Solution Manual for Trigonometry, 12th Edition

Chapter 2 Acute Angles and Right Triangles Section 2.1 Trigonometric Functions of Acute Angles 10. sin A  For Exercises 1–6, refer to the Function Values of Special Angles chart on page 170 of the text. cos A  1. C; sin 30  1 2 2 2. H; cos 45  2 3. B; tan 45  1 4. G; sec 60  1 1  2 cos 60 12 5. E; csc 60  1 1 2   3 sin 60 3 2 2 3 2 3    3 3 3 6. A; cot 30  7. cos 30  sin 30 3 2 1 2 side opposite 21 sin A   hypotenuse 29 side adjacent 20  hypotenuse 29 side opposite 21 tan A   side adjacent 20 cos A  8. 9. side opposite 45  hypotenuse 53 side adjacent 28 cos A   hypotenuse 53 side opposite 45 tan A   side adjacent 28 sin A  side opposite n  hypotenuse p side adjacent m cos A   hypotenuse p side opposite n tan A   side adjacent m sin A   3 2   3 2 1 side opposite k  hypotenuse z side adjacent y  hypotenuse z side opposite k tan A   side adjacent y 11. a = 5, b = 12 c 2  a 2  b 2  c 2  52  122  c 2  169  c  13 side opposite b 12 sin B    hypotenuse c 13 side adjacent a 5 cos B    hypotenuse c 13 side opposite b 12 tan B    side adjacent a 5 side adjacent a 5 cot B    side opposite b 12 hypotenuse c 13 sec B    side adjacent a 5 hypotenuse c 13 csc B    side opposite b 12 12. a = 3, b = 4 c 2  a 2  b 2  c 2  32  42  c 2  25  c5 side opposite b 4 sin B    hypotenuse c 5 side adjacent a 3 cos B    hypotenuse c 5 side opposite b 4 tan B    side adjacent a 3 side adjacent a 3 cot B    side opposite b 4 hypotenuse c 5 sec B    side adjacent a 3 hypotenuse c 5 csc B    side opposite b 4 Copyright © 2021 Pearson Education, Inc. 115 116 Chapter 2 Acute Angles and Right Triangles 13. a = 6, c = 7 c 2  a 2  b 2  7 2  62  b 2  49  36  b 2  13  b 2  13  b side opposite b 13   c hypotenuse 7 side adjacent a 6   cos B  c 7 hypotenuse side opposite b 13   tan B  side adjacent a 6 side adjacent a 6 cot B    side opposite b 13 6 13 6 13    13 13 13 hypotenuse c 7 sec B    side adjacent a 6 hypotenuse 7 c csc B    side opposite b 13 7 13 7 13    13 13 13 sin B  14. b = 7, c = 12 c 2  a 2  b 2  122  a 2  7 2  144  a 2  49  95  a 2  95  a side opposite b 7 sin B    hypotenuse c 12 side adjacent a 95 cos B    hypotenuse 12 c side opposite b 7 tan B    side adjacent a 95 7 95 7 95    95 95 95 side adjacent a 95   side opposite b 7 hypotenuse 12 c sec B    side adjacent a 95 12 95 12 95    95 95 95 hypotenuse c 12 csc B    side opposite b 7 cot B  15. a = 3, c = 10 c 2  a 2  b 2  102  32  b 2  100  9  b 2  91  b 2  91  b side opposite b 91   hypotenuse 10 c side adjacent a 3 cos B    hypotenuse c 10 sin B  side opposite b 91   side adjacent a 3 side adjacent a 3 cot B    side opposite b 91 3 91 3 91    91 91 91 hypotenuse c 10 sec B    side adjacent a 3 hypotenuse 10 c csc B    side opposite b 91 10 91 10 91    91 91 91 tan B  16. b = 8, c = 11 c 2  a 2  b 2  112  a 2  82  121  a 2  64  57  a 2  57  a side opposite b 8 sin B    hypotenuse c 11 side adjacent a 57 cos B    hypotenuse 11 c side opposite b 8 tan B    side adjacent a 57 8 57 8 57    57 57 57 side adjacent a 57   side opposite b 8 hypotenuse 11 c sec B    side adjacent a 57 11 57 11 57    57 57 57 hypotenuse c 11 csc B    side opposite b 8 cot B  17. a = 1, c = 2 c 2  a 2  b 2  22  12  b 2  4  1  b2  3  b2  3  b side opposite b 3   hypotenuse c 2 side adjacent a 1 cos B    hypotenuse c 2 side opposite b 3 tan B     3 side adjacent a 1 side adjacent a 1 cot B    side opposite b 3 1 3 3    3 3 3 sin B  Copyright © 2021 Pearson Education, Inc. (continued on next page) Section 2.1 Trigonometric Functions of Acute Angles (continued) hypotenuse c 2   2 side adjacent a 1 hypotenuse 2 c csc B    side opposite b 3 2 3 2 3    3 3 3 sec B  18. a  2, c  2 2 c 2  a 2  b 2  22  2  b 2  4  2  b2  2  b2  2  b side opposite b 2   hypotenuse 2 c side adjacent a 2 cos B    hypotenuse 2 c side opposite b 2 tan B    1 side adjacent a 2 side adjacent a 2 cot B    1 side opposite b 2 hypotenuse 2 c sec B    side adjacent a 2 2 2 2 2     2 2 2 2 2 c hypotenuse   csc B  side opposite b 2 2 2 2 2     2 2 2 2 sin B  19. b = 2, c = 5 c 2  a 2  b 2  52  a 2  2 2  25  a 2  4  21  a 2  21  a side opposite b 2 sin B    hypotenuse c 5 side adjacent a 21 cos B    hypotenuse 5 c side opposite b 2 tan B    side adjacent a 21 2 21 2 21    21 21 21 side adjacent a 21 cot B    side opposite b 2 hypotenuse 5 c sec B    side adjacent a 21 21 5 21 5    21 21 21 hypotenuse c 5 csc B    side opposite b 2 117 20. sin   cos 90    ; cos   sin 90    ; tan   cot 90    ; cot   tan 90    ; sec   csc 90    ; csc   sec 90    21. cos 30  sin 90  30  sin 60 22. sin 45  cos 90  45  cos 45 23. csc 60  sec 90  60  sec 30 24. cot 73  tan 90  73  tan17 25. sec 39  csc 90  39  csc 51 26. tan 25.4  cot 90  25.4  cot 64.6 27. sin 38.7  cos 90  38.7  cos 51.3 28. cos   20  sin 90    20  sin 70    29. sec   15  csc 90    15  csc 75    30. Using   50, 102, 248, and  26, we see that sin 90    and cos  yield the same values. For exercises 31–40, if the functions in the equations are cofunctions, then the equations are true if the sum of the angles is 90º. 31. tan   cot   10     10  90 2  10  90 2  80    40 32. cos   sin  2  30   2  30  90 3  30  90 3  120    40 Copyright © 2021 Pearson Education, Inc. 118 33. 34. 35. Chapter 2 Acute Angles and Right Triangles sin  2  10  cos 3  20 2  10  3  20  90 5  10  90 5  100    20 sec    10  csc  2   20  10      2   20  90 3  30  90 3  60    20 tan 3  4  cot 5  10 3  4  5  10  90 8  6  90 8   96    12 43. sin 46  cos 46 Using the cofunction identity, cos 46  sin 90  46  sin 44 . In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 46  sin 44  sin 46  cos 46 is false. 44. cos 28  sin 28 Using the cofunction identity, sin 28  cos 90  28  cos 62 . In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so cos 28  cos 62  cos 28  sin 28 is false. cot 5  2  tan  2  4 5  2  2  4  90 45. tan 41  cot 41 Using the cofunction identity, cot 41  tan 90  41  tan 49 . In the 37. sin   20  cos 2  5   20  2  5  90 3  15  90 3  105    35 46. cot 30  tan 40 Using the cofunction identity, cot 30  tan 90  30  tan 60 . In the 38. cos  2  50  sin  2  20 2  50  2  20  90 4  30  90 4  60    15 36. 7  6  90 7  84    12 39. sec 3  10  csc    8 3  10     8  90 4   18  90 4   72    18 40. csc    40  sec    20  40         20  90 2   20  90 2   70    35 interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan 41  tan 49  tan 41  cot 41 is true. interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan 60  tan 40  cot 30  cot 40 is false. 47. sec 60  sec 30 In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so the secant of the angle increases. Thus, sec 60  sec 30 is true. 48. csc 20  csc 30 In the interval from 0º to 90º, as the angle increases, sine of the angle increases, so cosecant of the angle decreases. Thus csc 20  csc 30 is false. Use the following figures for exercises 49–64. 41. sin 50  sin 40 In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 50º > sin 40º is true. 42. tan 28  tan 40 In the interval from 0º to 90º, as the angle increases, the tangent of the angle increases, so tan 40º > tan 28º  tan 28  tan 40 is true. side opposite 1  side adjacent 3 1 3 3    3 3 3 49. tan 30  Copyright © 2021 Pearson Education, Inc. Section 2.1 Trigonometric Functions of Acute Angles 50. cot 30  side adjacent 3   3 side opposite 1 side opposite 1  51. sin 30  hypotenuse 2 52. cos 30  65. Because sin 60  and 90°, A  60. 119 3 and 60° is between 0° 2 66. 0.7071067812 is a rational approximation for 2 (an irrational value). the exact value 2 side adjacent 3  hypotenuse 2 67. hypotenuse 2  side adjacent 3 2 3 2 3    3 3 3 53. sec 30  54. csc 30  hypotenuse 2  2 side opposite 1 55. csc 45  hypotenuse 2   2 side opposite 1 56. sec 45  hypotenuse 2   2 side adjacent 1 side adjacent 1  hypotenuse 2 1 2 2    2 2 2 58. cot 45  side adjacent 1  1 side opposite 1 59. tan 45  side opposite 1  1 side adjacent 1 side opposite 1  hypotenuse 2 1 2 2    2 2 2 3 x. 3   The line passes through 0, 0 and 1, 3 . 0, 0 . Let  x, y  be any other point on this side opposite 3   3 side adjacent 1 hypotenuse 2  side opposite 3 2 3 2 3    3 3 3 equation of the line is y  69. One point on the line y  3x is the origin side opposite 3  hypotenuse 2 side adjacent 1  62. cos 60  hypotenuse 2 64. csc 60  The slope is change in y over the change in x. 1 1 3 3    and the Thus, m  3 3 3 3 The slope is change in y over the change in x. 3 Thus, m   3 and the equation of the 1 line is y  3 x. 60. sin 45  63. tan 60   3, 1. 68. 57. cos 45  61. sin 60  The line passes through (0, 0) and line. Then, by the definition of slope, y0 y   3, but also, by the m x0 x definition of tangent, tan   3. Because tan 60  3, the line y  3 x makes a 60° angle with the positive x-axis (See exercise 68). Copyright © 2021 Pearson Education, Inc. 120 Chapter 2 Acute Angles and Right Triangles 3 x, is the origin 3 0, 0 . Let  x, y  be any other point on this 70. One point on the line y  72. (a) The diagonal forms two isosceles right triangles. Each angle formed by a side of the square and the diagonal measures 45º. line. Then, by the definition of slope, y0 y 3   , but also, by the m 3 x0 x 3 . Because 3 3 3 tan 30  , the line y  x makes a 30° 3 3 angle with the positive x-axis. (See Exercise 67). definition of tangent, tan   71. (a) Each of the angles of the equilateral triangle has measure 13 180  60 . (b) The perpendicular bisects the opposite side so the length of each side opposite each 30º angle is k. (b) By the Pythagorean theorem, k 2  k 2  c 2  2k 2  c 2  c  2 k . Thus, the length of the diagonal is 2 k . (c) In a 45º-45º right triangle, the hypotenuse has a length that is 2 times as long as either leg. 73. Apply the relationships between the lengths of the sides of a 30  60 right triangle first to the triangle on the left to find the values of y and x, and then to the triangle on the right to find the values of z and w. In the 30  60 right triangle, the side opposite the 30 angle (c) Let x = the length of the perpendicular. Then apply the Pythagorean theorem. is 12 the length of the hypotenuse. The longer leg is 3 times the shorter leg. x 2  k 2   2k   x 2  k 2  4k 2  2 x 2  3k 2  x  3k The length of the perpendicular is 3k . (d) In a 30º-60º right triangle, the hypotenuse is always 2 times as long as the shorter leg, and the longer leg has a length that is 3 times as long as that of the shorter leg. Also, the shorter leg is opposite the 30º angle, and the longer leg is opposite the 60º angle. Thus, we have 1 9 9 3 y  9  and x  y 3  2 2 2 9 9 3 3 3 y , y  z 3, so z   2   6 2 3 3 3 3  and w  2 z , so w  2  3 3  2  Copyright © 2021 Pearson Education, Inc. Section 2.1 Trigonometric Functions of Acute Angles 74. Apply the relationships between the lengths of the sides of a 30  60 right triangle first to the triangle on the left to find the values of a and b. In the 30  60 right triangle, the side opposite the 30 angle is 12 the length of the 3 times the hypotenuse. The longer leg is shorter leg. 1 24  12 and b  a 3  12 3 2 Apply the relationships between the lengths of the sides of a 45  45 right triangle next to the triangle on the right to find the values of d and c. In the 45  45 right triangle, the sides opposite the 45 angles measure the same. 121 Thus, we have r  q 3  q r 15 2 15 2 3     5 6 and 3 3 3 3   t  2q  2 5 6  10 6 76. Apply the relationships between the lengths of the sides of a 30  60 right triangle first to the triangle on the right to find the values of m and a. In the 30  60 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle). a The hypotenuse is 2 times the measure of a leg. d  b  12 3 and  c  d 2  12 3  2   12 6 75. Apply the relationships between the lengths of the sides of a 45  45 right triangle to the triangle on the left to find the values of p and r. In the 45  45 right triangle, the sides opposite the 45 angles measure the same. The hypotenuse is leg. 2 times the measure of a Thus, we have 7m 3m 7 7 3 7 3    and 3 3 3 3  7 3  14 3 a  2m  a  2   3  3  Apply the relationships between the lengths of the sides of a 45  45 right triangle next to the triangle on the left to find the values of n and q. In the 45  45 right triangle, the sides opposite the 45 angles measure the same. The hypotenuse is 2 times the measure of a leg. Thus, we have n  a  14 3 and 3  14 3  14 6 qn 2 . 2  3  3  Thus, we have p  15 and r  p 2  15 2 Apply the relationships between the lengths of the sides of a 30  60 right triangle next to the triangle on the right to find the values of q and t. In the 30  60 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle). 77. Because A  A 1 bh, we have 2 1 1 s2  s  s  s 2 or A  . 2 2 2 Copyright © 2021 Pearson Education, Inc. 122 Chapter 2 Acute Angles and Right Triangles 78. Let h be the height of the equilateral triangle. h bisects the base, s, and forms two 30°–60° right triangles. 81. 82. sin 45  The formula for the area of a triangle is 1 A  bh. In this triangle, b = s. The height h 2 of the triangle is the side opposite the 60° angle in either 30°–60° right triangle. The side s opposite the 30° angle is . The height 2 s s 3 So the area of the entire is 3  .  2 2 1  s 3  s2 3  . triangle is A  s  2  2  4 79. Graph the equations in the window [–1.5, 1.5] × [–1.2, 1.2]. The point of intersection is (0.70710678, 0.70710678). This  2 2 corresponds to the point  , . 2   2 y 2  y  4 sin 45  4  2 2 4 2 and cos 45  x 2  x  4 cos 45  4  2 2 4 2 83. The legs of the right triangle provide the   coordinates of P, 2 2, 2 2 . 84. y 3  y  2 sin 60  2   3 2 2 x 1 and cos 60   x  2 cos 60  2   1 2 2 The legs of the right triangle provide the sin 60  These coordinates are the sine and cosine of 45°. 80. Yes, the third angle can be found by subtracting the given acute angle from 90°, and the remaining two sides can be found using a trigonometric function involving the known angle and side.   coordinates of P. P is 1, 3 . Section 2.2 Trigonometric Functions of Non-Acute Angles 1. The value of sin 240° is negative because 240° is in quadrant III. The reference angle is 60°, and the exact value of sin 240° is  23 . 2. The value of cos 390° is positive because 390° is in quadrant I. The reference angle is 30°, and the exact value of cos 390° is Copyright © 2021 Pearson Education, Inc. 3 . 2 Section 2.2 Trigonometric Functions of Non-Acute Angles 7. A; 135  360  225 and 225  180 = 45° (225° is in quadrant III) 3. The value of tan (–150°) is positive because –150° is in quadrant III. The reference angle is 3 . 3 30° and the exact value of tan (–150°) is 123 8. B; 60  360  300 and 360  300  60 (300° is in quadrant IV) 4. The value of sec 135° is negative because 135° is in quadrant II. The reference angle is 45°, and the exact value of sec 135° is  2. 9. D; 750  2  360  30 (30° is in quadrant I) 5. C; 180  98  82 (98° is in quadrant II) 10. B; 480  360  120 and 180  120  60 (120° is in quadrant II) 6. F; 212  180 = 32° (212° is in quadrant III)  sin  cos  tan  cot  sec  csc  11. 30 1 2 3 2 3 3 3 2 3 3 2 12. 45 2 2 1 1 2 2 13. 60 3 2 14. 120 3 2 2 2 1 2 cos 120   cos 60 1  2 15. 135 2 2  2 2 16. 150 sin150  sin 30 1  2  3 2 3  3 tan135   tan 45  1  17. 210  1 2 cos 210   cos 30 3  2 18. 240 3  2  3 3 3 3 1 2 tan 240  tan 60  3 3 2 3 cot 120   cot 60 sec120   sec 60 3  2  3 cot135   cot 45  2  1 sec150 cot150   sec 30   cot 30 2 3   3 3 sec 210   sec 30 3 2 3  3 cot 240  cot 60 2 3  3 Copyright © 2021 Pearson Education, Inc. 2 3 3 2 3 3 2 2 2  2 3 3 124 Chapter 2 Acute Angles and Right Triangles 19. To find the reference angle for 300, sketch this angle in standard position. The reference angle is 360  300  60. Because 300° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 3 sin 300   sin 60   2 1 cos 300  cos 60  2 tan 300   tan 60   3 3 cot 300   cot 60   3 sec 300  sec 60  2 2 3 csc 300   csc 60   3 20. To find the reference angle for 315, sketch this angle in standard position. 21. To find the reference angle for 405°, sketch this angle in standard position. The reference angle for 405° is 405° – 360° = 45°. Because 405° lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 45° See the Function Values of Special Angles table on page 170.) 2 sin 405  sin 45  2 2 cos 405  cos 45  2 tan 405  tan 45  1 cot 405  cot 45  1 sec 405  sec 45  2 csc 405  csc 45  2 22. To find the reference angle for 420º, sketch this angle in standard position. The reference angle is 360  315  45. Because 315° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. sin 315   sin 45   2 2 2 2 tan 315   tan 45  1 cot 315   cot 45  1 sec 315  sec 45  2 csc 315   csc 45   2 cos 315  cos 45  The reference angle for 420º is 420º − 360º = 60º. Because 420º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. See the Function Values of Special Angles table on page 170.) 3 sin  420  sin 60  2 1 cos  420  cos 60 2 tan  420  tan 60  3 (continued on next page) Copyright © 2021 Pearson Education, Inc. Section 2.2 Trigonometric Functions of Non-Acute Angles 125 cot 495   cot 45  1 sec 495   sec 45   2 csc 495  csc 45  2 (continued) 3 3 sec  420  sec 60  2 cot  420  cot 60  25. To find the reference angle for 570º sketch this angle in standard position. 2 3 csc  420  csc 60 3 23. To find the reference angle for 480º, sketch this angle in standard position. 480º is coterminal with 480º − 360º = 120º. The reference angle is 180º − 120º = 60º. Because 480º lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. 3 sin  480  sin 60  2 1 cos  480   cos 60  2 tan  480   tan 60   3 cot  480   cot 60   sec  80   sec 60  2 3 3 570° is coterminal with 570º − 360º = 210º. The reference angle is 210º − 180º = 30º. Because 570º lies in quadrant III, the sine, cosine, secant, and cosecant are negative. 1 sin 570   sin 30   2 3 cos 570   cos 30   2 3 tan 570  tan 30  3 cot 570  cot 30  3 2 3 sec 570   sec 30   3 csc 570   csc 30  2 26. To find the reference angle for 750°, sketch this angle in standard position. 2 3 csc  480  csc 60 3 24. To find the reference angle for 495º, sketch this angle in standard position. 495º is coterminal with 495º − 360º = 135º. The reference angle is 180º − 135º = 45º. Because 495° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. 2 sin 495  sin 45  2 2 cos 495   cos 45   2 tan 495   tan 45  1 750° is coterminal with 30 because 750  2  360  750  720  30. Because 750 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 30. 1 sin 750  sin 30  2 3 cos 750  cos 30  2 3 tan 750  tan 30  3 (continued on next page) Copyright © 2021 Pearson Education, Inc. 126 Chapter 2 Acute Angles and Right Triangles (continued) cot 750  cot 30  3 2 3 sec 750  sec 30  3 csc 750  csc 30  2 27. 1305º is coterminal with 1305  3  360  1305  1080  225 . The reference angle is 225º – 180º = 45º. Because 1305º lies in quadrant III, the sine, cosine, and secant and cosecant are negative. 2 sin1305   sin 45   2 2 cos1305   cos 45   2 tan1305  tan 45  1 cot1305  cot 45  1 sec1305   sec 45   2 csc1305   csc 45   2 28. 1500º is coterminal with 1500  4  360  1500  1440  60 . Because 1500º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. 3 sin  420  sin 60  2 1 cos  420  cos 60 2 tan  420  tan 60  3 3 3 sec  420  sec 60  2 cot  420  cot 60  csc  420  csc 60 The reference angle for 300 is 300  360  60. Because 300 lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60°. See the Function Values of Special Angles table on page 170.) 3 sin   300  sin 60  2 1 cos   300  cos 60 2 tan   300  tan 60  3 3 3 sec   300  sec 60  2 cot   300  cot 60  csc   300  csc 60 2 3 3 30. –390° is coterminal with 390  2  360  390  720  330. The reference angle is 360  330  30. Because –390° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 1 sin  390   sin 30   2 3 cos  390  cos 30  2 3 tan  390   tan 30   3 cot  390   cot 30   3 2 3 3 csc  390   csc 30  2 sec  390  sec 30  2 3 3 29. To find the reference angle for 300, sketch this angle in standard position. 31. –510° is coterminal with 510  2  360  510  720  210. The reference angle is 210  180  30. Because –510° lies in quadrant III, the sine, cosine, and secant and cosecant are negative. 1 sin  510   sin 30   2 3 cos  510   cos 30   2 3 tan  510  tan 30  3 cot  510  cot 30  3 2 3 3 csc  510   csc 30  2 sec  510   sec 30   Copyright © 2021 Pearson Education, Inc. Section 2.2 Trigonometric Functions of Non-Acute Angles 32. –1020º is coterminal with 1020  3  360  1020  1080  60 . Because −1020º lies in quadrant I, the values of all of its trigonometric functions will be positive, so these values will be identical to the trigonometric function values for 60. 3 2 1 cos  420  cos 60 2 tan  420  tan 60  3 sin  420  sin 60  35. –1860º is coterminal with 1860  6  360  1860  2160  300 . The reference angle is 360º – 300º = 60º. Because –1860º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 3 sin   1860   sin 60   2 1 cos   1860  cos 60 2 tan   1860   tan 60   3 cot   1860   cot 60   3 cot  420  cot 60  3 sec  420  sec 60  2 sec   1860  sec 60  2 csc   1860   csc 60  2 3 csc  420  csc 60 3 33. –1290° is coterminal with 1290  4  360  1290  1440  150. The reference angle is 180  150  30. Because –1290° lies in quadrant II, the cosine, tangent, cotangent, and secant are negative. 1 sin 2670  sin 30  2 3 cos 2670   cos 30   2 3 tan 2670   tan 30   3 cot 2670   cot 30   3 2 3 sec 2670   sec 30   3 csc 2670  csc 30  2 34. –855º is coterminal with 855  3  360  855  1080  225 . The reference angle is 225º – 180º = 45º. Because –855º lies in quadrant III, the sine, cosine, and secant and cosecant are negative. 2 sin  855   sin 45   2 2 cos  855   cos 45   2 tan  855  tan 45  1 cot  855  cot 45  1 sec  855   sec 45   2 csc  855   csc 45   2 127 3 3 2 3 3 36. –2205º is coterminal with 2205  7  360  2205  2520  315 . The reference angle is 360º – 315º = 45º. Because –2205º lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 2 sin   2205   sin 45   2 2 cos   2205  cos 45 2 tan   2205   tan 45  1 cot   2205   cot 45  1 sec   2205  sec 45  2 csc   2205   csc 45  2 37. Because 1305 is coterminal with an angle of 1305  3  360  1305  1080  225, it lies in quadrant III. Its reference angle is 225  180  45. Because the sine is negative in quadrant III, we have 2 sin1305   sin 45   . 2 38. Because 1500 is coterminal with an angle of 1500  4  360  1500  1440  60, it lies in quadrant I. Because 1500 lies in quadrant I, the values of all of its trigonometric functions will be positive, so 3 sin1500  sin 60  . 2 Copyright © 2021 Pearson Education, Inc. 128 Chapter 2 Acute Angles and Right Triangles 39. Because 510 is coterminal with an angle of 510  2  360  510  720  210, it lies in quadrant III. Its reference angle is 210  180  30. The cosine is negative in quadrant III, so 3 cos  510   cos 30   . 2 40. Because 1020 is coterminal with an angle of 1020  3  360  1020  1080  60, it lies in quadrant I. Because 1020 lies in quadrant I, the values of all of its trigonometric functions will be positive, so tan  1020  tan 60  3. 41. Because –855º is coterminal with 855  3  360  855  1080  225 , it lies in quadrant III. Its reference angle is 225º – 180º = 45º. The cosecant is negative in quadrant III, so csc  855   csc 45   2. 42. Because −495º is coterminal with an angle of 495  2  360  495  720  225 , it lies in quadrant III. Its reference angle is 225  180  45. Tthe secant is negative in quadrant III, so sec  495   sec 45   2. 43. Because 3015º is coterminal with 3015  8  360  3015  2880  135 , it lies in quadrant II. Its reference angle is 180º – 135º = 45º. The tangent is negative in quadrant II, so tan 3015   tan 45  1. 44. Because 2280º is coterminal with 2280  6  360  2280  2160  120 , it lies in quadrant II. Its reference angle is 180º − 120º = 60º. The cotangent is negative in quadrant II, so 3 cot 2280   cot 60   . 3 2  3   1 2 45. sin 2 120  cos 2 120       2  2 3 1   1 4 4 2  2  2 46. sin 225  cos 225         2   2  2 2   1 4 4 2 2 47. 2 tan 2 120  3sin 2 150  cos 2 180  2  3 2   3  12   1 2 2 23 1  2 3  3    1  4 4 48. cot 2 135  sin 30  4 tan 45 1 1 9 2   1   4 1  1   4  2 2 2 49. sin 2 225  cos 2 270  tan 2 60 2  2 2    0   2   3   24  3  72 2 50. cot 2 90  sec2 180  csc 2 135  02   1  2  2   1  2  1 2 51. cos 2 60  sec2 150  csc2 210 2 2 2 1  2 3         2 2  3  1 4 29   4 4 3 12 52. cot 2 135  tan 4 60  sin 4 180  1  2  3   0  1  9  10 4 4 ? 53. cos 30  60  cos 30  cos 60 Evaluate each side to determine whether this statement is true or false. cos 30  60  cos 90  0 and cos 30  cos 60  0 3 1   2 2 3 1 2 3 1 , so the statement is false. 2 ? 54. sin 30  sin 60  sin 30  60 Evaluate each side to determine whether this statement is true or false. 1 3 1 3 and sin 30  sin 60    2 2 2 sin 30  60  sin 90  1 2 Because 1 3  1, the given statement is 2 false. Copyright © 2021 Pearson Education, Inc. Section 2.2 Trigonometric Functions of Non-Acute Angles ? 55. cos 60  2 cos 30 Evaluate each side to determine whether this statement is true or false.  3 1 and 2cos 30  2  cos 60   3 2  2  1 Because  3, the statement is false. 2 ? 56. cos 60  2 cos 2 30  1 Evaluate each side to determine whether this statement is true or false. 1 cos 60  and 2 2  3 3 2 cos 30  1  2    1  2  4   1 2   3 1  1  2 2 1 1 Because  , the statement is true. 2 2 2 ? 57. sin 2 45  cos 2 45  1 2 2  2  2 2 2  2    2   4  4 1     Because 1 = 1, the statement is true. 58. tan 2 60  1  sec 2 60 Evaluate each side to determine whether this statement is true or false. tan 2 60  1   3   1  3  1  4 and 2 sec 2 60  22  4 . Because 4 = 4, the statement is true. ? 59. cos  2  45   2 cos 45 Evaluate each side to determine whether this statement is true or false. cos  2  45   cos 90  0 and 3 1 3  2 sin 30  cos 30  2      2   2  2 Because Because 0  2, the statement is false. ? 60. sin 2  30  2 sin 30  cos 30 Evaluate each side to determine whether this statement is true or false. 3 sin 2  30  sin 60  2 3 3  , the statement is true. 2 2 1 2 Because sin  is positive,  must lie in quadrants I or II. Because one angle, namely 30, lies in quadrant I, that angle is also the reference angle,  . The angle in quadrant II will be 180     180  30  150. 61. sin   3 2 Because cos  is positive,  must lie in quadrants I or IV. One angle, namely 30, lies in quadrant I, so that angle is also the reference angle,  . The angle in quadrant IV will be 360     360  30  330. 62. cos   63. tan    3 Because tan  is negative,  must lie in quadrants II or IV. The absolute value of tan  is 3, so the reference angle,   must be 60. The quadrant II angle  equals 180     180  60  120, and the quadrant IV angle  equals 360     360  60  300. 64. sec    2 Because sec  is negative,  must lie in quadrants II or III. The absolute value of sec  is 2, so the reference angle,   must be 45. The quadrant II angle  equals 180     180  45  135, and the quadrant III angle  equals 180     180  45  225. 2 2 Because cos  is positive,  must lie in quadrants I or IV. One angle, namely 45, lies in quadrant I, so that angle is also the reference angle,  . The angle in quadrant IV will be 360     360  45  315. 65. cos    2 2 cos 45  2   2  2  129 Copyright © 2021 Pearson Education, Inc. 130 Chapter 2 Acute Angles and Right Triangles 3 3 Because cot  is negative,  must lie in quadrants II or IV. The absolute value of 3 cot  is , so the reference angle,   must 3 be 60. The quadrant II angle  equals 180     180  60  120, and the quadrant IV angle  equals 360     360  60  300. 66. cot    67. csc   2 Because csc  is negative,  must lie in quadrants III or IV. The absolute value of csc  is 2, so the reference angle,  , is 30 . The angle in quadrant III will be 180     180  30  210 , and the quadrant IV angle is 360     360  30  330 . 3 2 Because sin  is negative,  must lie in quadrants III or IV. The absolute value of 3 sin  is , so the reference angle,  , is 2 60°. The angle in quadrant III will be 180     180  60  240 , and the quadrant IV angle is 360     360  60  300 . 68. sin    3 3 Because tan  is positive,  must lie in quadrants I or III. One angle, namely 30 , lies in quadrant I, so that angle is also the reference angle,  . The angle in quadrant III will be 180     180  30  210 . 69. tan   1 2 Because cos  is negative,  must lie in quadrants II or III. The absolute value of 1 cos  is , so the reference angle,   must 2 be 60. The quadrant II angle  equals 180     180  60  120, and the quadrant III angle  equals 180     180  60  240. 70. cos    71. csc    2 Because csc  is negative,  must lie in quadrants III or IV. The absolute value of csc  is 2, so the reference angle,   must be 45. The quadrant III angle  equals 180     180  45  225. and the quadrant IV angle  equals 360     360  45  315. 72. cot   1 Because cot  is negative,  must lie in quadrants II or IV. The absolute value of cot  is 1, so the reference angle,   must be 45. The quadrant II angle  equals 180     180  45  135. and the quadrant IV angle  equals 360     360  45  315. 73. 150º is in quadrant II, so the reference angle is 180º − 150º = 30º. x 3 3 3 cos 30   x  r cos 30  6  2 r y 1 and sin 30   y  r sin 30  6   3 r 2 Because 150º is in quadrant II, the x- coordinate will be negative. The   coordinates of P are 3 3, 3 . 74. 225º is in quadrant III, so the reference angle is 225º – 180º = 45º. x 2 5 2 cos 45   x  r cos 45  10  2 r and y 2 5 2 sin 45   y  r sin 45  10  2 r Because 225º is in quadrant III, both the x- and y-coordinates will be negative. The   coordinates of P are 5 2, 5 2 . 75. For every angle  , sin 2   cos 2   1 . Because 0.8  0.6  0.64  0.36  1 , 2 2 there is an angle  for which cos   0.6 and sin   0.8 . Because cos   0 and sin   0 ,  lies in quadrant IV. 76. For every angle  , sin 2   cos 2   1 . Because 145  1 , there  34    23   169  94  144 2 2 is no angle  for which cos   23 and sin   34 . Copyright © 2021 Pearson Education, Inc. Section 2.3 Approximations of Trigonometric Function Values 77. If  is in the interval 90,180 , then 90    180  45   2  90 . Thus  lies in quadrant I, and cos 2  2 is positive.  2 lies in quadrant I, and sin  90 . Thus  2 86. The reference angle for 115° is 180° – 115° = 65°. Because 115° is in quadrant II the cosine is negative. Cos  decreases on the interval (90°, 180°) from 0 to –1. Therefore, cos 115° is closest to –0.4. 2 . Sine and 2 cosine are both positive in quadrant I and both negative in quadrant III. Because   180  45  180  225, 45º is the quadrant I angle, and 225º is the quadrant III angle. 87. When   45 , sin   cos   78. If  is in the interval 90,180 , then 90    180  45  131  2 is positive. 79. If  is in the interval 90,180 , then 90    180  270    180  360 . Thus   180 lies in quadrant IV, and sec   180 is positive. 80. If  is in the interval 90,180 , then 90    180  270    180  360 . Thus   180 lies in quadrant IV, and cot   180 is negative. 81. If  is in the interval 90,180 , then 90    180  90    180  180    90 Because 180º is coterminal with −180º + 360º = 180º and −90º is coterminal with −90º + 360º = 270º,  lies in quadrant III, and sin    is negative. 82. If  is in the interval 90,180 , then 90    180  90    180  180    90 Because 180º is coterminal with –180º + 360º = 180º and –90º is coterminal with –90º + 360º = 270º,  lies in quadrant III, and cos    is negative. 2 . Sine and 2 cosine are opposites in quadrants II and IV. Thus, 180    180  45  135 in quadrant II, and 360    360  45  315 in quadrant IV. 88. When   45 , sin   cos   Section 2.3 Approximations of Trigonometric Function Values For Exercises 1–10, be sure your calculator is in degree mode. 1. J 2. B 3. E 4. F 5. D 6. I 7. H 8. A 9. G 10. C For Exercises 11−40, be sure your calculator is in degree mode. If your calculator accepts angles in degrees, minutes, and seconds, it is not necessary to change angles to decimal degrees. Keystroke sequences may vary on the type and/or model of calculator being used. Screens shown will be from a TI-84 Plus C calculator. To obtain the degree (°) and (′) symbols, go to the ANGLE menu (2nd APPS). 83. When an integer multiple of 360° is added to  , the resulting angle is coterminal with  . The sine values of coterminal angles are equal. 84. When an integer multiple of 360° is added to  , the resulting angle is coterminal with  . The cosine values of coterminal angles are equal. 85. The reference angle for 115° is 180° – 115° = 65°. Because 115° is in quadrant II the cosine is negative. Sin  decreases on the interval (90°, 180°) from 1 to 0. Therefore, sin 115° is closest to 0.9. For Exercises 11–40, we include TI-84 screens only for those exercises involving cotangent, secant, and cosecant. 11. sin 3842  0.625243 12. cos 4124  0.750111 Copyright © 2021 Pearson Education, Inc. 132 Chapter 2 Acute Angles and Right Triangles 13. sec1315  1.027349 14. csc14545  1.776815 21. 1  tan 23.4  0.432739 cot 23.4 22. 1  cos14.8  0.966823 sec14.8 23. cos 77  cot 77  0.230868 sin 77 24. sin 33  tan 33  0.649408 cos 33 25. cot 90  4.72  tan 4.72  0.082566 26. cos 90  3.69  sin 3.69  0.064358 27. 15. cot18348  15.055723 28. 1  cos 51  0.629320 csc 90  51 1 1  tan 90  22 cot 22  tan 22  0.404026 29. tan   1.4739716   tan 1 1.4739716  55.845496 30. tan   6.4358841   tan 1 6.4358841  81.168073 16. tan 42130  1.841771 17. sin  31212   0.740805 18. tan  8006   5.729742 19. csc  31736   1.483014 31. sin   0.27843196   sin 1 0.27843196  16.166641 32. sin   0.84802194    sin 1 0.84802194  57.997172 33. cot   1.2575516 1    1.2575516    cot 1 1.2575516  tan 1   38.491580 20. cot  51220   1.907415 Copyright © 2021 Pearson Education, Inc. Section 2.3 Approximations of Trigonometric Function Values 133 34. csc   1.3861147 1    1.3861147    csc1 1.3861147   sin 1   46.173582 40. cot   0.21563481  1    0.21563481    cot 1 0.21563481  tan 1   77.831359 35. sec   2.7496222 1    2.7496222    sec 1 2.7496222  cos 1   68.673241 36. sec   1.1606249   sec 1 1.1606249  cos 1  30.502748 1 1.1606249 41. A common mistake is to have the calculator in radian mode, when it should be in degree mode (and vice verse). 42. If the calculator allows an angle  where 0    360 , then we need to find an angle within this interval that is coterminal with 2000º by subtracting a multiple of 360º: 2000  5  360  2000  1800  200 . Then find cos 200°. If the calculator is more restrictive on evaluating angles (such as 0    90 ), then reference angles need to be used. 43. Find tan 1 1.482560969 . 37. cos   0.70058013   cos 1 0.70058013  45.526434 A = 56º. 44. Find the sine of 22º. A = 0.3746065934º 45. sin 35 cos 55  cos 35 sin 55  1 46. cos100 cos 80  sin100 sin 80  1 38. cos   0.85536428   cos 1 0.85536428  31.199998 47. sin 2 36  cos 2 36  1 39. csc   4.7216543  48. 2 sin 2513 cos 2513  sin 5026  0 1    4.7216543    csc1  4.7216543  sin 1   12.227282 49. cos 7529 cos1431  sin 7529 sin1431  0 50. cos 2814 cos 6146  sin 2814 sin 6146  1 ? 51. sin10  sin10  sin 20 Using a calculator gives sin10  sin10  0.34729636 and sin 20  0.34202014. Thus, the statement is false. Copyright © 2021 Pearson Education, Inc. 134 Chapter 2 Acute Angles and Right Triangles ? 52. cos 40  2 cos 20 Using a calculator gives cos 40  0.76604444 and 2 cos 20  1.87938524. Thus, the statement is false. ? 60. tan 2 7225  1  sec 2 7225 Using a calculator gives tan 2 7225  1  10.9577102 and sec 2 7225  10.9577102. Thus, the statement is true. ? 53. sin 50  2 sin 25 cos 25 Using a calculator gives sin 50  0.76604444 and 2 sin 25 cos 25  0.76604444. Thus, the statement is true. ? 54. cos 70  2 cos 2 35  1 Using a calculator gives cos 70  0.34202014 and 2 1  2 sin 80  0.93969262. Thus, the statement is false. ? 56. 2 cos 3822  cos 7644 Using a calculator gives 2 cos 3822  1.56810939 and cos 7644  0.22948353. Thus, the statement is false. ? 57. sin 3948  cos 3948  1 Using a calculator gives sin 3948  cos 3948  1.40839322  1. Thus, the statement is false. 40 Thus, the statement is false. 2 Using a calculator gives cos 30  20  0.64278761 and 63. sin   0.92718385 sin  is positive in quadrants I and II. sin 1 0.92718385  68 The angle in quadrant II with the same sine is 180° – 68° = 112°. 64. sin   0.52991926 sin  is positive in quadrants I and II. sin 1 0.52991926  32 The angle in quadrant II with the same sine is 180° – 32° = 148°. 65. cos   0.71933980 cos  is positive in quadrants I and IV. cos 1 0.71933980  44 66. cos   0.10452846 cos  is positive in quadrants I and IV. cos 1 0.10452846  84 sin 12 40  0.34202014. ? 62. cos 30  20  cos 30  cos 20 The angle in quadrant IV with the same cosine is 360° – 44° = 316°. Using a calculator gives 1 sin 40  0.32139380 and 2 2 cos 30 cos 20  sin 30 sin 20  0.64278761. Thus, the statement is true. cos 30  cos 20  1.8057180. Thus, the statement is false. ? 55. cos 40  1  2 sin 2 80 Using a calculator gives cos 40  0.76604444 and ? 1 sin 40  sin 12 2 Using a calculator gives cos 30  20  0.64278761 and ? 2 cos 2 35  1  0.34202014. Thus, the statement is true. 58. ? 61. cos 30  20  cos 30 cos 20  sin 30 sin 20 59. 1  cot 42.5  csc 42.5 Using a calculator gives 1  cot 2 42.5  2.1909542 and csc2 42.5  2.1909542. Thus, the statement is true. The angle in quadrant IV with the same cosine is 360° – 84° = 276°. 67. tan   1.2348971 tan  is positive in quadrants I and III. tan 1 1.2348971  51 The angle in quadrant III with the same tangent is 180° + 51° = 231°. Copyright © 2021 Pearson Education, Inc. Section 2.3 Approximations of Trigonometric Function Values 68. tan   0.70020753 tan  is positive in quadrants I and III. tan 1 0.70020753  35 The angle in quadrant III with the same tangent is 180° + 35° = 215°. 69. F  W sin  F  2100 sin1.8  65.96  70 lb 70. F  W sin  F  2400 sin  2.4  100.5  100 lb F is negative because the car is traveling downhill. 71. 130  sin   2600 0.05  sin     sin 1  0.05  2.9 72. 3º 0.0523 0.0524 0.0524 3.5º 0.0610 0.0612 0.0611 4º 0.0698 0.0699 0.0698 180  180 . W  180 4  0.04 100 F  W tan   2000(0.04)  80 lb (c) tan   W  from part (b). Let 180   3.75 and W = 1800. 1800 (3.75) F  118 lb 180 F  W sin  (d) Use F  F  W sin  77. 45 mph = 66 ft/sec, V  66 ,   3, g  32.2, f  0.14 120 W  sin(2.7) W  2547  2500 lb 120  W sin(2.7)  V2 662  g  f  tan   32.2(0.14  tan 3)  703 ft R F  W sin  145 W  sin(3) W  2771  2800 lb 145  W sin(3)  75. F  W sin  F  2200 sin 2  76.77889275 lb F  2000 sin 2.2  76.77561818 lb The 2200-lb car on a 2° uphill grade has the greater grade resistance. 76. tan  (b) F  W sin   W tan   150  sin   3000 0.05  sin     sin 1 0.05  2.9 74. sin  sin   tan   150  3000 sin   73.   (a) From the table, we see that if  is small, F  W sin  130  2600 sin   135   sin  tan  0º 0.0000 0.0000 0.0000 0.5º 0.0087 0.0087 0.0087 1º 0.0175 0.0175 0.0175 1.5º 0.0262 0.0262 0.0262 2º 0.0349 0.0349 0.0349 2.5º 0.0436 0.0437 0.0436 180 78. There are 5280 ft in one mile and 3600 sec in one min. 1 hr 5280 ft 70 mph  70 mph  · 3600 sec 1 mi 2  102 ft per sec 3  102.67 ft per sec V  102.67 ,   3, g  32.2, f  0.14 V2 102.67 2  g  f  tan   32.2 0.14  tan 3  1701 ft R Copyright © 2021 Pearson Education, Inc. 136 Chapter 2 Acute Angles and Right Triangles 79. Intuitively, increasing  would make it easier to negotiate the curve at a higher speed much like is done at a race track. Mathematically, a larger value of  (acute) will lead to a larger value for tan  . If tan  increases, then the ratio determining R will decrease. Thus, the radius can be smaller and the curve sharper if  is increased. V2 662 R  g  f  tan   32.2 0.14  tan 4  644 ft V2 102.67 2 R  g  f  tan   32.2 0.14  tan 4  1559 ft As predicted, both values are less. 80. From Exercises 77 amd 78, V2 . Solving for V we have R g  f  tan   R (b) 1  39,  2  28, c1  3  108 m per sec c1 sin 1 c sin  2   c2  1  c2 sin  2 sin 1 82. (a) 1  40, c2  1.5  108 m per sec, and c1  3  108 m per sec c1 sin 1 c sin 1   sin  2  2  c2 sin  2 c1 3  108  1.5  108 sin 40    19  2  sin 1    3  108    V  Rg  f  tan    1150 32.20.14  tan 2.1  80.9 ft/sec 80.9 ft/sec · 3600 sec/hr · 1 mi/5280 ft  55 mph, The speed limit should be 55 mph. 81. (a) 1  46,  2  31, c1  3  108 m per sec c1  3  108 m per sec c1 sin 1 c sin 1   sin  2  2  c2 sin  2 c1 8 sin 46 Because c1 is only given to one significant digit, c2 can only be given to one significant digit. The speed of light in the second medium is about 2  108 m per sec. 2.6  10  sin 62  8 sin  2  3  108  2.6  108 sin 62    50  2  sin 1    3  108    8 2  (b) 1  62, c2  2.6  108 m per sec and c1 sin 1 c sin  2   c2  1  c2 sin  2 sin 1 3  10  sin 31  2  10 c  1.5  10  sin 40  8 sin  2  V  V 2  Rg  f  tan    g  f  tan   V  Rg  f  tan   8 sin 39 Because c1 is only given to one significant digit, c2 can only be given to one significant digit. The speed of light in the second medium is about 2  108 m per sec. 2 R  1150 ,   2.1, g  32.2, f  0.14 3  10  sin 28  2  10 8 c2   83. 1  90, c1  3  108 m per sec, and c2  2.254  108 c1 sin 1 c sin 1   sin  2  2 c2 sin  2 c1 2.254  10  sin 90 8 sin  2   3  108 2.254  108 1 8 3  10 1  2.254   2  sin   Copyright © 2021 Pearson Education, Inc. 3  2.254  3   48.7 Section 2.3 Approximations of Trigonometric Function Values 84. 1  90  29.6  60.4, c1  3  108 m per 8 8 sec, and c2  2.254  10 c1  3  10 m per sec c1 sin 1 c sin 1   sin  2  2 c2 sin  2 c1 D 200   1.05 V12  V2 2  64.4  K1  K 2  sin    1.05 1322  V2 2  64.4 0.4  0.02  sin  3.5   2.254  10  sin 60.4 sin   200  2.254 sin 60.4  3  2.254  2  sin 1  sin 60.4   40.8  3 Light from the object is refracted at an angle of 40.8° from the vertical. Light from the horizon is refracted at an angle of 48.7° from the vertical. Therefore, the fish thinks the object lies at an angle of 48.7° – 40.8° = 7.9° above the horizon. 4624  18, 295.2  1.05V2 2 13, 671.2  1.05V2 2 13, 671.2 V2 2  1.05 V2 2  13020.19048 V2  114.106 V2  114 ft/sec · 3600 sec/hr · 1 mi/5280 ft  78 mph 8 2 3  108  85. V1  55 mph  55 mph  1 hr 5280 ft · 3600 sec 1 mi 2 ft per sec  80.67 ft per sec, 3 1 hr 5280 ft · V2  30 mph  30 mph  3600 sec 1 mi  44 ft per sec   3.5, K1  0.4, K 2  0.02.  80 D   1.05 V12  V2 2  64.4  K1  K 2  sin    1.05 80.67 2  442  64.4 0.4  0.02  sin 3.5  155 ft 86. V1  80.67 ft per sec, V2  44 ft per sec,   2, K1  0.4, and K 2  0.02. D   1.05 V12  V2 2  64.4  K1  K 2  sin    1.05 80.67 2  442  64.4 0.4  0.02  sin  2  194 ft 87. Negative values of  require greater distances for slowing down than positive values. 1.05 1322  1.05V2 2 23.12 200  23.12  18, 295.2  1.05V2 2 89. For Auto A, calculate 70  cos10  68.94. Auto A’s reading is approximately 69 mph. For Auto B, calculate 70  cos 20  65.78. Auto B’s reading is approximately 66 mph. 90. The figure for this exercise indicates a right triangle. Because we are not considering the time involved in detecting the speed of the car, we will consider the speeds as sides of the r right triangle. Given angle  , cos   . a Thus, the speed that the radar detects is r  a cos  . This confirms the “cosine effect” that reduces the radar reading. 91. h = 1.9 ft,   0.9, 1  3,  2  4, S = 336 ft: L  4  (3) 3362 200 1.9  336 tan .9  132 ft per sec  550 ft 92. h = 1.9 ft,   1.5, 1  3,  2  4, S = 336 ft: L  4  (3) 3362 200 1.9  336 tan1.5 88. Using the values for K1 and K 2 from Exercise 73, determine V2 when D = 200,   3.5, V1  90 mph  90 mph  137 1 hr 5280 ft · 3600 sec 1 mi Copyright © 2021 Pearson Education, Inc.  369 ft 138 Chapter 2 Acute Angles and Right Triangles 5. 180º – 135º = 45º, so the reference angle is 45º. The original angle (135º) lies in quadrant II, so the sine and cosecant are positive, while the remaining trigonometric functions are negative. 2 2 ; cos135   sin135  2 2 tan135  1 ; cot135  1 sec135   2 ; csc135  2 Chapter 2 Quiz (Sections 2.1−2.3) sin A  1. side opposite 24 3   hypotenuse 40 5 side adjacent 32 4   hypotenuse 40 5 side opposite 24 3 tan A    side adjacent 32 4 side adjacent 32 4 cot A    side opposite 24 3 hypotenuse 40 5 sec A    side adjacent 32 4 hypotenuse 40 5 csc A    side opposite 24 3 cos A  2.  sin  cos  tan  cot  sec  csc  30 1 2 3 2 3 3 3 2 3 3 2 45 2 2 1 1 2 2 60 3 2 2 2 1 2 3 3 3 2 2 3 3 1 w  w  36 sin 30  36   18 36 2 3 x cos 30   x  36 cos 30  36   18 3 36 2 18 w tan 45   1   y  18 y y 2 18 36 w sin 45    z  18 2 2 z z 2 3. sin 30  4. The height of one of the six equilateral triangles from the solar cell is h sin    h  x sin  . x Thus, the area of each of the triangles is   12 bh  12 x 2 sin  . So, the area of the solar cell is   6  12 x 2 sin   3x 2 sin  . 6. −150º is coterminal with 360º – 150º = 210º. This lies in quadrant III, so the reference angle is 210º – 180º = 30º. In quadrant III, the tangent and cotangent functions are positive, while the remaining trigonometric functions are negative. 1 sin  150   sin 30   2 3 cos  150   cos 30   2 3 3 cot  150  cot 30  3 tan  150  tan 30  2 3 3 csc  150   csc 30  2 sec  150   sec 30   7. 1020º is coterminal with 1020º – 720º = 300º. This lies in quadrant IV, so the reference angle is 360º – 300º = 60º. In quadrant IV, the cosine and secant are positive, while the remaining trigonometric functions are negative. 3 sin1020   sin 60   2 1 cos1020  cos 60  2 tan1020   tan 60   3 3 cot1020   cot 60   3 sec1020  sec 60  2 2 3 csc1020   csc 60   3 3 2 Because sin  is positive,  must lie in quadrants I or II, and the reference angle,  , is 60°. The angle in quadrant I is 60º, while the angle in quadrant II is 180     180  60  120 . 8. sin   Copyright © 2021 Pearson Education, Inc. Section 2.4 Solutions and Applications of Right Triangles 9. sec    2 Because sec  is negative,  must lie in quadrants II or III. The absolute value of sec  is 2, so the reference angle,   must be 45. The quadrant II angle  equals 180     180  45  135, and the quadrant III angle  equals 180     180  45  225. 10. sin 4218  0.673013 11. sec  21212   1.181763 12. tan   2.6743210    69.497888 139 A  B  90  B  90  A  B  90  3620  8960  3620  5340 a a sin A   sin 36 20   964 c a  964 sin 3620  571m (rounded to three significant digits) b b cos A   cos 36 20   c 964 b  964 cos 36 20  777 m (rounded to three significant digits) 14. X = 47.8°, z = 89.6 cm 13. csc   2.3861147    24.777233 14. The statement is false. sin 60  30  sin 90  1 , while sin 60  sin 30  3 1   2 2 3 1 . 2 15. The statement is true. Using the cofunction identity, tan 90  35  cot 35 . Section 2.4 Solutions and Applications of Right Triangles 1. B 2. D 3. A 4. F 5. C 6. E Y  X  90  Y  90  X  Y  90  47.8  42.2 x x sin X   sin 47.8   89.6 z x  89.6 sin 47.8  66.4 cm (rounded to three significant digits) y y  cos 47.8   89.6 z y  89.6 cos 47.8  60.2 cm (rounded to three significant digits) cos X  7. 23.865 to 23.875 15. N = 51.2°, m = 124 m 8. 28,999.5 to 29,000.5 9. 8959.5 to 8960.5 10. No. Points scored in basketball are exact numbers and cannot take on values that are not whole numbers. 11. If h is the actual height of a building and the height is measure as 58.6 ft, then h  58.6  0.05. 12. If w is the actual weight of a car and the weight is measure as 1542 lb, then w  1542  0.5. 13. A  3620, c = 964 m M  N  90  M  90  N  M  90  51.2  38.8 n n tan N   tan 51.2   124 m n  124 tan 51.2  154 m (rounded to three significant digits) m 124 cos N   cos 51.2   p p 124 p  198 m (rounded to three cos 51.2 significant digits) Copyright © 2021 Pearson Education, Inc. 140 Chapter 2 Acute Angles and Right Triangles 16. A  3140, a  35.9 km a a  cos 68.5142   c 3579.42 a  3579.42 cos 68.5142  1311.04 m (rounded to six significant digits) cos B  19. a  12.5, b  16.2 A  B  90  B  90  A  B  90  3140  8960  3140  5820 a 35.9 sin A   sin 3140   c c 35.9 c  68.4 km (rounded to sin 3140 three significant digits) a 35.9 tan A   tan 3140   b b 35.9 b  58.2 km (rounded to tan 3140 three significant digits) 17. B  42.0892, b  56.851 A  B  90  A  90  B  A  90  42.0892  47.9108 b 56.851 sin B   sin 42.0892   c c 56.851 c  84.816 cm sin 42.0892 (rounded to five significant digits) b 56.851 tan B   tan 42.0892   a a 56.851 a  62.942 cm tan 42.0892 (rounded to five significant digits) 18. B  68.5142, c  3579.42 A  B  90  A  90  B  A  90  68.5142  21.4858 b b sin B   sin 68.5142  c 3579.42 b  3579.42 sin 68.5142  3330.68 m (rounded to six significant digits) Using the Pythagorean theorem, we have a 2  b 2  c 2  12.52  16.22  c 2  418.69  c 2  c  20.5 ft (rounded to three significant digits) a 12.5 tan A   tan A   b 16.2 12.5 A  tan 1  37.6540 16.2  37  0.6540  60  3739  3740 (rounded to three significant digits) b 16.2 tan B   tan B   12.5 a 16.2 B  tan 1  52.3460 12.5  52  0.3460  60  5221  5220 (rounded to three significant digits) 20. a  4.80, c  15.3 Using the Pythagorean theorem, we have a 2  b 2  c 2  4.802  b 2  15.32  4.802  b 2  15.32 b 2  15.32  4.802  211.05 b  14.5 m (rounded to three significant digits) a 4.80  sin A   sin A  b 15.3 4.80  18.2839 A  sin 1 15.3  18  0.2839  60  1817   1820 (rounded to three significant digits) Copyright © 2021 Pearson Education, Inc. (continued on next page) Section 2.4 Solutions and Applications of Right Triangles 141 26. B = 46.0°, c = 29.7 m (continued) a 4.80  cos B   c 15.3 4.80  71.7161 B  cos 1 15.3  71  0.7161  60  7143  7140 cos B  (rounded to three significant digits) 21. No. You need to have at least one side to solve the triangle. There are infinitely many similar right triangles satisfying the given conditions. 22. If we are given an acute angle and a side in a right triangle, the unknown part of the triangle requiring the least work to find is the other acute angle. It may be found by subtracting the given acute angle from 90°. 23. Answers will vary. If you know one acute angle, the other acute angle may be found by subtracting the given acute angle from 90°. If you know one of the sides, then choose two of the trigonometric ratios involving sine, cosine or tangent that involve the known side in order to find the two unknown sides. 24. If you know the lengths of two sides, the Pythagorean theorem can be used to find the length of the remaining side. Then an inverse trigonometric function can be used to find one of the acute angles. The other acute angle may be found by subtracting the calculated acute angle from 90°. 25. A = 28.0°, c = 17.4 ft A  B  90 B  90  A B  90  28.0  62.0 a a sin A   sin 28.0   c 17.4 a  17.4 sin 28.0  8.17 ft (rounded to three significant digits) b b cos A   cos 28.00   c 17.4 b  17.4 cos 28.00  15.4 ft (rounded to three significant digits) A  B  90 A  90  B A  90  46.0  44.0 a a cos B   cos 46.0   c 29.7 a  29.7 cos 46.0  20.6 m (rounded to three significant digits) b b sin B   sin 46.0   c 29.7 b  29.7 sin 46.0  21.4 m (rounded to three significant digits) 27. Solve the right triangle with B = 73.0°, b = 128 in. and C = 90° A  B  90  A  90  B  A  90  73.0  17.0 128 b tan B   tan 73.0   a a 128 a  a  39.1 in (rounded to tan 73.0 three significant digits) 128 b sin B   sin 73.0   c c 128 c  c  134 in (rounded to sin 73.0 three significant digits) 28. A = 62.5º, a = 12.7 m A  B  90  B  90  A  B  90  62.5  27.5 a 12.7 tan A   tan 62.5   b b 12.7 b  6.61 m (rounded to tan 62.5 three significant digits) (continued on next page) Copyright © 2021 Pearson Education, Inc. 142 Chapter 2 Acute Angles and Right Triangles (continued) a 12.7  sin 62.5   c c 12.7 c  14.3 m (rounded to sin 62.5 three significant digits) sin A  29. A = 61.0º, b = 39.2 cm A  B  90  B  90  A  B  90  61.0  29.0 a a tan A   tan 61.0   b 39.2 a  39.2 tan 61.0  70.7 cm (rounded to three significant digits) b 39.2 cos A   cos 61.0   c c 39.2 c  80.9 cm cos 61.0 (rounded to three significant digits) 39.2 cm c 2  a 2  b 2  222  132  b 2  484  169  b 2  315  b 2  b  18 m (rounded to two significant digits) We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other. 13 a  sin A   sin A  22 c  13  A  sin 1    36.2215  36 (rounded  22  to two significant digits) 13 b  cos B   cos B  22 c  13  B  cos 1    53.7784  54  22  (rounded to two significant digits) 32. b = 32 ft, c = 51 ft 30. B = 51.7º, a = 28.1 ft A  B  90  B  90  B  A  90  51.7  38.3 b b tan B   tan 51.7   a 28.1 b  28.1tan 51.7  35.6 ft (rounded to three significant digits) a 28.1 cos B   cos 51.7   c c 28.1 c  45.3 ft (rounded to cos 51.7 three significant digits) 31. a = 13 m, c = 22m c 2  a 2  b 2  512  a 2  322  2601  a 2  1024  1577  a 2  a  40 ft (rounded to two significant digits) We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other. b 32 cos A   cos A   c 51  32  A  cos 1    51.1377  51  51  (rounded to two significant digits) 32 b  sin B   sin B  51 c  32  B  sin 1    38.8623  39 (rounded  51  to two significant digits) Copyright © 2021 Pearson Education, Inc. Section 2.4 Solutions and Applications of Right Triangles We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other. a 958 tan A   tan A   b 489  958  A  tan 1   62.9585  489   63  0.9585  60  6258  6300 33. a = 76.4 yd, b = 39.3 yd c 2  a 2  b2  c  a 2  b2  143 76.42  39.32  5836.96  1544.49  7381.45  85.9 yd (rounded to three significant digits) We will determine the measurements of both A and B by using the sides of the right triangle. In practice, once you find one of the measurements, subtract it from 90 to find the other. a 76.4  tan A   tan A  39.3 b  76.4   62.7788 A  tan 1   39.3   62  0.7788  60  6247   6250 (rounded to three significant digits) b 39.3 tan B   tan B   a 76.4  39.3   27.2212 B  tan 1   76.4   27  0.2212  60  2713  2710 (rounded to three significant digits) 34. a = 958 m, b = 489 m (rounded to three significant digits) b 489 tan B   tan B   a 958  489  B  tan 1   27.0415  958   27  0.0415  60  2702  2700 (rounded to three significant digits) 35. a = 18.9 cm, c = 46.3 cm c 2  a 2  b 2  46.32  18.92  b 2  2143.69  357.21  b 2  1786.48  b 2  b  42.3 cm (rounded to three significant digits) a 18.9  sin A   c 46.3  18.9  A  sin 1   24.09227  46.3   24  0.09227  60  2406  2410 sin A  (rounded to three significant digits) a 18.9  cos B   cos B  c 46.3  18.9  B  cos 1   65.9077  46.3   65  0.9077  60  6554  6550 (rounded to three significant digits) c 2  a 2  b 2  c  a 2  b 2  9582  4892  917, 764  239,121  1,156,885  1075.565887  1080 m (rounded to three significant digits) Copyright © 2021 Pearson Education, Inc. 144 36. Chapter 2 Acute Angles and Right Triangles b = 219 cm, c = 647 m c2  a 2  b2 647 2  a 2  2192 418, 609  a 2  47, 961 370, 648  a 2 a  609 m (rounded to three significant digits) b 219  cos A   c 647  219  A  cos 1   70.2154  647   70  0.2154  60  7013  7010 cos A  (rounded to three significant digits) b 219  sin B   sin B  c 647  219  B  sin 1   19.7846  647   19  0.7846  60  1947   1950 (rounded to three significant digits) 37. A = 53º24′, c = 387.1 ft A  B  90 B  90  A B  90  5324  8960  5324  3636 a a  sin 5324   387.1 c a  387.1sin 5324  310.8 ft (rounded to four significant digits) b b cos A   cos 5324   387.1 c b  387.1cos 5324  230.8 ft (rounded to four significant digits) sin A  38. 39. B = 39º09′, c = 0.6231 m A  B  90 A  90  B A  90  3909  8960  3909  5051 b b  sin 3909   c 0.6231 b  0.6231sin 3909  0.3934 m (rounded to four significant digits) sin B  a a  cos 3909   c 0.6231 a  0.6231cos 3909  0.4832 m (rounded to four significant digits) cos B  40. B = 82º51′, c = 4.825 cm A  B  90  A  90  B  A  90  8251  8960  8251  709 b b sin B   sin 8251   c 4.825 b  4.825sin 8251  4.787 cm (rounded to four significant digits) a a  cos 8251   c 4.825 a  4.825 cos 8251  0.6006 cm (rounded to four significant digits) cos B  41. If B is a point above point A as shown in the figure, the angle of elevation from A to B is the acute angle formed by the horizontal line through A and the line of sight from A to B. A = 13º47′, c = 1285 m A  B  90  B  90  A  B  90  1347   8960  1347   7613 a a sin A   sin1347    1285 c a  1285sin1347   306.2 m (rounded to four significant digits) b b cos A   cos1347    1285 c b  1285 cos1347   1248 m (rounded to four significant digits) 42. No. An angle of elevation (and also an angle of depression) must be an acute angle. 43. Angles DAB and ABC are alternate interior angle formed by the transversal AB intersecting parallel lines AD and BC. Thus they have the same measure. 44. An angle of depression (and also an angle of elevation) is measured from a horizontal line to the line of sight. Angle CAB is formed by the line of sight and vertical line AC. Copyright © 2021 Pearson Education, Inc. Section 2.4 Solutions and Applications of Right Triangles d 13.5 d  13.5sin 4350 ‘  9.3496000 The ladder goes up the wall 9.35 m. (rounded to three significant digits) 45. sin 4350 ‘  145 48. Let x = the diameter of the sun. 46. T = 32° 10 and S = 5750 1 32   16. 2 We will use this angle, d, and half of the diameter to set up the following equation. 1 x 2  tan16  92, 919,800 x  2 92, 919,800 tan16   864, 943.0189 The diameter of the sun is about 864,900 mi. (rounded to four significant digits) The included angle is 32, so 53.1 m Because S  T  32 10  5750  8960  90, triangle RST is a right triangle. Thus, we have RS tan 3210  53.1 RS  53.1tan 3210  33.395727 The distance across the lake is 33.4 m. (rounded to three significant digits) 47. Let x represent the horizontal distance between the two buildings and y represent the height of the portion of the building across the street that is higher than the window. 49. The altitude of an isosceles triangle bisects the base as well as the angle opposite the base. The two right triangles formed have interior angles which have the same measure. The lengths of the corresponding sides also have the same measure. The altitude bisects the base, so each leg (base) of the right triangles is 42.36  21.18 in. 2 Let x = the length of each of the two equal sides of the isosceles triangle. 30.0 30.3 x  82.4 x tan 20.0 y tan 50.0   x  30.0  y  x tan 50.0   tan 50.0  tan 20.0  height = y  30.0  30.0  tan 50.0  30.0   tan 20.0   128.2295 The height of the building across the street is about 128 ft. (rounded to three significant digits) tan 20.0  21.18  x cos 38.12  21.18  x 21.18 x  26.921918 cos 38.12 The length of each of the two equal sides of the triangle is 26.92 in. (rounded to four significant digits) cos 38.12  Copyright © 2021 Pearson Education, Inc. 146 Chapter 2 Acute Angles and Right Triangles 50. The altitude of an isosceles triangle bisects the base as well as the angle opposite the base. The two right triangles formed have interior angles which have the same measure. The lengths of the corresponding sides also have the same measure. The altitude bisects the base, so each leg (base) of the right triangles are 184.2  92.10 cm. Each angle opposite to 2 the base of the right triangles measures 1 6844   3422 . 2 Let h = the altitude. In triangle ABC, 92.10  h tan 3422  92.10  tan 3422  h 92.10 h  134.67667 tan 3422 The altitude of the triangle is 134.7 cm. (rounded to four significant digits) 51. Let h represent the height of the tower. In triangle ABC we have h tan 34.6  40.6 h  40.6 tan 34.6  28.0081 The height of the tower is 28.0 m. (rounded to three significant digits) In triangle ABC, 252 sin 3230  d 252 d  469.0121 sin 3230 The distance from the top of the building to the point on the ground is 469 m. (rounded to three significant digits) 53. Let x = the length of the shadow. 5.75 tan 23.4  x x tan 23.4  5.75 5.75  13.2875 x tan 23.4 The length of the shadow is 13.3 ft. (rounded to three significant digits) 54. Let x = the horizontal distance that the plane must fly to be directly over the tree. 10, 500 x x tan1350  10, 500 10, 500 x  42, 641.2351 tan1350 The horizontal distance that the plane must fly to be directly over the tree is 42,600 ft. (rounded to three significant digits) tan1350  55. Let   the angle of depression. 52. Let d = the distance from the top B of the building to the point on the ground A. 39.82  39.82     tan 1   51.74  51.74   37.58  3735 tan   Copyright © 2021 Pearson Education, Inc. Section 2.4 Solutions and Applications of Right Triangles 56. Let x = the height of the taller building; h = the difference in height between the shorter and taller buildings; d = the distance between the buildings along the ground. 28.0  tan1410  28.0  d tan1410  d 28.0  110.9262493 m d tan1410 (We hold on to these digits for the intermediate steps.) To find h, solve h  tan 4640 d h  d tan 4640  110.9262493  tan 4640  117.5749 Thus, the value of h rounded to three significant digits is 118 m. x  h  28.0  118  28.0  146 m, so the height of the taller building is 146 m. 57. Let   the angle of elevation of the sun. 147 59. In order to find the angle of elevation,  , we need to first find the length of the diagonal of the square base. The diagonal forms two isosceles right triangles. Each angle formed by a side of the square and the diagonal measures 45˚. By the Pythagorean theorem, 7002  7002  d 2  2  7002  d 2  d  2  7002  d  700 2 Thus, length of the diagonal is 700 2 ft. To to find the angle,  , we consider the following isosceles triangle. 34.09  34.09     tan 1   37.62  37.62   42.18 tan   58. Let   the angle of elevation of the sun. The height of the pyramid bisects the base of this triangle and forms two right triangles. We can use one of these triangles to find the angle of elevation,  . 200  tan   350 2  200    tan 1   22.0017  350 2  Rounding this figure to two significant digits, we have   22. 55.20  55.20     tan 1   27.65  27.65   63.39 tan   Copyright © 2021 Pearson Education, Inc. 148 Chapter 2 Acute Angles and Right Triangles 60. Let y = the height of the spotlight (this measurement starts 6 feet above ground) y tan 30.0  1000 y  1000  tan 30.0  577.3502 Rounding this figure to three significant digits, we have y  577. However, the observer’s eye-height is 6 feet from the ground, so the cloud ceiling is 577  6  583 ft. 61. (a) Let x = the height of the peak above 14,545 ft. The diagonal of the right triangle formed is in miles, so we must first convert this measurement to feet. Because there are 5280 ft in one mile, we have the length of the diagonal is 27.0134 5280  142, 630.752. Find the value of x by x solving sin 5.82  . 142, 630.752 x  142, 630.752 sin 5.82  14, 463.2674 Thus, the value of x rounded to five significant digits is 14,463 ft. Thus, the total height is about 14,545 + 14,463 = 29,008 ≈ 29,000 ft. (b) The curvature of the earth would make the peak appear shorter than it actually is. Initially the surveyors did not think Mt. Everest was the tallest peak in the Himalayas. It did not look like the tallest peak because it was farther away than the other large peaks. 62. Let x = the distance from the assigned target. In triangle ABC, we have x tan 00 30  234, 000 x  234, 000 tan 00 30  34.0339 The distance from the assigned target is 34.0 mi. (rounded to three significant digits) Section 2.5 Further Applications of Right Triangles 1. C 2. D 3. A 4. G 5. B 6. H 7. F 8. J 9. I 10. E 11. (−4, 0) The bearing of the airplane measured in a clockwise direction from due north is 270°. The bearing can also be expressed as N 90° W, or S 90° W. 12. (5, 0) The bearing of the airplane measured in a clockwise direction from due north is 90°. The bearing can also be expressed as N 90° E, or S 90° E. Copyright © 2021 Pearson Education, Inc. Section 2.5 Further Applications of Right Triangles 13. (0, 4) 149 16. (–3, –3) The bearing of the airplane measured in a clockwise direction from due north is 0°. The bearing can also be expressed as N 0° E or N 0º W. The bearing of the airplane measured in a clockwise direction from due north is 225°. The bearing can also be expressed as S 45° W. 17. (2, –2) 14. (0, –2) The bearing of the airplane measured in a clockwise direction from due north is 180°. The bearing can also be expressed as S 0° E or S 0º W. The bearing of the airplane measured in a clockwise direction from due north is 135°. The bearing can also be expressed as S 45° E. 18. (2, 2) 15. (–5, 5) The bearing of the airplane measured in a clockwise direction from due north is 45°. The bearing can also be expressed as N 45° E. The bearing of the airplane measured in a clockwise direction from due north is 315°. The bearing can also be expressed as N 45° W. 19. Let x = the distance the plane is from its starting point. In the figure, the measure of angle ACB is 38  180  128  38  52  90°. Therefore, triangle ACB is a right triangle. (continued on next page) Copyright © 2021 Pearson Education, Inc. 150 Chapter 2 Acute Angles and Right Triangles (continued) Because d  rt , the distance traveled in 1.5 hr is (1.5 hr)(110 mph) = 165 mi. The distance traveled in 1.3 hr is (1.3 hr)(110 mph) = 143 mi. Using the Pythagorean theorem, we have x 2  1652  1432  x 2  27, 225  20, 449  x 2  47, 674  x  218.3438 The plane is 220 mi from its starting point. (rounded to two significant digits) 20. Let x = the distance from the starting point. In the figure, the measure of angle ACB is 27  180  117  27  63  90. Applying the Pythagorean theorem, we have x 2  27 2  392  x 2  729  1521  x 2  2250  x  2250  47.4342 The ships are 47 nautical mi apart (rounded to 2 significant digits). 22. Let x = distance the ships are apart. In the figure, the measure of angle BCA is 360  322  52  90. Therefore, triangle BCA is a right triangle. Because d  rt , the distance traveled by the first ship in 2.5 hr is (2.5 hr)(17 knots) = 42.5 nautical mi and the second ship is (2.5hr)(22 knots) = 55 nautical mi. Therefore, triangle ACB is a right triangle. Applying the Pythagorean theorem, we have x 2  552  1402  x 2  3025  19, 600  x 2  22, 625  x  22, 625  150.4161 The distance of the end of the trip from the starting point is about 150 km. (rounded to two significant digits) 21. Let x = distance the ships are apart. In the figure, the measure of angle CAB is 130  40  90. Therefore, triangle CAB is a right triangle. Because d  rt , the distance traveled by the first ship in 1.5 hr is (1.5 hr)(18 knots) = 27 nautical mi and the second ship is (1.5hr)(26 knots) = 39 nautical mi. x Applying the Pythagorean theorem, we have x 2  42.52  552  x 2  4831.25  x  4831.25  69.5072 The ships are 70 nautical mi apart (rounded to 2 significant digits). 23. Let b = the distance from dock A to the coral reef C. In the figure, the measure of angle CAB is 90  5822  3138, and the measure of angle CBA is 32822  270  5822 . Because 3138  5822  90, ABC is a right triangle. b cos A  2587 b cos 3138   b  2587 cos 3138 2587 b  2203 ft Copyright © 2021 Pearson Education, Inc. Section 2.5 Further Applications of Right Triangles 24. Let C = the location of the ship, and let c = the distance between the lighthouses. mBAC  180  12943  17960  12943  5017  5017   3943  90, so we have a right triangle. Thus, 3742 sin 3943   c sin 3943  3742  c 3742 c  5856.1020 sin 3943 The distance between the lighthouses is 5856 m (rounded to four significant digits). 151 The measure of angle CBA is 90  5340  8960  5340  3620 . The measure of angle CAB is 90  3620  8960  3620  5340 . A + B = 90°, so C = 90°. Thus, we have a a sin A   sin 5340   2.50 2.50 a  2.50 sin 5340  2.0140 The distance of the transmitter from B is 2.01 mi. (rounded to 3 significant digits) 27. Let b = the distance from A to C and let c = the distance from A to B. 25. Let x = distance between the two ships. The angle between the bearings of the ships is 180°   28°10 + 61°50   90°. The triangle formed is a right triangle. The distance traveled at 24.0 mph is (4 hr) (24.0 mph) = 96 mi. The distance traveled at 28.0 mph is (4 hr)(28.0 mph) = 112 mi. Applying the Pythagorean theorem we have x 2  962  1122  x 2  9216  12, 544  x 2  21, 760  x  21, 760  147.5127 The ships are 148 mi apart. (rounded to three significant digits) 26. Let C = the location of the transmitter; a = the distance of the transmitter from B. Because the bearing from A to B is N 84° E, the measure of angle ABD is 180° − 84° = 96°. The bearing from B to C is 38°, so the measure of angle ABC = 180° − (96° + 38°) = 46°. The bearing of A to C is 52°, so the measure of angle BAC is 180° − (52° + 84°) = 44°. The measure of angle C is 180° − (44° + 46°) = 90°, so triangle ABC is a right triangle. The distance from A to B, labeled c, is 2.4(250) = 600 miles. b b sin 46   c 600 b  600 sin 46  430 mi 28. The information in the example gives mDAC  64, mEAB  82, and mGBC  26. The sum of the measures of angles EAB and FBA is 180º because they are interior angles on the same side of a transversal. (continued on next page) Copyright © 2021 Pearson Education, Inc. 152 Chapter 2 Acute Angles and Right Triangles (continued) So mFBA  180  mEAB  180  82  98 . mCAB  180  64  82  34 and mABC  180  98  26  56 . Thus, angle C is a right angle. It takes 1.8 hours at 350 mph to fly from A to B, so AB  c  1.8  350  630 mi . To find the distance from B to C, use cos B . a cos ABC   a  630 cos 56  350 mi 630 29. Draw triangle WDG with W representing Winston-Salem, D representing Danville, and G representing Goldsboro. Name any point X on the line due south from D. The line from Atlanta to Macon makes an angle of 27° + 63°  90°, with the line from Macon to Augusta. Because d  rt , the distance from Atlanta to Macon is   60 1 14  75 mi. The distance from Macon to   Augusta is 60 1 34  105 mi. Use the Pythagorean theorem to find x: x 2  752  1052  x 2  5625  11, 025  x 2  16, 650  129.0349 The distance from Atlanta to Augusta is 130 mi. (rounded to two significant digits) 31. Let x = the distance from the closer point on the ground to the base of height h of the pyramid. 42° The bearing from W to D is 42° (equivalent to N 42 E), so angle WDX measures 42°. Because angle XDG measures 48°, the measure of angle D is 42° + 48° = 90°. Thus, triangle WDG is a right triangle. Using d = rt and the Pythagorean theorem, we have WG  WD 2   DG 2 2  65 1.1   65 1.8 2 WG  71.52  117 2  5112.25  13, 689  18,801.25  137 The distance from Winston-Salem to Goldsboro is approximately 140 mi. (rounded to two significant digits) 30. Let x = the distance from Atlanta to Augusta. In the larger right triangle, we have h tan 2110 ‘   h  135  x  tan 2110 ‘ 135  x In the smaller right triangle, we have h tan 3530 ‘   h  x tan 3530 ‘. x Substitute for h in this equation, and solve for x to obtain the following. 135  x  tan 2110 ‘  x tan 3530 ‘ 135 tan 2110 ‘ x tan 2110 ‘  x tan 3530 ‘ 135 tan 2110 ‘  x tan 3530 ‘ x tan 2110 ‘ 135 tan 2110 ‘  x  tan 3530 ‘ tan 2110 ‘ 135 tan 2110 ‘ x tan 3530 ‘ tan 2110 ‘ Substitute for x in the equation for the smaller triangle. 135 tan 2110 ‘ h tan 3530 tan 3530 ‘ tan 2110 ‘  114.3427 The height of the pyramid is 114 ft. (rounded to three significant digits) Copyright © 2021 Pearson Education, Inc. Section 2.5 Further Applications of Right Triangles 32. Let x = the distance traveled by the whale as it approaches the tower; y = the distance from the tower to the whale as it turns. 68.7  tan 3540  68.7  y tan 3540  y 68.7 68.7 y  tan1550 and x y tan 3540 68.7   x  y  tan1550 68.7 68.7 x y  x y tan1550 tan1550 68.7 68.7 x   146.5190 tan1550 tan 3540 The whale traveled 147 m as it approached the lighthouse. (rounded to three significant digits) 33. Let x = the height of the antenna; h = the height of the house. 153 In triangle ABC x y  7.00  y  7.00 tan1050 ‘  x y tan1050 ‘ 7.00 tan1050 ‘  x x  7.00 tan1050 ‘  y (2) tan1050 ‘ Setting equations 1 and 2 equal, we have x x  7.00 tan1050  tan 2240 tan1050 x tan1050  x tan 2240  7.00  tan1050  tan 2240  7.00  tan1050  tan 2240   x tan 2240  x tan1050 7.00  tan1050  tan 2240   x  tan 2240 ‘ tan1050 ‘ 7.00  tan1050  tan 2240  x tan 2240  tan1050 x  2.4725 The height of the top of Mt. Whitney above road level is 2.47 km. (rounded to three significant digits) tan1050 ‘  35. Algebraic solution: Let x = the side adjacent to 49.2º in the smaller triangle. In the smaller right triangle, we have h tan1810   h  28 tan1810 . 28 In the larger right triangle, we have xh tan 2710   x  h  28 tan 2710  28 x  28 tan 2710  h x  28 tan 2710  28 tan1810  5.1816 The height of the antenna is 5.18 m. (rounded to three significant digits) 34. Let x = the height of Mt. Whitney above the level of the road; y = the distance shown in the figure below. In triangle ADC, x tan 2240 ‘   y tan 2240 ‘  x  y x y . (1) tan 2240 ‘ In the larger right triangle, we have h tan 29.5   h  392  x  tan 29.5 . 392  x In the smaller right triangle, we have h tan 49.2   h  x tan 49.2 . x Substituting, we have x tan 49.2  392  x  tan 29.5 x tan 49.2  392 tan 29.5  x tan 29.5 x tan 49.2  x tan 29.5  392 tan 29.5 x  tan 49.2  tan 29.5  392 tan 29.5 392 tan 29.5 x tan 49.2  tan 29.5 Now substitute this expression for x in the equation for the smaller triangle to obtain h  x tan 49.2 392 tan 29.5  tan 49.2 h tan 49.2  tan 29.5  433.4762  433 ft (rounded to three significant digits. (continued on next page) Copyright © 2021 Pearson Education, Inc. 154 Chapter 2 Acute Angles and Right Triangles Graphing calculator solution: Graph y1   tan 41.2 x and (continued) Graphing calculator solution: Graph y1   tan 29.5 x and y2   tan 52.5 x  168 in the window y2   tan 29.5 x  392 in the window [0, 800] × [0, 600]. Then find the intersection. [0, 1000] × [0, 500]. Then find the intersection. The height of the triangle is 448 m (rounded to three significant digits. The height of the triangle is 433 ft (rounded to three significant digits. 37. Let x = the minimum distance that a plant needing full sun can be placed from the fence. 36. Algebraic solution: Let x = the side adjacent to 52.5º in the smaller triangle. 4.65  x tan 2320  4.65  x 4.65 x  10.7799 tan 2320 The minimum distance is 10.8 ft. (rounded to three significant digits) tan 2320  In the larger right triangle, we have h  h  168  x  tan 41.2 . tan 41.2  168  x In the smaller right triangle, we have h tan 52.5   h  x tan 52.5. x Substituting, we have x tan 52.5  168  x  tan 41.2 x tan 52.5  168 tan 41.2  x tan 41.2 x tan 52.5  x tan 41.2  168 tan 41.2 x  tan 52.5  tan 41.2  168 tan 41.2 168 tan 41.2 x tan 52.5  tan 41.2 Now substitute this expression for x in the equation for the smaller triangle to obtain h  x tan 52.5 168 tan 41.2  tan 52.5 h tan 52.5  tan 41.2  448.0432  448 m (rounded to three significant digits. 38. 1.0837  0.7261944649 1.4923 A  tan 1 0.7261944649  35.987  3559.2  3559 10 1.4923 tan B   1.377041617 1.0837 B  tan 1 1.377041617   54.013  5400.8  5400 50 tan A  Copyright © 2021 Pearson Education, Inc. Section 2.5 Further Applications of Right Triangles 39. Let h = the minimum height above the surface of Earth so a pilot at A can see an object on the horizon at C. Using the Pythagorean theorem, we have 4.00  10  h  4.00  10   125 2 3 3 2 2 4000  h2  40002  1252 4000  h2  16, 000, 000  15, 625 4000  h2  16, 015, 625 4000  h  16, 015, 625 h  16, 015, 625  4000  4001.9526  4000  1.9526 The minimum height above the surface of Earth would be 1.95 mi. (rounded to 3 significant digits) 40. Let y = the common hypotenuse of the two right triangles. cos 3050  198.4 198.4  y  231.0571948 y cos 3050 To find x, first find the angle opposite x in the right triangle: 5220  3050  5180  3050  2130 x x sin 2130   sin 2130   x  231.0571948 sin 2130  84.6827 y 231.0571948 The length x is approximate 84.7 m. (rounded) 41. (a) PQ  d   b  b   b cot  cot   cot  cot  2 2 2 2 2 2 2 (b) Using the result of part (a), let   37 48,   42 03, and b = 2.000  b   cot  cot   2 2 2 2.000  37 48 42 03  d  cot  cot   cot 0.315  cot 0.3504166667  345.3951 2 2 2  The distance between the two points P and Q is about 345.4 cm. d Copyright © 2021 Pearson Education, Inc. 155 156 Chapter 2 Acute Angles and Right Triangles 42. D  v 2 sin  cos   v cos  v sin  2  64h 32 All answers are rounded to four significant digits. (a) v  44 ft per sec and h  7 ft, so D  442 sin  cos   44 cos  44 sin  2  64  7 32 1936 sin 40 cos 40  44 cos 40  44 sin 40  448 2 If   40, D  32 1936 sin 42 cos 42  44 cos 42  44 sin 42  448  67.00 ft. 2 If   42, D  32 1936 sin 45 cos 45  44 cos 45  44 sin 45  448  67.14 ft 2 If   45, D  32 As  increases, D increases and then decreases.  66.84 ft v 2 sin 42 cos 42  v cos 42 v sin 42  64  7 2 (b) h  7 ft and   42, so D  32 43 sin 42 cos 42  43cos 42  43sin 42  448 2 2 If v  43, D  32 442 sin 42 cos 42  44 cos 42  44 sin 42  448  64.40 ft 2 If v  44, D  32 452 sin 42 cos 42  45 cos 42  45sin 42  448  67.14 ft 2 If v  45, D  32  69.93 ft As v increases, D increases. (c) The velocity affects the distance more. The shot-putter should concentrate on achieving as large a value of v as possible. 43. (a) If   37, then  2  37  18.5. 2 To find the distance between P and Q, d, we first note that angle QPC is a right angle. Hence, triangle QPC is a right triangle and we can solve d tan18.5  965 d  965 tan18.5  322.8845 The distance between P and Q, is 320 ft. (rounded to two significant digits) (b) We are dealing with a circle, so the distance between M and C is R. If we let x be the distance from N to M, then the distance from C to N will be R  x. Triangle CNP is a right triangle, so we can set up the following equation.  Rx   R cos  R  x  cos  2 R 2    x  R  R cos  x  R 1  cos   2 2 Copyright © 2021 Pearson Education, Inc. Section 2.5 Further Applications of Right Triangles 57.3S 57.3 336   32.088 R 600   d  R 1  cos   2  600 1  cos16.044  23.3702 ft The distance is 23 ft. (rounded to two significant digits) 44. (a)   57.3S 57.3  485   46.3175 600 R   d  R 1  cos   2  600 1  cos 23.15875  48.3488 The distance is 48 ft. (rounded to two significant digits) (b)   (c) The faster the speed, the more land needs to be cleared on the inside of the curve. 45. y  tan   x  a   y  tan 35  x  25 46. y  tan   x  a   y  tan15  x  5 47. 157 49. All points whose bearing from the origin is 240º lie in quadrant III. The reference angle,   , is 30º. For any point, x (x, y) on the ray   cos   and r y   sin  , where r is the distance from the r point to the origin. Let r = 2, so x   cos   r x   r cos    2 cos 30  2  23   3 y   sin   r y   r sin    2 sin 30  2  12  1   Thus, a point on the ray is  3, 1 . The ray contains the origin, the equation is of the form   y = mx. Substituting the point  3, 1 , we   have 1  m  3  The line that bisects quadrants I and III passes through the origin, so a = 0. In addition,   45 because the line bisects the angle formed by the axes. tan 45  1, so an equation of the line is y  tan 45  x  or y = x. m  1  1   3 3 3  33 . Thus, an equation of 3 the ray is y  33 x, x  0 (because the ray lies in quadrant III). 50. All points whose bearing from the origin is 150º lie in quadrant IV. 48. The line that bisects quadrants II and IV passes through the origin, so a = 0. In addition,   135 because the line bisects the angle formed by the axes and the reference angle   is 45°. tan135  1, so an equation of the line is y  tan135  x  or y = –x. The reference angle,   , is 60º. For any point, x y (x, y) on the ray  cos   and   sin  , r r where r is the distance from the point to the origin. (continued on next page) Copyright © 2021 Pearson Education, Inc. 158 Chapter 2 Acute Angles and Right Triangles 4. sec  2  10  csc  4  20 (continued) Let r = 2, so x  cos    x  r cos    2 cos 60  2  12  1 r y   sin   r y   r sin    2 sin 60  2  23   3   Thus, a point on the ray is 1,  3 . The ray contains the origin, so the equation is of the form y = mx. Substituting the point 1,  3  , we have  3  m 1  m   3 . Thus, an equation of the ray is y   3 x, x  0 (because the ray lies in quadrant IV). Chapter 2 Review Exercises side opposite 60  hypotenuse 61 side adjacent 11 cos A   hypotenuse 61 side opposite 60 tan A   side adjacent 11 side adjacent 11 cot A   side opposite 60 hypotenuse 61  sec A  side adjacent 11 hypotenuse 61  csc A  side opposite 60 1. sin A  Secant and cosecant are cofunctions, so the sum of the angles is 90º. 2  10  4  20  90 6  30  90 6  60    10 5. tan 5 x  11  cot 6 x  2 Tangent and cotangent are cofunctions, so the sum of the angles is 90º. 5 x  11  6 x  2  90 11x  13  90 11x  77  x  7  3   7  6. cos   11   sin   40   5   10  Sine and cosine are cofunctions, so the sum of the angles is 90º.  3   7   11     40   90     5 10  6   7   11     40   90    10  10 13 13  51  90   39 10 10 10   39   30 13 7. sin 46  sin 58 In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 46º < sin 58º is true. side opposite 40 20 2. sin A    hypotenuse 58 29 side adjacent 42 21   cos A  hypotenuse 58 29 side opposite 40 20   tan A  side adjacent 42 21 side adjacent 42 21 cot A    side opposite 40 20 hypotenuse 58 29   sec A  side adjacent 42 21 hypotenuse 58 29   csc A  side opposite 40 20 3. sin 4   cos 5  Sine and cosine are cofunctions, so the sum of the angles is 90º. 4   5   90  9   90    10 8. cos 47  cos 58 In the interval from 0º to 90º, as the angle increases, the cosine of the angle decreases, so cos 47  cos 58 is false. 9. tan 60  cot 40 Using the cofunction identity, cot 40  tan 90  40  tan 50 . In quadrant I, the tangent function is increasing. Thus cot 40  tan 50  tan 60, and the statement is true. 10. csc 22  csc 68 In quadrant I, the cosecant function is decreasing. Thus csc 22  csc 68, and the statement is false. Copyright © 2021 Pearson Education, Inc. Chapter 2 Review Exercises 11. cos A  bc and sin B  bc , so cos A  sin B. This is an example of equality of cofunctions of complementary angles. 12. If   135,    180  135  45. If   45,    45. If   300,    360  300  60. If   140,    180  140  40. Of these reference angles, 40° is the only one which is not a special angle, so D, tan 140°, is the only one which cannot be determined exactly using the methods of this chapter. 13. 1020º is coterminal with 1020  2  360  300 . The reference angle is 360  300  60. Because 1020° lies in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 3 sin1020   sin 60   2 1 cos1020  cos 60  2 tan1020   tan 60   3 3 cot1020   cot 60   3 sec1020  sec 60  2 2 3 csc1020   csc 60   3 14. A 120º angle lies in quadrant II, so the reference angle is 180  120  60. Because 120° is in quadrant II, the cosine, tangent, cotangent, and secant are negative. 3 sin120  sin 60  2 1 cos120   cos 60   2 tan120   tan 60   3 1 3  cot120   cot 60   3 3 sec120   sec 60  2 2 2 3  csc120  csc 60  3 3 159 15. –1470° is coterminal with 1470  5  360  330. This angle lies in quadrant IV. The reference angle is 360  330  30. Because –1470° is in quadrant IV, the sine, tangent, cotangent, and cosecant are negative. 1 sin(1470)   sin 30   2 3 cos(1470)  cos 30  2 3 tan(1470)   tan 30   3 cot(1470)   cot 30   3 2 3 sec(1470)  sec 30  3 csc(1470)   csc 30  2 16. –225° is coterminal with –225° + 360° = 135°. This angle lies in quadrant II. The reference angle is 180  135  45. Because –225° is in quadrant II, the cosine, tangent, cotangent, and secant are negative. 2 sin(225)  sin 45  2 2 cos(225)   cos 45   2 tan(225)   tan 45  1 cot(225)   cot 45  1 sec(225)   sec 45   2 csc(225)  csc 45  2 1 2 Because cos  is negative,  must lie in quadrants II or III. The absolute value of 1 cos  is , so the reference angle,   must 2 be 60. The quadrant II angle  equals 180     180  60  120, and the quadrant III angle  equals 180     180  60  240. 17. cos    1 2 Because sin  is negative,  must lie in quadrants III or IV. The absolute value of 1 sin  is , so the reference angle,  , is 30º. 2 The angle in quadrant III will be 180     180  30  210 , and the quadrant IV angle is 360     360  30  330 . 18. sin    Copyright © 2021 Pearson Education, Inc. 160 Chapter 2 Acute Angles and Right Triangles 2 3 3 Because sec  is negative,  must lie in quadrants II or III. The absolute value of 2 3 , so the reference angle,   sec  is 3 must be 30º. The quadrant II angle  equals 180     180  30  150, and the quadrant III angle  equals 180     180  30  210. 19. sec    20. cot   1 Because cot  is negative,  must lie in quadrants II or IV. The absolute value of cot  is 1, so the reference angle,   must be 45. The quadrant II angle  equals 180     180  45  135. and the quadrant IV angle  equals 360     360  45  315.  3  2   3  2 3  3 3  21. tan 2 120  2 cot 240   3  2 2 1 1 1 1 22. cos 60  2 sin 30   2      1   2 2 2 2 2  3 23. sec 300  2 cos 150  2  2     2  3 5  4  2 2 2 2 (b) 1,  3  The distance from the origin is r: r  x2  y 2  24. (a) (−3, −3) 2  1 3  4  2 y  3 3   r 2 2 x 1 cos    r 2 sin   tan   y  3   3 x 1 For the exercises in this section, be sure your calculator is in degree mode. 25. sec 22230  1  1.356342 cos 22230 26. sin 7230  0.953717 27. csc 7821  1  1.021034 sin 7821 28. cot 305.6  1  0.715930 tan 305.6 2 2 12   3  29. tan11.7689  0.208344 30. sec 58.9041  1  1.936213 cos 58.9041 31. sin   0.8254121   sin 1 0.8254121  55.673870 32. cot   1.1249386 The distance from the origin is r: r  x2  y 2  r  32  32  r  9  9  r  18  r  3 2 sin   y 3 1 2 2     2 r 3 2 2 2 x 3 1 2 2     2 r 3 2 2 2 y 3 tan    1 x 3 cos   1    1.1249386    cot 1 1.1249386  tan 1   41.635092 33. cos   0.97540415   cos 1 0.97540415  12.733938 34. sec   1.2637891 1    1.2637891    sec1 1.2637891  cos 1   37.695528 Copyright © 2021 Pearson Education, Inc. Chapter 2 Review Exercises 35. tan   1.9633124   tan 1 1.9633124  63.008286 161 44. We must find an angle having secant equal to 1 10, which means that its cosine is 10 . To find sec1 10, on a calculator, enter 36. csc   9.5670466 1    9.5670466    csc1 9.5670466  sin 1   5.9998273   1 cos 1 10 . 37. sin   0.73135370   sin 1 0.73135370  47 The value of sin  is positive in quadrants I and II, so the two angles in  0, 360 are 47° and 180° – 47° = 133°.   1 sec1 10  cos 1 10  84.26082952 38. tan   1.3763819   tan 1 1.3763819  54 The value of tan  is positive in quadrants I and III, so the two angles in  0, 360 are 54º and 180º + 54º = 234º. ? 39. sin 50  sin 40  sin 90 Using a calculator gives sin 50  sin 40  1.408832053 while sin 90º = 1. Thus, the statement is false. ? 40. 1  tan 2 60  sec 2 60 1  tan 2 60  1   3  1 3  4 2 46. a = 129.7, b = 368.1 sec 2 60  22  4 Thus, the statement is true. ? 41. sin 240  2 sin120 cos120 sin 240   sin 60   23 and 2 sin120 cos120  2 sin 60  cos 60 2      3 2 1 2 3 2 Thus, the statement is true. ? 45. A  58 30, c  748 A  B  90  B  90  A  B  90  5830  8960  5830  3130 a a  sin A   sin 5830  c 748 a  748sin 5830  638 (rounded to three significant digits) b b  cos A   cos 5830  c 748 b  748 cos 5830  391 (rounded to three significant digits) 42. sin 42  sin 42  sin 84 Using a calculator gives sin 42  sin 42  1.338261213 while sin 84º = 0.9945218954. Thus, the statement is false. 43. No, this will result in an angle having tangent equal to 25. The function tan 1 is not the reciprocal of the tangent (the cotangent), but is the inverse tangent function. To find cot 25°, the student must find the reciprocal of tan 25°. 1 cot 25   tan 1 25. tan 25 c  a 2  b 2  c  129.7 2  368.12  390.3 (rounded to four significant digits) a 129.7 tan A    b 368.1  129.70  A  tan 1   19.41  368.10   19  0.41  60  1925 b 368.1   a 129.7  368.1  B  tan 1   70.59  129.7   70  0.59  60  7035 tan B  Copyright © 2021 Pearson Education, Inc. 162 Chapter 2 Acute Angles and Right Triangles 47. A  39.72, b  38.97 m 49. Let x = the height of the tree. A  B  90  B  90  A  B  90  39.72  50.28 a a  tan A   tan 39.72  b 38.97 a  38.97 tan 39.72  32.38 m (rounded to four significant digits) b 38.97  cos 39.72   c c c cos 39.72  38.97  38.97 c  50.66 m cos 39.72 (rounded to five significant digits) c os A  tan 70  x  x  50 tan 70  137 ft 50 50. r  122  52  144  25  169  13 5 5 tan      tan 1    23  12  12 51. Let x = height of the tower. 48. B  4753, b  298.6 m x 93.2 x  93.2 tan 3820 x  73.6930 The height of the tower is 73.7 ft. (rounded to three significant digits) tan 3820  A  B  90  A  90  B  A  90  4753  8960  4753  4207  b 298.6  tan B   tan 4753  a a a tan 4753  298.6  298.6 a  270.0 m tan 4753 (rounded to four significant digits) b 298.6  sin B   sin 4753  c c c sin 4753  298.6  298.6 c  402.5 m sin 4753 (rounded to four significant digits) 52. Let h = height of the tower. h 36.0 h  36.0 tan 29.5  20.3678 The height of the tower is 20.4 m. (rounded to three significant digits) tan 29.5  Copyright © 2021 Pearson Education, Inc. Chapter 2 Review Exercises 53. Let x = length of the diagonal 15.24 x 15.24 x  18.7548 cos 35.65 The length of the diagonal is 18.75 cm (rounded to three significant digits). cos 35.65  54. Let x = the length of the equal sides of an isosceles triangle. Divide the isosceles triangle into two congruent right triangles 1 49.28  24.64 and 2 1   58.746  29.373 2 24.64 d  sin    sin 29.373  s s 24.64  50.2352 s sin 29.373 Each side is 50.24 m long (rounded to 4 significant digits). 163 Angle ACB = 344° – 254° = 90°, so ABC is a right triangle. Angle BAX = 32° because it is an alternate interior angle to 32°. Angle YCA = 360° – 344° = 16° Angle XAC = 16° because it is an alternate interior angle to angle YCA. Angle BAC = 32° + 16° = 48°. In triangle ABC, 780 AC cos A   cos 48   AB AB 780 AB cos 48  780  AB   1165.6917 cos 48 The distance from A to B is 1200 m. (rounded to two significant digits) 56. Draw triangle ABC and extend north-south lines from points A and B. Angle ABX is 55° (alternate interior angles of parallel lines cut by a transversal have the same measure) so Angle ABC is 55° + 35° = 90°. d 55. Draw triangle ABC and extend the north-south lines to a point X south of A and S to a point Y, north of C. Angle ABC is a right angle, so use the Pythagorean theorem to find the distance from A to C.  AC 2  812  742   AC 2  6561  5476   AC 2  12, 037  AC  12, 037  109.7133 It is 110 km from A to C. (rounded to two significant digits) 57. Suppose A is the car heading south at 55 mph, B is the car heading west, and point C is the intersection from which they start. After two hours, using d = rt, AC = 55(2) = 110. Angle ACB is a right angle, so triangle ACB is a right triangle. The bearing of A from B is 324º, so angle CAB = 360º – 324º = 36º. (continued on next page) Copyright © 2021 Pearson Education, Inc. 164 Chapter 2 Acute Angles and Right Triangles (continued) 110 AC  cos 36   AB AB 110 AB   135.9675 cos 36 The distance from A to B is about 140 mi (rounded to two significant digits). cos CAB  58. Let x = the leg opposite angle A x  x  k tan A and k hx tan B   x  k tan B  h . So, k k tan A  k tan B  h h  k tan B  k tan A  k  tan B  tan A tan A  59. Answers will vary. Sample answer: Find the value of x. 60. Answers will vary. Sample answer: Find the value of  . (b) Let R = 3955 mi, T = 30 min, P = 140 min.   1  h  R 1   cos 180T  P   1  h  3955  1   1103.6349  cos 18030  140 The height of the satellite is approximately 1104 mi.    62. (a) From the figure, we see that xQ  xP  xQ  xP  d sin  . sin   d Similarly, we have yQ  yP  yQ  yP  d cos  . cos   d (b) Let  xP , yP   123.62, 337.95 ,   1719 22, and d = 193.86 ft. xQ  xP  d sin   xQ  123.62  193.86 sin1719 22  181.3427 yQ  yP  d cos   yQ  337.95  193.86 cos1719 22  523.0170 Thus, the coordinates of Q are (181.34, 523.02), rounded to five significant digits. Chapter 2   (a) Let R = 3955 mi, T = 25 min, P = 140 min.   1 h  R  1   cos 180T  P   1 h  3955   1   715.9424  cos 18025  140 The height of the satellite is approximately 716 mi.     Test side opposite 12  hypotenuse 13 side adjacent 5 cos A   hypotenuse 13 side opposite 12 tan A   side adjacent 5 side adjacent 5 cot A   side opposite 12 hypotenuse 13 sec A   side adjacent 5 hypotenuse 13 csc A   side opposite 12 1. sin A    1 1  61. h  R    cos 180T  P  Copyright © 2021 Pearson Education, Inc. Chapter 2 Test 2. Apply the relationships between the lengths of the sides of a 30  60 right triangle first to the triangle on the right to find the values of y and w. In the 30  60 right triangle, the side opposite the 60 angle is 3 times as long as the side opposite to the 30 angle. The length of the hypotenuse is 2 times as long as the shorter leg (opposite the 30 angle). 165 5. A 240º angle lies in quadrant III, so the reference angle is 240  180  60. Because 240° is in quadrant III, the sine, cosine, secant, and cosecant are negative. 3 sin 240   sin 60   2 1 cos 240   cos 60   2 tan 240  tan 60  3 1 3  3 3 sec 240   sec 60  2 2 2 3 csc 240   csc 60    3 3 cot 240  cot 60  Thus, we have y  4 3 and w  2  4  8 . Apply the relationships between the lengths of the sides of a 45  45 right triangle next to the triangle on the left to find the values of x and z. In the 45  45 right triangle, the sides opposite the 45 angles measure the same. The hypotenuse is 2 times the measure of a leg. Thus, we have x  4 and z  4 2 3. sin   15  cos 2  30 Sine and cosine are cofunctions, so the sum of the angles is 90º. So,   15  2  30  90 3  45  90 3  45    15 4. (a) sin 24  sin 48 In the interval from 0º to 90º, as the angle increases, so does the sine of the angle, so sin 24º < sin 48º is true. (b) cos 24  cos 48 In the interval from 0º to 90º, as the angle increases, so the cosine of the angle decreases, so cos 24º < cos 48º is false. (c) cos 60  30  cos 60 cos 30  sin 60 sin 30 cos 60  30  cos 90  0 cos 60 cos 30  sin 60 sin 30 1 3 3 1     0  2  2  2 2 Thus, the statement is true. 6. –135° is coterminal with –135° + 360° = 225°. This angle lies in quadrant III. The reference angle is 225  180  45. Because –135° is in quadrant III, the sine, cosine, secant, and cosecant are negative. 2 sin(135)   sin 45   2 2 cos(35)   cos 45   2 tan(135)  tan 45  1 cot(135)  cot 45  1 sec(135)   sec 45   2 csc(135)   csc 45   2 7. 990º is coterminal with 990  2  360  270 , which is the reference angle. sin 990  sin 270  1 cos 990  cos 270  0 tan 990  tan 270 undefined cot 990  cot 270  0 sec 990  sec 270 undefined csc 990  csc 270  1 2 2 Because cos  is negative,  must lie in quadrant II or quadrant III. The absolute value 2 , so    45 . The quadrant II of cos  is 2 angle  equals 180     180  45  135, and the quadrant III angle  equals 180     180  45  225. 8. cos    Copyright © 2021 Pearson Education, Inc. 166 Chapter 2 Acute Angles and Right Triangles 2 3 3 Because csc  is negative,  must lie in quadrant III or quadrant IV. The absolute 2 3 value of csc  is , so    60 . The 3 quadrant III angle  equals 180     180  60  240, and the quadrant IV angle  equals 360     360  60  300. 9. csc    10. tan   1    45 or   225 11. tan   1.6778490 1 1 cot     tan   , so we can use tan  division or the inverse key (multiplicative inverse). A  B  90  B  90  A  B  90  5830  3130 a 748 tan A   tan 5830   b b 748  458 (rounded to three b tan 5830 significant digits a 748  sin A   sin 5830  c c 748  877 (rounded to three c sin 5830 significant digits 15. Let  = the measure of the angle that the guy wire makes with the ground. sin   71.3 77.4  71.3   67.1  6710  77.4    sin 1  16. Let x = the height of the flagpole. 12. (a) sin 7821  0.979399 (b) tan117.689  1.905608 (c) sec 58.9041  1  1.936213 cos 58.9041 13. sin   0.27843196   sin 1 0.27843196  16.166641 14. A  5830, a  748 x 24.7 x  24.7 tan 3210  15.5344 The flagpole is approximately 15.5 ft high. (rounded to three significant digits) tan 3210  17. Let h = the height of the top of mountain above the cabin. Then 2000 + h = the height of the mountain. h  h  6800 (rounded to two 14, 000 significant digits). Thus, the height of the mountain is about 6800 + 2000 = 8800 ft. tan 26  Copyright © 2021 Pearson Education, Inc. Chapter 2 Test 18. Let x = distance the ships are apart. In the figure, the measure of angle CAB is 122  32  90. Therefore, triangle CAB is a right triangle. 167 Angle ACB is a right angle, so use the Pythagorean theorem to find the distance from A to B.  AB 2  752  532   AB 2  5625  2809   AB 2  8434  AB  8434  91.8368 It is 92 km from the pier to the boat, rounded to two significant digits. 20. Let x = the side adjacent to 52.5° in the smaller triangle. Because d = rt, the distance traveled by the first ship in 2.5 hr is (2.5 hr)(16 knots) = 40 nautical mi and the second ship is (2.5hr)(24 knots) = 60 nautical mi. Applying the Pythagorean theorem, we have x 2  402  602  x 2  1600  3600  x 2  5200  x  5200  72.111 The ships are 72 nautical mi apart (rounded to 2 significant digits). 19. Draw triangle ACB and extend north-south lines from points A and C. Angle ACD is 62° (alternate interior angles of parallel lines cut by a transversal have the same measure), so Angle ACB is 62° + 28° = 90°. In the larger triangle, we have h tan 41.2   h  168  x  tan 41.2. 168  x In the smaller triangle, we have h tan 52.5   h  x tan 52.5. x Substitute for h in this equation to solve for x. 168  x  tan 41.2  x tan 52.5 168 tan 41.2  x tan 41.2  x tan 52.5 168 tan 41.2  x tan 52.5  x tan 41.2 168 tan 41.2  x  tan 52.5  tan 41.2 168 tan 41.2 x tan 52.5  tan 41.2 Substituting for x in the equation for the smaller triangle gives h  x tan 52.5 168 tan 41.2 tan 52.5 h  448.0432 tan 52.5  tan 41.2 The height of the triangle is approximately 448 m (rounded to three significant digits). Copyright © 2021 Pearson Education, Inc.