PLANE AND SPHERICAL TRIGONOMETRY BY GEORGE N. BAUER, PH.D. N AND W. E. BROOKE, C.E.,M.A. UNIVERSITY OF MINNESOTA SECOND REVISED EDITION D. C. HEATH & CO., PUBLISHERS BOSTON NEW YORK CHICAGO PHYSIOS COPYRIGHT, 1907 AND 1917, BY D. C. HEATH & Co. H7 PREFACE THE plan and scope of this work may be indicated briefly by the following characteristic features : Directed lines and Cartesian coordinates are introduced as a working basis. Each subject, when first introduced, is treated in a gen- eral manner and is presented as fully as the character of the work demands. This avoids the necessity of treating the same subjects several times for the purpose of modify- ing and extending certain conceptions. Statements in the form of theorems and problems are used freely to indicate the aim of various articles and to define the data clearly. Inverse functions are treated more fully than is customary. The general principles governing the solution of triangles, the solution of trigonometric equations, and the proof of identities are carefully presented. . Special attention is given to the arrangement of compu- tations. The sine and cosine series are obtained from De Moivre's Theorem, thus completing the line of development which leads to the calculation of the trigonometric functions. The work on spherical trigonometry contains the devel- opment of all the formulas that are generally used in practical astronomy. The right spherical triangle is treated from two points of view : as a special case of the oblique triangle, and directly iii iv PREFACE from geometric figures. The work is so arranged that either view may be presented independently. The solutions of the oblique spherical triangle by means of auxiliary quantities, characteristic of astronomy, are included as interesting mathematical problems and as preparation for astronomical work. PREFACE TO THE SECOND REVISED EDITION The second edition embodies such modifications, rear- rangements, and additions as have been suggested by experi- ence in the classroom. Due to numerous requests tables have been added. Simple three-place tables have been included. By their use the numerical work is reduced to a minimum, thus leaving the student free to give more attention to the principles in- volved in the solution of a triangle. Accordingly a few tri- angles, suitable for three-place table or slide rule computa- tion, have been introduced at the beginning of the list of problems under the right triangle and also under the oblique triangle. In order to give a wider range to computational work four- place tables as well as five-place tables have been included. G. N. B. W. E. B. MINNEAPOLIS, MINNESOTA, June, 1917. CONTENTS PLANE TRIGONOMETRY CHAPTER I RECTANGULAR COORDINATES AND ANGLES ABT. PAOB 1. Introduction 1 RECTANGULAR COORDINATES 2. Directed lines .1 3. Lines of reference 2 4. Quadrants 2 5. Coordinates of a point 2 6. Signs of coordinates . . 3 7. Exercises 3 ANGLES 8. Magnitude of angles 4 9. Direction of rotation. Positive and negative angles . . 4 10. Initial and terminal lines 6 11. Sign of terminal line 5 12. Algebraic sum of two angles 6 13. Measurement of angles 6 14. Circular or radian measure 7 15. Value of radian 7 16. Relation between degree and radian 7 17. Relation between angle, radius, and arc . . . .9 18. Linear and angular velocity 9 19. Examples 10 v CONTENTS CHAPTER II TRIGONOMETRIC FUNCTIONS ART. PAGE 20. Trigonometric functions introduced 12 21. Definitions of the trigonometric functions .... 12 22. Values of the trigonometric functions of 30, 45, and 60 . 13 23. Values of the trigonometric functions of 120, 135, and 150 15 24. Signs of the trigonometric ratios 17 25. Trigonometric functions are single valued .... 18 26. A given value of a trigonometric function determines an infinite number of angles 19 27. Examples . 20 CHAPTER III RIGHT TRIANGLES 28. Statement of problem 22 29. Application of the definitions of the trigonometric functions to the right triangle 22 30. Trigonometric tables . . 23 31. Formulas used in the solution of right triangles ... 24 32. Selection of formulas 25 33. Check formulas 25 34. Suggestions on solving a triangle 25 35. Illustrative examples 26 36. Examples .......... 29 37 . Oblique triangles 30 38. Applications 31 CHAPTER, IV VARIATIONS OF THE TRIGONOMETRIC FUNCTIONS REDUCTION OF FUNCTIONS OF n90 39. Values of functions at 0, 90, 180, 270, and 360 . . 34 40. Variation of the functions 35 41. Graphical representation 39 42. Periodicity of the trigonometric functions .... 40 43. Examples 41 CONTENTS vii ABT PAGE 44. Use of formulas 42 45. Functions of a in terms of functions of a . . . .42 46. Functions of 90 -f a in terms of functions of a . .43 47. Functions of 90 a in terms of functions of a . .44 48. Functions of 180 a in terms of functions of a . . .45 49. Laws of reduction . . .46 50. Examples 46 CHAPTER V FUNDAMENTAL KELATIONS. LINE VALUES 51. General statement 47 FUNDAMENTAL RELATIONS 52. Development of formulas 47 53. The use of exponents 48 54. Trigonometric identities 49 55. Trigonometric equations 49 56. Examples .51 LINE VALUES 57. Representation of the trigonometric functions by lines . 53 58. Variations of the trigonometric functions as shown by line values 55 59. Fundamental relations by line values 66 60. Examples 57 CHAPTER VI FUNCTIONS OF THE SUM OF TWO ANGLES DOUBLE ANGLES. HALF ANGLES 61. Statement of problem 61 62. The sine of the sum of two acute angles expressed in terms of the sines and cosines of the angles .... 61 63. The cosine of the sum of two acute angles .... 62 64. Importance of formulas ........ 62 65. Generalization of formulas 63 66. Tangent of the sum of two angles 65 67. Cotangent of the sum of two angles 65 68. Addition formulas . 65 viii CONTENTS ART. PAGB 69. Sine, cosine, tangent, and cotangent of the difference of two angles 66 70. Exercises .66 71. Double angles 67 72. Half angles 68 73. Sum and difference of two sines and of two cosines . . 69 74. Equations and identities 70 75. Examples . .71 CHAPTER VII INVERSE FUNCTIONS 76. Statement of problem 74 77. Fundamental idea of an inverse function .... 74 78. Multiple values of an inverse function . . . 75 79. Principal values .77 80. Interpretation of sin sin- 1 a and sin- 1 sin a . . . .77 81. Application of the fundamental relations to angles expressed as inverse functions 78 82. The value of any function of an inverse function ... 79 83. Some inverse functions expressed in terms of other inverse functions .81 84. Relations between inverse functions derived from the formulas for double angles, half angles, and the addition formulas 82 86. Examples 82 CHAPTER VIII OBLIQUE TRIANGLE 86. General statement 84 87. Law of sines 84 88. Law of tangents 86 89. Cyclic interchange of letters 85 90. Law of cosines 85 91. Sine of half angle in terms of sides of triangle ... 86 92. Cosine of half angle in terms of sides of triangle ... 87 93. Tangent of half angle in terms of sides 88 94. Area of plane triangle in terms of two sides and included angle 89 95. Area of triangle in terms of a side and two adjacent angles . 89 CONTENTS 96. Area in terms of sides 90 97. Formulas for solving an oblique triangle .... 90 98. Check formulas 91 99. Illustrative problems 92 100. The ambiguous case 95 101. Examples 98 102. MISCELLANEOUS EXERCISES 102 CHAPTER IX DE MOIVRE'S THEOREM WITH APPLICATIONS 103. Introduction to chapter 113 104. Geometric representation of a complex number . . . 113 105. De Moivre's theorem 114 106. Geometric interpretation 115 107. Applications of De Moivre's theorem 116 108. Cube roots of unity 116 109. Fifth roots of unity . . . ... . .117 110. Square root of a complex number 118 111. Any root of a complex number ...... 119 112. Sin n a and cos n a. expressed in terms of sin a and cos a . 120 113. Comparison of the values of sin a, a, and tan a, a being any acute angle 121 114. Value of sm a for small values of a 121 a 115. Sine and cosine series 122 116. Examples 124 SPHERICAL TRIGONOMETRY CHAPTER X FUNDAMENTAL FORMULAS 117. The spherical triangle 127 118. Law of sines 128 119. Law of cosines 129 120. Law of cosines extended 130 121. Relation between one side and three angles . . . 131 122. The sine-cosine law 131 CONTENTS 123. Relation between two sides and three angles . . . 132 124. Relation between two sides and two angles . . . .132 126. Formulas independent of the radius of the sphere . . 133 CHAPTER XI SPHERICAL RIGHT TRIANGLE 126. Definition of spherical right triangle 134 127. Formulas for the solution of right triangles .... 134 128. Direct geometric derivation of formulas .... 134 129. Sufficiency of formulas 137 130. Comparison of formulas of plane and spherical right triangles 137 131. Napier's rules 138 132. Side and angle opposite terminate in same quadrant . . 139 133. Two sides determine quadrant of a third .... 139 134. Two parts determine a triangle 139 135. The quadrant of any required part 140 136. Check formula 140 137. Solution of a right triangle 141 138. Two solutions 141 139. Examples 143 140. Quadrantal triangles 143 141. Isosceles triangles 143 142. Examples 144 CHAPTER XII OBLIQUE SPHERICAL TRIANGLE 143. Statement of aim of chapter 145 GENERAL SOLUTION 144. Angles determined from the three sides .... 145 145. Sides found from the three angles 146 146. Delambre's or Gauss's formulas . . . . . .147 147. Napier's analogies 147 148. Formulas collected . . 148 149. All formulas excepting law of sines determine quadrant . 148 CONTENTS xi ART. *AGE 150. Theorem to determine quadrant ...... 149 151. Second theorem to determine quadrant .... 149 152. Third theorem to determine quadrant ..... 150 153. Illustrative examples ........ 150 154. Two solutions ......... 153 155. Area of spherical triangle ....... 154 156. Examples .......... 154 SOLUTION WHEN ONLY ONE PART is REQUIRED 157. Statement of problem ....... .156 158. From two sides and the included angle, to find any one of the remaining parts . . . . . ' . . 156 169. Two parts required ........ 158 160. Problems .......... 159 161. Each unknown part found from two angles and the included side .......... 160 162. Each unknown part found from two sides and an angle opposite one of them ....... 161 163. Each unknown part found from two angles and a side opposite one of them ....... 163 164. The general triangle ........ 164 ANSWERS ........... 165 PLANE TRIGONOMETRY CHAPTER I RECTANGULAR COORDINATES AND ANGLES 1. Introduction. The word trigonometry is derived from the two Greek words for triangle (r/otytuvov) and measure- ment (/Aerpta). Originally trigonometry was concerned chiefly with the solution of triangles. At present this is but one part of the subject. Certain preliminary considerations, concerning directed lines and angles, are neqessary before introducing the fun- damental definitions of trigonometry. RECTANGULAR COORDINATES 2. Directed lines. A positive and a negative direction may be assigned arbitrarily to every line. If the direction from A to C is positive, the opposite direction from C to A is negative. If we let the order of the let- ^ ters indicate the direction in A c which a segment is measured, it is evident that AC and CA represent the same segment measured in opposite direc- tions; hence AC = - CA or -AC = CA. Also, if B be a third point on the line, the segments AB and BA are opposite in sign; likewise the segments BO and CB. Hence AB=-BA or -AB = BA, and BC= - CB or - BC= CB. 1 2 TRIGONOMETRY Then for all positions of A, B, and C on a line it follows that Thus for the following figures : % K_i_f Ti P Q 5 8 12 8 + f 4^ 12 -12 A -8 C B 4-_6 + (4-l(r)-4 C - B 4 \ A X' 3. Lines of reference. Two directed lines, perpendicular to each other, may be taken as lines of reference or axes. They are usually designated by X'X and F'F and are called the X-axis and the F-axis respectively. The X-axis is positive from left to right and negative from right to left. The F-axis is positive upward _j_ and negative downward. X The point of intersection of the axes is the origin. The origin serves as a convenient starting point from which to measure dis- tances. 4. Quadrants. The axes produced divide the plane into four parts called quadrants. The quadrants are designated by number. The first quadrant is indicated by XOF, the second by FOX', the third by X'OF', and the fourth by F'OX. 5. Coordinates of a point. From any given point P x in the plane, draw a line parallel to the F-axis intersecting the X-axis in some point A. Then the distance from the origin to the point of inter- section, or OA, is the abscissa of the given point P. RECTANGULAR COORDINATES AND ANGLES 3 X'C The distance from the point of intersection to the given point, or AP lf is the ordinate of the given point P x . The abscissa and the ordinate of the point P 2 are OC and CP 2 respec- tively. The abscissa and the ordinate of the point P 3 are OC and CP 3 respectively. The abscissa and the ordinate of the point P 4 are OA and AP 4 respectively. 6. Signs of coordinates. Abscissas are positive when measured from the origin to the right, and negative when measured from the origin to the left. Ordinates are positive when measured upward from the .X-axis and negative when measured downward. Thus OA, APv and CP 2 are positive; 0(7, CP 3 , and AP 4 are negative. 7. EXERCISES 1. Locate the point whose abscissa is 4 and whose ordi- nate is 7. This point is designated (4, 7). 2. Locate the point whose abscissa is 2 and whose ordi- nate is 5, i.e. the point (2, 5). 3. Locate the points (- 3, 4), (- 6, - 3), (5, - 1), (0, 4), (- 7, 0), and (- 8, 10). 4. Locate the points (1J, 3), (- |, 0), (m, n), (a?, y), (a, 0). 5. What is the locus of the points whose abscissas are 6 ? 6. What is the locus of the points whose ordinates are 3 ? 7. What is the locus of the points whose abscissas are twice their ordinates ? TRIGONOMETRY ANGLES 8. Magnitude of angles. In elementary geometry the angles considered are usually less than two right angles ; but in trigonometry it is necessary to introduce angles of any magni- tude, positive or negative. To extend the conception of an angle, suppose a line to revolve in a fixed plane about a fixed point 0, from the initial position OX to the successive positions OY } OX', and OP, generating a*, an angle greater than two right angles. If the line continues to revolve, making more than one complete rev- olution, it generates an angle a which is greater than four right angles. Evidently by continuing the rotation an angle of any magnitude may be generated. Thus the size of the angle a depends upon the amount of rotation of OP and is designated by an arc. 9. Direction of rotation. Positive and negative angles. As a positive and negative direction may be assigned arbitra- rily to a line, so a positive and nega- tive sense of generation may be assigned arbitrarily to an angle. The direction of rotation indicated by the arrows in the figures of Art. 8 is the positive direction ; the opposite direction, indicated by the arrow in the adjoining figure, is the negative direction. * Angles will usually be designated by Greek letters : a ... Alpha 5 ... Delta /3 . . . Beta 6 . . . Theta 7 ... Gamma ... Phi RECTANGULAR COORDINATES AND ANGLES 5 A positive angle is an angle generated by a line rotating in the positive direction. A negative angle is an angle generated by a line rotating in the negative direction. Negative angles, like positive angles, may have unlimited magnitude. 10. Initial and terminal lines. The fixed line from which both positive and negative angles are measured is the initial line. It usually coincides with that part of the X-axis lying to the right of the origin. Thus, in the last three figures, OX is the initial line. The final position of the revolving line, marking the ter- mination of the angle, is the terminal line. Thus, in the last three figures, OP is the terminal line. Two or more unequal angles may P N have the same terminal line. Thus the positive angle a and the nega- tive angles ft and y have the same terminal line. Angles having the same terminal line, and the same initial line, are called coterminal angles. 11. Sign of terminal line. The terminal line, drawn from Q f the origin in the direction of the ex- / tremity of the measuring arc, is posi- / tive ; the terminal line produced, drawn from the origin in the oppo- X site direction, is negative. Thus the terminal line OP, of the , angle a, is positive, and the terminal line produced, OQ, is negative. As the terminal line OP revolves it retains its positive sign. 12. The algebraic sum of two angles. To construct the algebraic sum of two angles, conceive OP to rotate from the 6 TRIGONOMETRY position OX } through the angle a, to OP'; then from this P f X a positive R positive O X a positive R negative. a negative positive position let it rotate through the angle fi, whether positive or negative, to OP. Then XOP is the desired angle a + ft. 13. Measurement of angles. Various units may be em- ployed in the measurement of angles. In elementary geom- etry the right angle is frequently used. Two other units in common use are the degree and the radian. The degree is generally used in practical problems involving numerical computations, while the radian is essential in many theoret- ical considerations. The degree is defined as one ninetieth of a right angle. The degree is divided into sixty equal parts called min- utes. The minute is divided into sixty equal parts called seconds. Then 60 seconds (60") = 1 minute. 60 minutes (60') = 1 degree. 90 degrees (90) = 1 right angle. The angle 26 degrees, 39 minutes, and 57 seconds is written 26 39' 57". RECTANGULAR COORDINATES AND ANGLES 7 14. Circular or radian measure. A radian is an angle of such magnitude that, if placed with its vertex at the center of any circle, it will intercept an arc equal in length to the radius of the circle. Thus, if the arc XP is equal to the radius OX, the angle XOP is a radian, or Z. XOP = l r , r being used to designate radian. 15. Value of the radian. Since, in the same circle or in equal circles, angles at the center are proportional to their intercepted arcs, it follows that /.XOP ZXOY one radian or one right angle i (2 irr) 2 Therefore, one radian = one right angle. 7T or TT radians = 180. It is clear that the value of the radian is independent of the radius, depending only upon the constant TT and the right angle, and hence is an invariable unit. 16. Relation between degree and radian. From the pre- ceding article it follows that i-=^, a) ' Also, from equation (1), r jf_ (3) 180' or 1 = 0.01745''. (4) Equations (1) and (2) are used to convert radians into 8 TRIGONOMETRY degrees, and equations (3) and (4) are used to convert degrees into radians. Thus, fromd), o Y TT from (2), 4 r = 4 (57 17' 44") = 229 10' 56", fro m( 3), 2 0= from (4), 3 = 3 (0.01745') = 0.05235 r . Using equation (3) to convert degrees into radians intro- duces TT into the numerical value of the angle. Thus TT becomes associated with radian measure. Since the radian is the angular unit with which TT is commonly associated, no ambiguity arises by omitting to state with each angle, expressed in terms of TT, that the radian is the unit. Thus, 90 = - radians, 180 = TT radians, 29 = radians are usually written 90 = -, 180 = TT, 29 = 2 loO It must be especially noted that when no unit is specified the radian is always understood. The constant TT is always equal to 3.1416, and can never equal 180, but TT radians are equal to 180. 17. Relation between angle, radius, and arc. It is evident from the definition of a radian that if any arc of a circle AB be divided by ^^ the radius, the quotient indicates the number of radians contained in the central angle subtended by the given arc, hence radius where angle AOB is expressed in radians. Representing RECTANGULAR COORDINATES AND ANGLES 9 the length of the arc AB by a, the radius by r, and the angle AOB by 0, the relation above may be written or a=r9 (1) It should be especially noted that the angle is expressed in terms of radians. In the above figure 9 is approximately equal to 21 radians. Equation (1) expresses the relation between a, r, and 0, and determines any one of these elements when the other two are known. PROBLEM. What is the length of the arc subtended by a central angle of 112 in a circle whose radius is 15 feet ? Solution. Reducing the angle to radians it is seen that 112 =?^- r = 1.95'". 45 Hence = 1.95. Substituting in Equation (1) we have a = 15 x 1.95 = 29.3 Hence the length of the arc is 29.3 ft. 18. Linear and angular velocity. Equation (1) of Art. 17 leads directly to the relation between linear velocity and angular velocity in uniform motion in a circle. Suppose a point P moves along the circumference of a circle at a constant velocity v, describing an arc a in time t ; then a/t is called the linear velocity and is represented by v. During the same time t, the angle is generated ; and 6/t is called the angular velocity, and is represented by the Greek letter o> (omega). 10 TRIGONOMETRY Dividing Equation (1), Art. 17, by t gives = ^or^ = r> t t i.e. the linear velocity is equal to r times the angular velocity. While in general angular velocity may be expressed in any units whatsoever, in the equation v = rco the angular velocity must be expressed in radians per unit of time. PROBLEM 1. If a point moves 26 feet in the arc of a circle of radius 7 feet in 3 seconds, what is its angular velocity ? Solution. The linear velocity of the point is %- feet per second. Substituting in equation (1) we have = 7 a, .-. w = |f = 1.238 radians per second. PROBLEM 2. A flywheel makes 200 revolutions per minute. Show that its angular velocity is 72,000 per minute or radians per second. If the radius of the flywheel is 3 feet, show that the velocity of a point on the rim is 42.84 miles per hour. 19. EXAMPLES 1. Construct the following angles : 30, 45, 60, 135, 300, - 60, - 90, - 390, - 420. 2. Construct approximately the following angles : 2 radians, 3|- radians, ^ radian, 4 radians, 9 radians. 3. Construct the following angles : 7T 7T 7T 5 7T 5 TT 2' ~3' 4' 1 ~T~' T' 4. Eeduce the following angles to radian or circular measure : 10, 30, 45, 60, 135, 225, - 270, - 12, - 18, 24 15', - 612 19' 25". RECTANGULAR COORDINATES AND ANGLES 11 5. Reduce to degree measure : 2 radians, 5 radians, 3 radians, ^ radian, ^ radian, b radians. 6. Reduce to degrees : TT 3_7r _5jr 3.1416 2 TT + 1 2 3' 4 ' " 3 ' 7T "' TT' 2 ' 7T + 3* 7. If an arc of 30 ft. subtends an angle of 4 radians, find the radius of the circle. 8. In a circle whose radius is 5, the length of an intercepted arc is 12. Find the angle (a) in radians, (6) in degrees. 9. If two angles of a plane triangle are respectively equal to 1 radian and ^ radian, express the third angle in degrees. 10. In a circle whose radius is 12 ft., find the length of the arc intercepted by a central angle of 16. 11. Find the angle between the tangents to a circle at two points whose distance apart measured on the arc of the circle is 378 ft., the radius of the circle being 900 ft. 12. An automobile whose wheels are 34 inches in diameter travels at the rate of 25 miles per hour. How many revolutions per minute does a wheel make ? What is its angular velocity in radians per second ? 13. Assuming the earth's orbit to be a circle of radius 92,000,000 miles, what is the velocity of the earth in its path in miles per second ? 14. The rotor of a steam turbine is two feet in diameter and makes 25,000 revolutions per minute. The blades of the turbine, situated on the circumference of the rotor, have one-half the velocity of the steam which drives them. What is the velocity of the steam in feet per second ? 15. A belt travels around two pulleys whose diameters are 3 feet and 10 inches respectively. The larger pulley makes 80 revolutions per minute. Find the angular velocity of the smaller pulley in radians per second, also the speed of the belt in feet per minute. CHAPTER II TRIGONOMETRIC FUNCTIONS 20. The trigonometric functions, upon which trigonom- etry is based, are functions of an angle. These functions are the sine, cosine, tangent, cotangent, secant, cosecant, versed sine, and coversed sine of an angle. For any angle a they are written sin a, cos a, tan a, cot a, sec a, esc a, vers a, and covers a. 21. Definitions of the trigonometric functions. Let OX and OP be the initial and terminal lines respectively of any angle a. Y AX X'A X A X' Let P be any point on the terminal line, OP or r the distance from the origin to the point P, OA or x the abscissa of the point P, and AP or y the ordinate of the point P. 12 TRIGONOMETRIC ^UNCTIONS 13 It should be noted that OP, OA, and AP are directed lines and hence r = OP and not PO x = OA and not AO y = AP and not PA. Then the trigonometric functions are defined as follows : sin a = - = ordinate r distance r distance Y ordinate tan a = ^ = : x abscissa = * = abscissa y ordinate = /: = distance x abscissa A* distance esc a = - = / ordinate vers a = 1 cos a covers a = 1 sin a It will be observed that each of the first six functions is defined as a ratio between two line segments. These are the fundamental trigonometric functions, and the ratios de- fining them are called the trigonometric ratios. The first six functions are called trigonometric ratios. The expression trigonometric functions is more general and embraces the versed sine, coversed sine, and the trigono- metric ratios. It is evident, from the definitions, that the values of the trigonometric functions are abstract numbers. 22. Values of the trigonometric functions of 30, 45, and 60 6 . A few concrete illustrations serve to show the nature of the trigonometric functions, to fix ideas, and to prepare the way for more general considerations. 14 TRIGONOMETRY Functions of 30. Let OPF be an equilateral triangle having its sides equal to 2 units. Place the triangle with a vertex at the origin so that OX bisects the angle POP. Then, by geometry, the angle A OP =30, the ordinate AP= 1, and the abscissa OA = -\/3. Hence, applying definitions, sin30 = i=.500 P cos 30 = =.866 X' o V3 jA X \j F V3 3 cot 30 = = V 1.732 vers30=l-iV3=.134 sec 30 =-^z = |V3 = 1.155 covers 30= 1 - - = i = .500 esc 30 = | = 2.000 2i 2i J- PROBLEM. Find the values of the trigonometric functions of 30, as above, taking OP= 1. Functions of 45. Let OAP be a right-angled isosceles triangle having its sides OA and AP each equal to 1. Then Z. AOP = 45 and X ' the distance OP = V2. Hence, applying definitions, sin 45 = = - V2 V2 2 cos 45 = = lV2 V2 2 1 1 A X sec 45=-^ = esc 45 = ^ = vers45 = l-i- cot45 = = covers 45 1-V2 TRIGONOMETRIC FUNCTIONS 15 PROBLEM. Find the values of the trigonometric functions of 45, taking OA = 3. Functions of 60. Let OPF y an equilateral triangle having each side equal to 2, be placed as in the figure. Then the abscissa and ordinate of P are 1 and V5, respectively. Hence, applying definitions, sin 60 = 1 A F X cos 60 = - 2 V3 2~2 cot 60 = -L = | V3 covers 60 = 1 - - V3 PROBLEM. Find the values of the functions of 60, as above, taking OP= 4 a. The values of the sines and cosines of 30, 45, and 60 are used frequently and should be memorized. The follow- ing table may be found helpful : sin 30 = i VI = cos 60 2 23. Values of the trigonometric functions of 120, 135, and 150. By using the magnitudes of the figures of Art. 22 and properly placing them with respect to the axes, the 16 TRIGONOMETRY values of the trigonometric functions of various angles may be obtained. Functions of 120. From the figure and definitions it is evident that V3 X' A -1 120 sin 120 = -^ -: 2 2 cos 120 = = -1 covers 120 = 1 V3 cotl20 = -=i = --V3 = - sec 120 = =-2 2 1 esc 120= 7==? A/3 Functions of 135. From the figure and definitions it is evident that V2 2" P 1 1 V2 :=l = _lV2 i V V2 2 1 X' A -1 _^ ~ "l"""" 1 Y X. sec 135 = = - V2 esc 135 = -^?= V2 covers 135 = l- TRIGONOMETRIC FUNCTIONS IT Functions of 150. From the figure and definitions it is evident that sin 150 = - cos 150 = tan 150 = cot 150 = sec 150 = -V3 -V3 : -V3 X A -Y3 3 V V3 ?V3 3 J50" csc 150 = = covers 150 = l- = w w PROBLEM. Find the values of the trigonometric functions of 210, 225, 240, 300, 315, and 330. 24. Signs of the trigonometric ratios. The signs of the trigonometric ratios of any angle depend upon the signs of the ordinate, the abscissa, and the distance of any point on its terminal line. As the terminal line passes from one quadrant to another, there is always a change of sign in either the abscissa or the ordinate of any point on that line, but the distance remains positive. When a coordinate changes its sign, every trigonometric ratio de- pendent upon it must also change its sign. The following table is constructed by taking account of the signs of the abscissa x and of the ordinate y, and re- membering that the distance r is always positive. It gives the sign of each trigonometric ratio of an angle terminating in any quadrant. 18 TRIGONOMETRY 1st Quadrant 2d Quadrant 3d Quadrant 4th Quadrant -t- + . + -"* + ~ + + cos a = + T = = jH tan a H _ 3=4- T = cot a |^ + ~ 3= + 4- _ sec a i^ + _ _ 1^ pep /y +. + . + + +- Hh A 2 A 25. Theorem. For every given angle there is one and only one value of each trigonometric function. The theorem is demonstrated for the sine of an angle. The same method is applicable to each of the remaining functions. - Let a be any angle. Eefer- ring to Art. 21, it is clear that if it be possible to obtain two or more values for sin a they must be obtained by taking dif- ferent points on the terminal line. Let Pj and P 2 be any two points on the terminal line. Then by definition X f But since the right triangles OA^ and OA 2 P 3 are similar, TRIGONOMETRIC FUNCTIONS 19 and have their corresponding sides in the same direction, it follows that OP, OP 2 Hence the value of sin a is independent of the position of the point chosen on the terminal line, but depends solely upon the position of the terminal line, i.e. upon the angle. The above theorem may be stated as follows : The trigo- nometric functions are single-valued functions of the angle. 26. Theorem. Every given value of a trigonometric func- tion determines an unlimited or infinite number of positive, and negative angles, among which there are always two positive angles less than 360. The theorem is demonstrated for a given tangent. A similar method is applicable to the remaining functions. Let tan a = where m n and n are positive numbers. Then -f- m m tan a n Locate the point P l5 whose abscissa is n, and whose ordinate is m. This point and the origin determine the terminal line of the angle a 1} whose tangent is n Likewise the point P 2 is located by using m and n as ordinate and abscissa respectively. Drawing the terminal line OP 2 , a second angle a 2 is found, which also has the given tangent. The angles a, and <*2> are evidently less than 360, and have the given tangent There is an unlimited num- ber of positive and negative angles coterminal with a, and 2, all of which have the given tangent. Hence the theorem. 20 TRIGONOMETRY 27. EXAMPLES In what quadrants does a terminate when 1. sin a = i 3. tan a = 5. 5. sin a = . L . . v T> .X' x ' x x X' J aj^V / Y y \/< ^ $* p' r y / Y \ / D ' Y Similar figures can be constructed for the other quadrants. Then in each quadrant sin (90 - a) = = - = cos a cos (90 -) = - = % = sin a r' r tan(90-- 3), (2) cos Y 4- cos 3 = 2 cos J (Y 4- 3) cos J (? 3), (3) cos Y cos d = 2 sin $(i + d) sin -J (n + Q.. 76 TRIGONOMETRY To find all the values o/cos" 1 ^. cos- 1 ^ 60 = -60 = 360 + 60 = 360 - 60 = 720 + 60 = 720 - 60 . = _ 360 + 60 = -360 -60 = _ 720 + 60 = _ 720 - 60 All these angles may be expressed by n 360 60, n being any positive or negative integer. Hence o In general, if a is an angle whose cosine is u, it may be shown that cos -i u = 2 nir a.. To find all the values o/sin- 1 !. sin- 1 i= 30 = 180 - 30 = 360 + 30 = 540 - 30 = 720 + 30 = 900 - 30 - 180 - 30 - 360 + 30 - 540 - 30 : - 720 + 30 INVERSE FUNCTIONS 77 All these angles may be expressed by n 180 + ( - 1)"30, n being any positive or negative integer. Hence In general, if a is an angle whose sine is w, it may be own that sin-' U = / nr +(-!)" a. In a similar manner it may be shown that sec" 1 u = 2 nir a, csc~ l u = nir + ( l) n a. 79. Principal values. The smallest numerical value of an inverse function is called its principal value, preference be- ing given to positive angles in case of ambiguity. The principal values of the inverse sine and the inverse cosecant lie between and - ; of the inverse cosine and 2 2 the inverse secant, between and ?r ; of the inverse tangent and the inverse cotangent, between - and - The principal values of an inverse function are sometimes distinguished from the general values by the use of a capital letter. Thus Sin- 1 -^-, while sin- 1 1 = mr + ( - l) w - 6 80. To interpret sin sin~ l u and sin~ l sin a. The expression sin sin" 1 !* is read: the sine of the angle whose sine is u. This sine is evidently u, hence sin sin' 1 u = u. 78 TRIGONOMETRY The expression sin" 1 sin a is read : the angle whose sine is the sine of a. This angle is evidently a, hence sin" 1 sin a == a, or more generally, sin" 1 sin a = nir -\- ( l) n . Similar relations exist between any direct function and the corresponding inverse function. Thus cos cos" 1 u = u ; cos" 1 cos a = a, or cos" 1 cos a = 2 n-n a ; tan tan" 1 u = u ; tan" 1 tan a = a, or tan" 1 tan = nir + a, etc. 81. Application of the fundamental relations to angles ex- pressed as inverse functions. The fundamental relations, being true for all angles, must necessarily be true when the angles are expressed as inverse functions. Thus, letting a = tan" 1 u in the identity sin 2 a + cos 2 a = 1, we have (sin tan" 1 u) 2 -f- (cos tan" 1 u) 2 = 1. Similarly (sin esc" 1 u) 2 -f- (cos esc" 1 u) 2 = 1 ; (tan sec" 1 u) 2 + 1 = (sec sec" 1 u) 2 j sin cos" 1 u = esc cos" 1 u By expressing the angle of the fundamental relations as an inverse function, we may develop relations between the inverse functions. 82. Given an angle, expressed as an inverse function of u, to find the value of any function of the angle in terms of u. By the application of one or more of the fundamental relations, it is always possible to solve the stated problem. INVERSE FUNCTIONS 79 Several illustrations are given below. The method em- ployed can be readily applied to the other functions. 1. To find the value of tan cos" 1 u in terms of u. If tan cos" 1 u is expressed in terms of the cosine of cos" 1 u, the problem is solved, since cos cos" 1 u = u. This may be done as follows : tan cos" 1 u sin cos" 1 u _ Vl (cos cos" 1 u) 2 __ Vl u 2 COS COS" 1 U U U This result may be obtained geometrically. Since u is given, it is evident that cos" 1 ^ represents, among others, two positive angles, a and o, each less than 360. Let us assume u positive and let us construct these angles defined by cos" 1 u. Then from the figure and the definition of the tangent, and Since the tangent of any angle Co- terminal with ! is -t- , and the tangent of any angle coterminal with 0% is , and since cos" 1 ^ u represents all angles coterminal with either ai or #2, we have tan cos" 1 u = Vl - u* 2. To find the value of sec cot 1 u in terms of u. To solve this problem it is only necessary to express sec cot" 1 u in terms of cot cot" 1 u. 80 TRIGONOMETRY Thus sec coir 1 u = Vl 4- (tan cot" 1 u) 2 = -v/1 4- (cot cot" 1 w) 2 This result may be obtained geometrically. Construct the angles given by cot" 1 u. Let us assume in this problem that 16 is negative and hence that u is positive. If the cotangent of an ^angle is nega- tive the angle must terminate in either the second or fourth quadrant. Since OA and OB are the terminal lines of otj and 2 respectively, and since the terminal line is always positive, we have OA = OB = u u or by considerations similar to those in the previous ex- ample, we have + Vl 4- u 2 Vl 4- w 2 sec cot" 1 16=- 16 U 3. To find the value of sin cos" 1 u in terms ofu. We have sin cos" 1 u = Vl (cos cos" 1 16) 2 = Vl i* 2 . 4. To find the value of cot vers" 1 16 in terms of u. We have 1-u Vl (1 vers vers" 1 w) 2 1-u ^/2u u* INVERSE FUNCTIONS 81 83. Some inverse functions expressed in terms of other in- verse functions. 1. To express cos" 1 u in terms of an inverse tangent. From tan cos" 1 u = , (Art. 82, prob. 1) by u taking the inverse tangent of each member (Art. 80), there results yi _ ^2 cos" 1 u = tan" 1 u 2. To express the cot" 1 u in terms of an inverse secant. From sec cot" 1 u = + M> (Art. 82, prob. 2) there u results 3. Similarly from sin cos" 1 u = Vl w 2 there results cos" 1 u = sin" 1 ( Vl w 2 ). 4. Also from cot vers" 1 M = ~ there results V2 u - u* 1-u vers" 1 u = cot" 1 V2 u - u 2 By the method exemplified in Arts. 82 and 83 it is possi- ble to express any inverse function in terms of any other inverse function. In applying the above formulas care must be exercised in selecting the angles, since each inverse function repre- sents an infinite number of angles and one member of the equation may represent angles not represented by the other. For example, in problem 1, if u be positive the cos" 1 u represents angles terminating in the first and fourth quadrants ; but tan" 1 -5^- -^- represents angles terminat- ing in the second and third quadrants as well as angles terminating in the first and fourth quadrants. 82 TRIGONOMETRY 84. Some relations between inverse functions derived from the formulas for double angles, half angles, and the addition formulas. The general method applicable to this class of problems will be illustrated by a few examples. 1. To express cos (2 sec" 1 u) in terms of u. cos (2 sec" 1 u) = 2 (cos sec" 1 u) z - 1 Art. 71, Eq. 4. 1=1-1. (sec sec" 1 u) 2 u 2 From this relation it follows that, (2 \ 1 ). u 2 ) 2. To express tan (1 cos" 1 u) in terms of u. lY^^L COS Art. 72, Eq. 3. From this relation it follows that, l , *. ,, _i_ /I if 2 3. To express sin (sin" 1 u + cos" 1 v) in terms of u and v. sin (sin" 1 u + cos" 1 v) = sin sin" 1 u cos cos" 1 v + cos sin" 1 u sin cos" 1 v Art. 62, Eq. 1. From this relation it follows that, sin" 1 u + cos" 1 v = sin" 1 (uv Vl w 2 Vl v 2 ). 85. EXAMPLES Find the value of each of the following : 1. sin-^VS. 3. tan" 1 !. 2. cos-^-iVS). 4. tan cot" 1 4. 5. sin cot" 1 4. INVERSE FUNCTIONS 83 Express the following in terms of u and v: 6. cos cot" 1 w. 17. cos (2 cos' 1 w). 7. sec cot" 1 u. 18. cos (2 skr 1 u). 8. esc cot" 1 u. 19. cos (2 tan" 1 u). 9. cos sin" 1 u. 20. sin (sin" 1 u -f sin" 1 v). 10. cos tan" 1 u. 21. cos (sin" 1 w 4- sin" 1 v). 11. cossec^w. 22. tan (tan" 1 u + cot" 1 v). 12. cos esc" 1 u. 23. tan (sec" 1 u + sec" 1 v). 13. sin (2 cos" 1 w). 24. cos (see" 1 w + osc" 1 v). 14. tan (2 tan- 1 w). 25. sin ( cos- 1 w) 15. tan (2 sec- 1 w). 26. cos (J cos' 1 u). 16. tan (2 cos- 1 w). 27. sin (i sec' 1 w). 28. sin (| tan- 1 w). Find x in terms of a. 29. tan" 1 a; = cot" 1 a. 30. sin - 1 x = tan" 1 a. 31. cos" 1 x = 2 sin"" 1 a. 32. cos" 1 # = sin" 1 a + tan" 1 a. 33. sin" 1 x = -J- sec" 1 a. Find a; in terms of a and &. 34. tan" 1 x = sin" 1 a + sin" 1 6. 35. cos" 1 x = sec" 1 a sec" 1 b. 36. sin" 1 x = 2 cos" 1 a -f- cos" 1 6. CHAPTER VIII OBLIQUE TRIANGLE 86. In the present chapter we develop the formulas by means of which a triangle may be completely solved when any three independent parts are given. 87. Law of sines. In a plane triangle any two sides are to each other as the sines of the opposite angles. c c Let a, b, c be the sides of a triangle and a, ft, y the angles opposite these sides, respectively. From the vertex of y draw h perpendicular to the side c, or c produced. Then, for each figure, sin a = - , b and siny3=-< a Art. 21 Art. 29 Dividing the first equation by the second, we have a _ sin a b sin/? In a similar manner, dropping a perpendicular from the vertex of a, it is seen that c sny The last two equations may be written a b c sin a sin p sin -y 84 OBLIQUE TRIANGLE 85 88. Law of tangents. The tangent of half the difference of two angles is to the tangent of half their sum, as the difference of the corresponding opposite sides is to their sum. From the law of sines we have a _ sin a b sin/3 By division and composition, this becomes a b _ sin a sin ft a + b sin a + sin/?' which reduces to a- 6 2cos -( + sina ff) a + 6 2 sin 1 (a + J3) cos | (a - ft = cot i (a + J3) tan *- ( - J3) tana + a + Singly tanHP-V) ^_g, (2) tani(p+Y> 6+* and tanj^^ L o) = c--a > (3) tan i (-y + a) c + a 89. Cyclic interchange of letters. Each formula pertaining to the oblique triangle gives rise to two other formulas of the same type by a cyclic interchange of letters. A cyclic interchange of letters may be accomplished by arranging the letters around the circumference of a circle as in the figure, and then replacing each letter by the next in order as indicated by the arrows. Thus by this cyclic interchange of letters formula (1) of Art. 88 gives rise to formula (2) ; likewise formula (2) gives rise to formula (3). 90. Law of cosines. The square of any side of a plane tri- angle is equal to the sum of the squares of the other sides minus twice the product of those sides into the cosine of the included angle. 86 TRIGONOMETRY c D B DA B For each figure AB = AD + DB, or DB = AB- AD. Art. 2 But AB = c, ^4Z> = 6 cos a, hence DB = c b cos a. Also D(7 = b sin a. From the right triangle CDB we have BC 2 = DC 2 -\- DB 2 Substituting values, this equation becomes a 2 = (6 sin a) 2 + (c 6 cos a) 2 = 6 2 sin 2 a + c 2 2 fcccos a + 6 2 cos 2 a = 6 2 (sin 2 a + cos 2 a) + c 2 2 6c cos a. Hence a 2 = 6 2 + c 2 - 2 be cos a. Similarly 6 2 = c 2 + a 2 and c == fl ~f~ o 91. To find the sine of half an angle of a plane triangle in terms of the sides of the triangle. From equation (1), Art. 72, cos whence From the cosine law . cos a. = 52 + c2 _ 2 be (1) (2) OBLIQUE TRIANGLE 87 From equations (1) and (2) 2 be 2 be (a 6 2 be Let a + b -f c = 2 s. (4) Subtracting 2 a, 26, and 2 c from each member of (4) we have respectively -a + 6 + c = 2( S -a) a -f- 6 c = 2 (s c). Then equation (3) becomes a Similarly sin = -j- ca -82. ^/in ter and 6in? = + J(* -")(*-*). 2 * a0 In these formulas the positive sign is given to the radical, since it is known that half an angle of any plane triangle is less than 90. The same applies to the corresponding for- mulas for the cosine and the tangent of half an angle. To find the cosine of half an angle of a plane triangle 'n terms of the sides of the triangle. 88 TRIGONOMETRY From equation (2), Art. 72, we have 2cos 2 - = l + cosa. 2 Then 1 . fr 2 + c 2 - a 2 l+___ (b + c) 2 - a 2 2 be 26c and Hence Similarly and 93. To find the tangent of half an angle of a plane tri- s angle in terms of the sides of the triangle. Since Similarly and sin? tan"- 2 - (1) (2) (3} HUI A , a tan a + % fe _ )($ _ c ) 2 "" \ 5(5 - a) tan P 4- A /( 5 - c)(s a) 2 * s(s - b) ,n Y - ^fe -a)(s-b) t OBLIQUE TRIANGLE Formula (1) may be written 89 Letting Similarly and tan" 1 ,/(-)(*-&)(* - c >- (4) (5) (6) (7) Ln 2 s-a\ r _:,_.J(-)(*-W*-c) * * tan a - r 2 s -a tan p - r 2~-6' tan'' r 2 s-o 94. To find the area of a plane triangle in terms of two sides and the included angle. Let A be the area and h the altitude. Then in each figure A = ^ ch, and h = b sin a. Therefore A = \ be sin a. Similarly A = J ca sin 0, and A = \ab sin -y. 95. To find the area of a plane triangle in terms of a side and two adjacent angles. From Art. 94, A = % be sin a. 90 TRIGONOMETRY From the sine law , _ csin/? siny . _ c 2 sin a sin J3 2 siny Then since a + ft + y = 180, M _c 2 sin a sin p ~2sin(a + p)' 96. To find the area of a plane triangle in terms of the three sides. From Art. 94. A = bc sin a. Since sin a = 2 sin - cos - . 2 2' A be sin - cos - 2 2 Substituting the values of sin ^ and cos ^ as found in Arts. 91 and 92, we have, after reduction, -a)( S -b)(s-c). 97. Formulas for solving an oblique triangle. The formu- las developed in the present chapter are sufficient to solve a plane triangle when three independent parts are given. The law of sines a b c sin a sin ft sin y is used when two of the given parts are an angle and the opposite side. The law of tangents tan i (a- ft) = ^| tan 1 (a + 0) a + 6 is used when two sides and the included angle are given. OBLIQUE TRIANGLE 91 If a, 6, and y are the given parts, \ (a + ft) is obtained from the relation a + /? + y = 180. The formula then gives the value of -J (a /3), which, united with the value of % (a gives a and /?. TTie ta;s of half angles s (s a) s a are used when the three sides are given. The last formula is the most accurate, since the tangent varies more rapidly than either the sine or the cosine. The formula involving r is advantageous when all the angles are to be computed. TJie law of cosines may be used to determine the third side when two sides and an angle are given. It may also be used to determine an angle when the three sides are given. This formula is used with natural functions, not being adapted to logarithmic computation. 98. Check formulas. Any formula which was not used in the solution of the triangle may be used as a check formula. The relation a + (3 + y = 180 cannot be used as a check when a problem has been solved by the law of tangents, since the law of tangents involves this relation. When two equations from the sine law have been used to find two elements of a triangle, the third equation from the sine law cannot be used as a check, since the first two equations involve the third. 92 TRIGONOMETRY 99. Illustrative problems. 1. Given c = 127.32 a = 71 58' 22" J3 =52 19' 40" SOLUTION. Construction and estimates. A A\ 7 V to find a b y . a = 140 6 = 120 7 = 60 A Outline c sin a tan * fa ^ a C%ecfc _ 5 ian 2 ^ p; sm 7 a + 6 c a ft a-6 a + 6 y log a b ) colog (a + 6) log sin a log tan (a + /3) logc colog sin 7 log tan (a /3) log a a !>+ , ~ sin 7 r, more compactly, log sin /3 logc colog sin 7 log& 6 log sin a . w> j colog sin 7 ^ [ log sin /3 log a log& a 6 OBLIQUE TRIANGLE 93 Filling in the above outline, the completed work appears as follows c sin a sin 7 c a V 127.32 c , , 71 58' 22" 52 19' 40" tan*Ca /3) a ~ & tan *fa + /3) 124 18' 2" a + 6 55 41' 58" a-6 24.57 268.55 62 9' 1" log sin a logc colog sin 7 log a a 9.97814-10 a + b 2.10490 ?( + ) 0.08297 log (a -6) 1.39041 7.57098 0.27708 2.16601 colog (a + 6) 146.56 log tan (a + /3) 3 sin/* logtanKa-0) 9.23847-10 9 49' 28" 62 9' 1" 71 58' 29" 52 19' 33" Shl7 Ha -I-/9) log sin ft logc colog sin 7 log 6 6 Or, in the 9.89846-10 ' r ' a 2.10490 ,3 0.08297 2.08633 121.99 more compact form, (log sin a 9.97814-10 f logc 2.10490 "S)j colog sin 7 0.08297 -2 1 log sin ft 9.89846 log a 2.16601 log b 2.08633 a 146.56 b 121.99 2. Given a = 1674.3 c = 1021.7 = 28 44' 39" to find a 7 b. Estimates a = 120 7 = 30 6 = 900. 94 TRIGONOMETRY a -f c T _ c sin /3 sin 7 a c /s a c a + c a + 7 log (a - c) colog (a + c) log tan (a + 7) log tan H-7) 1674.3 1021.7 28 44' 39" 652.6 2696.0 151 15' 21" 75 37' 40" logo log sin /3 colog sin 7 log 6 3.00932 9.68205-10 0.27268 2.96405 920.66 Check 2.81465 6.56928-10 0.59135 b = sin a 9.97528-10 43 22' 12" 75 37' 40" 118 59' 52" 32 15' 28" log a log sin colog sin a log 6 3.22384 9.68205-10 0.05817 2.96406 3. Given a = 1.4932 b = 2.8711 = 1.9005 to find a Estimates ft a= 25 /3 = 120 y- 7 = 35 r tin 7 r 2 S - C a 1.4932 6 2.8711 c 1.9005 2s 6.2648 s 3.1324 s a 1.6392 s- 6 0.2613 * c 1.2319 s 3.1324 log (s - a) 0.21463 log 0-6) 9.41714 - 10 log (s - c) 0.09058 colog s 9.50412 - 10 logr 2 19.22647 - 20 logr 9.61324 - 10 log tan J 9.39861 - 10 log tan | 0.19610 log tan J 9.52266 - 10 2 14 3' 26" e 2 57 31' 2" 2 2 18 25' 34" a 28 6' 52" ft 115 2' 4" 7 36 51' 8" CAecfc : a + + 7 180 00' 4" OBLIQUE TRIANGLE 95 4. Given 6 = .0060041 c = .0093284 = 44 47' 58" to find Estimates j8= 30 a = 105 a = .012. Check sin/3 6 sin 7 tin "" ("ft "v} a c tan 1 (OL 1 V^ c a + c b c 7 .0060041 .0093284 44 47' 58" a c a c a + c a + 7 K + 7) .012574 .0093284 .0032456 .021902 153 1'48" 76 30' 54" log& colog c log sin 7 log sin /3 a a 7.77845 - 10 2.03019 9.84796 - 10 9.65660 - 10 26 58' 12" 71 46' 10" 108 13' 50" csina log(a-c) colog (a + c log tan K a + 7) log tan ( - 7) U-7) 7 7.51129 - 10 1.65952 0.62015 9.79096 - 10 31 42' 51 76 30' 54" 108 13' 45" 44 48' 3" sin 7 logc colog sin 7 log sin a log a a 7.96981 - 10 0.15204 9.97764-10 8.09949-10 0.012574 100. The ambiguous case. When two sides and an angle opposite one of them are given, the triangle may admit of no solution, of one solution, or of two solutions. Let a, b, a be given. The formula sin/?; determines 96 TRIGONOMETRY If the calculated sin ft is greater than 1, there can be no solu- tion. a < b sin a If the calculated sin ft equals 1, ft = 90 and there is one solution. .5 1 a a = 6 sin a If the calculated sin ft is less than 1, two supplementary values of ft are deter- mined, giving two so- lutions unless the larger value of ft plus a is equal to or greater than 180. a > b sin a a etc., respectively. (4) 118 TRIGONOMETRY Substituting the values from (4) in equation (1), and remembering that r = 1, we have j-> 2 -v/1 = 1 for n = 0, P, ^ il + isini? f or n = 2, o o 5 5 = cos + isin forn = 4. 5 5 110. To extract the square root of a + ib. Let Va + ib = r (cos a + i sin a). (1) Squaring, a + ib = r 2 (cos 2 a + i sin 2 a). (2) Also a + ib = ^[003(2 a 2nir)+i sin (2 a 2 rnr)], (3) where n is any integer. Equating the real and imaginary parts, ^008(2 a - 2tt7r)= a, (4) r 2 sin(2a-2ri7r) = 6. (5) Squaring and adding, we have r 4 =a 2 + & 2 , (6) and r = Va 2 + & 2 . (8) Equation (8) gives the value of r in terms of the known numbers a and 6. From equations (4) and (7) whence 2 a 2 nir cos' the quadrant of 2 a 2 n-ir being determined by (4) and (5). DE MOIVRE'S THEOREM WITH APPLICATIONS 119 Then a = cos- 1 a forn = 0, (9) Va 2 + b 2 and = TT + cos" 1 a for n = 1. (10) The values of a for n = 2, 3, 4, etc., are coterminal with the values of a in equations (9) or (10). Hence there are only two values for sin a and cos a ; namely those for which n = and n = l. Substituting these values of a in equation (1), we have, finally, Va + ib = \/a 2 -f 6 2 [~cos / cos- 1 CT and Va-M& = -\/a 2 + 6 2 |cos f + 4 cos' 1 a L V Va 2 + + i sin f TT 4 i cos- 1 - - - \\ . (12) V Va 2 + 6V J These values of Va + 16, while more complicated than Va H- ib itself, are nevertheless in the standard form of complex numbers, r (cos a + i sin a). 111. To extract the Mh root ofa + ib. Let -fya + ib = r (cos a + i sin a). (1) Then a + 6 = r* (cos fca + i sin fca) (2) = r*[cos (Tea - 2 rnr) + i sin(fca - 2 WTT)], (3) where n is an integer. Equating the real and imaginary parts of equation (3), we have r* cos (ka 2 HIT) = a, (4) and i* sin (& 2 WTT) = 6. (5) 120 TRIGONOMETRY -Squaring and adding equations (4) and (5), we have r 2 * = a 2 + b 2 . (6) Then r* = Va 2 + 6 2 , (7) and r= ^a 2 + 6 2 . (8) From equations (4) and (7), we have cos (ka- or fca 2 nir = cos' 1 the quadrant of ka 2 mr being determined by the signs of (4) and (5). Therefore = ?^Z[ + ! cos" 1 a k k whence a = =^ + icos- 1 k k 4?r 1 c^-r+'-cos- 1 k k Va 2 + b* Substituting the values for r and a as found in equations (8), (10), (11), (12), etc., in equation (1), we have the several kih roots of a + ib. When A; is an integer there are k roots. Consecutive values of a, as given by (9), differ by , hence all the fc values of a after the fcth are coterminal with one of the first k values. 112. To express sin na and cos na in terms of sin a and cosa. We have cos na + i sin na = (cos a + i sin ) n . DE MOIVRE'S THEOREM WITH APPLICATIONS 121 Expanding the second member by the binomial theorem and equating the real and imaginary parts, the problem is solved. Thus, for sin 4 a and cos 4 a we have cos 4 a 4- i sin 4 a = (cos a -f i sin a) 4 = cos 4 a, -h 4 i cos 3 a sin a 6 cos 2 a sin 2 a 4 i cos a sin 3 a + sin 4 a. Therefore sin 4 a = 4 cos 3 a sin a 4 cos a sin 3 a. and cos 4 a = cos 4 a 6 cos 2 a sin 2 a -f- sin 4 a. 113. Comparison of the values of sin a, a, and tan a, a being an acute angle. Let a be any acute angle expressed in ra- dians. With the vertex as a center and any radius OB, describe the arc EC. Draw AG and BD perpendicular to OB, and join B with O. The area of the triangle OBC is less than the area of the sector OBC, and the sector OBC is less than the triangle OBD. But since AC= OC sin a, BD = OB tan a, and arc BC = OB a, Art. 17, the area of the triangle OBC is equal to | OB OC sin a, the area of the sector OBC is equal to ^ OB OB - a, and the area of the triangle OBD is equal to |- OB - OB tan . Hence or sin a < a < tan a. 114. Value of for small values of a. In Art. 40 we a saw that sin a approaches as a approaches 0. The value of therefore approaches . as a approaches 0. a (J 122 TRIGONOMETRY But from the previous article sin a. < a < tan a. Dividing by sin a, we have a 1 sin a cos a sin a or 1 > > cos a. a sin a , . . sin a Since lies between 1 and cos a, must approach a a. 1 as a approaches 0, since cos a approaches 1. Then for very small angles sin a may be replaced by , expressed in radians. The error thus introduced is so small that it may be neglected in many problems. Thus, to five decimal places, sin 1 = 0.01745 1 = 0.01745 radians sin 2 = 0.03490 2 = 0.03491 radians sin 3 = 0.05234 3 = 0.05236 radians sin 4 = 0.06976 4 = 0.06981 radians 115. To develop sin a and cos a in terms of a. By De Moivre's theorem, cos nO + i sin nO = (cos -f i sin 0) n . On expanding the second member by the binomial theorem, we have cos nO + * sin n& = cos n -{-in cos"" 1 sin 6. ^j2 -cos n ~ 2 sin 2 ?! j^ ' cos n ~ 3 sin 3 | n(n-l)(n~2)(n-8) cos ,_ 4gsin4g | ... (1) 11 Equating the imaginary parts, we have ' sin n$ = n cos"" 1 sin r^ cos w ~ 3 sin 3 tt(n-l)(n-2)(n-3)(tt-4) + _A & A A / cos n l Let nO = a. Then equation (2) may be written DE MOIVRE'S THEOREM WITH APPLICATIONS 123 sin a = ? cos"- 1 sin - ^ ^ ^ cos w ~ 3 sin 3 |3 ---- " " Let a remain constant while n increases indefinitely. Then necessarily decreases indefinitely, since n0 = a, a constant. By Art. 114, when approaches 0, ^ ap- proaches 1, and cos 6 approaches 1. Making these substi- tutions in equation (4), we have a 3 a 5 a 7 Equating the real parts of equation (1), we have cos n6 = cos" - n ^~ 1 ) cos"- 2 6 sin 2 6 e n(n- l)(n-2)(tt-3) **-.* /K N 4. _1 - d\_^ /i -- ? C os n ~ 4 6 sm 4 ^ -. (5) By the same process as above, equation (5) becomes The series for sin a, and cos a are convergent for all finite values of a* They enable us to compute the sine and cosine of any angle. It is then possible to construct a table of natural functions, from which the logarithmic functions may be obtained. In using these series a must, of course, be expressed in radian measure. * See any College Algebra on the convergency of series. 124 TRIGONOMETRY 116. EXAMPLES 1. Find the four fourth roots of unity by De Moivre's theorem. 2. Find the six sixth roots of unity by De Moivre's theorem. 3. Find the square root of 5 3 i. SOLUTION. Let V5 3 i = r (cos a + I'sin a). (1) Then 5 - 3 1 = r 2 (cos 2 a + i sin 2 ) = r 2 [cos (2 a - 2 WTT) +isin(2a 2nir)]. Equating the real and the imaginary parts, r 2 cos (2 a - 2 WTT) = 5, (2) and r 2 sin (2 a - 2 WTT) = - 3. (3) Squaring and adding (2) and (3), we have r* = 34, . . r 2 = V&, and r = ^34. (4) Then cos (2 a 2 nir) = V34 and 2 a - 2 nir = 329 2', (5) the quadrant being determined by (2) and (3). When n = 0, a = 164 31' ; (6) = !, = 844 31'. (7) Substituting from (4) and (6) in (1), we have \/5^3l = - 2.3271 + .6446 L Substituting from (4) and (7) in (1), we have V5^3l = 2.3271 -.6446i. 4. Find the square root of 3 -f 4 i. 5. Find the square root of 3 4 i. 6. Find the square root of 1 + 2 1. 7. Find the square root of i. 8. Find the square root of i. 9. Find the cube root of 2 3 i. SOLUTION. Let \/2 - 3 i - r (cos a + i sin a). (1) Then 2 - 3 i =r s (cos 3 a + i sin 3 a) (3a-2n7r)]. DE MOIVRE'S THEOREM WITH APPLICATIONS 125 Equating the real and imaginary parts, r 3 cos (3 a - 2 WTT) = 2 ; (2) and r 8 sin (3 a - 2 nir) = - 3. (3) Squaring and adding (2) and (3), we have i* = 13, .-. r 8 = Vl3~ and r = \/13T (4) 2 Then cos (3 a - 2 WTT) =- , and 3 a - 2 nir = 303 41' ; (5) the quadrant being determined by (2) and (3). When n = 0, a = 101 14' ; (6) n = l, a = 221 14'; (7) n = 2, a = 341 14'. (8) Substituting from (4) and (6) in (1), we have v / 2^3l = - .2987 + 1.5041 i. Substituting from (4) and (7) in (1), we have v 2-3i = - 1.1530 - 1.0107 *. Substituting from (4) and (8) in (1), we have 2/2 - 3 i = 1.4518 - .4933 i. We have thus found the three cube roots of 2 3 i. 10. Find the cube root of 1 + i. 11. Find the cube root of 1 +i. 12. Find the cube root of 2 -f 3 i. 13. Find the values of sin 3x and cos 3 a; in terms of sin a? and cos a;. 14. Find the values of sin 5 a; and cos 5 a; in terms of sin x and cos x. 15. Prove by De Moivre's theorem that sin a = 2 sin ~ cos ~> also cos a = cos 2 ~ sin 2 - 2, L & & I a . a\2 SUGGESTION, cos a + i sin a = I cos - + i sm - 1 . f 16. Show that cos a = cos 3 ^ 3 cos f sin 2 1, o o o sin a = 3 cos 2 ^ sin ^ sin 3 o o o SPHERICAL TRIGONOMETRY CHAPTER X FUNDAMENTAL FORMULAS 117. The spherical triangle.* Spherical trigonometry treats of the relations between the various parts of a spherical triangle and of the methods of solving the spheri- cal triangle. The sides of a spherical triangle are always arcs of great circles. Having given a spherical triangle ABC, situated upon a sphere S, a triedral angle 0- ABC may be formed by passing planes through 0, the center of the sphere, and through the sides of the triangle. It is known from geometry that the arc AB and the angle AOB contain the same number of degrees, and that the angle CAB and the diedral angle C-AO-B contain the same number of degrees. The sides and angles of a spherical triangle may have any values between and 360. A triangle having one or more of its parts greater than 180 is called a general spherical triangle. A triangle having each of its parts less than 180 is called a spherical triangle. We shall consider only those triangles whose parts are each less than 180. * For a course on the right spherical triangle read Arts. 117 and 126 and from Art. 128 to end of Chapter XI. 127 128 TRIGONOMETRY 118. Law of sines. To find the relation between two sides of a spherical triangle and the angles opposite. /3 > 90, 6 > 90. Given a spherical triangle and its accompanying triedral angle ; through the vertex P pass planes perpendicular to OR and OQ, intersecting in PF a line perpendicular to the plane ORQ. Then sin ft w Also sin a -- *-=& therefore Hence Likewise sin a RP sin b sin a QP sin a sin/3 sin 6 sn y sn c Uniting these equations, we have sing _ sin b _ sine sin a sin p sin Y FUNDAMENTAL FORMULAS 129 This demonstration applies to similar figures drawn for all possible cases,* hence the theorem is always true. To find the relation between the three A J v 119. Law of cosines. sides and an angle. Given a spherical tri- angle and its accompany- ing triedral angle ; pass a plane through the ver- tex A perpendicular to OA, intersecting the planes of the triedral angle in the lines AB, AC, and BC. b < 90, c < 90 ; a < 180, a < 180. Then AB = r tan c, OB = r sec c, AC = r tan b, OC = r sec b. From the triangle OBC, by Art. 90, B& = (r sec b) 2 + (r sec c) 2 2(r sec &) (r sec c) cos a. (1) Likewise from the triangle ABC, BC 2 = (r tan b) 2 + (r tan c) 2 - 2 (r tan 6) (r tan c) cos a. (2) Subtracting (2) from (1), we have = r 2 (sec 2 6 - tan 2 6) + r 2 (sec 2 c - tan 2 c) 2 r 2 sec b sec c cos a + 2 r 2 tan 6 tan c cos <*, which reduces to == 1 sec 6 sec c cos a + tan b tan c cos , or cos a = cos b cos c + sin b sin c cos a. (3) Also cos b = cos c cos a -f- sin c sin a cos p, (4) and cos c = cos a cos 6 + sin a sin 6 cos y. (5) * The following seven cases can arise : (1) 3 sides < 90, 3 angles < 90 (4) 1 side < 90, 2 angles < 90 (2) 3 sides < 90, 2 angles < 90 (5) 1 side < 90, 1 angle < 90 (8) 2 sides < 90, 2 angles < 90 (6) 1 side < 90, angle < 90 (7) side < 90, angle <90. It is to be understood that all parts not mentioned are greater than 90. 130 TRIGONOMETRY 120- To extend the law of cosines. In the derivation of the formula cos a = cos b cos c -f- sin b sin c cos a, b and c were less than 90, while a and a were less than 180. To show that the formula is true in general, it is necessary to consider two additional cases : 1st. Both b and c greater than 90. 2d. Either b or c greater than 90. Since ft and y do not enter the formula, they may have any value consistent with the above conditions. First. Given the triangle ABC in which b > 90 and c > 90. Extend the sides 6 and c of the tri- angle ABC, forming the lune whose angle is a. Then in the triangle A'BC, the sides A'B and A'C are each less than 90, hence by Art. 119 cos a = cos (180 - b) cos (180 - c) + sin (180 - 6) sin (180 - c) cos , or cos a = cos b cos c + sin b sin c cos a. Hence the law of cosines holds when both b and c are greater than 90. Second. Given the triangle ABC } in which b < 90 and c > 90. Extend the sides a and c of the triangle ABC, forming the lune whose angle is ft. Then in the tri- angle AB'C, the sides AB' and AG are each less than 90, and the angle B'AC is equal to 180 - a. Then, by Art. 119, cos (180 - a) = cos b cos (180 - c) + sin b sin (180 - c) cos (180 - a), or cos a = cos b cos c -f sin b sin c cos a. FUNDAMENTAL FORMULAS 131 Hence the law of cosines holds when either b or c is greater than 90. The law of cosines is therefore true in general. 121. To find the relation between one side and the three angles. Let a, b, and c be the sides of any spherical triangle, and a', b' } c' the sides of its polar triangle. Applying the law of cosines to the polar triangle, we have cos a' = cos b' cos c' + sin b' sin c' cos a'. But a' = 180 - a, b'= 180 - /?, etc. Therefore cos (180 - a) = cos (180 - /?) cos (180 - y) + sin (180 - ) sin (180 - y) cos (180 - a), or cos a = cos /3 cos y + sin /? sin y cos a. (1) Also cos (3 = cos y cos a + sin y sin a cos 6, (2) and cos y = cos a cos ft + sin a sin /? cos c. (3) 122. The sine- cosine law. To find the relation between three sides and two angles. We have cos a = cos 6 cos c + sin b sin c cos a, (1) and cos b = cos c cos a -f sin c sin a cos /?. (2) Eliminating cos a by substitution, cos b = cos b cos 2 c + sin b sin c cos c cos a + sin c sin a cos /?. Transposing and factoring, cos b (1 cos 2 c) = sin b sin c cos c cos a -f- sin a sin c cos /?. Eeplacing (1 cos 2 c) by sin 2 c, and dividing by sin c, we have cos b sin c = sin b cos c cos a + sin a cos ft. Kearranging terms, sin a cos /3 = cos 6 sin c sin 6 cos c cos . (3) 132 TRIGONOMETRY Also sin 6 cos y = cos c sin a sin. c cos a cos ft, (4) and sin c cos a = cos a sin b sin a cos 6 cos y. (5) Interchanging ft and y and consequently b and c, we have from (3) sin a cos y = cos c sin b sin c cos 6 cos a. (6) Similarly from (4) sin 6 cos a = cos a sin c sin a cos c cos /?, (7) and from (5) sin c cos ft = cos 6 sin a sin & cos a cos y. (8) 123. To find the relation between two sides and the three angles. Applying the sine-cosine law to the polar of the given triangle, we have sin a' cos ft' = cos b f sin c' sin b ! cos c r cos ex! . But a'= 180 - a, 0' = 180- 6, etc. Then sin (180 -a) cos (180-&)=cos (180 -0) sin (180 -y) -sin (180- /?) cos (180- y) cos (180- a). Therefore sin a cos 6 = cos ft sin y -f- sin ft cos y cos a. (1) Also sin /? cos c = cos y sin a + sin y cos a cos &, (2) sin y cos a = cos a sin /3 -f sin a cos /? cos c, (3) sin a cos c = cos y sin ft + sin y cos /? cos a, (4) sin ft cos a = cos a sin y 4- sin a cos y cos b, (5) sin y cos b = cos /3 sin a + sin /? cos a cos c. (6) 124. To find the relation between two sides and two angles t one of the angles being included between the given sides. From Art. 122 sin a cos ft = cosb sin c sin 6 cos c cos a. Dividing this equation by sin a sin ft = sin b sin a, Art. 118 FUNDAMENTAL FORMULAS 133 member by member, we have , Q _ cot b sin c cos c cos a sin a Therefore sin a cot (3 = cot b sin c cos c cos a. , (1) Similarly sin ft cot y = cot c sin a cos a cos ft (2) and sin y cot a = cot a sin b cos 6 cos y. (3) Interchanging a and ft and consequently a and 6, we have from (1) sin /? cot a = cot a sin c cos c cos ft (4) Similarly from (2) sin y cot ft = cot b sin a cos a cos y, (5) and from (3) sin a cot y = cot c sin 6 cos b cos a. (6) 125. Formulas independent of the radius of the sphere. It will be noticed that r, the radius of the sphere, does not enter any of the formulas thus far developed ; hence they are independent of the radius of the sphere, and may be applied, without modification, to triangles on any sphere. The fact is serviceable in problems of Astronomy and Ge- odesy where the formulas are applied to triangles situated upon the celestial and terrestrial spheres. CHAPTER XI SPHERICAL RIGHT TRIANGLE 126. The spherical right triangle is a spherical triangle one of whose angles is a right angle. The other parts may have any values between and 180. In the work that follows the angle y will be taken as the right angle. 127. Formulas for the solution of right triangles. The formulas for the solution of any spherical right triangle are obtained from the general formulas of Chapter X by letting y = 90. We thus have from eq. (1) Art. 118 sin a = sin c sin a from-eq. (1) Art. 118 sin b = sin c sin /3 from eq. (5) Art. 119 cos c = cos a cos b from eq. (1) Art. 121 cos a = sin ft cos a from eq. (2) Art. 121 cos ft = sin a cos b from eq. (3) Art. 121 cos c = cot a cot ft from eq. (2) Art. 124 cos ft = tan a cot c from eq. (3) Art. 124 sin b = cot a tan a from eq. (5) Art. 124 sin a =cot ft tan b from eq. (6) Art. 124 cos a = tan b cot c. 128. Direct geometric derivation of formulas. Let a and b be the sides of a given spherical right triangle, a and ft the angles opposite, and c its hypotenuse. Let 0-ABC be its accompanying triedral angle. Through the vertex B pass a plane perpendicular to OA, intersecting the planes of the triedral angle in AB } BC, and CA. 134 SPHERICAL RIGHT TRIANGLE 135 Then Z BAO=a, Z BOC=a, Z AOC=b, and Z.AOB=c. Also Z jB(L4, Z BOO, ZCAO,^BAO are each a right angle. From the triangle ABC we have OB ; sina ? (1) OB AC AC OA tan 6 cos a. = = -pr- = , AB AB tan c OA CB CB OC tana tana = __ = ^ = ___. OC By interchanging a and /? and consequently a and 6, or by passing a plane through D JL to OB and proceeding as above, we have (2) (3) o cos/? = sin c tana - -- , tan c sin a Also from the figure. OA OA OC cos b = OB = 0^ = seca OC Dividing (1) by (5) and reducing by (7) we have , COS and by interchange of letters as before (5) (6) (7) (8) sin (3 cos a cos a 136 TRIGONOMETRY Substituting the values of cos a and cos b from (8) and (9) in (7), we have, after reduction, cos c = cot a cot ft. (10) In the demonstration of formulas (1) to (10) the parts a, ft, a, b, and c were assumed less than 90. To show that these formulas are true in general, it is necessary to con- sider two additional cases : viz. (1) when one side and the hypotenuse are each greater than 90, and (2) when the two sides are each greater than 90. 1. Wlien a > 90 and c > 90. Let ABC be the given spherical right triangle. Draw the lune BB'. Then in the right triangle AB'C a each part is less than 90. and formulas (1) to (10) are applicable. sm(180-a) = sin ( 180 - a ) ' sin (180 - c) sin a From (1) or sin a = sine which shows that (1) holds when a and c are each greater than 90. From (2) cos (180 - a) = tan b or cos a = tan (180 -c) tan b tanc' which shows that (2) also holds in this case. Similarly it may be shown that formulas (3) to (10) hold when a and c are each greater than 90. 2. When a > 90 and b > 90. Let ABC be the given spherical right triangle. Draw the lune (7(7. Then in the right triangle ABC' c each part is less than 90 and for- mulas (1) to (10) are applicable. SPHERICAL RIGHT TRIANGLE 137 From (1) sin (180 - a) = Bin (180 -a) sine or sin a = sin a sin G which shows that (1) holds when a and b are each greater than 90. From (2) cos (180 - ) = tan b tan c or cos a = tan c' which shows that (2) also holds in this case. Similarly it may be shown that formulas (3) to (10) hold when a and 6 are each greater than 90. 129. Sufficiency of formulas. It will be noticed that the ten formulas of Arts. 127 and 128 contain all possible com- binations of the five parts of a spherical right triangle, taken three at a time ; hence they are sufficient to solve any spheri- cal right triangle directly from two given parts. 130. Comparison of formulas of plane and spherical right triangles. By rearranging the formulas of the previous article, the analogy between the formulas of the plane and the spherical right triangles is made apparent. IN PLANE RIGHT TRIANGLES* a sin a = cos a = - c tan=- o sin /?=* C cos/3 = - c tan/?=- a sin a = cos ft sin ft = cos a c 2 = a 2 + b 2 IN SPHERICAL RIGHT TRIANGLES cos a = sin a sin c tan 6 tanc tana sin b sin ft = cos ft = sin b sin c tana tanc tan b sin a sn = 1 = cot a cot ft. *The above comparison is taken from Chauvenet's Spherical Trigonometry." cos b cos a cos c = cos a cos b cos c = cot a cot /?. Plane and 138 TRIGONOMETRY 131. Napier's Rules. The ten formulas used in the solution of spherical right triangles can all be expressed by means of two rules, known as Napier's rules of circular parts. Napier's circular parts are the sides a and 5, the complements of the angles opposite or 90 a, 90 /?, and the complement of the hypot- enuse or 90 c. They are usually written c o a a, b, GO a, co/3, coc. It will be noticed that the right angle is not one of the circular parts. Let the five circular parts be placed in the sectors of a circle in the order in which they occur in the triangle. Whenever any three parts are considered, it is always possible to select one of them in such a manner that the other two parts will either be adjacent to this part, or opposite this part. The part having the other two parts adjacent to it or opposite it is called the middle part. Thus let co a, 6, and a be the parts under consideration. Then b is the middle part and co a and a are adjacent parts. If co c, co /?, and b are the parts under consideration, b is the middle part and co c and co y8 are opposite parts. Napier's rules may now be stated as follows : The sine of the middle part is equal to the product of the cosines of the opposite parts. The sine of the middle part is equal to the product of the tangents of the adjacent parts.* * To associate cosine with opposite and tangent with adjacent, it may be noticed that the words cosine and opposite have the same vowels ; likewise the words tangent and adjacent. SPHERICAL RIGHT TRIANGLE 139 132. Theorem. In a spherical right triangle, a side and the angle opposite terminate in the same quadrant. From the equation cos a = cos a sin ft it is seen that cos a and cos a must always have the same sign, since sin j3 is always positive. Hence a and a termi- nate in the same quadrant. 133. Theorem. Of the three parts a, b, c, if any two ter- minate in the same quadrant, the third terminates in the first quadrant ; if any two terminate in different quadrants, the third terminates in the second quadrant. This follows directly from the equation cos c = cos a cos b by noticing that if any two of the quantities cos a, cos 6, and cos c have like signs, the third is positive ; if any two have unlike signs, the third is negative. 134. Two parts determine a triangle. In order to solve a spherical right triangle two parts, in addition to the right angle, must be given. Each of the required parts should be obtained directly from the given parts. Thus, given 6 = 50, c = 110 we have cos a = - , cos 50 ' cos a = tan 50 cot 110, using the formulas for spherical right triangles, or Napier's Rules. If, in the solution of a problem, the sine of any required part is found to be negative, no triangle is possible, since no part of a spherical triangle can be greater than 180. Likewise if the logarithmic sine or cosine of any required part is found to be greater than zero, the triangle is impossi- ble, since no sine or cosine is numerically greater than unity. 140 TRIGONOMETRY 135. The quadrant of any required part. Since the parts of a spherical triangle may have any value between and 180, it is always necessary to determine whether the required parts are greater or less than 90. This can be done by the theorems of Arts. 132 and 133. Thus, given 6 = 50, c=110, we have a > 90, a > 90, and ft < 90. The quadrant in which any required part terminates may also be determined from the formula used in calculating that part, by observing the signs of the functions involved. But when the unknown part is determined from its sine, the part terminates in both quadrants, giving two solutions, unless limited by the theorems of Arts. 132 and 133. Thus, given 6 = 50, c = 110, by writing the signs of each function above the function, we have cos 110 cos a = - , cos 50 cos a = tan 50 cot 110, sin 110 Then a > 90 and a > 90, since their cosines are negative, and (3 < 90 by Art. 132. 136. Check formula. The formula containing the three computed parts may always be used as a check formula. Thus, having given 6 = 50, c = 110, the check formula is cos a = cos a sin . SPHERICAL RIGHT TRIANGLE 141 137. Solution of a right triangle. Given b = 77 35' 16" and a = 112 19' 42" tan a = ^=-^- cota 6 a log sin b log cot a log tan a a 77 35' 16" 112 19' 42'' 9.98973 - 10 9.61353 - 10 0.37620 67 11' 30" 112 48' 30" cos a + tan b log cos a log tan b log cot c c 9.57969 - 10 0.65740 8.92229 - 10 85 13' 13'' 94 46' 47" cos /3 = cos & sin a log cos 6 log sin a log cos j3 9.33233 - 10 9.96615 - 10 9.29848 - 10 78 31' 53" Check + cos /3 = tan a cot c log tan a log cot c log cos /3 0.37620 8.92229 - 10 9.29849 - 10 138. Two solutions or the ambiguous case. Whenever a side and the angle opposite are given, there are two solutions. Thus if a and a are given, the only formulas by which 6, /3, and c can be determined are sin 6 = tan a cot a, sin/?.; cos a cos a 3 sin a Since the unknown parts are obtained from their sines, each may have two values, giving two solutions, the theorems of Arts. 132 and 133 not restricting the values of the parts to one solution. Having found the two values for each part, the theorems 142 TRIGONOMETRY of Arts. 132 and 133 determine the values that belong to each solution. Thus, given a = 155 27' 45" to find b a = 100 21' 50" c ft SOLUTION sin b = tan a cot a a a log tan a log cot a log sin b b 155 27' 45" 100 21' 50" 9.65945 - 10 9.26217 - 10 8.92162 - 10 4 47' 20" and 175 12' 40" stn cos a P _ cos a log cos a log cos a log sin /3 18 9.25503 - 10 9.95889 - 10 9.29614 - 10 11 24' 22" and 168 35' 38" sin c = sing -4- sina log sin a log sin a log sin c c 9.61835 9.99286 - 10 -10 9.62549 24 58' 155 1' - 10 18" and 42" Check sin b = sin c sin /3 log sin c log sin /3 log sin 6 9.62549 - 10 9.29614 - 10 8.92163 - 10 FIRST SOLUTION &!= 4 47' 20" ft= 11 24' 22" C = 155 1'42" SECOND SOLUTION 6 2 = 17512'40" A = 168 35' 38" c 2 = 24 58' 18" SPHERICAL RIGHT TRIANGLE 143 139. EXAMPLES Solve the following spherical right triangles, right- angled at y. 9. a = 132 25' a = 107 30' 1. 6= 10 32' a= 12 3' 2. a = 25 18' b= 32 41' 3. c = 12037' ft = 9 49' 4. c= 46 40' a= 20 50' 5. a = 115 6' b = 123 14' 6. a = 112 43' 30" c= 85 10' 10" 7. a = 15 18' 20" c= 21 30' 40" 8. 6 = 168 13' 45" c = 150 9' 20" 140. Quadrantal triangles. A quadrantal triangle is a triangle one of whose sides is 90. Its polar triangle is then a right triangle. The solution of a quadrantal tri- angle is effected through the solution of its polar triangle. 141. Isosceles triangles. The solution of an isosceles triangle is effected by solving the two equal right triangles formed by dropping a perpendicular from the vertex to the base. 10. c= 80 3' 20" = 135 16' 30" 11. 6=171 3' 15" c= 12 20' 30" 12. a= 35 54' 20" a= 47 6' 10" 13. b= 15 2' 30" J3= 20 11' 40" 14. = 20 26' 20" = 84 41' 40" 15. a= 25 41' 30" a= 34 25' 40" 144 TRIGONOMETRY 142. EXAMPLES Solve the following triangles, 1. a = 117 54' 30" 3. =153 16' b= 95 42' 20" a= 19 3' c= 90 c= 90 2. a = 69 45' 4. a = 15933'40" (3= 94 40' 6= 95 18' 20" c= 90 c= 90 5. The base of an isosceles triangle is 51 8'. The equal angles are each 41 57'. Find the equal sides and the angle at the vertex. 6. The base angles of an isosceles triangle are each 100 12' 30", the vertical angle is 50 19' 40". Find the equal sides and the base. CHAPTER XII OBLIQUE SPHERICAL TRIANGLE 143. In the present chapter the general formulas of spherical trigonometry, already developed (Chap. X) are transformed into standard formulas adapted to logarithmic computation; and the problem of the solution of the spheri- cal triangle is discussed. GENEEAL SOLUTION 144. To find the angles when the three sides are given. From Art. 119, we have cos a = cos b cos c + sin b sin c cos a. Therefore cos = cos a .- cos b cos c . (1) sin b sin c But 2 sin 2 i a = 1 cos a _ sin b sin c cos a -f- cos b cos c sin b sin c cos a cos (6 c) sin b sin c Applying formula (4) of Art. 73, we have sin b sin c Letting 2s = a + b-\-c, we have 8in i a= sin '6 sin c 145 146 TRIGONOMETRY Similarly uniting (1) with 2 cos 2 1 a = 1 4- cos a, we have 2 cos 2 4 = sin 6 sin c + cos -cos 6 cos c sin b sin c cos (b 4- c) cos CT sin 6 sin c 2 sin.a + b c sin- a sin 6 sin c Therefore cos^ a=J sin * h (*-). (3) * sin b sin c Uniting (2) and (3), ten 1 q = J 8in ~ *) Sin (' ~ C ) = tan ** (4) ^ sin 5 sin (s a) sin (s a) Similarly tan * p = J sin ( J ~ c > f < ~ ">= . ^ n \. (5) \ sin . sin fx h\ sin <^.c A^ ^ ' and tan H = J 8in (^ ~ ") 8in ( ^ sin 5 sin (s 6) sin (s b) tan sin 5 sin (5 (?) sin (5 c) where tan r = /sin (s - a) sin (s - fr) sin (3 - c) . v sins 145. To find the sides when the three angles are given. Following the method of the last article, the equation cos a= cos ft cos y 4- sin ft sin y cos a gives . - sin 6 sin Y and cos J a = J(*-P) eo. (J- ^ sin p sin -y and tan J a = J ~' ct s ^ Q cos ( ^~ a ) = tan /? cos (5 -a) ^- - ctQ cos(S-P) OBLIQUE SPHERICAL TRIANGLE 147 where 2# = a tan R 146. Delambre's or Gauss's formulas express relations be- tween the six parts. By Art. 68 sin i (a + (3) = sin i a cos i'/3 + cos ^ a sin i. /?. Substituting for sin \ a, cos 1 a, sin 1 y8, and cos | /8 their values in terms of the sides of the triangle, and simplifying, we have sin i (a 1 ff) = sin ( s ~ 6 ) + sin (* ~ a ^ A sln S Sln ( .' sin c ^ sin a si 2 sin c cos c sin b *r- Then sin \ (a + p) = cos i (a -6) . CQs ^ ^ (1) cos^c Similarly sin (a - p) = sin ^ g ~^ . cos i Y , (2) sin^c and cos J- (a + p) = . 8in J Y , (3) COS C and co8i(a-p) = .si n | Y . (4) 147. Napier's analogies express relations between five parts of a triangle. They are easily obtained from Gauss's formulas. Dividing (1) by (2), Art. 146, 8 in|(a_p) tan|(a-6) Dividing (3) by (4), . tan c cos|(a-p) 148 TRIGONOMETRY Dividing (4) by (2), cotjH Dividing (3) by (1), 148. Formulas collected. The following formulas are suf- ficient to solve a spherical triangle when any three parts are given : c 1nr sin s sin (s a) sin (s a) ) = tan sin !(<* + /?) _ tan^c sin i (a ft) tan 1 (a 6) cos - + ) tan c cos i (a ./8) tan 1 (a + 6) sin 1 (a + 6) _ cot ^- y sin i (a 6) tani( a ^) 1-6) tanf sin a _ sin b sin a sin /? VII Formula I is used to determine an angle when the three sides are given. Formula II is used to determine a side when the three angles are given. Formulas III and IV are used when two angles and the * These formulas are typical. Other formulas of the same type are obtained by a cyclic change of letters. OBLIQUE SPHERICAL TRIANGLE 149 included side are given. Formula III determines ^ (a b) and formula IV determines -j- (a -j- b), from which a and b are obtained. Either formula may also be used to deter- mine the side c when the other two sides and their opposite angles are given. Formulas V and VI are used when two sides and the in- cluded angle are given. Formula V determines ^ (a ft) and formula VI determines J (a + (3), from which a and ft are obtained. Either formula may also be used to deter- mine the angle y when the other two angles and their op- posite sides are given. Formula VII is used when an angle and the side opposite are among the given parts. 149. Whenever the formulas I to VI are employed in the solution of a spherical triangle as indicated above, the quadrant in which any part terminates may always be deter- mined by noticing the signs of the functions involved. But when the law of sines is employed, two values are found for the required part. This leads to two solutions unless limited to one solution by the following principles. 150. Theorem. Half the sum of any two angles is in the same quadrant as half the sum of the sides opposite. This follows from a consideration of the signs of the functions involved in cos \(a + ft) _ tan^- c cos %(a ft) tan |(a -f b) ' Since each part is less than 180, tan 1 c and cos ^( ft) are always positive. Hence cos|( + /?) an( i tan i (a + 6) must always have the same sign. Hence i(a + ft) and |-(a + b) terminate in the same .quadrant. 151. Theorem. A side which differs more from 90 than an- other side, terminates in the same quadrant as its opposite angle. 150 TRIGONOMETRY We have from Art. 119 cos a cos b cos c cos a sin b sin c If a differs more from 90 than 6, cos a is numerically greater than cos b. Cos a is also greater than cos b cos c, since cos c is not greater than unity. Hence the numerator of this fraction has the same sign as cos a. The denominator being always positive, cos a and cos a have the same sign. Hence a and a terminate in the same quadrant. The negative of this theorem is not true. Thus given, a = 165, b = 120, ft = 135, a terminates in the second quadrant, since a differs more from 90 than 6. 152. Theorem. An angle which differs more from 90 than another angle, terminates in the same quadrant as its opposite side. This follows from sff cosy sin ft sin y by considerations similar to those of the previous article. Thus given, a = 80, y = 140, and a = 120, c terminates in the second quadrant, since y differs more from 90 than a. 153. Illustrative examples. 1. Given the three sides, a = 105 27' 20", 6 c = 96 53' 10", to find a, ft, and y. 83 14' 40", a 105 27' 20" b 83 14' 40" c 96 53' 10" 2s 284 94' 70" s 142 47' 35" s a 37 20' 15" s-b 59 32 '55" s c 45 54' 25" Check-: s 142 47' 35" tan r = J sin (-) sin (s-&)sin Q-c) * sins tan \ a = tanr tan \ 8 = sin (s a) tanr sin (s b) tan r sin (s c) OBLIQUE SPHERICAL TRIANGLE 151 log sin (s a) 9.78284-10 log sin (s 6) 9.93553-10 log sin (s c) 9.85623-10 colog sin s 0.21846 2 log tan r 9.79306-10 log tan r 9.89653-10 log tan A a 0.11369 log tan A /3 9.96100-10 log tan A 7 0.04030 i 52 24' 55" i 42 25' 51" IT 47 39' 17" 0E 104 49' 50" 84 51' 42" T 95 18' 34" Check ^ U2/ siiiK-&) a + b a-b *(' + ) *(*-&) K-0) 188 42' 22 12' 40' 94 21' 0' 11 6' 20' 9 59' 4' log sin (a + &) log tan (-) colog sin A (a 6) log cot 7 log tan 1 7 9.99875 - 10 9.24563 - 10 0.71531 9.95969 - 10 0.04031 2. Given two sides and the included angle, a = 29 18', b = 37 30', y = 51 52', to find a, (3, and c. *>=!!! J ( ,g + ??*K6-) b 37 30' a 29 18' y 51 52' b-a 8 12' b + a 66 48' 4 (ft - ) 4 6' i (& + a ) 33 24' 17 25 56' log sin |(6 a) 8.85429 -10 log cot A 7 0.31310 colog sin i(6 + a) 0.25926 log tan ( cc) 9.42665 - 10 1 03 - a) 14 57' 14" log cos A (& _ a) log cot \ 7 colog cos \ (6 + a) log tan A (|8 + ) 103 + 0) a. 9.99889 - 10 0.31310 0.07839 0.39038 67 51' 8" 14 57' 14" 82 48' 22" 52 53' 54" log tan \ (b - a) log sin A (|8 + a) colog sin A (^3 ) log tan A c c 8.85540 - 10 9.96671 - 10 0.58831 9.41042 - 10 14 25' 43" 28 51' 26" 152 TRIGONOMETRY Check sin a _ sin 6 _ sin c sin a sin j8 sin 7 log sin a log sin a 9.68965 9.90177 -10 -10 log sin 6 log sin /3 9.78445-10 0.99656 - 10 log sin c log sin 7 9.68361 9.89574 -10 -10 9.78788 -10 9.78789 - 10 9.78787 -10 3. Given two sides, and an angle opposite one of them, a = 63 24' 50", b = 17 36' 40", a = 44 48' 20", to find ft y, and c. sin a taDic = *'"}( g + ^tanK-6) a 63 24' 50" log sin J (a -f &) 9.71445 - 10 a 44 48' 20" log tan K - /3) 9.57083 - 10 b 17 36' 40" colog sin \ (a ft) 0.62876 a + b 62 25' 0" log cot \ 7 9.91404 - 10 a-b 27 11' 40" i 7 50 38' 0" \ (a + 6) 31 12' 30" 7 101 16' 0' ( a _ 5) 13 35' 50" log sin a 9.95147 - 10 log sin b 9.48081 - 10 log sin (a + ) 9.83375 - 10 colog sin a 0.15200 log tan (a &) 9.38359 - 10 log sin /3 9.58428 - 10 colog sin \ (a /3) 0.45735 /3 22 34' 44" . log tan , c 9.67469 - 10 a /3 40 50' 6" j C 25 18' 20" + /3 85 59' 34" c 50 36' 4(X !(*-/) 20 25' 3" i (a + /3) 42 59' 47" Cftecfc * sin ft _ sin c sin /3 sin 7 log sin 6 9.48081 - 10 log sin c 9.88810 - 10 log sin 9.58428- 10 log sin 7 9.99155 - 10 9.89653 - 10 9.89655 - 10 OBLIQUE SPHERICAL TRIANGLE 153 154. Two solutions. There are two solutions, if any, when- ever two sides and an angle opposite one of them, or two an- gles and a side opposite one of them, are given, unless limited to one solution by the principles of Arts. 150, 151, and 152. Thus, having given = 45 15' 12", b = 56 49' 46", a = 68 52' 48", to find a, y, and c. sin 6 a b 68 52' 48" 56 49' 46" 45 15' 12" a a + P a-p ci + b a-b fo-&) 52 19' 33" 97 34' 45" 7 4' 21" 48 47' 22" 3 32' 10" 125 42' 34" 12 3' 2" 62 51' 17" 6 1'31" 127 40' 27" 172 55' 39" 82 25' 15" 86 27' 50" 41 12' 38" log sin p log sin a colog sin b . log sin a a 9.85140 9.96980 0.07725 9.89845 52 19' 33", or 127 40' 27" log sin ( + ) log tan $ (a - 6) colog sin & (a /9) log tan c I c 9.87639 9.02346 1.20984 9.99917 9.02346 0.18124 0.10969 52 9' 35" 104 19' 10'' 9.20387 9 5' 7" 18 10' 14" log sin (a + 6) logtanO-) colog sin } (a ft) log cot 1 7 i7 7 9.94932 8.79099 0.97895 9.94932 9.94238 0.97895 9.71926 6221'1" 124 42' 2" 0.87065 7 40' 16" 15 20' 32" 154 TRIGONOMETRY Check cosJ(a-/8) _ ; logcos|(a + j3) log tan %(a + 6) cologcos(ot /3) log tan \ c 9.81877 0.29012 0.00083 8.79013 0.29012 0.12361 0.10972 &) tan^r 9.20386 a 4- B) cos J(a - b) FIRST SOLUTION = 52 19' 33" c = 104 19' 10" = 12442' 2" SECOND SOLUTION a = 127 40' 27" c = 18 10' 14" y = 15 20' 32" 155. Area of spherical triangle. Representing the area of a sphere, S, by 720 spherical degrees, it is demonstrated in geometry that the area of a spherical triangle, A, in terms of spherical degrees, is equal to its spherical excess ; or >f=(a + p+y 180) spherical degrees -' log cos ( a + 6) log tan ( + ) cologcos (o &) log cot 7 9.65920 0.05762 0.00241 9.65920 0.20904 0.00241 9.71923 0.87065 Let A' and S' represent the area of the triangle and the sphere respectively, in terms of the unit in which r is ex- pressed. Since the ratio of the area of the spherical tri- angle to the area of the sphere is independent of the units used, we have A' __A Therefore But the area of the sphere expressed in terms of r is 4 ?rr 2 , therefore the area of the triangle is given by A' = 180 156. EXAMPLES Solve the following spherical triangles and check the results : OBLIQUE SPHERICAL TRIANGLE 155 1. 6= 10 O'lO" c = 114 40' 40" a= 92 28' 20" 11. = 61 8' J3= 59 12' y= 78 25' 2. b= 85 4' 19" c = 139 58' 25" = 12 20' 31" 12. a= 76 43' 15" b= 83 35' 27" c= 98 26' 38" 3. c= 82 3' 4" a= 70 14' 12" = 84 20' 9" 13. = 11035' ^8 = 135 42' y = 146 8' 4. a=, 95 3' 30" b = 128 38' 50" y = 170 52' 20" 5. o= 29 18' 6= 37 30' y = 51 52' 6. a= 11 21' 10" y = 19 0'20" 6= 66 19' 30" 7. y = 179 22' 11" a = 148 17' 17" 6= 25 39' 34" 8. a = 108 5' 18" 6 = 170 30' 46" c= 85 50' 22" 9. a = 105 27' 20" 6= 83 14' 40" c= 96 53' 10" 10. a= 34 19' 30" 6= 28 37' 10" c= 22 44' 40" 14. y= 11 34' 10" 6= 82 56' 30" c= 27 9' 40" 15. a= 32 4' 10" /? = 12856'20" a= 39 50' 30" 16. /?= 80 40' 2" c= 75 54' 0" 6 = 100 21' 28" 17. a= 21 I'lO" y= 17 22' 50" a= 14 13 '30" 18. )8= 77 44' 55" y= 92 17 '24" a= 26 29' 39" 19. a = 114 23' 9" 0= 88 41' 11" y= 79 0' 4" 20. 0= 99 4' 12" y = 106 0' 9" a = 161 2' 10" 156 TRIGONOMETRY 21. c = 100 10' 40" 23. b= 42 15' 20" a= 65 20' 30" c = 127 3' 30" y= 94 30' 10" (3= 31 44' 20" 22. a = 103 19' 50" 24. y = 127 4' 10" /?= 92 37' 30" a= 88 12' 0" y = 128 54' 20" c = 141 20' 30" SOLUTION WHEN ONLY ONE PART IS REQUIRED 157. In many problems of astronomy and geodesy, it is re- quired to find only one or two of the unknown parts of a spher- ical triangle, the other unknown parts being of no importance in the problem. It then becomes desirable to have a method whereby the required parts can be computed without being under the necessity of first computing any part not desired. It has already been shown that any angle can be found directly from three given sides (Art. 144), and that any side can be found directly from three given angles (Art. 145). It is evident that any part can be found from any three given parts by the use of the general formulas containing the required part and the three given parts. By the intro- duction of auxiliary quantities, these formulas will now be adapted to logarithmic computation. 158. Given two sides and the included angle, to find any one of the remaining parts. Let a, b, y be the given parts. First. To find c. The relation between a, 6, y, and c is (Art. 119), cos c = cos a cos 6 -f sin a sin b cos y. (1) To adapt this formula to logarithmic computation, let m sin M= sin b cos y, (2) and m cos M = cos b. (3) OBLIQUE SPHERICAL TRIANGLE 157 Then eliminating b by uniting (1), (2), and (3), we have cos c = m (cos a cos M+ sin a sin M ), or cos c = m cos (a M ). From (2) and (3) tan Jf = tan 6 cos y, and from (3) and (4) cos b cos (a M") cos c = = t cos (4) (5) (6) Equations (5) and (6) enable us to find c. ILLUSTRATION. Given o= 75 38' 20", & = 54 54' 38", and = 30 17' 43". tan M = tan & cos 7 a 75 38' 20" 6 54 54' 38" 7 30 17' 43" log tan 6 log cos 7 log tan Jf M 0.15333 9.93623 0.08956 50 51' 58" a- M 24 46' 22" cos c = cos 5 cos (a -Jf) cos Jf log cos b log cos (a M) colog cos M log cos c c 9.75956 9.95808 0.19987 9.91751 34 12' 27" Second. To find ft. The relation between a, b, y, and ft is given by Art. 124, (equation 5), from which we have , Q _ cot b sin a cos a cos y x^ siny Multiplying numerator and denominator by sin b, , ~_ cos b sin a sin b cos a cos y sin b siiiy Again using equations (2) and (3) we have m (cos M sin a sin Jf cos a) sin 6 sin y (8) , cot = or sin b sm y (9) (10) 158 TRIGONOMETRY As before tan M = tan 6 cos y. (11) From (2) and (10) . (12) . sin M Equations (11) and (12) enable us to find ft. Third. To find y. By interchanging a and 6 and consequently a and ft in (11) and (12) we have, calling the auxiliary angle N 9 tan JV= tan a cos y (13) and cota = tysin(6-^) (14) sm^" to determine a. 159. Two parts required. It will be noticed that the same auxiliary quantity M is used to find both c and ft. We thus have a convenient method, much used in astronomy, for finding a side and an angle when two sides and the included angle are given. For finding c and ft we have, collecting our formulas, tan M = tan 6 cos y cos M cot/?= cotysin(q-Jf) t sin M Dividing equation (10) of Art. 158 by equation (4), we have cot/?_tan (a Jf) cos c sin b sin y ' which serves as a check upon c and ft. Similarly, for finding c and a, we have OBLIQUE SPHERICAL TRIANGLE tan N= tan a cos y cos a cos (b N} cos c = 5s L cos N 159 sin N 160. PROBLEMS An arc of l r on the earth's surface is equal to one English geographical mile. 1. Find the distance between Boston, latitude 42 21' N., longitude 71 41' W., and San Francisco, latitude 37 48' N. and longitude 122 28' W. P SOLUTION. Let APD be the me- ridian of Greenwich from which longi- tude is measured, ABCD the equator, and P the north pole. Let the positions of San Francisco and Boston be represented by E and F respectively. The desired distance is EF. Then But Letting angle DPE = 122 28', angle DPF = 71 41', arc EB= 37 48', arc CF= 42 21', arc PB = arc PC = 90. PE=PB-EB = 52 12', PF = PC - FC = 47 39', Z FPE = /. DPE - Z DPF = 50 47'. Z FPE = 7, PE = a, PF= &, we may find EF or c by the formulas tan M = tan 6 cos 7, cos c = cos&cos ( cos M -M) 160 TRIGONOMETRY _ cos b cos (a M ) cos M a b y 52 12' 47 39' 50 47' log cos b log cos (a M ) colog cos M log cos c c 9.82844 - 10 9.97951 - 10 0.08526 log tan b log cos y log tan M M a- M 0.04023 9.80089 - 10 9.89321 - 10 38 33' 20", or 2313 miles 9.84112 - 10 34 44' 23" 17 27' 37'' 2. Find the distance between New York (lat. 40 43' K, long. 74 0' W.) and San Francisco (lat. 37 48' K, long. 122 28' W.). 3. Find the distance between Calcutta (lat. 22 33 f N., long. 88 19' E.) and Greenwich (lat. 51 29' N.). 4. Find the distance between Baltimore (lat. 39 17' N., long. 76 37' W.) and Calcutta. 161. Given two angles and the included side, to find any one of the rema ining parts. Let a, ft c be the given parts. First. To find y. The relation between a, ft c, and y is, Art. 121, eq. (3), cos y = cos a cos ft + sin a sin ft cos c. - (1) Let ra sin M = cos a, (2) and m cos M = sin a cos c. (3) Uniting (1), (2), and (3) cos y = m ( sin M cos /? -f cos M sin /?), or cos y = m sin (ft M). (4) From (2) and (3) cot M = tan a cos c, (5) OBLIQUE SPHERICAL TRIANGLE 161 and from (2) and (4) cos a sin (8 sinM Equations (5) and (6) enable us to find y. (6) Second. To find a. The relation between a } (3, c, and a is given by Art. 124, equation (4), from which sin 8 cot a + cos c cos 8 ^ cot a = . ( 1 1 sine , _ sin ft cos a + sin a cos c cos ft t ,^ sin a sin c Uniting equations (2), (3), and (8), cot a = ^ " "" ^ ' , (9) sin a sin c sin a sin c From (2), (3), and (10) cot M = tan a cos c (11) cot c cos (8 Jf) /., \ and cota = ^ L , (12) cos Jf from which a is found. To find b. From (11) and (12) by interchanging a and /?, and conse- quently a and 6, we have, calling the auxiliary quantity JV, cot N= tan cos c (13) and cot6 = - (14) cos N to determine &. 162. Given two sides and an angle opposite one of them, to find any one of the remaining parts. Let a, bj a be the given parts. 162 TRIGONOMETRY First. To find c. The relation between a, b } a and c is cos a = cos b cos c + sin 6 sin c cos a. (1) Let m sin Jf = sin b cos a, (2) and m cos Jf = cos 6. (3) Then cos a = m cos (c M ). (4) From equations (2), (3), and (4), tan M = tan b cos a (5) and cos (c - M) = Cos a cos ^ (6) cos b Equation (5) determines M and equation (6) determines c M. Adding these values, we have c. In general there are 2 solutions for c. We may limit M to positive values less than 180. By equation (6) c M may have two values, numerically equal but opposite in sign, giving two values for c unless the sum M -\- (c M ) is greater than 180 or negative, in which case there is but one solution. Second. To find y. The relation between a, 6, a, and y is sin y cot a = cot a sin b cos b cos y. Multiplying by sin a and rearranging, we have sin y cos a + cos y sin cos b = sin a cot a sin 6. (7) Let n cos JV= cos a, (8) and n sin ^= sin a cos 6. (9) From (7), (8), and (9) we have n sin (y -f- JV) = sin a cot a sin 6. (10) Then from (8), (9), and (10), tan N== tan a cos 6, (11) and sin (y -f JV) = sin N cot a tan 6. (12) Equation (12) determines y + N and equation (11) deter- mines H. Subtracting the second value from the first gives y. OBLIQUE SPHERICAL TRIANGLE 163 In general there are two solutions, since y + N may have two values. Third. To find ft. The angle ft is found from . sin 6 sin a ,* o\ sm ft = : , (lo) sin a which in general gives two values. 163. Given two angles and a side opposite one of them, to find any one of the remaining parts. Let a, ft, a be the given parts. First. To find y. The relation between a, ft, a, and y is cos a = cos ft cos y + sin ft sin y cos a. (1) Let m sin M = cos ft, (2) and m cos M = sin ft cos a. (3) Then cos a = msin(y M). (4) From (2), (3), and (4) cot M = tan ft cos a, (5) -, . / , f ^ cos a sin J/" //>. and sm(y M)= (6) cosp Equations (5) and (6) determine M and y M, from which y is found. Two solutions may be possible, as in Art. 162. Second. To find c. The relation between a, ft, a, and c is sin ft cot a = cot a sin c cos c cos /?. (7) Multiplying by sin a and transposing, cos a sin c sin a cos c cos /3 = sin /? sin a cot a. (8) Let n cos JV= cos a, (9) and w sin JV= sin a cos ft. (10) 164 TRIGONOMETRY Then uniting (8), (9), and (10) n sin (c JV) = sin ft sin a cot a. (11) From (9), (10), and (11) tan N= tan a cos /?, (12) and sin (c N) = tan ft cot a sin N. (13) Equations (12) and (13) determine N and c JV, from which c is found. There may be two solutions. Third. To find b. We have sin b = sin ? sin * (14) Bin There are two values of b unless restricted to one solution by the principles of Arts. 150, 151, and 152. 164. The general triangle. The parts of the general spheri- cal triangle are not restricted to values less than 180. It can be shown that all the formulas developed for the oblique spherical triangle are true for the general spherical triangle if the double sign is introduced in the formulas of Arts. 144, 145, and 146. ANSWERS Art. 19 ; Page 10 5. 114 35' 28", 286 28' 40", etc. 6. 60, 135, - 300, 57 17' 44", 36 28' 31", etc. 7. 7ift. 8. 2f radians, 137 .30' 34". 12. 247.16 E. P. M. 25.882. 9. 94 3' 24". 13. 18.33 mi. per sec. 10. ifTrft. 14. 5236. 11. 2.7216 radians. 15. 9.6 TT 240 it ft. per min. Art. 27 ; Page 20 1. 3d and 4th. 2. 1st and 4th. 3. 1st and 3d 7. 2d. 8. 3d. 20. -?^. 21. -^. o o 28. sin r = -g% V85, cos ^ = ^ V85, cot a 1 = ^ ) sec ! = ^ V85, esc 06! = ^- V85, sin 2 = -^- V85, cos 2 = -g 7 - A/ 85, cot 02 = -J-, sec 2 = -J- V85, esc c^ = 1 V85. 35. if Art. 36 ; Page 29 1. 6 = 16.5 3. c-=.869 5. a = 27 c = 17.5 a = .225 ft = 63 ft = 70. a = 15. b = 72.6 2. a = 1.44 4. c = 65 6. a = 18 b = 2.05 a = 23 ft = 71| = 55. = 67 & = . 00867 165 166 TRIGONOMETRY 7. a = 346 8. a = 27 9. a = .029 10. a c = 507 ft = 63 b = .089 a = 8.11 = 47 a = 3.50 0=72 6 = 9.48 11. 6 = 5161 c = 5489 =70 5' 14. c=. 00006294 a = 72 26' = 17 34' 12. a = .1384 6 = .2878 ft = 64 19' 13. a = 1.446 c = 1.719 a = 57 17' 15. b = 810.80 16. 17. etc. a = 47 31' 32" check your = 42 28' 28" results. 31. Base 1331.1, vertical angle 149* 19' 10". 32. Base angles 39 23' 56", base 1477.0. 33. Equal sides 1622.9, base angles 37 59' 37". 34. Equal sides 219.75, base angles 68 27' 19". 35. 37.504. Art. 38; Page 31 1. 290.83 ft. 7. 2nrtai i 12. 54.775 mi. 2. 405.24 ft. n 13. 153.72 Ibs. 3. 263.92 ft. 8. 130.99 ft. 38 31' 46". 4. 289.93 21.442 ft. ft. 9. 10. 226.11 2572.5 ft. ft. 14. 739.38 hi. mi. per 5. 2 nr sii 180 11. 21.360 22.638 in. in. 15. 16 ft. l& o Z in. n 6. 132.52 ft. 69 26' 36". 16. 35 16'. Art. 50; Page 46 2. - cos 10, - sin 80. 5. cos 20, sin 70. 3. - cot 20, - tan 70. 6. - tan 80, - cot 10. 4. -.cot 20, - tan 70. 7. - sin 60, - cos 30. 8. cos B. 9. -tan B. 10. tan0. 11. -cos ANSWERS 16T 11. 12. 15. 18. 20. 21. 35. 37. 33. 39. 41. Art. 56 ; Page 51 _, tan = - V5, cot0 = -V5, sec = | V5, esc = f . sin = ^VUfl, cos = T |^ V149, tan = J^, sec = I V149, esc = T V VI49. z = 30. 16. = 45. 17. z=0, 60. = 45, 60. 19. sina = ^-i. + V5. Identity. 22. = 120. 24. Identity. x = 60, 120. 23. = 30. 25. y = 30, 150. a = 30, 150. 36. a = 30, 60, 120, 150. x = 45, 135. 38. x = 45. Art. 60; Page 57 tan 2 a-l. 34. sinzTVl-sin 2 *. 35. 1 + COS X tan 2 # + l 1 cos 2 a; ' vr r o 2 ' _i_->/1 __ /S 1 1 -, sec0 a Vl - a 2 COS0 cot0 CSC0 = , sec0 COS0' 43. 50. 55. 58. 59. 60. ? 47. 0, 90. 48. 135. 49. 0, 30, 60, 150. 2 30, 150. 51. 45. 54. 36 52'. (See tables.) 46 24', 90. 57. 65 54', 114 6'. 0, _ 30, - 150. tanu=|V5, iV2. 61. 195, 345. j, |V2. 62. 54, 234. 168 TRIGONOMETRY Art. 70 ; Page 66 2. ^V2(V3 - 1). 3. iV2( V3 - 1). 4. 3+V3 1). 5. 1. 2. 3. 32. 34. 35. 1. 3. 5. 7. 10. 13. 14. 20. 3-V3 7. cos ft. 8. cot a. 9. cos a. Art. 75 ; Page 71 * V2 - V2, fV2 + V2, V3 - 2V2, V3 + 2 V2. |V15, - |, | V15, 20 /O O /?^O O/W H fr .raj;;- 400 65 !; 2 30, 90, 150. 42. 0, 30, 90, 150. 0, 120. 43. 0, 90, 120. Art. 85 ; Page 82 60, 120, 420, etc. 2. 150, 210, - 150, etc. 45, 225, - 135, etc. 4. A. 6. u Vl 4- % 2 2 ?^ Vl - Vw 2 + 1 9. vr^: u. i u 15. 2 Vw 2 - 1 v*v^/l u 2 . 21. Vl u 2 Vl v 2 uv. ANSWERS 169 - V 2 ' V 2u 2Q SoT TTTTOV T 1 1 30 z a - cot 31. 1 1 34. .: V cot" 1 a a rt 2 VI + a 2 32 VT=rf-aV .4 a . \Xl-62_j_5Vl-a 2 35. -i (1 + Va 2 1 a& V6 2 -l). Art. 101 ; Page 98 1. 6 = 24.5 c = 31.8 y=78. 2. a & c = 50 = 54.9 = 58.6 3. a = 61.4 c = 47.7 y = 48. 4. /? y c = 48 20' = 62 40' = 76.1 5. = 68 22' y = 56 58' a = 107 6. c = 40 48' = 104 52' = 141. 7. a = 69 22' = 38 38' c = 42.7 8. ft y c = 38 37' = 31 23' = 173. 9. a = 44 42' = 60 20' y = 74 54' 10. a y = 72 21' = 49 38' = 58 V 11. /3 = 34 44' or y = 125 16' or c = 35.8 or 146 16' 12. a 13 44' ft 10.4 6 = 51 44' or 128 16' = 84 26' or 7 54' = .431 or .059 170 TRIGONOMETRY Art. 101 ; Page 98 13. y = 39 49' 50" 14. a = 13.081 a = 41.581 c = 13.620 c = 41.432 /? = 972'15" 15. a = 90 20' 34" 16. a = 126 59' 18" ft = 65 17' 34" ft = 27 29' 38" y= 24 21' 50" y = 2531'4" 17. a = 1.9555 18. y = 77 22' 16" a = 43 36' 35" ft = 51 57' 20" ft = 15 37' 7" a = 83.732 19. a = 87 33' 58" 20. a = 78 40' 32" y = 18 6' 24" ft = 41 20' 47" b = 1.4033 y = 59 58' 41'^ 21. a = 32 24' 0" 22. Impossible. ft = 55 0' 28" y = 92 35' 32" 23. a = 142 28' 9" or 21 31' 11" ft = 29 31' 31" or 150 28' 29" a = .080746 or .04862 24. y = 76 50' 20" or 103 9' 40" a =69 39' 35" or 43 20' 15" a = 17.405 or 12.739 25. c = 502.28 26. a = 96 9' 32" b = 300.25 . ft = 41 11' 10" y = 23 7' 3" y = 42 39' 18" 27. .4 = 150. 30. ^ = 108.61 33. A = 368.91. 35. 172.8 ft. 36. 106.1ft. 37. 3.710 mi. 38. 97.14 124.59 39. 60.1ft. 178.64 40. 57.93 mi. per Lr. -41. 3888.0ft. 42. 6328.7 ft. 43. 239600 mi. ANSWERS 171 Art. 102 ; Page 102 1. 4. 4258 / 18 // . 5. 35 48' 35". 6. 13.754 in. 7. . 8. 2. 60',72M35'. 3. , , , 7T 9* 7T 7T 7T 6' 3' 2 55 7T 73 7T 144 144 9. 4 7T 9 ' 3' 9 12. 50. 51. 54. 56. 57. 59. 60. 61. 62. 63. 64. 66. 67. 68. 69. 73. 11 tE 10?r 16 22 TT 28 TT 34 TT 40 * ' 77 ' 77 ' 77 ' 77 ' 77 ' 77 ' 77 ' 9fin. 49. 0, 180. 30, 150, 210, 330, 45, 135, 225, 315. 30, 150. 52. tan-\(2 V3). 53. 0, 180, cos' 1 f 45, 135, 225, 215, sin' 1 Vf 55. 60, 300, 90, 270. 60, 120, 240, 300, 45, 135, 225, 315. tan- 1 !. 58. 90, 270, sin- 1 (-|i). 0, 45, 90, 135, 180, 225 a , 270, 315. 0, 45, 60, 90, 120, 180, 225, 240, 270, 300, 315. 7i, 37i, 67$., 97i, 127^, 1571, etc. 30, 150, 210, 330. 90, 270, 70, 110, 190, 230, 310, 350. 90, 180. 65. 210, 330. 45, 215, 671, 1571, 247f, 337^. 60, 120, 240, 300, 18, 54, 90, 126, 162, etc. 60, 90, 120, 240, 270, 300. 70. 0, V|. 71. 172 TRIGONOMETRY 99. 532. 100. 50.0. 101. 1478.5. 102. 93,470,000 mi. 103. 51.9 ft. 104. 27.925. 105. 112.36. 106. - 107. 129.90 ft. 106. f 108. 8.2596. 109. 70 31' 43". 111. 196. 112. 89.431. 115. 90 ft., 40 ft. 116. 13.66. 118. 820.54. 119. 535.4. 121. 567.3. 122. 132.6. 124. 641. 125. 962.605. 128. a. 129. 4009. 110. 5296 ft., 251 ft. 114. V2. 117. 103.97. 120. 25.43. 123. 36.7 ft. 127. 16.33 N. 75 36' E. 130. 1674.3. Art. 116 ; Page 124 1. 1, i, 1, i. 2.1, i 4. (2 + 0- 6. (1.272 + .786 1). 7. 8. _ 10. 1.0842 + .29051 i, - .79370 + .79370 1, - .29051 - 1.0842 1. 5. (l-2t). Art. 139 ; Page 143 1. /3 = 789'22" c = 10 45' 55" a = 2 14' 5" 3. 6 = 8 26' 14" a = 120 59' 19" a = 95 2' 10" 2 ? a= 41 11' 53" = 56 19' 56" c = 40 27' 11" 4. ^ = 75 21' 53" 6 = 44 43' 49" a = 14 59' 33" ANSWERS 173 5. a = 111 23' 47'.' 6. ft =101 38' 28" = 120 40' 56" a = 112 13'. 48" c = 76 33' 24" b = 102 35' 26" 7. = 46r28" 8. a = 68 42' 11" b = 15 18' 0" = 15548'0" a = 46 2' 40" a = 27 37' 26" 9. p = 153 31' 29" or 26 28' 31" c = 50 43' 22" or 129 16' 38" b = 159 48' 44" or 20 11' 16" Art. 142; Page 144 1. a = 117 45' 28" 2. y = 88 23' 11" = 9627'1" a = 69 48' 42" y = 930'61" b = 94 22' 46" 3. a = 8 49' 46" 4. a =160 13' 48" y = 283'4" = 105 21' 16" b = 10.6 56' 53" y = 104 25' 45" 5. Sides, 32 45' 6" 6. Sides, 112 32' 20" angles, 105 49' 32" base, 46 15' 12" Art. 156 ; Page 154 1. = 110'47" 2. = 14 53' 47" y=928'27" y = 170 26' 51" a = 114 42' 50" a = 55 56' 0" 3. = 71r23" 6. a = 24 35' 10" y = 84 22' 25" c = 43 29' 48" 6 = 821'30" = 154 19' 20" 8. a = 133 28' 34" 9. a = 104 49' 50" = 169 38' 12" = 84 51' 42" y = 132 6' 14" y = 95 18' 24" 10. a = 84 57' 8" = 57 47' 44" v = 434'36" 174 TRIGONOMETRY 14. a = 155 14' 24" or 21 52' 40" = 25 50' 58" or 154 9' 2" a = 107 34' 50" or 58 0' 44" 15. Two solutions. 24. No solution. Art. 160 ; Page 159 2. 2229| miles. 3. 4291J miles. ^ -.. 1 .L Til ffil5 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. f, 40?MC* Si f - 1 ;JAN 15 1959 ^1 REC'D LD 24Jan'6SJP -0 NOV 30 1956 """ + LI 14 Dec'56RK REC'D LD REC'D UD JAN25'65-4PM Jf DEC 13 1956 V, H ! SjanSSKK > r - ^5("" .f"ril L MB- / REG D LD ;" '-/ DEC 2 7 1957 29 0a'59t~l LD 21-100m-6 '56 .General library VB 17137 T- , o o ..^ >;