 •4772 
 ■4797 
 .4823 
 
 .4848 
 
 .4874 
 .4899 
 ■4924 
 •495° 
 ■4975 
 .5000 
 
 Cos. 
 
 9239 
 9228 
 9216 
 
 9205 
 
 9194 
 9182 
 9171 
 9159 
 9H7 
 
 9135 
 
 9124 
 9112 
 9100 
 9088 
 9075 
 
 9063 
 
 9051 
 9038 
 9026 
 9013 
 9001 
 
 8975 
 8962 
 
 8949 
 8936 
 
 8923 
 
 8910 
 
 8897 
 
 8870 
 8857 
 8843 
 
 8829 
 
 8816 
 8802 
 8788 
 
 8774 
 8760 
 
 8746 
 
 8732 
 8718 
 8704 
 
 8675 
 8660 
 
 Sin. 
 
 30' 
 20' 
 10' 
 67° 
 
 40' 
 30' 
 20' 
 10' 
 66° 
 
 40' 
 30' 
 20' 
 10' 
 65° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 64 c 
 
 50; 
 
 40' 
 
 3°' 
 20' 
 10' 
 63 c 
 5C 
 40' 
 30' 
 20' 
 10' 
 62° 
 5°' 
 4°' 
 30' 
 20' 
 10' 
 61° 
 50; 
 40' 
 30' 
 20' 
 10' 
 60° 
 
 30° 
 
 10' 
 20' 
 30' 
 40' 
 5°' 
 31° 
 10' 
 20' 
 30' 
 40' 
 5°' 
 32° 
 10' 
 20' 
 30' 
 40' 
 S<* 
 33° 
 10' 
 20' 
 
 3C 
 40' 
 
 5°' 
 34° 
 
 10' 
 20' 
 30' 
 40' 
 
 5^ 
 35° 
 
 10' 
 20' 
 30' 
 
 4 d 
 5°' 
 
 36° 
 
 10' 
 
 2& 
 30' 
 40' 
 50' 
 
 37° 
 10' 
 20' 
 
 3°' 
 
 5 02 5 
 5°5° 
 5075 
 5100 
 5125 
 
 5150 
 
 5'75 
 5200 
 5225 
 5250 
 5275 
 
 5000 
 
 8660 
 
 8646 
 8631 
 8616 
 8601 
 8587 
 
 5299 
 
 5324 
 5348 
 5373 
 5398 
 5422 
 
 5446 
 
 547' 
 5495 
 5519 
 5544 
 5568 
 
 5592 
 
 5616 
 5640 
 5664 
 5688 
 5712 
 
 5736 
 
 5760 
 
 5783 
 5807 
 
 5831 
 5854 
 
 5878 
 
 59oi 
 5925 
 5948 
 5972 
 5995 
 
 6018 
 
 6041 
 6065 
 6088 
 
 8572 
 
 8557 
 8542 
 8526 
 8511 
 8496 
 
 8480 
 
 8465 
 8450 
 
 8434 
 8418 
 8403 
 
 8387 
 
 8371 
 8355 
 8339 
 8323 
 8307 
 
 8290 
 
 8274 
 8258 
 8241 
 8225 
 8208 
 
 8192 
 
 8i75 
 8158 
 8141 
 8124 
 8107 
 
 8090 
 
 8073 
 8056 
 8039 
 8021 
 8004 
 
 7986 
 
 7969 
 795i 
 7934 
 
 Cos. Sin. 
 
 60° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 59° 
 50; 
 40' 
 30' 
 20' 
 10' 
 
 58° 
 
 so; 
 40 
 30' 
 20' 
 10' 
 
 57° 
 
 50; 
 40' 
 30* 
 20' 
 10' 
 56° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 55° 
 50' 
 40' 
 30' 
 20' 
 10' 
 54° 
 
 5 °; 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 53° 
 
 so; 
 40' 
 30' 
 
 A. 
 
 40' 
 
 5°' 
 38° 
 10' 
 20* 
 30' 
 40' 
 50' 
 
 39° 
 
 10' 
 20' 
 30' 
 40' 
 5°' 
 40° 
 10' 
 20' 
 30' 
 40' 
 5°' 
 41° 
 10' 
 20' 
 30' 
 40' 
 50' 
 42° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 5°' 
 
 43° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 S o' 
 
 44c 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 45° 
 
 .6088 
 .61 1 1 
 6134 
 
 ,6157 
 
 ,6180 
 .6202 
 6225 
 .6248 
 ,6271 
 
 .6293 
 
 ,6316 
 
 ■6338 
 ,6361 
 
 6383 
 6406 
 
 .6428 
 
 6450 
 .6472 
 .6494 
 .6517 
 ■6539 
 
 .6561 
 
 ■6583 
 .6604 
 .6626 
 .6648 
 .6670 
 
 .6691 
 
 6841 
 ,6862 
 6884 
 
 6905 
 .6926 
 
 6947 
 
 .6967 
 
 7009 
 7030 
 7°5° 
 7071 
 
 •7934 
 .7916 
 
 ■7898- 
 
 .7880 
 
 .7862 
 
 ■7844 
 .7826 
 .7808 
 •7790 
 
 .7771 
 
 •7753 
 ■7735 
 .7716 
 .7698 
 .7679 
 
 .7660 
 
 .7642 
 .7623 
 .7604 
 
 •7585 
 •7566 
 
 ■7547 
 
 •7528 
 •75°9 
 .7490 
 .7470 
 
 •745 1 
 
 •7431 
 
 .7412 
 •7392 
 •7373 
 ■7353 
 ■7333 
 
 •73>4 
 
 •7294 
 ■7274 
 •7254 
 •7234 
 .7214 
 
 •7'93 
 
 •7173 
 •7'53 
 •7'33 
 .7112 
 .7092 
 
 •7071 
 
 30' 
 
 20' 
 
 10' 
 
 52° 
 
 50' 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 51° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 50° 
 
 5^ 
 40' 
 30' 
 20' 
 10' 
 49° 
 
 5C 
 40' 
 
 3°' 
 20' 
 10' 
 
 48° 
 
 40' 
 30' 
 20' 
 10' 
 
 47° 
 
 4°' 
 30' 
 20' 
 id 
 46° 
 
 5o' 
 4d 
 30' 
 20' 
 10' 
 45° 
 
 Cos. Sin. 
 
22 
 
 NATURAL TANGENTS AND COTANGENTS. 
 
 A. 
 
 Tim. 
 
 Cot. 
 
 
 A. 
 
 Tan. 
 
 Cot. 
 
 
 A. 
 
 Tan. 
 
 Cot. 1 
 
 0° 
 
 10' 
 20' 
 30' 
 40' 
 5<* 
 1° 
 10' 
 20' 
 30' 
 40' 
 50' 
 2° 
 10' 
 20' 
 30' 
 40' 
 50' 
 3° 
 10' 
 20' 
 30' 
 40' 
 
 5°' 
 
 40 
 
 10' 
 20' 
 30' 
 
 4°' 
 50' 
 
 5° 
 10' 
 20' 
 30' 
 40' 
 50' 
 6° 
 10' 
 20' 
 30' 
 40' 
 5°' 
 7° 
 10' 
 20' 
 30' 
 
 .OOOOOO 
 
 00 
 
 90° 
 
 40' 
 30' 
 20' 
 10' 
 89° 
 
 40' 
 30' 
 20' 
 10' 
 88° 
 50; 
 40 
 30' 
 20' 
 10' 
 
 87° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 86° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 
 85° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 84° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 83° 
 
 5o' 
 40' 
 30' 
 
 30' 
 40' 
 
 50' 
 8° 
 10' 
 20' 
 30' 
 40' 
 50' 
 9° 
 10' 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 10° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 50' 
 
 11° 
 
 10' 
 
 20' 
 
 30 7 
 
 40' 
 
 5°' 
 
 12° 
 
 10' 
 20' 
 
 30' 
 40' 
 
 SO' 
 
 13° 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 14° 
 
 IO 1 
 
 2d 
 yd 
 40' 
 5°' 
 15° 
 
 •1317 
 .1346 
 
 •■376 
 
 7-5958 
 7.4287 
 7.2687 
 
 30' 
 20' 
 10' 
 82° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 81° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 
 80° 
 
 5o' 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 79° 
 
 50; 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 78° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 77° 
 50' 
 40' 
 30' 
 20' 
 10' 
 76° 
 
 s°; 
 40' 
 
 30' 
 20' 
 10' 
 
 75° 
 
 15° 
 
 10' 
 20/ 
 30' 
 40' 
 5o' 
 16° 
 10' 
 20' 
 30' 
 40' 
 50' 
 17° 
 10' 
 20' 
 30' 
 40' 
 5o' 
 18° 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 19° 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 20° 
 10' 
 20' 
 30' 
 40' 
 
 5°' 
 21° 
 
 10' 
 
 20' 
 
 30/ 
 
 40' 
 
 50' 
 
 22° 
 
 10' 
 
 20' 
 
 30' 
 
 .2679 
 
 3-7321 
 
 75° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 74 c 
 
 50' 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 73° 
 
 50' 
 
 40' 
 
 30' 
 
 20' 
 
 10' 
 
 72° 
 
 S°> 
 40' 
 
 30' 
 20' 
 10' 
 
 71° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 70° 
 
 50' 
 40- 
 30' 
 20 1 
 10' 
 69° 
 
 5o' 
 40' 
 3°' 
 20' 
 10' 
 68° 
 
 5°' 
 40' 
 
 30' 
 
 .OO2909 
 .005818 
 .008727 
 
 .on 636 
 
 .014545 
 
 343-7737 
 
 171.8854 
 
 114.5887 
 
 85.9398 
 
 68.7501 
 
 .27II 
 .2742 
 
 • 2 773 
 .2805 
 .2836 
 
 3.6891 
 
 3.647 
 3-6059 
 3-5656 
 3.5261 
 
 .1405 
 
 7-H54 
 
 ••435 
 .1465 
 
 •1495 
 .1524 
 
 • I 554 
 
 6.9682 
 6.8269 
 6.6912 
 6.5606 
 6.4348 
 
 ■OI745S 
 
 57.2900 
 
 .2867 
 
 34874 
 
 .02036 
 .02328 
 .02619 
 .O29IO 
 .O32OI 
 
 49.1039 
 42.9641 
 38.1885 
 34-3678 
 31.2416 
 
 .2899 
 .2931 
 .2962 
 .2994 
 .3026 
 
 3-4495 
 3-4124 
 3-3759 
 3-3402 
 3-3052 
 
 .1584 
 
 6.3138 
 
 .1614 
 .1644 
 •'673 
 •i7°3 
 •1733 
 
 6.1970 
 6.0844 
 
 5-975 8 
 5.8708 
 
 5-7694 
 
 .03492 
 
 28.6363 
 
 ■3057 
 
 3.2709 
 
 •037 8 3 
 .04075 
 .04366 
 .04658 
 .04949 
 
 26.4316 
 24.5418 
 22.9038 
 21.4704 
 20.2056 
 
 .3089 
 .3121 
 •3153 
 ■3185 
 •3217 
 
 3-.237 1 
 3.2041 
 3.1716 
 
 3-1397 
 3.1084 
 
 •1763 
 
 5-6713 
 
 •1793 
 .1823 
 
 •1853 
 .1883 
 .1914 
 
 5-5764 
 5-4845 
 5-3955 
 5 -3°93 
 5-2257 
 
 .05241 
 
 19.081 1 
 
 •3249 
 
 3-0777 
 
 •05533 
 ■05824 
 .06116 
 .06408 
 .067OO 
 
 18.0750 
 17.1693 
 16.3499 
 15.6048 
 14.9244 
 
 .3281 
 ■33H 
 ■3346 
 ■3378 
 ■341 1 
 
 3-0475 
 3.0178 
 2.9887 
 2.9600 
 2.9319 
 
 ■1944 
 
 5.1446 
 
 •1974 
 .2004 
 .2035 
 .2065 
 .2095 
 
 5.0658 
 4.9894 
 4.9152 
 4.8430 
 4.7729 
 
 .06993 
 
 14.3007 
 
 •3443 
 
 2.9042 
 
 .07285 
 
 •07578 
 .07870 
 .08163 
 .08456 
 
 13.7267 
 13.1969 
 12.7062 
 12.2505 
 11.8262 
 
 •3476 
 .3508 
 •3541 
 •3574 
 .3607 
 
 2.8770 
 2.8502 
 2.8239 
 2.7980 
 2.7725 
 
 .2126 
 
 4.7046 
 
 .2156 
 .2186 
 .2217 
 .2247 
 .2278 
 
 4.6382 
 
 4.5736 
 4.5107 
 4.4494 
 4-3897 
 
 .08749 
 
 11.4301 
 
 .3640 
 
 2 -7475 
 
 .09042 
 
 •09335 
 .09629 
 .09923 
 .IO216 
 
 11.0594 
 10.7119 
 10.3854 
 10.0780 
 9.7882 
 
 ■3673 
 .3706 
 
 •3739 
 •377 2 
 .3805 
 
 2.7228 
 2.6985 
 2.6746 
 2.65 1 1 
 2.6279 
 
 .2309 
 
 4.33I5 
 
 •2339 
 .2370 
 .2401 
 .2432 
 .2462 
 
 4-2747 
 4-2193 
 4-1653 
 4.1 126 
 4.061 1 
 
 .IO510 
 
 9-5!44 
 
 •3839 
 
 2.6051 
 
 .10805 
 .IIO99 
 
 •I 1394 
 .11688 
 .II983 
 
 9- 2 553 
 9.0098 
 8.7769 
 8-5555 
 8.3450 
 
 .3872 
 .3906 
 ■3939 
 •3973 
 .4006 
 
 2.5826 
 2.5605 
 2.5386 
 2.5172 
 2.4960 
 
 •2493 
 
 4.0108 
 
 .2524 
 
 •2555 
 ■2586 
 .2617 
 .2648 
 
 3-9617 
 3-9136 
 3.8667 
 3.8208 
 3.7760 
 
 .12278 
 
 8.1443 
 
 .4040 
 
 2.4751 
 
 •12574 
 .12869 
 .13165 
 
 7-9530 
 7.7704 
 7-5958 
 
 .4074 
 .4108 
 .4142 
 
 2-4545 
 2.4342 
 2.4142 
 
 .2679 
 
 3-7321 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 
NATURAL TANGENTS AND COTANGENTS. 
 
 23 
 
 A. 
 
 Tan. 
 
 Cot. 
 
 
 A. 
 
 Tan. 
 
 Cot. 
 
 
 A. 
 
 Tan. 
 
 Cot. 
 
 
 30' 
 40' 
 5°' 
 
 4142 
 4176 
 4210 
 
 2.4142 
 2-3945 
 2-375° 
 
 30' 
 20' 
 10' 
 67° 
 
 4°' 
 30' 
 20' 
 10' 
 66° 
 
 40' 
 30' 
 20' 
 10' 
 65° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 64° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 63° 
 5c/ 
 40' 
 30' 
 20' 
 10' 
 62° 
 50; 
 40' 
 30' 
 20' 
 10' 
 61° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 60° 
 
 30° 
 
 10' 
 20' 
 30' 
 40' 
 5°' 
 31° 
 10' 
 20' 
 30' 
 40' 
 50' 
 32° 
 10' 
 20' 
 30' 
 40' 
 
 s& 
 
 33° 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 34° 
 10' 
 20' 
 30' 
 40' 
 50' 
 35° 
 10' 
 20' 
 30' 
 40' 
 5°' 
 36° 
 10' 
 20' 
 3 of 
 40' 
 50' 
 37° 
 10' 
 20' 
 3° r 
 
 •5774 
 
 I.7321 
 
 60° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 59° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 58° 
 
 50; 
 40' 
 30' 
 20' 
 10' 
 
 57° 
 
 5°' 
 40' 
 3 o> 
 20' 
 10' 
 56° 
 50' 
 40' 
 30' 
 20' 
 10' 
 55° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 54° 
 
 5°' 
 40' 
 30' 
 20' 
 io 7 
 53° 
 50' 
 40' 
 30' 
 
 3 c/ 
 40' 
 5°' 
 38° 
 10' 
 20' 
 30' 
 40' 
 5* 
 39° 
 10' 
 20' 
 30' 
 40' 
 50' 
 40° 
 10' 
 20' 
 
 3°' 
 40' 
 
 5°' 
 41° 
 
 10' 
 20' 
 30' 
 40' 
 5°' 
 
 42° 
 10' 
 20' 
 30' 
 4°' 
 5°' 
 
 43° 
 
 10' 
 
 20' 
 
 30' 
 
 40' 
 
 5°' 
 
 44° 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 45° 
 
 •7673 
 .7720 
 .7766 
 
 1.3032 
 1.2954 
 1.2876 
 
 30' 
 20' 
 10' 
 
 52° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 51° 
 
 50' 
 40' 
 30' 
 20' 
 10' 
 50° 
 
 5^ 
 40' 
 30' 
 20' 
 10' 
 49° 
 
 5°; 
 40' 
 30' 
 20' 
 10' 
 
 48° 
 
 40' 
 30' 
 20' 
 10' 
 
 47° 
 
 40' 
 30' 
 20' 
 10' 
 46° 
 
 5°' 
 40' 
 30' 
 20' 
 10' 
 45° 
 
 .5812 
 
 •5851 
 .5890 
 
 •593° 
 ■5969 
 
 I.7205 
 I.7090 
 I.6977 
 I.6864 
 I-6753 
 
 23° 
 
 4245 
 
 2-3559 
 
 •78l3 
 
 1.2799 
 
 10' 
 20' 
 30' 
 
 40' 
 5° 
 
 4279 
 43H 
 4348 
 4383 
 4417 
 
 2.33 6 9 
 2-3183 
 2.2998 
 2.2817 
 2.2637 
 
 .7860 
 .7907 
 
 ■7954 
 .8002 
 •8050 
 
 I.2723 
 1.2647 
 1.2572 
 1.2497 
 1.2423 
 
 .6009 
 
 1.6643 
 
 .6048 
 .6088 
 .6128 
 .6168 
 .6208 
 
 1-6534 
 1 .6426 
 I.6319 
 1. 621 2 
 1. 6107 
 
 24° 
 
 •445 2 
 
 2.2460 
 
 .8098 
 
 1-2349 
 
 10' 
 20' 
 30' ■ 
 
 4 <y 
 
 50' 
 
 •4487 
 .4522 
 
 •4557 
 ■4592 
 4628 
 
 2.2286 
 2.2113 
 2.1943 
 2-1775 
 2.1609 
 
 .8146 
 
 .8195 
 .8243 
 .8292 
 .8342 
 
 1.2276 
 1.2203 
 1-2131 
 1.2059 
 1. 1988 
 
 .6249 
 
 I.6003 
 
 .6289 
 •633° 
 ■6371 
 .6412 
 
 •6453 
 
 I.59OO 
 I.5798 
 I-5697 
 
 !-5597 
 1-5497 
 
 25° 
 
 4663 
 
 2.1445 
 
 ■8391 
 
 1.1918 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 4699 
 
 4734 
 4770 
 4806 
 4841 
 
 2.1283 
 2.1 123 
 2.0965 
 2.0809 
 2.0655 
 
 .8441 
 .8491 
 .8541 
 .8591 
 .8642 
 
 1.1847 
 1.1778 
 1. 1708 
 1. 1640 
 1.1571 
 
 .6494 
 
 1-5399 
 
 .6536 
 
 ■6577 
 .6619 
 .6661 
 .6703 
 
 1.5301 
 1.5204 
 1.5108 
 1.5013 
 I-49I9 
 
 26° 
 
 4877 
 
 2.0503 
 
 •8693 
 
 1. 1504 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 4913 
 4950 
 4986 
 5022 
 5°59 
 
 2 -°353 
 2.0204 
 2.0057 
 1.9912 
 1.9768 
 
 •8744 
 .8796 
 .8847 
 .S899 
 .8952 
 
 1. 1436 
 1.1369 
 
 i-i3°3 
 1.1237 
 1.1171 
 
 •6745 
 
 1.4826 
 
 .6787 
 .6830 
 
 ■6873 
 .6916 
 .6959 
 
 1-4733 
 1-464? 
 
 i-455° 
 1.4460 
 
 i-437° 
 
 27° 
 
 5°95 
 
 1.9626 
 
 .9004 
 
 1.1106 
 
 10' 
 20 1 
 30' 
 40' 
 5°' 
 
 5'32 
 5169 
 5206 
 
 •5 2 43 
 5280 
 
 1 .9486 
 
 1-9347 
 1.9210 
 1.9074 
 1.8940 
 
 .9057 
 .9110 
 ■9163 
 .9217 
 .9271 
 
 1.1041 
 1.0977 
 
 i°9i3 
 1.0850 
 1.0786 
 
 .7002 
 
 1. 4281 
 
 .7046 
 .7089 
 
 •7'33 
 .7177 
 .7221 
 
 1 .41 93 
 1.4106 
 1.4019 
 
 1-3934 
 1.3848 
 
 28° 
 
 •5317 
 
 1.8807 
 
 •9325 
 
 1.0724 
 
 it/ 
 20' 
 30' 
 40' 
 50' 
 
 •5354 
 •5392 
 543° 
 .5467 
 
 5505 
 
 1.8676 
 1.8546 
 1.8418 
 1.8291 
 1.8165 
 
 .9380 
 
 •9435 
 •949° 
 
 •9545 
 .9601 
 
 1. 066 1 
 1.0599 
 1.0538 
 1.0477 
 1. 041 6 
 
 .7265 
 
 I-3764 
 
 •73i° 
 ■7355 
 .7400 
 
 •7445 
 •749° 
 
 1.3680 
 1-3597 
 I-35H 
 1-3432 
 I-335I 
 
 29 c 
 
 5543 
 
 1.8040 
 
 •9657 
 
 ^ss 
 
 10' 
 20' 
 30' 
 40' 
 50' 
 
 558i 
 5619 
 
 5658 
 5696 
 
 5735 
 
 1. 7917 
 1.7796 
 I-7675 
 1-7556 
 1-7437 
 
 •97'3 
 
 .9770 
 
 ■9827 
 .9884 
 .9942 
 
 1.0295 
 1.0235 
 1.0176 
 1.0117 
 1.0058 
 
 •7536 
 
 1.3270 
 
 •758i 
 .7627 
 
 •7673 
 
 1.3190 
 1.3m 
 1.3032 
 
 30° 
 
 5774 
 
 1.7321 
 
 1.0000 
 
 1. 0000 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 
 
 Cot. 
 
 Tan. 
 
 A. 
 

 2 
 

 1 
 
 : : llilLli'ii'l'i-i 1 
 
 
 "'" j;| 
 
 .