Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 8 Applications of Trigonometric Functions 11. opposite = 2; adjacent = 3; hypotenuse = ? (hypotenuse) 2 = 22 + 32 = 13 Section 8.1 1. a = 52 − 32 = 25 − 9 = 16 = 4 hypotenuse = 13 sin θ = 1 , 0 < θ < 90° 2 1 θ = tan −1 ≈ 26.6o 2 2. tan θ = opp 2 2 13 2 13 = = ⋅ = hyp 13 13 13 13 adj 3 3 13 3 13 = = ⋅ = hyp 13 13 13 13 opp 2 tan θ = = adj 3 cos θ = 1 , 0 < θ < 90° 2 1 θ = sin −1 = 30° 2 3. sin θ = csc θ = hyp 13 = opp 2 hyp 13 = adj 3 adj 3 cot θ = = opp 2 sec θ = 4. False; sin 52° = cos 38° 5. True 6. angle of elevation 12. opposite = 3; adjacent = 3; hypotenuse = ? (hypotenuse) 2 = 32 + 32 = 18 7. True hypotenuse = 18 = 3 2 8. False 9. opposite = 5; adjacent = 12; hypotenuse = ? (hypotenuse)2 = 52 + 122 = 169 sin θ = hypotenuse = 169 = 13 opp 5 hyp 13 sin θ = csc θ = = = hyp 13 opp 5 adj 12 hyp 13 cos θ = sec θ = = = hyp 13 adj 12 opp 5 adj 12 tan θ = cot θ = = = adj 12 opp 5 cos θ = opp 3 3 2 2 = = ⋅ = hyp 3 2 3 2 2 2 adj 3 3 2 2 = = ⋅ = hyp 3 2 3 2 2 2 opp 3 tan θ = = =1 adj 3 csc θ = hyp 3 2 = = 2 opp 3 hyp 3 2 = = 2 adj 3 adj 3 = =1 cot θ = opp 3 sec θ = 10. opposite = 3; adjacent = 4, hypotenuse = ? (hypotenuse) 2 = 32 + 42 = 25 hypotenuse = 25 = 5 opp 3 hyp 5 sin θ = csc θ = = = hyp 5 opp 3 adj 4 hyp 5 cos θ = sec θ = = = hyp 5 adj 4 opp 3 adj 4 tan θ = cot θ = = = adj 4 opp 3 791 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 13. adjacent = 2; hypotenuse = 4; opposite = ? (opposite) 2 + 22 = 42 (opposite)2 = 16 − 4 = 12 opposite = 12 = 2 3 opp 2 3 3 = = hyp 4 2 adj 2 1 = = cos θ = hyp 4 2 sin θ = tan θ = opp 2 3 = = 3 adj 2 opp 2 = = 2 adj 1 csc θ = hyp 3 = = opp 2 sec θ = hyp 3 = = 3 adj 1 cot θ = adj 1 1 2 2 = = ⋅ = opp 2 2 2 2 3 2 6 ⋅ = 2 2 2 16. opposite = 2; adjacent = (hypotenuse) = 2 + 2 hyp 4 4 3 2 3 = = ⋅ = opp 2 3 2 3 3 3 hyp 4 sec θ = = =2 adj 2 csc θ = cot θ = tan θ = 2 ( 3) 3 ; hypotenuse = ? 2 =7 hypotenuse = 7 adj 2 2 3 3 = = ⋅ = opp 2 3 2 3 3 3 14. opposite = 3; hypotenuse = 4; adjacent = ? 32 + (adjacent)2 = 42 (adjacent)2 = 16 − 9 = 7 adjacent = 7 sin θ = opp 2 2 7 2 7 = = ⋅ = hyp 7 7 7 7 cos θ = adj 3 3 7 21 = = ⋅ = hyp 7 7 7 7 tan θ = opp 2 2 3 2 3 = = ⋅ = adj 3 3 3 3 csc θ = hyp 7 = opp 2 sin θ = opp 3 = hyp 4 sec θ = hyp 7 7 3 21 = = ⋅ = adj 3 3 3 3 cos θ = adj 7 = hyp 4 cot θ = adj 3 = opp 2 opp 3 3 7 3 7 = = ⋅ = adj 7 7 7 7 hyp 4 = csc θ = opp 3 tan θ = 17. opposite = 1; hypotenuse = 12 + (adjacent)2 = (adjacent)2 = 5 − 1 = 4 hyp 4 4 7 4 7 = = ⋅ = sec θ = adj 7 7 7 7 cot θ = adjacent = 4 = 2 sin θ = adj 7 = opp 3 15. opposite = ( 2) 2 adj 2 2 5 2 5 = = ⋅ = hyp 5 5 5 5 opp 1 tan θ = = adj 2 + 12 = 3 hypotenuse = 3 opp sin θ = = hyp 2 = 3 opp 1 1 5 5 = = ⋅ = hyp 5 5 5 5 cos θ = 2 ; adjacent = 1; hypotenuse = ? (hypotenuse) 2 = ( 5) 5 ; adjacent = ? 2 csc θ = 2 3 6 ⋅ = 3 3 3 hyp 5 = = 5 opp 1 hyp 5 = adj 2 adj 2 cot θ = = =2 opp 1 sec θ = adj 1 1 3 3 cos θ = = = ⋅ = hyp 3 3 3 3 792 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.1: Right Triangle Trigonometry; Applications 18. adjacent = 2; hypotenuse = (opposite) + 2 = 2 2 ( 5) 5 ; opposite = ? 25. tan 20º − 2 (opposite) = 5 − 4 = 1 2 opposite = 1 = 1 sin θ = =0 opp 1 1 5 5 = = ⋅ = hyp 5 5 5 5 26. cot 40º − adj 2 2 5 2 5 cos θ = = = ⋅ = hyp 5 5 5 5 opp 1 tan θ = = adj 2 csc θ = 27. cos 35º ⋅ sin 55º + cos 55º ⋅ sin 35º = cos 35º ⋅ cos(90º −55º ) + sin(90º −55º ) ⋅ sin 35º = cos 35º ⋅ cos 35º + sin 35º ⋅ sin 35º = cos 2 35º + sin 2 35º =1 19. sin 38º − cos 52º = sin 38º − sin(90º − 52º ) 28. sec35º ⋅ csc 55º − tan 35º ⋅ cot 55º = sin 38º − sin 38º = sec 35º ⋅ sec(90º −55º ) − tan 35º ⋅ tan(90º −55º ) = sec 35º ⋅ sec35º − tan 35º ⋅ tan 35º =0 20. tan12º − cot 78º = tan12º − tan(90º − 78º ) = sec 2 35º − tan 2 35º = tan12º − tan12º = (1 + tan 2 35º ) − tan 2 35º =1 =0 22. sin 50º cos(90º − 50º ) = cot 40º − sin 40º sin 40º cos 40º = cot 40º − sin 40º = cot 40º − cot 40º =0 hyp 5 = = 5 opp 1 hyp 5 sec θ = = adj 2 adj 2 cot θ = = =2 opp 1 21. cos 70º sin(90º − 70º ) = tan 20º − cos 20º cos 20º sin 20º = tan 20º − cos 20º = tan 20º − tan 20º cos10º sin(90º − 10º ) sin 80º = = =1 sin 80º sin 80º sin 80º 29. b = 5, B = 20º b a 5 tan ( 20º ) = a tan B = cos 40º sin(90º − 40º ) sin 50º = = =1 sin 50º sin 50º sin 50º a= 23. 1 − cos 2 20º − cos 2 70º = 1 − cos 2 20º − sin 2 (90º −70º ) = 1 − cos 2 20º − sin 2 (20º ) ( = 1 − cos 2 20º + sin 2 (20º ) 5 5 ≈ ≈ 13.74 tan ( 20º ) 0.3640 b c 5 sin ( 20º ) = c ) sin B = = 1−1 =0 c= 24. 1 + tan 2 5º − csc 2 85º = sec2 5º − csc 2 85º = sec2 5º − sec2 (90º − 85º ) 5 5 ≈ ≈ 14.62 sin ( 20º ) 0.3420 A = 90º − B = 90º − 20º = 70º = sec2 5º − sec2 5º =0 793 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 30. b = 4, B = 10º 33. b = 4, A = 10º b a 4 tan (10º ) = a a b a tan (10º ) = 4 a = 4 tan (10º ) ≈ 4 ⋅ (0.1763) ≈ 0.71 tan B = a= tan A = 4 4 ≈ ≈ 22.69 tan (10º ) 0.1763 b c 4 cos (10º ) = c cos A = b sin B = c 4 sin (10º ) = c c= 4 4 c= ≈ ≈ 23.04 sin (10º ) 0.1736 4 4 ≈ ≈ 4.06 0.9848 cos (10º ) B = 90º − A = 90º − 10º = 80º A = 90º − B = 90º − 10º = 80º 34. b = 6, A = 20º 31. a = 6, B = 40º b a b tan ( 40º ) = 6 b = 6 tan ( 40º ) ≈ 6 ⋅ (0.8391) ≈ 5.03 a b a tan ( 20º ) = 6 a = 6 tan ( 20º ) ≈ 6 ⋅ (0.3640) ≈ 2.18 a c 6 cos ( 40º ) = c b c 6 cos ( 20º ) = c tan A = tan B = cos B = c= cos A = 6 6 ≈ ≈ 7.83 cos ( 40º ) 0.7660 c= A = 90º − B = 90º − 40º = 50º B = 90º − A = 90º − 20º = 70º 32. a = 7, B = 50º b tan B = a b tan ( 50º ) = 7 b = 7 tan ( 50º ) ≈ 7 ⋅ (1.1918) ≈ 8.34 35. a = 5, A = 25º b cot A = a b cot ( 25º ) = 5 b = 5cot ( 25º ) ≈ 5 ⋅ ( 2.1445) ≈ 10.72 a c 7 cos ( 50º ) = c cos B = c= 6 6 ≈ ≈ 6.39 cos ( 20º ) 0.9397 c a c csc ( 25º ) = 5 c = 5csc ( 25º ) ≈ 5 ⋅ ( 2.3662 ) ≈ 11.83 csc A = 7 7 ≈ ≈ 10.89 cos ( 50º ) 0.6428 A = 90° − B = 90° − 50° = 40° B = 90° − A = B = 90º − A = 90° − 25° = 65° 794 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.1: Right Triangle Trigonometry; Applications 36. a = 6, A = 40º a cot A = b b cot ( 40º ) = 6 b = 6 cot ( 40º ) ≈ 6 ⋅ (1.1918) ≈ 7.15 39. a = 5, b = 3 c 2 = a 2 + b 2 = 52 + 32 = 25 + 9 = 34 c = 34 ≈ 5.83 a 5 = b 3 5 A = tan −1 ≈ 59.0° 3 tan A = c a c csc ( 45º ) = 6 c = 6 csc ( 40º ) ≈ 6 ⋅ (1.5557 ) ≈ 9.33 csc A = B = 90° − A ≈ 90° − 59.0° = 31.0° 40. a = 2, b = 8 c 2 = a 2 + b 2 = 22 + 82 = 4 + 64 = 68 B = 90° − A = 90° − 40° = 50° c = 68 ≈ 8.25 37. c = 9, B = 20º b sin B = c b sin ( 20º ) = 9 b = 9sin ( 20º ) ≈ 9 ⋅ (0.3420) ≈ 3.08 a 2 1 = = b 8 4 1 A = tan −1 ≈ 14.0° 4 tan A = B = 90° − A ≈ 90° − 14.0° = 76.0° 41. a = 2, c = 5 a c a cos ( 20º ) = 9 a = 9 cos ( 20º ) ≈ 9 ⋅ (0.9397) ≈ 8.46 cos B = c2 = a2 + b2 b 2 = c 2 − a 2 = 52 − 22 = 25 − 4 = 21 b = 21 ≈ 4.58 a 2 = c 5 2 A = tan −1 ≈ 23.6º 5 sin A = A = 90° − A = 90° − 20° = 70° 38. c = 10, A = 40º a c a sin ( 40º ) = 10 a = 10sin ( 40º ) ≈ 10 ⋅ (0.6428) ≈ 6.43 sin A = B = 90° − A ≈ 90° − 23.6° = 66.4° 42. b = 4, c = 6 c2 = a 2 + b2 a 2 = c 2 − b2 = 62 − 42 = 36 − 16 = 20 b cos A = c b cos ( 40º ) = 10 b = 10 cos ( 40º ) ≈ 10 ⋅ (0.7660) ≈ 7.66 a = 20 ≈ 4.47 b 4 2 = = c 6 3 2 A = tan −1 ≈ 48.2º 3 cos A = B = 90° − A = 90° − 40° = 50° B = 90º − A ≈ 90º − 48.2º = 41.8º 795 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 43. c = 5, a = 2 Case 2: θ = 25º , b = 5 2 5 sin A = c 2 A = sin ≈ 23.6° 5 B = 90º − A ≈ 90º − 23.6º = 66.4º The two angles measure about 23.6° and 66.4º . a −1 25º 5 cos ( 25º ) = c= 44. c = 3, a = 1 B = 90º − A ≈ 90º − 19.5º = 70.5º The two angles measure about 19.5° and 70.5º . b. 45. c = 8, θ = 35º 5 5 ≈ ≈ 5.52 in. cos ( 25º ) 0.9063 There are two possible cases because the given side could be adjacent or opposite the given angle. 48. Case 1: θ = 8 5 c a π , a=3 8 a. 35º b a 8 a = 8sin ( 35º ) b 8 b = 8cos ( 35º ) ≈ 8(0.5736) ≈ 8(0.8192) ≈ 4.59 in. ≈ 6.55 in. sin ( 35º ) = cos ( 35º ) = c 8 b π 3 sin = 8 c 3 3 ≈ ≈ 7.84 m. c= π 0.3827 sin 8 46. c = 10, θ = 40º 10 a 40º Case 2: θ = π , b=3 8 c a b a 10 a = 10sin ( 40º ) b 10 b = 10 cos ( 40º ) ≈ 10(0.6428) ≈ 10(0.7660) ≈ 6.43 cm. ≈ 7.66 cm. sin ( 40º ) = cos ( 40º ) = _π 8 a. b. There are two possible cases because the given side could be adjacent or opposite the given angle. 5 25º b sin ( 25º ) = c= 3 π 3 cos = 8 c 3 3 ≈ ≈ 3.25 m. c= π 0.9239 cos 8 47. Case 1: θ = 25º , a = 5 c 3 _π 49. tan ( 35º ) = 5 c AC 100 AC = 100 tan ( 35º ) ≈ 100(0.7002) ≈ 70.02 feet 5 5 ≈ ≈ 11.83 in. sin ( 25º ) 0.4226 796 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.1: Right Triangle Trigonometry; Applications 56. opposite side = 10 feet, adjacent side = 35 feet 10 tan θ = 35 10 θ = tan −1 ≈ 15.9º 35 AC 50. tan ( 40º ) = 100 AC = 100 tan ( 40º ) ≈ 100(0.8391) ≈ 83.91 feet 51. Let x = the height of the Eiffel Tower. x tan ( 85.361º ) = 80 x = 80 tan ( 85.361º ) ≈ 80(12.3239) ≈ 985.91 feet 57. a. 52. Let x = the distance to the shore. 100 feet The truck is traveling at 111.96 ft/sec, or 111.96 ft 1 mile 3600 sec ⋅ ⋅ ≈ 76.3 mi/hr . sec 5280 ft hr 25º x tan ( 25º ) = x= 100 x b. 100 100 ≈ ≈ 214.45 feet tan ( 25º ) 0.4663 c. 20º x 50 x 50 50 ≈ ≈ 137.37 meters tan ( 20º ) 0.3640 54. Let x = the distance up the building 22 feet x 22 x = 22sin ( 70º ) ≈ 22(0.9397) ≈ 20.67 feet sin ( 70º ) = 300 =6 50 A = tan −1 6 ≈ 80.5º The angle of elevation of the sun is about 80.5º . 30 30 ≈ ≈ 82.42 feet tan ( 20º ) 0.3640 A ticket is issued for traveling at a speed of 60 mi/hr or more. 60 mi 5280 ft 1hr ⋅ ⋅ = 88 ft/sec. hr mi 3600 sec 30 , the trooper should issue a If tan θ < 88 30 ticket. Now, tan −1 ≈ 18.8° , so a ticket 88 is issued if θ < 18.8º . 58. If the camera is to be directed to a spot 6 feet above the floor 12 feet from the wall, then the “side opposite” the angle of depression is 3 feet. (see figure) 12 3 1 = tan A = A 12 4 3 −1 1 A = tan ≈ 14.0° 4 The angle of depression should be about 14.0° . x 70º 55. 30 x The truck is traveling at 82.42 ft/sec, or 82.42 ft 1 mile 3600 sec ⋅ ⋅ ≈ 56.2 mi/hr . sec 5280 ft hr 50 meters x= tan ( 20º ) = x= 53. Let x = the distance to the base of the plateau. tan ( 20º ) = Let x represent the distance the truck traveled in the 1 second time interval. 30 tan (15º ) = x 30 30 x= ≈ ≈ 111.96 feet tan (15º ) 0.2679 tan A = 300 feet A 50 feet 797 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 59. a. ( ) 61. Let h = the height of the monument. 4.22 ⋅ 5.9 × 1012 = 24.898 ×1012 h 789 h = 789 tan ( 35.1º ) tan ( 35.1º ) = = 2.4898 ×1013 Proxima Centauri is about 2.4898 ×1013 miles from Earth. b. ≈ 789(0.7028) ≈ 554.52 ft Construct a right triangle using the sun, Earth, and Proxima Centauri as shown. The hypotenuse is the distance between Earth and Proxima Centauri. b Sun a 35.1º 789 ft 62. The elevation change is 11200 − 9000 = 2200 ft . Let x = the length of the trail. Proxima Centauri θ h x Parallax 2200 ft 17º c 2200 x 2200 2200 x= ≈ ≈ 7524.67 ft. sin (17º ) 0.2924 sin17º = Earth a 9.3 × 107 = c 2.4898 × 1013 sin θ ≈ 0.000003735 sin θ = 63. Begin by finding angle θ = ∠BAC : (see figure) 40º The parallax of Proxima Centauri is 0.000214° . 60. a. ( 11.14 ⋅ 5.9 × 10 12 i E Α ) = 65.726 ×10 12 C 1 =2 0.5 θ = 63.4º ∠DAC = 40º + 63.4º = 103.4º ∠EAC = 103.4º − 90º = 13.4º Now, 90º −13.4º = 76.6º The control tower should use a bearing of S76.6˚E. tan θ = = 6.5726 ×10 61 Cygni is about 6.5726 ×1013 miles from Earth. Construct a right triangle using the sun, Earth, and 61 Cygni as shown. The hypotenuse is the distance between Earth and 61 Cygni. b Sun 61 Cygni θ 64. Find ∠AMB and subtract from 80˚ to obtain θ (see figure). 80º M a 1m θ 13 b. B i D 0.5 m θ ≈ sin −1 0.000003735 θ ≈ 0.000214° Parallax Α 15 c θ 30 C Earth a 9.3 × 107 sin θ = = c 6.5726 × 1013 sin θ ≈ 0.000001415 B ∠CMA = 80° 30 =2 15 ∠AMB = tan −1 2 ≈ 63.4º θ = 80º − 63.4º = 16.6º The bearing of the ship from port is S16.6ºE . tan ∠AMB = θ ≈ sin −1 0.000001415 θ ≈ 0.00008° The parallax of 61 Cygni is 0.00008° . 798 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.1: Right Triangle Trigonometry; Applications 65. Let y = the height of the embankment. .8 y 36o2’ x θ = 36°2 ' ≈ 36.033° 140º 40º sin ( 40º ) = x= z= 43 27 34 a= o x 2593 x = 2593 ⋅ cos 34° ≈ 2150 The surveyor is located approximately 2150 feet from the building. cos 34° = b. c. d. b= 1 a 1 ≈ 1.1918 mi tan ( 40º ) tan ( 50º ) = x 1 z 1 ≈ 1.3054 mi sin ( 50º ) tan ( 40º ) = h 1 x 1 ≈ 1.5557 mi sin ( 40º ) sin ( 50º ) = a 93 b 3 mi The length of the highway = x + y + z Let h = the height of the building, a = the height of the antenna, and x = the distance between the surveyor and the base of the building. 25 z 50º 130º a y sin ( 36.033° ) = 51.8 y = 51.8sin ( 36.033° ) ≈ 30.5 meters The embankment is about 30.5 meters high. 66. a. 1 mi y 1 mi 51 67. Let x, y, and z = the three segments of the highway around the bay (see figure). 1 b 1 ≈ 0.8391 mi tan ( 50º ) a+ y+b = 3 y = 3− a −b h sin 34° = 2593 h = 2593 ⋅ sin 34° ≈ 1450 The building is about 1450 feet high. ≈ 3 − 1.1918 − 0.8391 = 0.9691 mi The length of the highway is about: 1.5557 + 0.9691 + 1.3054 ≈ 3.83 miles . 68. Let x = the distance from George at which the camera must be set in order to see his head and feet. Let θ = the angle of inclination from the surveyor to the top of the antenna. 2150 cos θ = 2743 2150 θ = cos−1 ≈ 38.4° 2743 x 20º 4 ft tan ( 20º ) = h + a 1450 + a = 2743 2743 2743 ⋅ sin θ = 1450 + a a = 2743 ⋅ sin θ − 1450 sin θ = 4 x 4 ≈ 10.99 feet tan ( 20º ) If the camera is set at a distance of 10 feet from George, his feet will not be seen by the lens. The camera would need to be moved back about 1 additional foot (11 feet total). x= 2150 a = 2743sin cos −1 − 1450 2743 a ≈ 253 The antenna is about 253 feet tall. 799 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 69. We construct the figure below: 32º h 23º 500 ft 32º 71. Let h represent the height of Lincoln's face. 23º y x tan ( 32º ) = 500 x b tan ( 23º ) = 500 x= tan ( 32º ) 32º 500 y 800 feet b 800 b = 800 tan ( 32º ) ≈ 499.90 tan ( 32º ) = 500 y= tan ( 23º ) b+h 800 b + h = 800 tan ( 35º ) ≈ 560.17 Thus, the height of Lincoln’s face is: h = (b + h) − b = 560.17 − 499.90 ≈ 60.27 feet tan ( 35º ) = Distance = x + y 500 500 = + tan ( 32º ) tan ( 23º ) ≈ 1978.09 feet 70. Let h = the height of the balloon. 54º 35º 72. Let h represent the height of tower above the Sky Pod. 61º h h h b 61º 20.1º 54º 4000 feet x − 100 100 ft x tan ( 54º ) = x= b 4000 b = 4000 tan ( 20.1º ) ≈ 1463.79 tan ( 20.1º ) = h x b+h 4000 b + h = 4000 tan ( 24.4º ) ≈ 1814.48 Thus, the height of tower above the Sky Pod is: h = (b + h) − b = 1814.48 − 1463.79 ≈ 350.69 feet tan ( 24.4º ) = h tan ( 54º ) h x − 100 h = ( x − 100) tan ( 61º ) tan ( 61º ) = 73. Let x = the distance between the buildings. h h= − 100 tan ( 61º ) tan ( 54º ) h− 24.4º tan(61°) tan(54°) h = −100 tan ( 61° ) 1451 o 10.3 tan(61°) h 1 − = −100 tan ( 61° ) tan(54°) h= x 1451 x 1451 x= ≈ 7984 tan (10.3° ) tan θ = −100 tan ( 61° ) ≈ 580.61 tan ( 61° ) 1 − tan ( 54° ) Thus, the height of the balloon is approximately 580.61 feet. The two buildings are about 7984 feet apart. 800 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.1: Right Triangle Trigonometry; Applications 74. tan (α + β ) = x +2 x +8 = 2 4 4( x + 2) = 2( x + 8) 4 x + 8 = 2 x + 16 2x = 8 x=4 Thus, the hypotenuse of the smaller triangle is θ 2 1 4 + 2 = 6 . Now, sin = = , so we have 2 6 3 2070 630 2070 630 2070 α = tan −1 −β 630 2070 −1 67 = tan −1 − cot 630 55 ≈ 33.69° Let x = the distance between the Arch and the boat on the Missouri side. x tan α = 630 x = 630 tan α = 630 tan ( 33.69° ) α + β = tan −1 that: x = 420 Therefore, the Mississippi River is approximately 2070 − 420 = 1650 feet wide at the St. Louis riverfront. 78. a. 75. The height of the beam above the wall is 46 − 20 = 26 feet. 26 tan θ = = 2.6 10 θ = tan −1 2.6 ≈ 69.0º The pitch of the roof is about 69.0º . c. 3960 d cos = 3960 + h 7920 e. 2 θ/2 x 2 22 4 4 3960 θ cos = 2 3960 + h d = 3960 θ 76. tan A = 77. A line segment drawn from the vertex of the angle through the centers of the circles will bisect θ (see figure). Thus, the angle of the two θ right triangles formed is . 1 = sin −1 3 1 θ = 2 ⋅ sin −1 ≈ 38.9° 3 2 b. d. 10 − 6 4 = 15 15 −1 4 A = tan ≈ 14.9º 15 The angle of elevation from the player’s eyes to the center of the rim is about 14.9º . θ 3960 2500 cos = 3960 + h 7920 3960 0.9506 ≈ 3960 + h 0.9506(3960 + h) ≈ 3960 3764 + 0.9506h ≈ 3960 0.9506h ≈ 196 h ≈ 206 miles 3960 3960 d cos = = 7920 3960 + 300 4260 d 3960 = cos −1 7920 4260 3960 d = 7970 cos −1 4260 ≈ 2990 miles Let x = the length of the segment from the vertex of the angle to the smaller circle. Then the hypotenuse of the smaller right triangle is x + 2 , and the hypotenuse of the larger right triangle is x + 2 + 2 + 4 = x + 8 . Since the two right triangles are similar, we have that: 801 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 79. Adding some lines to the diagram and labeling special points, we obtain the following: b. 5 D C B α 1.8 α Let y = the difference in height between Freedom Tower and the office building. Together with the result from part (a), we get the following diagram y 20° 2633 E opposite tan θ = adjacent y tan 20° = 2633 y ≈ 958 The Freedom Tower is about 958 feet taller than the office building. Therefore, the office building is 1776 − 958 = 818 feet tall. 1.2 1.5 A 1 3 If we let x = length of side BC, we see that, in 3 ΔABC , tan α = . Also, in ΔEDC , x 1.8 tan α = . Therefore, we have 5− x 3 1.8 = x 5− x 15 − 3x = 1.8 x 15 = 4.8 x 15 x= = 3.125 ft 4.8 1 + 3.125 = 4.125 ft The player should hit the top cushion at a point that is 4.125 feet from upper left corner. 80. a. 81 – 82. Answers will vary. 83. Let θ = the central angle formed by the top of the lighthouse, the center of the Earth and the point P on the Earth’s surface where the line of sight from the top of the lighthouse is tangent to 362 miles. the Earth. Note also that 362 feet = 5280 The distance between the buildings is the length of the side adjacent to the angle of elevation in a right triangle. 1776 θ = cos−1 3960 o ≈ 0.33715 + 3960 362 / 5280 34° x opposite and we know the adjacent angle measure, we can use the tangent to find the distance. Let x = the distance between the buildings. This gives us 1776 tan 34° = x 1776 x= tan 34° x ≈ 2633 The office building is about 2633 feet from the base of the tower. Since tan θ = Verify the airplane information: Let β = the central angle formed by the plane, the center of the Earth and the point P. 3960 β = cos −1 ≈ 1.77169° + 3960 10,000 / 5280 Note that d d tan θ = 1 and tan β = 2 3690 3690 d1 = 3960 tan θ d 2 = 3960 tan β So, d1 + d 2 = 3960 tan θ + 3960 tan β ≈ 3960 tan(0.33715°) + 3960 tan(1.77169°) ≈ 146 miles 802 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines To express this distance in nautical miles, we express the total angle θ + β in minutes. That is, Section 8.2 θ + β ≈ ( 0.33715o + 1.77169o ) ⋅ 60 ≈ 126.5 1. sin A cos B − cos A sin B nautical miles. Therefore, a plane flying at an altitude of 10,000 feet can see the lighthouse 120 miles away. 2. cos θ = 3 2 3 π = = 30° 2 6 θ = cos −1 Verify the ship information: The solution set is {π } . 6 = 30° 3. For similar triangles, corresponding pairs of sides occur in the same ratio. Therefore, we can write: x 5 15 = or x = 3 2 2 15 The missing length is . 2 Let α = the central angle formed by 40 nautical miles then. α = 40 2 = 60 3 4. oblique o 5. 3960 3960 + x 3960 cos ( (2 / 3)° − 0.33715° ) = 3960 + x ( 3960 + x ) cos ( (2 / 3)° − 0.33715° ) = 3960 cos(α − θ ) = 3960 + x = x= sin A sin B sin C = = a b c 6. False 7. False: You must have at least one angle opposite one side. 3960 cos ( (2 / 3)° − 0.33715° ) 8. ambiguous case 9. c = 5, B = 45º , C = 95º A = 180º − B − γ = 180º − 45º − 95º = 40º 3960 − 3960 cos ( (2 / 3)° − 0.33715° ) sin A sin C = a c sin 40º sin 95º = 5 a 5sin 40º a= ≈ 3.23 sin 95º ≈ 0.06549 miles ft ≈ ( 0.06549 mi ) 5280 mi ≈ 346 feet Therefore, a ship that is 346 feet above sea level can see the lighthouse from a distance of 40 nautical miles. sin B sin C = b c sin 45º sin 95º = 5 b 5sin 45º b= ≈ 3.55 sin 95º 803 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 10. c = 4, A = 45º , B = 40º C = 180º − A − B = 180º − 45º − 40º = 95º 13. b = 7, A = 40º , B = 45º C = 180º − A − B = 180º − 40º − 45º = 95º sin A sin C = a c sin 45º sin 95º = 4 a 4sin 45º ≈ 2.84 a= sin 95º sin A sin B = a b sin 40º sin 45º = 7 a 7 sin 40º ≈ 6.36 a= sin 45º sin B sin C = b c sin 40º sin 95º = 4 b 4 sin 40º b= ≈ 2.58 sin 95º sin C sin B = c b sin 95º sin 45º = 7 c 7 sin 95º ≈ 9.86 c= sin 45º 11. b = 3, A = 50º , C = 85º B = 180º − A − C = 180º − 50º − 85º = 45º 14. c = 5, A = 10º , B = 5º C = 180º − A − B = 180º − 10º − 5º = 165º sin A sin B = a b sin 50º sin 45º = 3 a 3sin 50º a= ≈ 3.25 sin 45º sin A sin C = a c sin10º sin165º = 5 a 5sin10º a= ≈ 3.35 sin165º sin C sin B = c b sin 85º sin 45º = 3 c 3sin 85º c= ≈ 4.23 sin 45º sin B sin C = b c sin 5º sin165º = 5 b 5sin 5º b= ≈ 1.68 sin165º 12. b = 10, B = 30º , C = 125º A = 180º − B − C = 180º − 30º − 125º = 25º 15. b = 2, B = 40º , C = 100º A = 180º − B − C = 180º − 40º − 100º = 40º sin A sin B = a b sin 25º sin 30º = 10 a 10 sin 25º ≈ 8.45 a= sin 30º sin A sin B = a b sin 40º sin 40º = 2 a 2 sin 40º a= =2 sin 40º sin C sin B = c b sin125º sin 30º = 10 c 10sin125º c= ≈ 16.38 sin 30º sin C sin B = c b sin100º sin 40º = 2 c 2 sin100º c= ≈ 3.06 sin 40º 804 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 16. b = 6, A = 100º , B = 30º C = 180º − A − B = 180º − 100º − 30º = 50º 19. B = 70º , C = 10º , b = 5 A = 180º − B − C = 180º − 70º − 10º = 100º sin A sin B = a b sin100º sin 30º = 6 a 6 sin100º a= ≈ 11.82 sin 30º sin A sin B = a b sin100º sin 70º = 5 a 5sin100º a= ≈ 5.24 sin 70º sin C sin B = c b sin 50º sin 30º = 6 c 6sin 50º c= ≈ 9.19 sin 30º sin C sin B = c b sin10º sin 70º = 5 c 5sin10º c= ≈ 0.92 sin 70º 17. A = 40º , B = 20º , a = 2 C = 180º − A − B = 180º − 40º − 20º = 120º 20. A = 70º , B = 60º , c = 4 C = 180º − A − B = 180º − 70º − 60º = 50º sin A sin B = a b sin 40º sin 20º = 2 b 2sin 20º b= ≈ 1.06 sin 40º sin A sin C = a c sin 70º sin 50º = 4 a 4 sin 70º a= ≈ 4.91 sin 50º sin C sin A = c a sin120º sin 40º = 2 c 2sin120º c= ≈ 2.69 sin 40º sin B sin C = b c sin 60º sin 50º = 4 b 4sin 60º b= ≈ 4.52 sin 50º 18. A = 50º , C = 20º , a = 3 B = 180º − A − C = 180º − 50º − 20º = 110º 21. A = 110º , C = 30º , c = 3 B = 180º − A − C = 180º − 110º − 30º = 40º sin A sin B = a b sin 50º sin110º = 3 b 3sin110º b= ≈ 3.68 sin 50º sin A sin C = a c sin110º sin 30º = 3 a 3sin110º a= ≈ 5.64 sin 30º sin C sin A = c a sin 20º sin 50º = 3 c 3sin 20º c= ≈ 1.34 sin 50º sin C sin B = c b sin 30º sin 40º = 3 b 3sin 40º b= ≈ 3.86 sin 30º 805 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 22. B = 10º , C = 100º , b = 2 A = 180º − B − C = 180º − 10º − 100º = 70º 25. a = 3, b = 2, A = 50º sin B sin A = b a sin B sin ( 50º ) = 2 3 2sin ( 50º ) ≈ 0.5107 sin B = 3 sin A sin B = a b sin 70º sin10º = 2 a 2sin 70º a= ≈ 10.82 sin10º B = sin −1 ( 0.5107 ) B = 30.7º or B = 149.3º The second value is discarded because A + B > 180º . C = 180º − A − B = 180º − 50º − 30.7º = 99.3º sin C sin B = c b sin100º sin10º = 2 c 2sin100º c= ≈ 11.34 sin10º sin C sin A = c a sin 99.3º sin 50º = 3 c 3sin 99.3º c= ≈ 3.86 sin 50º One triangle: B ≈ 30.7º , C ≈ 99.3º , c ≈ 3.86 23. A = 40º , B = 40º , c = 2 C = 180º − A − B = 180º − 40º − 40º = 100º sin A sin C = a c sin 40º sin100º = 2 a 2sin 40º a= ≈ 1.31 sin100º 26. b = 4, c = 3, B = 40º sin B sin C = b c sin 40º sin C = 4 3 3sin 40º sin C = ≈ 0.4821 4 sin B sin C = b c sin 40º sin100º = 2 b 2sin 40º b= ≈ 1.31 sin100º C = sin −1 ( 0.4821) C = 28.8º or C = 151.2º The second value is discarded because B + C > 180º . A = 180º − B − C = 180º − 40º − 28.8º = 111.2º 24. B = 20º , C = 70º , a = 1 A = 180º − B − C = 180º − 20º −70º = 90º sin A sin B = a b sin 90º sin 20º = 1 b 1sin 20º b= ≈ 0.34 sin 90º sin B sin A = b a sin 40º sin111.2º = 4 a 4 sin111.2º a= ≈ 5.80 sin 40º One triangle: A ≈ 111.2º , C ≈ 28.8º , a ≈ 5.80 sin C sin A = c a sin 70º sin 90º = 1 c 1sin 70º ≈ 0.94 c= sin 90º 806 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 27. b = 5, c = 3, B = 100º 29. a = 4, b = 5, A = 60º sin B sin C = b c sin100º sin C = 5 3 3sin100º sin C = ≈ 0.5909 5 sin B sin A = b a sin B sin 60º = 5 4 5sin 60º sin B = ≈ 1.0825 4 There is no angle B for which sin B > 1 . Therefore, there is no triangle with the given measurements. C = sin −1 ( 0.5909 ) C = 36.2º or C =143.8º The second value is discarded because B + C > 180º . A = 180º − B − C = 180º − 100º − 36.2º = 43.8º 30. b = 2, c = 3, B = 40º sin B sin C = b c sin 40º sin C = 2 3 3sin 40º sin C = ≈ 0.9642 2 C = sin −1 ( 0.9642 ) sin B sin A = b a sin100º sin 43.8º = 5 a 5sin 43.8º a= ≈ 3.51 sin100º One triangle: A ≈ 43.8º , C ≈ 36.2º , a ≈ 3.51 C1 = 74.6º or C2 = 105.4º For both values, B + C < 180º . Therefore, there are two triangles. 28. a = 2, c = 1, A = 120º sin C sin A = c a sin C sin120º = 1 2 1sin120º sin C = ≈ 0.4330 2 A1 = 180º − B − C1 = 180º − 40º − 74.6º = 65.4º sin B sin A1 = b a1 sin 40º sin 65.4º = 2 a1 C = sin −1 ( 0.4330 ) a1 = C = 25.7º or C =154.3º The second value is discarded because A + C > 180º . B = 180º − A − C = 180º − 120º − 25.7º = 34.3º 2sin 65.4º ≈ 2.83 sin 40º A2 = 180º − B − C2 = 180º − 40º − 105.4º = 34.6º sin B sin A2 = b a2 sin B sin A = b a sin 34.3º sin120º = 2 b 2sin 34.3º b= ≈ 1.30 sin120º sin 40º sin 34.6º = 2 a2 a2 = 2 sin 34.6º ≈ 1.77 sin 40º Two triangles: A1 ≈ 65.4º , C1 ≈ 74.6º , a1 ≈ 2.83 or One triangle: B ≈ 34.3º , C ≈ 25.7º , b ≈ 1.30 A2 ≈ 34.6º , C2 ≈ 105.4º , a2 ≈ 1.77 807 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 31. b = 4, c = 6, B = 20º 34. b = 4, c = 5, B = 95º sin B sin C = b c sin 20º sin C = 4 6 6sin 20º sin C = ≈ 0.5130 4 C = sin −1 ( 0.5130 ) sin C sin B = c b sin C sin 95º = 5 4 5sin 95º sin C = ≈ 1.2452 4 There is no angle C for which sin C > 1 . Thus, there is no triangle with the given measurements. C1 = 30.9º or C2 = 149.1º For both values, B + C < 180º . Therefore, there are two triangles. A1 = 180º − B − C1 = 180º − 20º − 30.9º = 129.1º 35. a = 2, c = 1, C = 25º sin A sin C = a c sin A sin 25º = 2 1 2sin 25º sin A = ≈ 0.8452 1 A = sin −1 ( 0.8452 ) sin B sin A1 = b a1 sin 20º sin129.1º = 4 a1 a1 = 4sin129.1º ≈ 9.08 sin 20º A1 = 57.7º or A2 = 122.3º For both values, A + C < 180º . Therefore, there are two triangles. A2 = 180º − B − C2 = 180º − 20º − 149.1º = 10.9º sin B sin A2 = b a2 B1 = 180º − A1 − C = 180º − 57.7º − 25º = 97.3º sin 20º sin10.9º = 4 a2 sin B1 sin C = b1 c 4 sin10.9º ≈ 2.21 sin 20º sin 97.3º sin 25º = b1 1 Two triangles: A1 ≈ 129.1º , C1 ≈ 30.9º , a1 ≈ 9.08 or b1 = a2 = A2 ≈ 10.9º , C2 ≈ 149.1º , a2 ≈ 2.21 1sin 97.3º ≈ 2.35 sin 25º B2 = 180º − A2 − γ = 180º − 122.3º − 25º = 32.7º 32. a = 3, b = 7, A = 70º sin B2 sin C = b2 c sin B sin 70º = 7 3 7 sin 70º sin B = ≈ 2.1926 3 There is no angle B for which sin B > 1 . Thus, there is no triangle with the given measurements. sin 32.7º sin 25º = b2 1 b2 = 1sin 32.7º ≈ 1.28 sin 25º Two triangles: A1 ≈ 57.7º , B1 ≈ 97.3º , b1 ≈ 2.35 or 33. a = 2, c = 1, C = 100º A2 ≈ 122.3º , B2 ≈ 32.7º , b2 ≈ 1.28 sin C sin A = c a sin100º sin A = 1 2 sin A = 2 sin100º ≈ 1.9696 There is no angle A for which sin A > 1 . Thus, there is no triangle with the given measurements. 808 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 36. b = 4, c = 5, B = 40º sin 60º sin 65º = b 150 150 sin 60º b= ≈ 143.33 miles sin 65º Station Able is about 143.33 miles from the ship, and Station Baker is about 135.58 miles from the ship. sin B sin C = b c sin 40º sin C = 4 5 5sin 40º sin C = ≈ 0.8035 4 C = sin −1 ( 0.8035 ) b. C1 = 53.5º or C2 = 126.5º For both values, B + C < 180º . Therefore, there are two triangles. A1 = 180º − B − C1 = 180º − 40º − 53.5º = 86.5º a 135.6 = ≈ 0.68 hours 200 r min 0.68 hr ⋅ 60 ≈ 41 minutes hr t= 38. Consider the figure below. C sin B sin A1 = b a1 sin 40º sin 86.5º = 4 a1 b 4 sin 86.5º a1 = ≈ 6.21 sin 40º a A2 = 180º − B − C2 = 180º − 40º − 126.5º = 13.5º c B A We are given that A = 49.8974° , B = 180° − 49.9312° = 130.0688° , and c = 300 km . Using angles A and B, we can find C = 180° − 130.0688° − 49.8974° sin B sin A2 = b a2 sin 40º sin13.5º = 4 a2 a2 = 4 sin13.5º ≈ 1.45 sin 40º = 0.0338° Using the Law of Sines, we can determine b and a. sin B sin C = b c c ⋅ sin B b= sin C 300 ⋅ sin130.0688° = sin 0.0338° ≈ 389,173.319 sin A sin C = a c c ⋅ sin A a= sin C 300 ⋅ sin 49.8974° = sin 0.0338° ≈ 388,980.139 At the time of the measurements, the moon was about 389,000 km from Earth. Two triangles: A1 ≈ 86.5º , C1 ≈ 53.5º , a1 ≈ 6.21 or A2 ≈ 13.5º , C2 ≈ 126.5º , a2 ≈ 1.45 37. a. Find C ; then use the Law of Sines (see figure): B 60º a C 150 mi 55º b A C = 180º − 60º − 55º = 65º sin 55º sin 65º = a 150 150 sin 55º a= ≈ 135.58 miles sin 65º 809 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 39. ∠QPR = 180º − 25º = 155º h h = x 1126.57 h = (1126.57 ) sin 69.2º ≈ 1053.15 feet The bridge is about 1053.15 feet high. sin 69.2º = ∠PQR = 180º − 155º − 15º = 10º Let c represent the distance from P to Q. sin15º sin10º = c 1000 1000 sin15º c= ≈ 1490.48 feet sin10º 43. Let h = the height of the tree, and let x = the distance from the first position to the center of the tree (see figure). 40. From Problem 39, we have that the distance from P to Q is 1490.48 feet. Let h represent the distance from Q to D. h sin 25º = 1490.48 h = 1490.48sin 25º ≈ 629.90 feet 10º 60º h 20º 30º 40 ft Using the Law of Sines twice yields two equations relating x and h. 41. Let h = the height of the plane and x = the distance from Q to the plane (see figure). Equation 1: R h 40º P x Equation 2: 35º 1000 ft Q ∠PRQ = 180º − 40º − 35º = 105º h h = x 665.46 h = ( 665.46 ) sin 35º ≈ 381 .69 feet The plane is about 381.69 feet high. 42. Let h = the height of the bridge, x = the distance from C to point A (see figure). 880 ft x sin 20º sin 70º = h x + 40 h sin 70º x= − 40 sin 20º h sin 60º h sin 70º = − 40 sin 30º sin 20º sin 60º sin 70º h − = − 40 sin 30º sin 20º −40 h= sin 60º sin 70º − sin 30º sin 20º ≈ 39.4 feet sin 35º = 69.2º sin 30º sin 60º = h x h sin 60º x= sin 30º Set the two equations equal to each other and solve for h. sin 40º sin105º = x 1000 1000sin 40º x= ≈ 665.46 feet sin105º A x The height of the tree is about 39.4 feet. B 44. Let x = the length of the new ramp (see figure). 65.5º x h 10 ft 12º 168º h 18º Using the Law of Sines: sin162º sin12º = x 10 10sin162º x= ≈ 14.86 feet sin12º The new ramp is about 14.86 feet long. C ∠ACB = 180º − 69.2º − 65.5º = 45.3º sin 65.5º sin 45.3º = x 880 880sin 65.5º x= ≈ 1126.57 feet sin 45.3º 810 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 45. Note that ∠KOS = 57.7° − 29.6° = 28.1° (See figure) 79.4o K sin10º sin162.875º = 70 PQ 70sin162.875º PQ = ≈ 118.7 sin10º S 29.6o es mil 1 . 461 57.7 118.67 = 0.478 hour 250 The trip should have taken 0.4748 hour but, because of the incorrect course, took 0.48 hour. Thus, the trip took 0.0052 hour, or about 18.7 seconds, longer. t= o 28.1o O 47. Let h = the perpendicular distance from R to PQ, (see figure). R From the diagram we find that ∠KSO = 180° − 79.4° − 57.7° = 42.9° and ∠OKS = 180° − 28.1° − 42.9° = 109.0° . We can use the Law of Sines to find the distance between Oklahoma City and Kansas City, as well as the distance between Kansas City and St. Louis. sin 42.9° sin109.0° = OK 461.1 461.1sin 42.9° ≈ 332.0 OK = sin109.0° 184.5 ft 60º P 123 ft Q sin R sin 60º = 123 184.5 123sin 60º ≈ 0.5774 sin R = 184.5 R = sin −1 ( 0.5774 ) ≈ 35.3º sin 28.1° sin109.0° = KS 461.1 461.1sin 28.1° ≈ 229.7 KS = sin109.0° ∠RPQ = 180º − 60º − 35.3º ≈ 84.7º h 184.5 h = 184.5sin 84.7º ≈ 183.71 feet sin 84.7º = Therefore, the total distance using the connecting flight is 332.0 + 229.7 = 561.7 miles. Using the connecting flight, Adam would receive 561.7 − 461.1 = 100.6 more frequent flyer miles. 48. Let θ = ∠AOP sin θ sin15º = 9 3 9sin15º ≈ 0.7765 sin θ = 3 θ = sin −1 ( 0.7765 ) ≈ 50.94º 46. The time of the actual trip was: 50 + 70 120 = = 0.48 hour t= 250 250 Q or θ ≈ 180º − 50.94º = 129.06º A = 114.06º or A = 35.94º sin114.06º sin15º = 3 a 3sin114.06º a= ≈ 10.58 inches sin15º sin 35.94º sin15º or = 3 a 3sin 35.94º a= ≈ 6.80 inches sin15º The approximate distance from the piston to the center of the crankshaft is either 6.80 inches or 10.58 inches. 70 mi 10º P h 50 mi R RQ = 70, PR = 50, P = 10º Solve the triangle: sin10º sin Q = 70 50 50sin10º sin Q = ≈ 0.1240 70 Q = sin −1 ( 0.1240 ) ≈ 7.125º R = 180º − 10º − 7.125º = 162.875º 811 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 49. A = 180º − 140º = 40º ; B = 180º − 135º = 45º ; C = 180º − 40º − 45º = 95º b. Use the Law of Sines: sin 50º sin 75º = 3 QR 3sin 75º ≈ 3.78 miles QR = sin 50º The ship is about 3.78 miles from lighthouse Q. c. Use the Law of Sines: sin 90º sin 75º = 3.2 d 3.2sin 75º ≈ 3.10 miles d= sin 90º The ship is about 3.10 miles from the shore. A 140º D 0.125 mi C 2 mi 0.125 mi E B 135º sin 40º sin 95º = 2 BC 2sin 40º BC = ≈ 1.290 mi sin 95º sin 45º sin 95º = 2 AC 2sin 45º AC = ≈ 1.420 mi sin 95º BE = 1.290 − 0.125 = 1.165 mi AD = 1.420 − 0.125 = 1.295 mi For the isosceles triangle, 180º − 95º ∠CDE = ∠CED = = 42.5º 2 sin 95º sin 42.5º = DE 0.125 0.125sin 95º ≈ 0.184 miles DE = sin 42.5º The approximate length of the highway is AD + DE + BE = 1.295 + 0.184 + 1.165 ≈ 2.64 mi. 51. Determine other angles in the figure: 50º 40º 65º 88 in 65º 65º Using the Law of Sines: sin(65 + 40)° sin 25° = L 88 88sin 25° L= ≈ 38.5 inches sin105° The awning is about 38.5 inches long. 52. The tower forms an angle of 95˚ with the ground. Let x be the distance from the ranger to the tower. B 50. Let PR = the distance from lighthouse P to the ship, QR = the distance from lighthouse Q to the ship, and d = the distance from the ship to the shore. From the diagram, ∠QPR = 75º and ∠PQR = 55º , where point R is the ship. 45º 100 ft A a. x 40º 5º C ∠ABC = 180º − 95º − 40º = 45º P 3 mi 95º 15º d 35º sin 40º sin 45º = x 100 100sin 45º ≈ 110.01 feet x= sin 40º The ranger is about 110.01 feet from the tower. R Q Use the Law of Sines: sin 50º sin 55º = PR 3 3sin 55º ≈ 3.21 miles PR = sin 50º The ship is about 3.21 miles from lighthouse P. 812 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 53. Let h = height of the pyramid, and let x = distance from the edge of the pyramid to the point beneath the tip of the pyramid (see figure). h 40.3º 46.27º 100 ft 100 ft x Using the Law of Sines twice yields two equations relating x and y: sin 46.27º sin(90º − 46.27º ) Equation 1: = h x + 100 ( x + 100)sin 46.27º = h sin 43.73º x sin 46.27º +100sin 46.27º = h sin 43.73º h sin 43.73º −100sin 46.27º sin 46.27º sin 40.3º sin(90º − 40.3º ) = h x + 200 ( x + 200 ) sin 40.3º = h sin 49.7º x= Equation 2: x sin 40.3º +200sin 40.3º = h sin 49.7º h sin 49.7º − 200sin 40.3º x= sin 40.3º Set the two equations equal to each other and solve for h. h sin 43.73º −100sin 46.27º h sin 49.7º − 200sin 40.3º = sin 46.27º sin 40.3º h sin 43.73º ⋅ sin 40.3º − 100sin 46.27º ⋅ sin 40.3º = h sin 49.7º ⋅ sin 46.27º − 200sin 40.3º ⋅ sin 46.27º h sin 43.73º ⋅ sin 40.3º −h sin 49.7º ⋅ sin 46.27º = 100sin 46.27º ⋅ sin 40.3º − 200sin 40.3º ⋅ sin 46.27º 100sin 46.27º ⋅ sin 40.3º − 200sin 40.3º ⋅ sin 46.27º h= sin 43.73º ⋅ sin 40.3º − sin 49.7º ⋅ sin 46.27º ≈ 449.36 feet The current height of the pyramid is about 449.36 feet. _________________________________________________________________________________________________ 54. Let h = the height of the aircraft, and let x = the distance from the first sensor to a point on the ground beneath the airplane (see figure). 5º 70º 15º 700 ft Equation 2: Set the equations equal to each other and solve for h. h h sin 70º h sin 75º = − 700 sin 20º sin15º sin 70º sin 75º − h = − 700 sin 20º sin15º − 700 h= sin 70º sin 75º − sin 20º sin15º ≈ 710.97 feet 20º x Using the Law of Sines twice yields two equations relating x and h. Equation 1: sin15º sin 75º = h x + 700 h sin 75º − 700 x= sin15º sin 20º sin 70º = h x h sin 70º x= sin 20º The height of the aircraft is about 710.97 feet. 813 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 56. Using the Law of Sines: 55. Using the Law of Sines: V enus βB Mercury βB x 57,910,000 γC x γ C Sun 15º Earth 149,600,000 Earth sin15º sin B = 57,910, 000 149, 600, 000 149, 600, 000 ⋅ sin15º sin B = 57,910,000 14,960 ⋅ sin15º = 5791 14,960 ⋅ sin15º o B = sin −1 ≈ 41.96 5791 or 10 º 108,200,000 Sun 149,600,000 sin10º sin B = 108, 200, 000 149, 600, 000 149, 600, 000 ⋅ sin10º sin B = 108, 200, 000 1496 ⋅ sin10º = 1082 1496 ⋅ sin10º o B = sin −1 ≈ 13.892 1082 or B ≈ 138.04o B ≈ 166.108o C ≈ 180° − 13.892° − 10° = 156.108° or C ≈ 180° − 166.108° − 10° = 13.892° C ≈ 180o − 41.96o − 15º = 123.04o or C ≈ 180o − 138.04o − 15º = 26.96o sin10º sin C = 108, 200, 000 x 108, 200, 000 ⋅ sin C x= sin10º 108, 200, 000 ⋅ sin156.108° x= sin10º ≈ 252,363, 760.4 km or 108, 200, 000 ⋅ sin 3.892° x= sin10º ≈ 42, 293, 457.3 km So the approximate possible distances between Earth and Venus are 42,300,000 km and 252,400,000 km. sin15º sin C = 57,910, 000 x 57,910, 000 ⋅ sin C x= sin15° 57,910, 000 ⋅ sin123.04o = sin15º ≈ 187,564,951.5 km or 57,910, 000 ⋅ sin 26.96o x= sin15º ≈ 101, 439,834.5 km So the possible distances between Earth and Mercury are approximately 101,440,000 km and 187,600,000 km. 57. Since there are 36 equally spaced cars, each car 360° is separated by = 10° . The angle between 36 the radius and a line segment connecting 170° consecutive cars is = 85° (see figure). If 2 we let r = the radius of the wheel, we get sin10° sin 85° = 22 r 22sin 85° ≈ 126 r= sin10° The length of the diameter of the wheel is approximately d = 2r = 2 (126 ) = 252 feet. 814 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.2: The Law of Sines 58. 59. a + b a b sin A sin B + = + = c c c sin C sin C sin A + sin B = sin C A+ B A− B 2sin cos 2 2 = C C 2sin cos 2 2 π C A− B sin − cos 2 2 2 = C sin ( C ) cos 2 C A− B cos cos 2 2 = C C sin cos 2 2 A− B 1 cos cos 2 ( A − B ) 2 = = C 1 sin sin C 2 2 60. a = b sin A b sin [180º − ( B + γ ) ] = sin B sin B b = sin( B + C ) sin B b = ( sin B cos C + cos B sin C ) sin B b sin C = b cos C + cos B sin B = b cos C + c cos B a−b a−b = c 61. a+b a+b c 1 sin ( A − B) 2 1 cos C 2 = 1 cos ( A − B) 2 1 sin C 2 1 1 sin ( A − B ) sin C 2 ⋅ 2 = 1 1 cos C cos ( A − B) 2 2 1 1 = tan ( A − B) tan C 2 2 1 1 = tan ( A − B) tan ( π − ( A + B) ) 2 2 a − b a b sin A sin B sin A − sin B − = = − = sin C c c c sin C sin C A− B A+ B 2sin cos 2 2 = C sin 2 ⋅ 2 A B − cos A + B 2sin 2 2 = C C 2sin cos 2 2 A− B π C sin cos − 2 2 2 = C C sin cos 2 2 A− B C sin sin 2 2 = C C sin cos 2 2 A− B 1 sin sin 2 ( A − B ) 2 = = 1 C cos cos C 2 2 π A + B 1 = tan ( A − B) tan − 2 2 2 1 A+ B = tan ( A − B) cot 2 2 1 tan ( A − B) 2 = 1 tan ( A + B ) 2 62. sin B = sin ( ∠PQR ) = sin ( ∠PP ' R ) = b 2r sin B 1 = b 2r The result follows from the Law of Sines. 63 – 65. Answers will vary. 815 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions c 2 = a 2 + b 2 − 2ab cos C Section 8.3 ( x2 − x1 ) 1. d = 2 2 θ = 45° or + ( y2 − y1 ) 2 a 2 + b2 − c 2 2ab 2.052 + 32 − 42 −2.7975 = cos C = 2(2.05)(3) 12.3 cos C = 2 2. cos θ = −2.7975 C = cos −1 ≈ 103.1º 12.3 π 4 The solution set is {45°} or {π4 } B = 180º − A − C ≈ 180º −30º −103.1º = 46.9º 3. Cosines 11. a = 2, b = 3, C = 95º 4. Sines c 2 = a 2 + b 2 − 2ab cos C 5. Cosines c 2 = 22 + 32 − 2 ⋅ 2 ⋅ 3cos 95º = 13 − 12 cos 95º c = 13 − 12 cos 95º ≈ 3.75 6. False: Use the Law of Cosines 7. False a 2 = b 2 + c 2 − 2bc cos A 8. True b2 + c2 − a 2 2bc 32 + 3.752 − 22 19.0625 = cos A = 2(3)(3.75) 22.5 cos A = 9. a = 2, c = 4, A = 45º b 2 = a 2 + c 2 − 2ac cos B b 2 = 22 + 42 − 2 ⋅ 2 ⋅ 4 cos 45º 19.0625 A = cos −1 ≈ 32.1º 22.5 2 = 20 − 16 ⋅ 2 = 20 − 8 2 B = 180º − A − C ≈ 180º −32.1º −95º = 52.9º 12. a = 2, c = 5, B = 20º b = 20 − 8 2 ≈ 2.95 b 2 = a 2 + c 2 − 2ac cos B a 2 = b 2 + c 2 − 2bc cos A b 2 = 22 + 52 − 2 ⋅ 2 ⋅ 5cos 20º = 29 − 20cos 20º 2bc cos A = b 2 + c 2 − a 2 b = 29 − 20 cos 20º ≈ 3.19 b2 + c2 − a2 2bc 2.952 + 42 − 22 20.7025 = cos A = 2(2.95)(4) 23.6 cos A = a 2 = b 2 + c 2 − 2bc cos A cos A = 20.7025 A = cos −1 ≈ 28.7º 23.6 b2 + c2 − a 2 2bc ( cos A = C = 180º − A − B ≈ 180º −28.7º −45º ≈ 106.3º 29 − 20 cos 20º ) 2 + 52 − 2 2 2( 29 − 20 cos 20º )(5) ≈ 0.97681 A = cos −1 ( 0.97681) ≈ 12.4º 10. b = 3, c = 4, A = 30º C = 180º − A − B ≈ 180º −12.4º −20º = 147.6º a 2 = b 2 + c 2 − 2bc cos A a 2 = 32 + 42 − 2 ⋅ 3 ⋅ 4 cos 30º 3 = 25 − 24 2 = 25 − 12 3 a = 25 − 12 3 ≈ 2.05 816 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.3: The Law of Cosines 13. a = 6, b = 5, c = 8 16. a = 4, b = 3, c = 4 a 2 = b 2 + c 2 − 2bc cos A cos A = a 2 = b 2 + c 2 − 2bc cos A b 2 + c 2 − a 2 52 + 82 − 62 53 = = 2bc 2(5)(8) 80 cos A = 53 A = cos −1 ≈ 48.5º 80 9 A = cos −1 ≈ 68.0º 24 b 2 = a 2 + c 2 − 2ac cos B b 2 = a 2 + c 2 − 2ac cos B a 2 + c2 − b2 2ac 62 + 82 − 52 75 cos B = = 2(6)(8) 96 cos B = cos B = C = 180º − A − B ≈ 180º −68.0º −44.0º = 68.0º 17. a = 3, b = 4, C = 40º C = 180º − A − B ≈ 180º −48.5º −38.6º = 92.9º c 2 = a 2 + b 2 − 2ab cos C 14. a = 8, b = 5, c = 4 c 2 = 32 + 42 − 2 ⋅ 3 ⋅ 4 cos 40º = 25 − 24 cos 40º a = b + c − 2bc cos A 2 2 c = 25 − 24 cos 40º ≈ 2.57 23 b 2 + c 2 − a 2 52 + 4 2 − 82 cos A = = =− 2bc 2(5)(4) 80 a 2 = b 2 + c 2 − 2bc cos A 23 A = cos −1 − ≈ 125.1º 80 cos A = a 2 + c 2 − b2 82 + 42 − 52 55 = = 2ac 2(8)(4) 64 B = 180º − A − C ≈ 180º −48.6º −40º = 91.4º 55 B = cos−1 ≈ 30.8º 64 18. a = 2, c = 1, B = 10º b 2 = a 2 + c 2 − 2ac cos B C = 180º − A − B ≈ 180º −125.1º −30.8º = 24.1º b 2 = 22 + 12 − 2 ⋅ 2 ⋅1cos10º = 5 − 4 cos10º b = 5 − 4 cos10º ≈ 1.03 15. a = 9, b = 6, c = 4 a 2 = b 2 + c 2 − 2bc cos A cos A = a 2 = b 2 + c 2 − 2bc cos A 29 b 2 + c 2 − a 2 62 + 4 2 − 9 2 = =− 2bc 2(6)(4) 48 cos A = 29 A = cos −1 − ≈ 127.2º 48 1.9391 b 2 + c 2 − a 2 1.032 + 12 − 22 = =− 2bc 2(1.03)(1) 2.06 1.9391 A = cos −1 − ≈ 160.3º 2.06 C = 180º − A − B ≈ 180º −160.3º −10º = 9.7º b 2 = a 2 + c 2 − 2ac cos B cos B = b 2 + c 2 − a 2 42 + 2.57 2 − 32 13.6049 = = 2bc 2(4)(2.57) 20.56 13.6049 A = cos −1 ≈ 48.6º 20.56 b 2 = a 2 + c 2 − 2ac cos B cos B = a 2 + c 2 − b 2 42 + 42 − 32 23 = = 2ac 2(4)(4) 32 23 B = cos−1 ≈ 44.0º 32 75 B = cos −1 ≈ 38.6º 96 2 9 b 2 + c 2 − a 2 32 + 42 − 42 = = 2bc 2(3)(4) 24 a 2 + c 2 − b 2 92 + 42 − 62 61 = = 2ac 2(9)(4) 72 19. b = 1, c = 3, A = 80º a 2 = b 2 + c 2 − 2bc cos A 61 B = cos ≈ 32.1º 72 −1 a 2 = 12 + 32 − 2 ⋅1 ⋅ 3cos 80º = 10 − 6 cos80º a = 10 − 6 cos 80º ≈ 2.99 C = 180º − A − B ≈ 180º −127.2º −32.1º = 20.7º 817 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 23. a = 2, b = 2, C = 50º b2 = a 2 + c 2 − 2ac cos B cos B = c 2 = a 2 + b 2 − 2ab cos C a 2 + c 2 − b 2 2.992 + 32 − 12 16.9401 = = 2ac 2(2.99)(3) 17.94 c 2 = 22 + 22 − 2 ⋅ 2 ⋅ 2 cos 50º = 8 − 8 cos 50º 16.9401 B = cos −1 ≈ 19.2º 17.94 c = 8 − 8 cos 50º ≈ 1.69 a 2 = b 2 + c 2 − 2bc cos A C = 180º − A − B ≈ 180º −80º −19.2º = 80.8º cos A = 20. a = 6, b = 4, C = 60º 2.8561 A = cos−1 ≈ 65.0º 6.76 c 2 = a 2 + b 2 − 2ab cos C c 2 = 62 + 42 − 2 ⋅ 6 ⋅ 4 cos 60º = 28 B = 180º − A − C ≈ 180º −65.0º −50º = 65.0º c = 28 ≈ 5.29 24. a = 3, c = 2, B = 90º a 2 = b 2 + c 2 − 2bc cos A cos A = b 2 = a 2 + c 2 − 2ac cos B 4 + 5.29 − 6 7.9841 b +c −a = = 2bc 2(4)(5.29) 42.32 2 2 2 2 2 2 b 2 = 32 + 22 − 2 ⋅ 3 ⋅ 2 cos 90º = 13 7.9841 A = cos −1 ≈ 79.1º 42.32 b = 13 ≈ 3.61 a 2 = b 2 + c 2 − 2bc cos A B = 180º − A − C ≈ 180º −79.1º −60º = 40.9º cos A = 21. a = 3, c = 2, B = 110º b = 3 + 2 − 2 ⋅ 3 ⋅ 2 cos110º = 13 − 12 cos110º 2 2 C = 180º − A − B ≈ 10º −56.3º −90º = 33.7º b = 13 − 12 cos110º ≈ 4.14 25. a = 12, b = 13, c = 5 c 2 = a 2 + b 2 − 2ab cos C cos C = a 2 = b 2 + c 2 − 2bc cos A 3 + 4.14 − 2 22.1396 a +b −c = = 2ab 2(3)(4.14) 24.84 2 2 2 2 b 2 + c 2 − a 2 ( 13)2 + 22 − 32 = ≈ 0.55470 2bc 2( 13)(2) A = cos −1 ( 0.55470 ) ≈ 56.3º b 2 = a 2 + c 2 − 2ac cos B 2 b 2 + c 2 − a 2 22 + 1.692 − 22 2.8561 = = 2bc 2(2)(1.69) 6.76 2 2 cos A = 22.1396 C = cos −1 ≈ 27.0º 24.84 b 2 + c 2 − a 2 132 + 52 − 122 50 = = 2bc 2(13)(5) 130 50 A = cos−1 ≈ 67.4º 130 A = 180º − B − C ≈ 180º −110º −27.0º = 43.0º b 2 = a 2 + c 2 − 2ac cos B 22. b = 4, c = 1, A = 120º cos B = a 2 = b 2 + c 2 − 2bc cos A a 2 + c 2 − b 2 122 + 52 − 132 = =0 2ac 2(12)(5) B = cos −1 0 = 90º a 2 = 42 + 12 − 2 ⋅ 4 ⋅1cos120º = 21 C = 180º − A − B ≈ 180º −67.4º −90º = 22.6º a = 21 ≈ 4.58 c 2 = a 2 + b 2 − 2ab cos C cos C = a 2 + b 2 − c 2 4.582 + 42 − 12 35.9764 = = 2ab 2(4.58)(4) 36.64 35.9764 C = cos −1 ≈ 10.9º 36.64 B = 180º − A − C ≈ 180º −120º −10.9º = 49.1º 818 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.3: The Law of Cosines 26. a = 4, b = 5, c = 3 29. a = 5, b = 8, c = 9 a 2 = b 2 + c 2 − 2bc cos A cos A = a 2 = b 2 + c 2 − 2bc cos A b 2 + c 2 − a 2 52 + 32 − 42 = = 0.6 2bc 2(5)(3) cos A = A = cos −1 0.6 ≈ 53.1º 120 A = cos −1 ≈ 33.6º 144 b 2 = a 2 + c 2 − 2ac cos B cos B = b 2 = a 2 + c 2 − 2ac cos B a 2 + c 2 − b2 42 + 32 − 52 = =0 2ac 2(4)(3) cos B = B = cos −1 0 = 90º 27. a = 2, b = 2, c = 2 C = 180o − A − B ≈ 180o − 33.6o − 62.2o = 84.2o a 2 = b 2 + c 2 − 2bc cos A b 2 + c 2 − a 2 2 2 + 22 − 2 2 = = 0.5 2bc 2(2)(2) 30. a = 4, b = 3, c = 6 a 2 = b 2 + c 2 − 2bc cos A −1 A = cos 0.5 = 60º cos A = b 2 = a 2 + c 2 − 2ac cos B cos B = b 2 = a 2 + c 2 − 2ac cos B C = 180º − A − B ≈ 180º −60º −60º = 60º cos B = 28. a = 3, b = 3, c = 2 b 2 + c 2 − a 2 32 + 22 − 32 1 = = 2bc 2(3)(2) 3 C = 180o − A − B ≈ 180o − 36.3o − 26.4o = 117.3o 1 A = cos −1 ≈ 70.5º 3 31. a = 10, b = 8, c = 5 a 2 = b 2 + c 2 − 2bc cos A b = a + c − 2ac cos B cos B = 2 2 cos A = 3 + 2 −3 1 a +c −b = = 2ac 2(3)(2) 3 2 2 2 2 a 2 + c 2 − b 2 42 + 62 − 32 43 = = 2ac 48 2 ( 4 )( 6 ) 43 B = cos −1 ≈ 26.4º 48 a 2 = b 2 + c 2 − 2bc cos A 2 b 2 + c 2 − a 2 32 + 62 − 42 29 = = 2bc 2(3)(6) 36 29 A = cos−1 ≈ 36.3º 36 a 2 + c 2 − b 2 22 + 22 − 2 2 = = 0.5 2ac 2(2)(2) B = cos −1 0.5 = 60º cos A = a 2 + c 2 − b2 52 + 92 − 82 42 = = 2ac 90 2 ( 5 )( 9 ) 42 B = cos −1 ≈ 62.2o 90 C = 180º − A − B ≈ 180º −53.1º −90º = 36.9º cos A = b 2 + c 2 − a 2 82 + 92 − 52 120 = = 2bc 2(8)(9) 144 2 2 11 b 2 + c 2 − a 2 82 + 52 − 102 = =− 2bc 2(8)(5) 80 11 A = cos −1 − ≈ 97.9º 80 1 B = cos ≈ 70.5º 3 −1 b 2 = a 2 + c 2 − 2ac cos B C = 180º − A − B ≈ 180º −70.5º −70.5º = 39.0º cos B = 61 a 2 + c 2 − b 2 102 + 52 − 82 = = 2ac 2(10)(5) 100 61 B = cos −1 ≈ 52.4º 100 C = 180o − A − B ≈ 180o − 97.9o − 52.4o = 29.7 o 819 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 32. a = 9, b = 7, c = 10 b 2 = a 2 + c 2 − 2ac cos B a 2 = b 2 + c 2 − 2bc cos A cos A = a 2 + c 2 − b2 2ac 2 6 + 9 2 − 82 53 = = 2 ( 6 )( 9 ) 108 cos B = 68 b 2 + c 2 − a 2 72 + 102 − 92 = = 2bc 2(7)(10) 140 68 A = cos−1 ≈ 60.9º 140 B = cos −1 b = a + c − 2ac cos B 2 2 cos B = 2 c 2 = a 2 + b 2 − 2ab cos C a 2 + c 2 − b2 92 + 102 − 72 132 = = 2ac 2(9)(10) 180 a 2 + b2 − c2 2ab 2 6 + 82 − 92 19 = = 2 ( 6 )( 8) 96 cos C = 132 B = cos −1 ≈ 42.8º 180 C = 180o − A − B ≈ 180o − 60.9o − 42.8o = 76.3o C = cos −1 33. B = 20° , C = 75° , b = 5 sin B sin C = b c b sin C 5sin 75° = ≈ 14.12 c= sin B sin 20° 19 ≈ 78.6° 96 36. a = 14 , b = 7 , A = 85° sin 85° sin B = 14 7 sin 85° sin B = ≈ 0.49810 2 B = sin −1 ( 0.49810 ) ≈ 29.9° or 150.1° The second value is discarded since A + B > 180° . Therefore, B ≈ 29.9° . C = 180° − 29.9° − 85° = 65.1° sin 85° sin 65.1° = 14 c 14 ⋅ sin 65.1° c= ≈ 12.75 sin 85° A = 180° − 20° − 75° = 85° sin A sin B = a b b sin A 5sin 85° = ≈ 14.56 a= sin B sin 20° 34. A = 50° , B = 55° , c = 9 C = 180° − 50° − 55° = 75° sin C sin B = c b c sin B 9sin 55° = ≈ 7.63 b= sin C sin 75° 37. B = 35° , C = 65° , a = 15 A = 180° − 35° − 65° = 80° sin C sin A = c a c sin A 9 sin 50° = ≈ 7.14 a= sin C sin 75° sin A sin C = a c a sin C 15sin 65° = ≈ 13.80 c= sin A sin 35° 35. a = 6 , b = 8 , c = 9 sin A sin B = a b a sin B 15sin 35° = ≈ 8.74 b= sin A sin 80° a 2 = b 2 + c 2 − 2bc cos A b2 + c2 − a 2 2bc 2 8 + 92 − 62 109 = = 2 ( 8)( 9 ) 144 cos A = A = cos −1 53 ≈ 60.6° 108 109 ≈ 40.8° 144 820 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.3: The Law of Cosines 38. a = 4 , c = 5 , B = 55° 41. b = 5 , c = 12 , A = 60° b 2 = a 2 + c 2 − 2ac cos B a 2 = b 2 + c 2 − 2bc cos A b 2 = 42 + 52 − 2 ( 4 )( 5) cos 55° a 2 = 52 + 122 − 2 ⋅ 5 ⋅12 ⋅ cos 60° = 109 b 2 = 41 − 40 cos 55° a = 109 ≈ 10.44 b = 41 − 40 cos 55° ≈ 4.25 sin 60° sin B 5 109 5sin 60° sin B = ≈ 0.414751 109 sin 55° sin A = 4 41 − 40 cos 55° 4 sin 55° sin A = ≈ 0.77109 41 − 40 cos 55° = B = sin −1 ( 0.414751) ≈ 24.5° or 155.5° A = sin −1 ( 0.77109 ) ≈ 50.5° or 129.5° We discard the second value since A + B > 180° . Therefore, A ≈ 50.5° . We discard the second value because it would give A + B > 180° . Therefore, B ≈ 24.5° . C = 180° − 60° − 24.5° = 95.5° C = 180° − 55° − 50.5° = 74.5° 42. a = 10 , b = 10 , c = 15 39. a = 3 , b = 10 , A = 10° sin10° sin B = 3 10 10 sin10° sin B = ≈ 0.578827 3 B = sin −1 ( 0.578827 ) ≈ 35.4° or 144.6° a 2 = b 2 + c 2 − 2bc cos A b2 + c 2 − a 2 2bc 102 + 152 − 102 225 3 = = = 2 (10 )(15) 300 4 cos A = Since both values yield A + B < 180° , there are two triangles. B1 = 35.4° and B2 = 144.6° A = cos −1 3 ≈ 41.4° 4 b 2 = a 2 + c 2 − 2ac cos B C1 = 180° − 10° − 35.4° = 134.6° a2 + c2 − b2 2ac 102 + 152 − 102 225 3 = = = 2 (10 )(15) 300 4 cos B = C2 = 180° − 10° − 144.6° = 25.4° Using the Law of Cosines we get c1 = 32 + 102 − 2 ⋅ 3 ⋅10 ⋅ cos134.6° ≈ 12.29 B = cos −1 c2 = 32 + 102 − 2 ⋅ 3 ⋅10 ⋅ cos 25.4° ≈ 7.40 3 ≈ 41.4° 4 c 2 = a 2 + b 2 − 2ab cos C a 2 + b2 − c 2 2ab 102 + 102 − 152 −25 1 = = =− 2 (10 )(10 ) 200 8 40. A = 65° , B = 72° , b = 7 sin 72° sin 65° = 7 a 7 sin 65° a= ≈ 6.67 sin 72° cos C = 1 C = cos −1 − ≈ 97.2° 8 C = 180° − 65° − 72° = 43° sin 43° sin 72° = c 7 7 sin 43° c= ≈ 5.02 sin 72° 821 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 124,148.39 A = cos −1 − ≈ 153.6º 138,570 The captain needs to turn the ship through an angle of 180° − 153.6° = 26.4° . 43. Find the third side of the triangle using the Law of Cosines: a = 150, b = 35, C = 110º c 2 = a 2 + b 2 − 2ab cos C = 1502 + 352 − 2 ⋅150 ⋅ 35 cos110º = 23, 725 − 10,500 cos110º b. c = 23, 725 − 10,500 cos110º ≈ 165 The ball is approximately 165 yards from the center of the green. 44. a. The angle inside the triangle at Sarasota is 180° − 50° = 130° . Use the Law of Cosines to find the third side: a = 150, b = 100, C = 130º 46. a. c 2 = a 2 + b 2 − 2ab cos C After 15 minutes, the plane would have flown 220(0.25) = 55 miles. Find the third side of the triangle: a = 55, b = 330, γ = 10º c 2 = a 2 + b 2 − 2ab cos C = 1502 + 1002 − 2 ⋅150 ⋅100 cos130º = 32,500 − 30, 000 cos130º = 552 + 3302 − 2 ⋅ 55 ⋅ 330cos10º = 111,925 − 36,300 cos10º c = 32,500 − 30, 000cos130º ≈ 227.56 mi b. 461.9 nautical miles ≈ 30.8 hours are 15 knots required for the second leg of the trip. (The total time for the trip will be about 40.8 hours.) t= c = 111,925 − 36,300cos10º ≈ 276 Use the Law of Sines to find the angle inside the triangle at Ft. Myers: sin A sin130º = 100 227.56 100sin130º sin A = 227.56 100sin130º A = sin −1 ≈ 19.7º 227.56 Since the angle of the triangle is 19.7˚, the pilot should fly at a bearing of N 19.7° E . Find the measure of the angle opposite the 330-mile side: a 2 + c2 − b2 cos B = 2ac 552 + 2762 − 3302 29, 699 = =− 2(55)(276) 30,360 29, 699 B = cos −1 − ≈ 168.0º 30,360 The pilot should turn through an angle of 180° − 168.0° = 12.0° . b. 45. After 10 hours the ship will have traveled 150 nautical miles along its altered course. Use the Law of Cosines to find the distance from Barbados on the new course. a = 600, b = 150, C = 20º c 2 = a 2 + b 2 − 2ab cos C = 6002 + 1502 − 2 ⋅ 600 ⋅150 cos 20º = 382,500 − 180, 000 cos 20º If the total trip is to be done in 90 minutes, and 15 minutes were used already, then there are 75 minutes or 1.25 hours to complete the trip. The plane must travel 276 miles in 1.25 hours: 276 r= = 220.8 miles/hour 1.25 The pilot must maintain a speed of 220.8 mi/hr to complete the trip in 90 minutes. c = 382,500 − 180, 000cos 20º ≈ 461.9 nautical miles a. Use the Law of Cosines to find the angle opposite the side of 600: b2 + c2 − a 2 cos A = 2bc 2 150 + 461.92 − 6002 124,148.39 =− cos A = 2(150)(461.9) 138,570 822 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.3: The Law of Cosines 47. a. b. Find x in the figure: 3rd 2nd 60 .5 ft y Home 7200 = (46 + y )2 90 ft β x 45° 46 + y = 7200 ≈ 84.85 y ≈ 38.85 feet It is about 38.85 feet from the pitching rubber to second base. 1st 90 ft Use the Pythagorean Theorem to find y in the figure: 602 + 602 = (46 + y )2 x 2 = 60.52 + 902 − 2(60.5)(90) cos 45º 2 = 11, 760.25 − 10,980 2 c. = 11, 760.25 − 5445 2 x = 11, 760.25 − 5445 2 ≈ 63.7 feet It is about 63.7 feet from the pitching rubber to first base. b. Use the Pythagorean Theorem to find y in the figure: 902 + 902 = (60.5 + y ) 2 Find B in the figure by using the Law of Cosines: 462 + 42.582 − 602 329.0564 cos B = = 2(46)(42.58) 3917.36 329.0564 B = cos −1 ≈ 85.2º 3917.36 The pitcher needs to turn through an angle of 85.2˚ to face first base. 49. a. Find x by using the Law of Cosines: 16, 200 = (60.5 + y ) 2 60.5 + y = 16, 200 ≈ 127.3 y ≈ 66.8 feet It is about 66.8 feet from the pitching rubber to second base. c. y Find B in the figure by using the Law of Cosines: 60.52 + 63.72 − 902 382.06 cos B = =− 2(60.5)(63.7) 7707.7 x = 260, 000 − 100, 000cos 80º ≈ 492.6 ft The guy wire needs to be about 492.6 feet long. b. 46 ft y Home 45° 60 ft Use the Pythagorean Theorem to find the value of y: y 2 = 1002 + 2502 = 72,500 y = 269.3 feet The guy wire needs to be about 269.3 feet long. 2nd β 10° x 2 = 5002 + 1002 − 2(500)(100) cos80º = 260, 000 − 100, 000 cos80º Find x in the figure: 3rd 80° 100 ft 382.06 B = cos −1 − ≈ 92.8º 7707.7 The pitcher needs to turn through an angle of about 92.8˚ to face first base. 48. a. 250 ft 500 ft x 60 ft x 1st x 2 = 462 + 602 − 2(46)(60) cos 45º 2 = 5716 − 5520 = 5716 − 2760 2 2 x = 5716 − 2760 2 ≈ 42.58 feet It is about 42.58 feet from the pitching rubber to first base. 823 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 50. Find x by using the Law of Cosines: 53. Use the Law of Cosines: 45o x y 10 10 500 ft 95° 85° 5° c c 2 = a 2 + b2 − 2ab cos C x 2 = 5002 + 1002 − 2(500)(100) cos 85º = 260, 000 − 100, 000 cos85º = 102 + 102 − 2 (10 )(10 ) cos ( 45° ) x = 260, 000 − 100, 000 cos 85º ≈ 501.28 feet The guy wire needs to be about 501.28 feet long. 2 = 100 + 100 − 200 2 Find y by using the Law of Cosines: y 2 = 5002 + 1002 − 2(500)100 cos 95º ( = 200 − 100 2 = 100 2 − 2 ( = 260, 000 − 100, 000 cos 95º ) ) c = 100 2 − 2 = 10 2 − 2 ≈ 7.65 y = 260, 000 − 100, 000 cos 95º ≈ 518.38 feet The guy wire needs to be about 518.38 feet long. The footings should be approximately 7.65 feet apart. 51. Find x by using the Law of Cosines: 54. Use the Law of Cosines: A x L r 400 ft 45° O 90 ft θ B x L2 = x 2 + r 2 − 2 x r cos θ x 2 = 4002 + 902 − 2(400)(90) cos ( 45º ) x 2 − 2 x r cos θ + r 2 − L2 = 0 = 168,100 − 36, 000 2 Using the quadratic formula: x = 168,100 − 36, 000 2 ≈ 342.33 feet It is approximately 342.33 feet from dead center to third base. x= 52. Find x by using the Law of Cosines: x= x x= 280 ft 45° 2r cos θ + (2r cos θ )2 − 4(1)(r 2 − L2 ) 2(1) ( 2r cos θ + 4r 2 cos 2 θ − 4 r 2 − L2 2 ( 2r cos θ + 4 r cos 2 θ − r 2 + L2 2 2 2r cos θ + 2 r cos 2 θ − r 2 + L2 2 2 60 ft x= x 2 = 2802 + 602 − 2(280)(60) cos ( 45º ) x = r cos θ + r 2 cos 2 θ + L2 − r 2 = 82, 000 − 16,800 2 x = 82, 000 − 16,800 2 ≈ 241.33 feet It is approximately 241.33 feet from dead center to third base. 824 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall ) ) Section 8.4: Area of a Triangle 55. Use the Law of Cosines to find the length of side d: d 2 = r 2 + r 2 − 2 ⋅ r ⋅ r ⋅ cos θ = 2r 2 − 2r 2 cos θ = 2r 2 (1 − cos θ ) 1 − cos θ 1 − cos θ = 4r 2 = 2r 2 2 θ = 2r sin 2 If s = rθ is the length of the arc subtended by θ θ , then d < s , and we have 2r sin < rθ or 2 θ θ θ 2sin < θ . Given sin θ = 2sin cos and 2 2 2 θ θ <θ . 2 2 Therefore, sin θ < θ for any angle θ > 0 . cos 56. cos ≤ 1 , we have sin θ ≤ 2sin 58. C 1 + cos C = 2 2 = 1+ a 2 + b2 − c2 2ab 2 = 2ab + a 2 + b 2 − c 2 4ab = ( a + b) 2 − c 2 4ab = = 4s( s − c) = 4ab = s( s − c) ab = 4( s − b)( s − a) 4ab = ( s − a)( s − b) ab cos A cos B cos C + + a b c 2 2 2 2 b +c −a a + c2 − b2 a 2 + b2 − c2 = + + 2bca 2acb 2abc 2 2 2 2 2 2 2 b + c − a + a + c − b + a + b2 − c2 = 2abc a 2 + b2 + c2 = 2abc 1 bh 2 2. K = 1 ab sin C 2 s ( s − a)( s − b)( s − c) ; K= = = −(a 2 − 2ab + b2 − c 2 ) 4ab = (2s − 2b)(2s − 2a) 4ab 1 (a + b + c) 2 5. a = 2, c = 4, B = 45º 2ab − a 2 − b2 + c 2 4ab ( = 4. True a2 + b2 − c2 2ab 2 − ( a − b) 2 − c 2 (a − b + c)(b + c − a) 4ab 1. K = 3. C 1 − cos C 57. sin = 2 2 1− = Section 8.4 4ab 2s (2 s − c − c) 4ab −(a − b + c)(a − b − c) 4ab 59 – 63. Answers will vary. ( a + b + c )( a + b − c ) = = 1 1 ac sin B = (2)(4)sin 45º ≈ 2.83 2 2 6. b = 3, c = 4, A = 30º 1 1 K = bc sin α = (3)(4) sin 30º = 3 2 2 7. a = 2, b = 3, C = 95º ) K= 1 1 ab sin C = (2)(3) sin 95º ≈ 2.99 2 2 4ab 825 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 8. a = 2, c = 5, B = 20º K= 16. a = 6, b = 4, C = 60º 1 1 ac sin B = (2)(5) sin 20º ≈ 1.71 2 2 K= 9. a = 6, b = 5, c = 8 17. a = 3, c = 2, B = 110º 1 1 19 ( a + b + c ) = 2 ( 6 + 5 + 8) = 2 2 K = s ( s − a)( s − b)( s − c) s= K= 1 1 ac sin B = (3)(2)sin110º ≈ 2.82 2 2 18. b = 4, c = 1, A = 120º 3591 19 7 9 3 = = ≈ 14.98 2 2 2 2 16 1 1 K = bc sin A = (4)(1) sin120º ≈ 1.73 2 2 10. a = 8, b = 5, c = 4 19. a = 12, b = 13, c = 5 1 1 17 s = ( a + b + c ) = (8 + 5 + 4 ) = 2 2 2 K = s ( s − a)( s − b)( s − c) 1 1 ( a + b + c ) = 2 (12 + 13 + 5 ) = 15 2 K = s ( s − a)( s − b)( s − c) s= 1071 17 1 7 9 = = ≈ 8.18 16 2 2 2 2 = (15 )( 3)( 2 )(10 ) = 900 = 30 20. a = 4, b = 5, c = 3 11. a = 9, b = 6, c = 4 1 1 ( a + b + c ) = 2 ( 4 + 5 + 3) = 6 2 K = s ( s − a)( s − b)( s − c) s= 1 1 19 s = ( a + b + c ) = (9 + 6 + 4) = 2 2 2 K = s ( s − a)( s − b)( s − c ) = 1463 19 1 7 11 = = ≈ 9.56 16 2 2 2 2 ( 6 )( 2 )(1)( 3) = 36 = 6 21. a = 2, b = 2, c = 2 1 1 ( a + b + c ) = 2 ( 2 + 2 + 2) = 3 2 K = s ( s − a)( s − b)( s − c) s= 12. a = 4, b = 3, c = 4 1 1 11 ( a + b + c ) = 2 ( 4 + 3 + 4) = 2 2 K = s ( s − a)( s − b)( s − c) s= = 495 11 3 5 3 = = ≈ 5.56 2 2 2 2 16 ( 3)(1)(1)(1) = 3 ≈ 1.73 22. a = 3, b = 3, c = 2 1 1 ( a + b + c ) = 2 (3 + 3 + 2) = 4 2 K = s ( s − a)( s − b)( s − c) s= 13. a = 3, b = 4, C = 40º K= 1 1 ab sin C = (6)(4) sin 60º ≈ 10.39 2 2 1 1 ab sin C = (3)(4) sin 40º ≈ 3.86 2 2 = 14. a = 2, c = 1, B = 10º ( 4 )(1)(1)( 2 ) = 8 ≈ 2.83 23. a = 5, b = 8, c = 9 1 1 K = ac sin B = (2)(1)sin10º ≈ 0.17 2 2 1 1 a + b + c ) = ( 5 + 8 + 9 ) = 11 ( 2 2 K = s ( s − a)( s − b)( s − c) s= 15. b = 1, c = 3, A = 80º = 1 1 K = bc sin A = (1)(3)sin 80º ≈ 1.48 2 2 (11)( 6 )( 3)( 2 ) = 396 ≈ 19.90 826 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.4: Area of a Triangle 30. A = 70º , B = 60º , c = 4 C = 180º − A − B = 180º − 70º − 60º = 50º 24. a = 4, b = 3, c = 6 1 1 13 ( a + b + c ) = 2 ( 4 + 3 + 6) = 2 2 K = s ( s − a)( s − b)( s − c) s= 13 5 7 1 = = 2 2 2 2 K= 455 ≈ 5.33 16 31. A = 110º , C = 30º , c = 3 B = 180º − A − C = 180º − 110º − 30º = 40º K= sin A sin B 25. From the Law of Sines we know . = a b a sin B . Thus, Solving for b, so we have that b = sin A 1 1 a sin B K = ab sin C = a sin C 2 2 sin A a 2 sin B sin C = 2sin A K= π 7π = 180 18 1 7π 112π 2 ASector = ⋅ 82 ⋅ = ft 2 18 9 1 ATriangle = ⋅ 8 ⋅ 8sin 70º = 32 sin 70º ft 2 2 112π ASegment = − 32sin 70º ≈ 9.03 ft 2 9 34. Area of a sector = 1 2 r θ where θ is in radians. 2 π 2π = 180 9 1 2π 25π 2 ASector = ⋅ 52 ⋅ = in 2 9 9 1 25 ATriangle = ⋅ 5 ⋅ 5sin 40º = sin 40º in 2 2 2 25π 25 ASegment = − sin 40º ≈ 0.69 in 2 9 2 θ = 40 ⋅ a 2 sin B sin C 22 sin 20º ⋅ sin120º = ≈ 0.92 2sin A 2 sin 40º 28. A = 50º , C = 20º , a = 3 B = 180º − A − C = 180º − 50º − 20º = 110º 35. Find the area of the lot using Heron's Formula: a = 100, b = 50, c = 75 a sin B sin C 3 sin110º ⋅ sin 20º = ≈ 1.89 2 sin A 2 sin 50º 2 s= 29. B = 70º , C = 10º , b = 5 A = 180º − B − C = 180º − 70º − 10º = 100º K= 1 2 r θ where θ is in radians. 2 θ = 70 ⋅ 27. A = 40º , B = 20º , a = 2 C = 180º − A − B = 180º − 40º − 20º = 120º K= b 2 sin A sin C 22 sin 70º ⋅ sin100º = ≈ 10.66 2sin B 2sin10º 33. Area of a sector = sin A sin B sin C = = , we have that a b c b sin C c sin A c= and a = . Thus, sin B sin C 1 1 b sin C K = bc sin A = b sin A 2 2 sin B b 2 sin A sin C = 2sin B 1 1 c sin A K = ac sin B = c sin B 2 2 sin C c 2 sin A sin B = 2sin C 2 c 2 sin A sin B 32 sin110º ⋅ sin 40º = ≈ 5.44 2sin C 2 sin 30º 32. B = 10º , C = 100º , b = 2 A = 180º − B − C = 180º − 10º − 100º = 70º 26. From K= c 2 sin A sin B 42 sin 70º ⋅ sin 60º = ≈ 8.50 2sin C 2sin 50º 1 1 225 ( a + b + c ) = 2 (100 + 50 + 75) = 2 2 b 2 sin A sin C 52 sin100º ⋅ sin10º = ≈ 2.27 2sin B 2sin 70º 827 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions The triangle is a right triangle. Find the other leg: 82 + b 2 = 102 K = s ( s − a)( s − b)( s − c) 225 25 125 75 = 2 2 2 2 b 2 = 100 − 64 = 36 b = 36 = 6 1 ATriangle = ⋅ 8 ⋅ 6 = 24 in 2 2 AShaded region = 12.5π − 24 ≈ 15.27 in 2 52, 734,375 16 ≈ 1815.46 Cost = ( $3)(1815.46 ) = $5446.38 = 39. The area is the sum of the area of a triangle and a sector. 1 1 ATriangle = r ⋅ r sin(π − θ ) = r 2 sin(π − θ ) 2 2 1 2 ASector = r θ 2 36. Diameter of canvas is 24 feet; radius of canvas is 12 feet; angle is 260˚. 1 Area of a sector = r 2θ where θ is in radians. 2 π 13π = θ = 260 ⋅ 180 9 1 13 π 936π ASector = ⋅122 ⋅ = = 104π ≈ 326.73 ft 2 2 9 9 K= = 37. Divide home plate into a rectangle and a triangle. 12” 12” = 17” 8.5” = 8.5” = 17” ARectangle = lw = (17)(8.5) = 144.5 in 2 1 2 1 r sin ( π − θ ) + r 2θ 2 2 1 2 r ( sin(π − θ ) + θ ) 2 1 2 r ( sin π cos θ − cos π sin θ + θ ) 2 1 2 r ( 0 ⋅ cos θ − (−1)sin θ + θ ) 2 1 2 r (θ + sin θ ) 2 40. Use the Law of Cosines to find the lengths of the diagonals of the polygon. x 2 = 352 + 802 − 2 ⋅ 35 ⋅ 80 cos15º Using Heron’s formula we get ATriangle = s ( s − a)( s − b)( s − c) = 7625 − 5600 cos15º 1 1 ( a + b + c ) = (12 + 12 + 17 ) = 20.5 2 2 Thus, ATriangle = (20.5)(20.5 − 12)(20.5 − 12)(20.5 − 17) s= x = 7625 − 5600 cos15º ≈ 47.072 feet The interior angle of the third triangle is: 180° − 100° = 80° . y 2 = 452 + 202 − 2 ⋅ 45 ⋅ 20 cos 80º = (20.5)(8.5)(8.5)(3.5) = 2425 − 1800 cos80º = 5183.9375 y = 2425 − 1800 cos80º ≈ 45.961 feet ≈ 72.0 sq. in. So, ATotal = ARectangle + ATriangle Find the area of the three triangles: 1 ( 35 + 80 + 47.072 ) = 81.036 2 s1 − a1 = 81.036 − 35 = 46.036 = 144.5 + 72.0 s1 ≈ = 216.5 in 2 The area of home plate is about 216.5 in 2 . s1 − b1 = 81.036 − 80 = 1.036 s1 − c1 = 81.036 − 47.072 = 33.964 38. Find the area of the shaded region by subtracting the area of the triangle from the area of the semicircle. Area of the semicircle 1 1 25 ASemicircle = π r 2 = π(5)2 = π in 2 2 2 2 K1 ≈ 81.036(46.036)(1.036)(33.964) ≈ 362.31 ft 2 828 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.4: Area of a Triangle 1 ( 40 + 45.961 + 47.072 ) = 66.5165 2 s2 − a2 = 66.5165 − 40 = 26.5165 s2 ≈ Area ΔOAC = b. Area ΔOCB = s2 − b2 = 66.5165 − 45.961 = 20.5555 s2 − c2 = 66.5165 − 47.072 = 19.4445 K 2 ≈ 66.5165(26.5165)(20.5555)(19.4445) ≈ 839.62 ft 2 1 (45 + 20 + 45.961) = 55.4805 2 s3 − a3 = 55.4805 − 45 = 10.4805 s3 ≈ 1 OC ⋅ AC 2 AC 1 OC = ⋅ ⋅ 2 1 1 1 = cos α sin α 2 1 = sin α cos α 2 43. a. s3 − b3 = 55.4805 − 20 = 35.4805 s3 − c3 = 55.4805 − 45.961 = 9.5195 1 OC ⋅ BC 2 OC BC 2 1 = ⋅ OB ⋅ ⋅ 2 OB OB 1 OB 2 1 = OB 2 = K3 ≈ 55.4805(10.4805)(35.4805)(9.5195) ≈ 443.16 ft 2 The approximate area of the lake is 362.31 + 839.62 + 443.16 = 1645.09 ft 2 c. 41. Use Heron’s formula: a = 87 , b = 190 , c = 173 1 1 s = ( a + b + c ) = ( 87 + 190 + 173) = 225 2 2 K = s ( s − a )( s − b )( s − c ) = 225 (138 )( 35)( 52 ) d. = 56,511, 000 ≈ 7517.4 The building covers approximately 7517.4 square feet of ground area. e. 42. Use Heron’s formula: a = 1028 , b = 1046 , c = 965 1 1 s = ( a + b + c ) = (1028 + 1046 + 965 ) = 1519.5 2 2 cos β sin β 2 sin β cos β 1 BD ⋅ OA 2 1 = BD ⋅1 2 BD 1 = ⋅ OB ⋅ 2 OB Area ΔOAB = = = 225 ( 225 − 87 )( 225 − 190 )( 225 − 173) 2 1 OB sin(α + β ) 2 OC OA OC OB cos A = = ⋅ = OB cos B 1 OC OC OB Area ΔOAB = Area ΔOAC + Area ΔOCB 1 1 1 OB sin(α + β ) = sin α cos α + OB 2 2 2 2 sin β cos β cosα cos 2 α sin(α + β ) = sin α cosα + sin β cos β cos β cos 2 β cos β cosα sin(α + β ) = sin α cos α + sin β cos β cos α cos β sin(α + β ) = sin α cos β + cos α sin β K = s ( s − a )( s − b )( s − c ) = 1519.5 ( 491.5 )( 473.5 )( 554.5) ≈ 442,816 The area of the Bermuda Triangle is approximately 442,816 square miles. 44. a. Area of ΔOBC = 1 sin θ ⋅1 ⋅1 ⋅ sin θ = 2 2 b. Area of ΔOBD = 1 tan θ sin θ ⋅1 ⋅ tan θ = = 2 2 2 cos θ 829 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions < Area ΔOBD Area ΔOBC < Area OBC 1 1 sin θ sin θ < θ < 2 2 2 cos θ sin θ sin θ < θ < cos θ sin θ sin θ θ < < sin θ sin θ sin θ cos θ 1 θ 1< < sin θ cos θ c. Thus, the total grazing area is: 23,561.94 + 2(3499.66) + 2(292.19) ≈ 31,146 ft 2 46. We begin by dividing the grazing area into five regions: three sectors and two triangles (see figure). Region A1 is a sector representing three-fourths of a circle with radius 100 feet: 3 2 Thus, A1 = π (100 ) = 7500π ≈ 23,561.9 ft 2 . 4 To find the areas of regions A2 , A3 , A4 , and 45. The grazing area must be considered in sections. Region A1 represents three-fourth of a circle with radius 100 feet. Thus, 3 2 A1 = π (100 ) = 7500π ≈ 23,561.94 ft 2 4 A5 , we first position the rectangular barn on a rectangular coordinate system so that the lower right corner is at the origin. The coordinates of the corners of the barn must then be O(0, 0) , P(−20, 0) , Q(−20,10) , and R(0,10) . Angles are needed to find regions A2 and A3 : (see the figure) T y C A2 D B 10 ft R 90 ft A A4 A5 U x Q(−20,10) and radius 90 feet. The equation of the circle then is ( x + 20)2 + ( y − 10) 2 = 902 . Likewise, region A5 is a sector of a circle with center O(0, 0) and radius 80 feet. The equation ∠DAC = 130.49º − 90º = 40.49º of this circle then is x 2 + y 2 = 802 . We use a graphing calculator to find the intersection point 1 (10)(90)sin 40.49º ≈ 292.19 ft 2 2 S of the two sectors. Let Y1 = 802 − x 2 and The angle for the sector A2 is 90º − 40.49º = 49.51º . A2 = A3 Now, region A2 is a sector of a circle with center ∠BAC = 180º − 45º − 4.51º = 130.49º A3 = 80 ft 10 ft Q P O 20 ft A1 In ΔABC , ∠CBA = 45º , AB = 10, AC = 90 . Find ∠BCA : sin ∠CBA sin ∠BCA = 90 10 sin 45º sin ∠BCA = 90 10 10 sin 45º sin ∠BCA = ≈ 0.0786 90 10sin 45º ∠BCA = sin −1 ≈ 4.51º 90 S 90 ft Y2 = 902 − ( x + 20)2 + 10 . 112 1 2 π ≈ 3499.66 ft 2 90 ) 49.51 ⋅ ( 2 180 Since the cow can go in either direction around the barn, both A2 and A3 must be doubled. −64 124 −12 830 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.4: Area of a Triangle K = s ( s − a )( s − b )( s − c ) The approximate coordinates are S (57.7,55.4) . Now, consider ΔQRS (i.e. region A3 ). The “base” of this triangle is 20 feet and the “height” is approximately 55.4 − 10 = 45.4 feet (the ycoordinate of the intersection point minus the side of the barn). Thus, the area of region A3 is = 18 (18 − 9 )(18 − 10 )(18 − 17 ) = 18 ( 9 )( 8 )(1) = 1296 = 36 Since the perimeter and area are numerically equal, the given triangle is a perfect triangle. 1 A3 ≈ ⋅ 20 ⋅ 45.4 = 454 ft 2 . Likewise, consider 2 ΔORS (i.e. region A4 ). The “base” of this triangle is 10 feet and the “height” is about 57.7 feet (the x-coordinate of the intersection point). 1 Thus, A4 ≈ ⋅10 ⋅ 57.7 = 288.5 ft 2 . 2 b. Area: 1 1 s = ( a + b + c ) = ( 6 + 25 + 29 ) = 30 2 2 To find the area of sectors A2 and A5 , we must determine their angles: ∠TQS and ∠SOU , K = s ( s − a )( s − b )( s − c ) respectively. Now, we know A3 ≈ 454 ft 2 . Also, we know A3 = Thus, = 30 ( 30 − 6 )( 30 − 25 )( 30 − 29 ) 1 ⋅ 90 ⋅ 20sin ( ∠SQR ) . 2 = 30 ( 24 )( 5)(1) 1 ⋅ 90 ⋅ 20 sin ( ∠SQR ) ≈ 454 2 sin ( ∠SQR ) ≈ 0.5044 = 3600 = 60 Since the perimeter and area are numerically equal, the given triangle is a perfect triangle. ∠SQR ≈ 0.5287 rad. Since ∠TQR is a right angle, we have ∠TQS ≈ π 2 1 2K 2K h a , so h1 = . Similarly, h 2 = 2 1 a b 2K . Thus, and h 3 = c 1 1 1 a b c + + = + + h1 h 2 h 3 2 K 2 K 2 K 48. K = − .5287 ≈ 1.0421 rad. So, 1 2 1 r θ ≈ ⋅ 902 ⋅1.0421 ≈ 4220.5 ft 2 . 2 2 1 Similarly, ⋅ 80 ⋅10sin ( ∠SOR ) ≈ 288.5 2 sin ( ∠SOR ) ≈ 0.72125 A2 = So, ∠SOU ≈ A5 ≈ π 2 a + b + c 2s = 2K 2K s = K = ∠SOR ≈ 0.8056 rad. − .8056 ≈ 0.7652 rad. and 1 ⋅ 802 ⋅ 0.7652 ≈ 2448.6 ft 2 2 1 1 ah and K = ab sin C , which 2 2 means h = b sin C . From the Law of Sines, we sin A sin B a sin B = , so b = . Therefore, know sin A a b a sin B sin C a sin B h= . sin C = A sin sin A 49. We know K = Thus, the total grazing area is: 23,561.9 + 4220.5 + 454 + 288.5 + 2448.6 = 30,973 ft 2 47. a. Perimeter: P = a + b + c = 6 + 25 + 29 = 60 Perimeter: P = a + b + c = 9 + 10 + 17 = 36 Area: 1 1 s = ( a + b + c ) = ( 9 + 10 + 17 ) = 18 2 2 831 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 52. Use the result of Problem 51: A B C s −a s −b s −c + + cot + cot + cot = 2 2 2 r r r s −a + s −b+ s −c = r 3s − ( a + b + c ) = r 3s − 2 s = r s = r a sin B sin C where h is the altitude to side a. sin A In ΔOPQ , c is opposite ∠POQ . The two 50. h = adjacent angles are A B and . Then 2 2 A B sin 2 2 . Now, ∠POQ = π − A + B , r= sin ( ∠POQ ) 2 2 so A B sin ( ∠POQ ) = sin π − + 2 2 c sin 53. K = Area ΔPOQ + Area ΔPOR + Area ΔQOR 1 1 1 = rc + rb + ra 2 2 2 1 = r (a + b + c) 2 = rs A B = sin + 2 2 A+ B = sin 2 π A + B = cos − 2 2 π − ( A + B) = cos 2 C = cos 2 A B c sin sin 2 2 . Thus, r = C cos 2 C cos C 2 = 51. cot = 2 sin C 2 Now, K = s ( s − a)( s − b)( s − c) , so rs = s ( s − a)( s − b)( s − c) s ( s − a )( s − b)( s − c) s ( s − a)( s − b)( s − c) = s r= 54 – 56. Answers will vary. A B ⋅ sin 2 2 r ( s − a)( s − b) ab c ⋅ sin Section 8.5 1. 5 = 5 ; ( s − b)( s − c) ( s − a)( s − c) c⋅ bc ac = ( s − a)( s − b) r ab c ( s − b)( s − c) ( s − a )( s − c) ab = ( s − a )( s − b) r bc ac 2π π = 4 2 2. simple harmonic; amplitude 3. simple harmonic; damped 4. True 5. d = − 5cos ( πt ) c ab( s − a )( s − b)( s − c) 2 = r abc 2 ( s − a)( s − b) 2π 6. d = −10 cos t 3 c ( s − c) 2 c s − c = ⋅ r r c c2 s−c = r = 7. d = − 6 cos ( 2t ) 8. d = − 4 cos ( 4 t ) 832 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves 9. d = − 5sin ( π t ) 18. d = − 2 cos(2t ) a. Simple harmonic 2π 10. d = −10sin t 3 11. d = − 6sin ( 2t ) 12. d = − 4sin ( 4 t ) 5 meters c. 2π seconds 3 d. 3 oscillation/second 2π c. π seconds d. 1 oscillation/second π b. 2 meters c. 1 second d. 1 oscillation/second 20. d = 4 + 3sin(πt ) a. Simple harmonic 14. d = 4sin(2t ) a. Simple harmonic b. 2 meters 19. d = 6 + 2 cos(2πt ) a. Simple harmonic 13. d = 5sin(3t ) a. Simple harmonic b. b. 4 meters π seconds 1 oscillation/second d. π c. b. 3 meters c. 2 seconds d. 1 oscillation/second 2 21. d ( t ) = e −t / π cos ( 2t ) 15. d = 6 cos(πt ) a. Simple harmonic b. 6 meters c. 2 seconds d. 1 oscillation/second 2 π 16. d = 5cos t 2 a. Simple harmonic b. 5 meters c. 4 seconds d. 1 oscillation/second 4 22. d ( t ) = e −t / 2π cos ( 2t ) 1 17. d = − 3sin t 2 a. Simple harmonic b. 3 meters c. 4π seconds d. 1 oscillation/second 4π 833 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 23. d ( t ) = e −t / 2π cos t 27. f ( x ) = x − sin x 28. f ( x ) = x − cos x 29. f ( x ) = sin x + cos x 30. f ( x ) = sin ( 2 x ) + cos x 24. d ( t ) = e −t / 4π cos t 25. 26. f ( x ) = x + cos x f ( x ) = x + cos ( 2 x ) 834 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves 31. f ( x ) = sin x + sin ( 2 x ) b. 35. a. 32. G ( x ) = cos ( 4 x ) cos ( 2 x ) 1 [cos(4 x − 2 x) + cos(4 x + 2 x)] 2 1 = [ cos(2 x) + cos(6 x) ] 2 f ( x ) = cos ( 2 x ) + cos x = b. 33. a. f ( x ) = sin ( 2 x ) sin x 1 [ cos(2 x − x) − cos(2 x + x)] 2 1 = [ cos( x) − cos(3 x)] 2 = 36. a. b. h ( x ) = cos ( 2 x ) cos ( x ) 1 [cos(2 x − x) + cos(2 x + x) ] 2 1 = [ cos( x) + cos(3x)] 2 = b. 34. a. F ( x ) = sin ( 3x ) sin x 1 [ cos(3x − x) − cos(3x + x)] 2 1 = [ cos(2 x) − cos(4 x) ] 2 = 835 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 37. a. H ( x ) = 2sin ( 3 x ) cos ( x ) 40. a. 1 = ( 2 ) [sin(3 x + x) + sin(3x − x) ] 2 = sin(4 x) + sin(2 x) b. 2π 2 (0.75)2 d = −15e −0.75 t / 2(20) cos − 6 4(20)2 π2 0.5625 d = −15e −0.75 t / 40 cos t − 9 1600 15 b. 30 0 −15 41. a. 38. a. g ( x ) = 2sin ( x ) cos ( 3 x ) 1 = ( 2 ) [sin( x + 3 x) + sin( x − 3x) ] 2 = sin(4 x) + sin(−2 x) π 2 (0.6)2 d = −18e −0.6 t / 2(30) cos − 2 4(30) 2 π2 0.36 d = −18e −0.6 t / 60 cos t − 4 3600 t 18 b. 0 = sin(4 x) − sin(2 x) t 20 −18 42. a. 2π 2 (0.65)2 d = −16e−0.65 t / 2(15) cos − 5 4(15) 2 4π2 0.4225 d = −16e−0.65 t / 30 cos t − 25 900 16 b. 39. a. 2π 2 (0.7)2 d = −10e−0.7t / 2(25) cos − 5 4(25) 2 4π2 0.49 d = −10e−0.7t / 50 cos t − 25 2500 0 t −16 43. a. 10 b. 0 25 25 2π 2 (0.8)2 d = −5e−0.8 t / 2(10) cos − 3 4(10)2 4π2 0.64 d = −5e−0.8 t / 20 cos t − 9 400 −10 836 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall t t Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves 5 b. 0 15 e. −5 44. a. 2π 2 (0.7) 2 d = −5e−0.7 t / 2(10) cos − 3 4(10)2 4π2 0.49 − d = −5e−0.7 t / 20 cos t 9 400 t 47. a. b. 15 35 −30 The displacement of the bob at the start of the second oscillation is about 28.47 meters. e. The displacement of the bob approaches zero, since e −0.6 t / 80 → 0 as t → ∞ . Damped motion with a bob of mass 20 kg and a damping factor of 0.7 kg/sec. 20 meters leftward 20 c. 0 25 48. a. −20 d. 30 d. −5 b. 30 meters leftward 0 0 45. a. Damped motion with a bob of mass 40 kg and a damping factor of 0.6 kg/sec. c. 5 b. The displacement of the bob approaches zero, since e −0.8 t / 40 → 0 as t → ∞ . b. The displacement of the bob at the start of the second oscillation is about 18.33 meters. Damped motion with a bob of mass 35 kg and a damping factor of 0.5 kg/sec. 30 meters leftward 30 c. 0 e. 46. a. b. −30 The displacement of the bob approaches zero, since e −0.7 t / 40 → 0 as t → ∞ . d. The displacement of the bob at the start of the second oscillation is about 29.15 meters. e. The displacement of the bob approaches zero, since e −0.5 t / 70 → 0 as t → ∞ . Damped motion with a bob of mass 20 kg and a damping factor of 0.8 kg/sec. 20 meters leftward 20 c. 0 25 49. a. −20 d. 20 b. The displacement of the bob at the start of the second oscillation is about 18.10 meters. Damped motion with a bob of mass 15 kg and a damping factor of 0.9 kg/sec. 15 meters leftward 837 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 15 c. position at time t = 0 . Since a cos ( ωt ) peaks at 0 t = 0 if a > 0 and is at its lowest if a < 0 , we select a cosine model and need a = −82.5 . Using the model d = a cos (ωt ) + b , we have 30 d = −82.5cos ( wt ) + b . When t = 0 , the rider should be 15 feet above the ground. That is, 15 = −82.5cos ( 0 ) + b −15 d. The displacement of the bob at the start of the second oscillation is about 12.53 meters. 15 = −82.5 + b 97.5 = b Therefore, the equation that describes the rider’s motion is d = 97.5 − 82.5cos ( 3.2π t ) . e. 50. a. b. The displacement of the bob approaches zero, since e −0.9 t / 30 → 0 as t → ∞ . 53. The maximum displacement is the amplitude so we have a = 0.01 . The frequency is given by Damped motion with a bob of mass 25 kg and a damping factor of 0.8 kg/sec. f = movement of the tuning fork is described by the equation d = 0.01sin ( 880π t ) . 10 meters leftward 10 c. 0 54. The maximum displacement is the amplitude so we have a = 0.025 . The frequency is given by 15 ω = 329.63 . Therefore, ω = 659.26π and 2π the movement of the tuning fork is described by the equation d = 0.025sin ( 659.26π t ) . f = −10 d. ω = 440 . Therefore, ω = 880π and the 2π The displacement of the bob at the start of the second oscillation is 9.53 meters. 55. a. V 1 e. 1 The displacement of the bob approaches zero, since e −0.8 t / 50 → 0 as t → ∞ . 2 3 t −1 51. The maximum displacement is the amplitude so we have a = 0.80 . The frequency is given by b. ω f = = 520 . Therefore, ω = 1040π and the 2π On the interval 0 ≤ t ≤ 3 , the graph of V touches the graph of y = e −t / 3 when t = 0, 2 . The graph of V touches the graph motion of the diaphragm is described by the equation d = 0.80 cos (1040π t ) . of y = − e −t / 3 when t = 1, 3 . c. 52. If we consider a horizontal line through the center of the wheel as the equilibrium line, then 165 = 82.5 . The wheel the amplitude is a = 2 2π 1 completes 1.6 so the period is = and ω 1.6 ω = 3.2π . We want the rider to be at the lowest We need to solve the inequality −0.4 < e −t / 3 cos (π t ) < 0.4 on the interval 0 ≤ t ≤ 3 . To do so, we consider the graphs of y = −0.4, y =e −t / 3 cos (π t ) , and y = 0.4 . On the interval 0 ≤ t ≤ 3 , we can use the INTERSECT feature on a calculator to determine that y = e −t / 3 cos (π t ) intersects 838 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 8.5: Simple Harmonic Motion; Damped Motion; Combining Waves y = 0.4 when t ≈ 0.35, t ≈ 1.75 , and t ≈ 2.19, y = e −t / 3 Let Y1 = c. cos (π t ) intersects y = −0.4 when t ≈ 0.67 and t ≈ 1.29 and the graph shows that −0.4 < e −t / 3 cos (π t ) < 0.4 when t = 3 . 1 Therefore, the voltage V is between –0.4 and 0.4 on the intervals 0.35 < t < 0.67 , 1.29 < t < 1.75 , and 2.19 < t ≤ 3 . 1.25 0 1.25 0 3.2 0 −1.25 1.25 d. f ( x ) = 3.2 3.2 0 4 −1 −1.25 1.25 0 1 1 sin ( 2π ) + sin ( 4π x ) 2 4 1 1 + sin ( 8π x ) + sin (16π x ) 8 16 1 1 1 sin ( 2π x ) + sin ( 4π x ) + sin ( 8π x ) 2 4 8 1 1 + sin (16π x ) + sin ( 32π x ) 16 32 57. Let Y1 = sin ( 2π ( 852 ) x ) + sin ( 2π (1209 ) x ) . 2.5 3.2 0 −1.25 1.25 −1.25 1.25 0 3.2 0 −1.25 56. a. −2.5 58. The sound emitted by touching * is y = sin ( 2π ( 941) t ) + sin ( 2π (1209 ) t ) . 3.2 Let Y1 = sin ( 2π ( 941) x ) + sin ( 2π (1209 ) x ) . −1.25 2.5 1 1 Let Y1 = sin ( 2π ) + sin ( 4π x ) . 2 4 0 1 2 −2.5 4 0 2 59 – 60. CBL Experiments −1 61. Let Y1 = 1 1 1 b. Let Y1 = sin ( 2π ) + sin ( 4π x ) + sin ( 8π x ) . 2 4 8 1 1 0 4 sin x . x 0 5π −0.3 −1 As x approaches 0, sin x approaches 1. x 839 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 62. y = x sin x 1 y = 3 sin x x 3 0.05 2π 0 0 −5 −0.015 y = x 2 sin x Possible observation: As x gets larger, the graph 1 of y = n sin x gets closer to y = 0 . x 64. Answers will vary. 5 2π 0 3π −25 y = x3 sin x Chapter 8 Review Exercises 10 1. opposite = 4; adjacent = 3; hypotenuse = ? (hypotenuse)2 = 42 + 32 = 25 0 hypotenuse = 25 = 5 2π opp 4 = hyp 5 adj 3 cos θ = = hyp 5 opp 4 tan θ = = adj 3 sin θ = −125 Possible observations: The graph lies between the bounding curves y = ± x, y = ± x 2 , y = ± x3 , respectively, touching them at odd multiples of π . The x-intercepts of each graph are the 2 multiples of π . 2. opposite = 3; adjacent = 5; hypotenuse = ? (hypotenuse)2 = 32 + 52 = 34 1 63. y = sin x x hypotenuse = 34 1 opp 3 = = hyp 34 adj 5 = = cos θ = hyp 34 opp 3 = tan θ = adj 5 sin θ = 0 5π −0.3 1 y= 2 x 0.1 0 hyp 5 = opp 4 hyp 5 sec θ = = adj 3 adj 3 cot θ = = opp 4 csc θ = sin x csc θ = 3 34 5 34 ⋅ ⋅ 34 34 34 34 hyp 34 = opp 3 hyp 34 = adj 5 adj 5 = cot θ = opp 3 sec θ = 7π −0.06 840 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall = 3 34 34 = 5 34 34 Chapter 8 Review Exercises 3. adjacent = 2; hypotenuse = 4; opposite = ? (opposite)2 + 22 = 42 7. sec55º csc(90º −55º ) csc 35º = = =1 csc35º csc 35º csc 35º 8. tan 40º tan 40º tan 40º = = =1 cot 50º tan(90º − 50º ) tan 40º (opposite) 2 = 16 − 4 = 12 opposite = 12 = 2 3 opp 2 3 3 = = hyp 4 2 adj 2 1 cos θ = = = hyp 4 2 sin θ = tan θ = 9. cos 2 40º + cos 2 50º = sin 2 (90º −40º ) + cos 2 50º = sin 2 50º + cos 2 50º =1 opp 2 3 = = 3 adj 2 ( hyp 4 4 3 2 3 = = ⋅ = csc θ = opp 2 3 2 3 3 3 hyp 4 = =2 sec θ = adj 2 cot θ = ( 2) 2 = −1 + sec 40º − sec (90º −50º ) 2 11. c = 10, B = 20º b c b sin 20º = 10 b = 10sin 20º ≈ 3.42 sin B = 2 ; hypotenuse = 2; adjacent = ? + (adjacent)2 = 22 (adjacent)2 = 4 − 2 = 2 a c a cos 20º = 10 a = 10 cos 20º ≈ 9.40 cos B = opposite = 2 sin θ = opp 2 = hyp 2 cos θ = adj 2 = hyp 2 tan θ = opp = adj csc θ = hyp 2 2 2 2 2 = = ⋅ = = 2 opp 2 2 2 2 sec θ = hyp 2 2 2 2 2 = = ⋅ = = 2 adj 2 2 2 2 cot θ = adj = opp 2 2 2 2 2 = −1 + sec 2 40º − sec2 40º = −1 adj 2 2 3 3 = = ⋅ = opp 2 3 2 3 3 3 4. opposite = ) 10. tan 2 40º − csc 2 50º = −1 + sec2 40º − csc2 50º A = 90º − B = 90º − 20º = 70º 12. a = 5, A = 35º =1 a c 5 sin 35º = c sin A = c= =1 5 ≈ 8.72 sin 35º a b 5 tan 35º = b tan A = 5. cos 62º − sin 28º = cos 62º − cos(90º −28º ) = cos 62º − cos 62º =0 b= 5 ≈ 7.14 tan 35º B = 90º − A = 90º − 35º = 55º 6. tan15º − cot 75º = tan15º − tan(90º − 75º ) = tan15º − tan15º =0 841 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 13. b = 2, c = 5 sin C sin B = c b sin 40º sin130º = 2 b 2sin130º ≈ 2.38 b= sin 40º c 2 = a 2 + b2 a 2 = c 2 − b 2 = 52 − 22 = 25 − 4 = 21 a = 21 ≈ 4.58 b 2 = c 5 2 B = sin −1 ≈ 23.6º 5 sin B = B = 180º − A − C = 180º − 10º − 40º = 130º 17. A = 100º , c = 2, a = 5 sin C sin A = c a sin C sin100º = 2 5 2 sin100º ≈ 0.3939 sin C = 5 2sin100º C = sin −1 5 C ≈ 23.2º or C ≈ 156.8º The value 156.8º is discarded because A + C > 180º . Thus, α ≈ 23.2º . A = 90º − B ≈ 90º − 23.6º = 66.4º 14. b = 1, a = 3 c2 = a 2 + b2 c 2 = 32 + 12 = 9 + 1 = 10 c = 10 ≈ 3.16 b 1 = a 3 1 B = tan −1 ≈ 18.4º 3 tan B = B = 180º − A − C ≈ 180º − 100º − 23.2º = 56.8º A = 90º − B ≈ 90º −18.4º = 71.6º sin B sin A = b a sin 56.8º sin100º = b 5 5sin 56.8º ≈ 4.25 b= sin100º 15. A = 50º , B = 30º , a = 1 C = 180º − A − B = 180º − 50º − 30º = 100º sin A sin B = a b sin 50º sin 30º = 1 b 1sin 30º ≈ 0.65 b= sin 50º 18. a = 2, c = 5, A = 60º sin C sin A = c a sin C sin 60º = 5 2 5sin 60º sin C = ≈ 2.1651 2 No angle C exists for which sin C > 1 . Therefore, there is no triangle with the given measurements. sin C sin A = c a sin100º sin 50º = c 1 1sin100º ≈ 1.29 c= sin 50º 16. A = 10º , C = 40º , c = 2 19. a = 3, c = 1, C = 110º sin C sin A = c a sin 40º sin10º = 2 a 2sin10º ≈ 0.54 a= sin 40º sin C sin A = c a sin110º sin A = 1 3 3sin110º sin A = ≈ 2.8191 1 No angle A exists for which sin A > 1 . Thus, there is no triangle with the given measurements. 842 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Review Exercises 20. a = 3, c = 1, C = 20º 23. a = 2, b = 3, c = 1 sin C sin A = c a sin 20º sin A = 1 3 3sin 20º sin A = ≈ 1.0260 1 There is no angle A for which sin A > 1 . Thus, there is no triangle with the given measurements. a 2 = b 2 + c 2 − 2bc cos A cos A = A = cos −1 1 = 0º No triangle exists with an angle of 0˚. 24. a = 10, b = 7, c = 8 a 2 = b 2 + c 2 − 2bc cos A 21. a = 3, c = 1, B = 100º cos A = b = a + c − 2ac cos B 2 2 2 = 32 + 12 − 2 ⋅ 3 ⋅1cos100º b 2 + c 2 − a 2 72 + 82 − 102 13 = = 2bc 2(7)(8) 112 13 A = cos −1 ≈ 83.3º 112 = 10 − 6 cos100º b = 10 − 6 cos100º ≈ 3.32 b 2 = a 2 + c 2 − 2ac cos B a 2 = b 2 + c 2 − 2bc cos A cos A = b 2 + c 2 − a 2 32 + 12 − 22 = =1 2bc 2(3)(1) cos B = b 2 + c 2 − a 2 3.322 + 12 − 32 3.0224 = = 2bc 2(3.32)(1) 6.64 a 2 + c 2 − b2 102 + 82 − 7 2 115 = = 2ac 2(10)(8) 160 115 B = cos −1 ≈ 44.0º 160 C = 180º − A − B ≈ 180º − 83.3º − 44.0º ≈ 52.7º 3.0224 A = cos −1 ≈ 62.9º 6.64 C = 180º − A − B ≈ 180º −62.9º −100º = 17.1º 25. a = 1, b = 3, C = 40º c 2 = a 2 + b 2 − 2ab cos C 22. a = 3, b = 5, B = 80º = 12 + 32 − 2 ⋅1⋅ 3cos 40º = 10 − 6 cos 40º sin A sin B = a b sin A sin 80º = 3 5 3sin 80º ≈ 0.5909 sin A = 5 3sin 80º A = sin −1 5 A ≈ 36.2º or A ≈ 143.8º The value 143.8º is discarded because A + B > 180º . Thus, A ≈ 36.2º . c = 10 − 6 cos 40º ≈ 2.32 a 2 = b 2 + c 2 − 2bc cos A cos A = b 2 + c 2 − a 2 32 + 2.322 − 12 13.3824 = = 2bc 2(3)(2.32) 13.92 13.3824 A = cos−1 ≈ 16.0º 13.92 B = 180º − A − C ≈ 180º −16.0º −40º = 124.0º 26. a = 4, b = 1, C = 100º C = 180º − A − B ≈ 180º − 36.2º − 80º = 63.8º c 2 = a 2 + b 2 − 2ab cos C sin C sin B = c b sin 63.8º sin 80º = 5 c 5sin 63.8º c= ≈ 4.56 sin 80º c 2 = 42 + 12 − 2 ⋅ 4 ⋅1cos100º = 17 − 8cos100º c = 17 − 8cos100º ≈ 4.29 a 2 = b 2 + c 2 − 2bc cos A cos A = b 2 + c 2 − a 2 12 + 4.292 − 42 3.4041 = = 2bc 2(1)(4.29) 8.58 3.4041 A = cos −1 ≈ 66.6º 8.58 B = 180º − A − C ≈ 180º −66.6º −100º = 13.4º 843 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 27. a = 5, b = 3, A = 80º sin A sin C2 = a c2 sin B sin A = b a sin B sin 80º = 3 5 3sin 80º ≈ 0.5909 sin B = 5 3sin 80º B = sin −1 5 B ≈ 36.2° or B ≈ 143.8° The value 143.8° is discarded because A + B > 180º . Thus, B ≈ 36.2° . sin 20º sin10.9º = c2 2 c2 = 2sin10.9º ≈ 1.11 sin 20º Two triangles: B1 ≈ 30.9º , C1 ≈ 129.1º , c1 ≈ 4.54 or B2 ≈ 149.1º , C2 ≈ 10.9º , c2 ≈ 1.11 1 4 , c= 2 3 a 2 = b 2 + c 2 − 2bc cos A 29. a = 1, b = C = 180º − A − B ≈ 180º − 80º − 36.2º = 63.8º 2 2 1 + 4 − 12 27 b2 + c2 − a 2 2 3 = = cos A = 2bc 37 1 4 2 2 3 27 A = cos −1 ≈ 39.6º 37 sin C sin A = c a sin 63.8º sin 80º = 5 c 5sin 63.8º c= ≈ 4.56 sin 80º b 2 = a 2 + c 2 − 2ac cos B 28. a = 2, b = 3, A = 20º 2 2 4 1 12 + − 3 2 91 a 2 + c 2 − b2 = = cos B = 2ac 96 4 2(1) 3 91 B = cos−1 ≈ 18.6º 96 sin B sin A = b a sin B sin 20º = 3 2 3sin 20º ≈ 0.5130 sin B = 2 3sin 20º B = sin −1 2 B1 = 30.9º or B2 =149.1º For both values, A + B < 180º . Therefore, there are two triangles. C = 180º − A − B ≈ 180º − 39.6º − 18.6º ≈ 121.8º 30. a = 3, b = 2, c = 2 a 2 = b 2 + c 2 − 2bc cos A cos A = C1 = 180º − A − B1 ≈ 180º − 20º − 30.9º ≈ 129.1º 1 b 2 + c 2 − a 2 22 + 22 − 32 = =− 2bc 2(2)(2) 8 1 A = cos −1 − ≈ 97.2º 8 sin A sin C1 = a c1 sin 20º sin129.1º = c1 2 b 2 = a 2 + c 2 − 2ac cos B cos B = 2sin129.1º ≈ 4.54 c1 = sin 20º a 2 + c 2 − b2 32 + 22 − 22 9 = = 2ac 2(3)(2) 12 9 B = cos −1 ≈ 41.4º 12 C = 180º − A − B ≈ 180º − 97.2º − 41.4º ≈ 41.4º C2 = 180º − A − B2 ≈ 180º − 20º − 149.1º ≈ 10.9º 844 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Review Exercises 31. a = 3, A = 10º , b = 4 33. c = 5, b = 4 , A = 70º a 2 = b 2 + c 2 − 2bc cos A sin B sin A = b a sin B sin10º = 4 3 4sin10º sin B = ≈ 0.2315 3 4sin10º B = sin −1 3 B1 ≈ 13.4º or B2 ≈ 166.6º For both values, A + B < 180º . Therefore, there are two triangles. a 2 = 42 + 52 − 2 ⋅ 4 ⋅ 5cos 70º = 41 − 40cos 70º a = 41 − 40 cos 70º ≈ 5.23 c 2 = a 2 + b 2 − 2ab cos C a 2 + b2 − c 2 2ab 5.232 + 42 − 52 18.3529 = = 2(5.23)(4) 41.48 cos C = 18.3529 C = cos −1 ≈ 64.0º 41.48 B = 180º − A − C ≈ 180º − 70º − 64.0º ≈ 46.0º C1 = 180º − A − B1 ≈ 180º − 10º − 13.4º ≈ 156.6º sin A sin C1 = a c1 34. a = 1, b = 2, C = 60º c 2 = a 2 + b 2 − 2ab cos C sin10º sin156.6º = 3 c1 c1 = = 12 + 22 − 2 ⋅1⋅ 2 cos 60º = 3 3sin156.6º ≈ 6.86 sin10º c = 3 ≈ 1.73 a 2 = b 2 + c 2 − 2bc cos A C2 = 180º − A − B2 = 180º − 10º − 166.6º ≈ 3.4º 6 3 b 2 + c 2 − a 2 22 + ( 3 ) − 12 cos A = = = = 2bc 2 4 3 2(2) ( 3 ) 3 A = cos −1 = 30º 2 B = 180º − A − C = 180º − 30º − 60º = 90º 2 sin A sin C2 = a c2 sin10º sin 3.4º = 3 c2 c2 = 3sin 3.4º ≈ 1.02 sin10º 35. a = 2, b = 3, C = 40º Two triangles: B1 ≈ 13.4º , C1 ≈ 156.6º , c1 ≈ 6.86 K= or B2 ≈ 166.6º , C2 ≈ 3.4º , c2 ≈ 1.02 1 1 ab sin C = (2)(3) sin 40º ≈ 1.93 2 2 36. b = 5, c = 5, A = 20º 32. a = 4, A = 20º , B = 100º C = 180º − A − B = 180º − 20º − 100º = 60º 1 1 K = bc sin A = (5)(5)sin 20º ≈ 4.28 2 2 sin A sin B = a b sin 20º sin100º = 4 b 4sin100º b= ≈ 11.52 sin 20º 37. b = 4, c = 10, A = 70º 1 1 K = bc sin A = (4)(10)sin 70º ≈ 18.79 2 2 38. a = 2, b = 1, C = 100º K= sin C sin A = c a sin 60º sin 20º = c 4 4sin 60º c= ≈ 10.13 sin 20º 1 1 ab sin C = (2)(1)sin100º ≈ 0.98 2 2 39. a = 4, b = 3, c = 5 1 1 (a + b + c) = (4 + 3 + 5) = 6 2 2 K = s ( s − a)( s − b)( s − c) s= = ( 6 )( 2 )( 3)(1) = 36 = 6 845 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 46. c = 12 feet, a = 8 feet. We need to find A and B (see figure). 40. a = 10, b = 7, c = 8 1 1 (a + b + c) = (10 + 7 + 8) = 12.5 2 2 K = s ( s − a)( s − b)( s − c) s= = B 12 ft (12.5)( 2.5 )( 5.5)( 4.5 ) A = 773.4375 ≈ 27.81 sin A = 41. a = 4, b = 2, c = 5 1 1 s = (a + b + c) = (4 + 2 + 5) = 5.5 2 2 K = s ( s − a)( s − b)( s − c) = 8 8 A = sin −1 ≈ 41.8° 12 12 B = 180° − 90° − A = 180° − 90° − 41.8° = 48.2° 47. Let x = the distance across the river. x tan(25°) = 50 x = 50 tan(25°) ≈ 23.32 Thus, the distance across the river is 23.32 feet. ( 5.5 )(1.5 )( 3.5)( 0.5 ) = 14.4375 ≈ 3.80 48. Let h = the height of the building. h tan(25°) = 80 h = 80 tan(25°) ≈ 37.30 Thus, the height of the building is 37.30 feet. 42. a = 3, b = 2, c = 2 1 1 s = (a + b + c) = (3 + 2 + 2) = 3.5 2 2 K = s ( s − a)( s − b)( s − c) = 8 ft ( 3.5 )( 0.5 )(1.5 )(1.5) 49. Let x = the distance the boat is from shore (see figure). Note that 1 mile = 5280 feet. = 3.9375 ≈ 1.98 1454 ft 43. A = 50º , B = 30º , a = 1 5º C = 180º − A − B = 180º − 50º − 30º = 100º K= 5280 ft a 2 sin B sin C 12 sin 30º ⋅ sin100º = ≈ 0.32 2sin A 2sin 50º 1454 x + 5280 1454 x + 5280 = tan(5°) 1454 − 5280 x= tan(5°) ≈ 16, 619.30 − 5280 = 11,339.30 Thus, the boat is approximately 11,339.30 feet, 11,339.30 ≈ 2.15 miles, from shore. or 5280 tan(5°) = 44. A = 10º , C = 40º , c = 3 B = 180º − A − C = 180º − 10º − 40º = 130º K= x c 2 sin A sin B 32 sin10º ⋅ sin130º = ≈ 0.93 2sin C 2sin 40º 45. To find the area of the segment, we subtract the area of the triangle from the area of the sector. 1 1 π ≈ 15.708 in 2 ASector = r 2θ = ⋅ 62 50 ⋅ 2 2 180 1 1 A Triangle = ab sin θ = ⋅ 6 ⋅ 6sin 50º ≈ 13.789 in 2 2 2 ASegment = 15.708 − 13.789 ≈ 1.92 in 2 846 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Review Exercises 52. Let θ = the inclination (grade) of the trail. The “rise” of the trail is 4100 − 5000 = −900 feet (see figure). 50. Let x = the length of the lake, and let y = the distance from the edge of the lake to the point on the ground beneath the balloon (see figure). 500 ft 25º y 900 4100 900 θ = sin −1 ≈ 12.7º 4100 The trail is inclined about 12.7º from the lake to the hotel. sin θ = 500 x 500 x= tan ( 65º ) tan ( 65º ) = 500 x+ y 500 x+ y = tan ( 25º ) tan ( 25º ) = 53. Let h = the height of the helicopter, x = the distance from observer A to the helicopter, and γ = ∠AHB (see figure). H 500 −y tan ( 25º ) h 40º 25º A B 100 ft γ = 180º − 40º − 25º = 115º ≈ 1072.25 − 233.15 = 839.10 Thus, the length of the lake is approximately 839.10 feet. sin 40º sin115º = x 100 100 sin 40º x= ≈ 70.92 feet sin115º 51. Let x = the distance traveled by the glider between the two sightings, and let y = the distance from the stationary object to a point on the ground beneath the glider at the time of the second sighting (see figure). h h ≈ x 70.92 h ≈ 70.92 sin 25º ≈ 29.97 feet The helicopter is about 29.97 feet high. sin 25º = 40º 10º γ x 500 500 = − tan ( 25º ) tan ( 65º ) 54. ∠ACB = 12º + 30º = 42º ∠ABC = 90º − 30º = 60º ∠CAB = 90º − 12º = 78º 200 ft y −900 ft 65º x x= 4100 ft θ 2 mi A x y 200 y = 200 tan(10°) x+ y tan(40°) = 200 x + y = 200 tan(40°) tan(10°) = B P x b 12º a 30º C a. x = 200 tan(40°) − y = 200 tan(40°) − 200 tan(10°) sin 60º sin 42º = b 2 2sin 60º b= ≈ 2.59 miles sin 42º ≈ 167.82 − 35.27 = 132.55 The glider traveled 132.55 feet in 1 minute, so the speed of the glider is 132.55 ft/min. 847 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 56. a. sin 78º sin 42º = a 2 2sin 78º a= ≈ 2.92 miles sin 42º b. Find the third side of the triangle to determine the distance from the island: x 2.59 x = 2.59 cos12º ≈ 2.53 miles 15º 55. α = 180º − 120º = 60º ; β = 180º − 115º = 65º ; γ = 180º − 60º − 65º = 55º D 3 mi β BWI c c 2 = a 2 + b 2 − 2ab cos C c 2 = 722 + 2002 − 2 ⋅ 72 ⋅ 200 cos15º = 45,184 − 28,800 cos15º c ≈ 131.8 miles The sailboat is about 131.8 miles from the island. 0.25 mi γ C 0.25 mi E b. B 115º sin 60º sin 55º = BC 3 3sin 60º BC = ≈ 3.17 mi sin 55º Find the measure of the angle opposite the 200 side: a 2 + c 2 − b2 cos B = 2ac 2 72 + 131.82 − 2002 17, 444.76 cos B = =− 2(72)(131.8) 18,979.2 17, 444.76 B = cos −1 − ≈ 156.8º 18,979.2 The sailboat should turn through an angle of 180° − 156.8° = 23.2° to correct its course. sin 65º sin 55º = AC 3 3sin 65º AC = ≈ 3.32 mi sin 55º c. BE = 3.17 − 0.25 = 2.92 mi AD = 3.32 − 0.25 = 3.07 mi For the isosceles triangle, ∠CDE = ∠CED = 72 mi B a = 72, b = 200, C = 15º 120º Aα 200 mi ST cos12º = c. After 4 hours, the sailboat would have sailed 18(4) = 72 miles. 180º − 55º = 62.5º 2 sin 55º sin 62.5º = 0.25 DE 0.25sin 55º DE = ≈ 0.23 miles sin 62.5º The original trip would have taken: 200 t= ≈ 11.11 hours . The actual trip takes: 18 131.8 t = 4+ ≈ 11.32 hours . The trip takes 18 about 0.21 hour, or about 12.6 minutes longer. 57. Find the third side of the triangle using the Law of Cosines: a = 50, b = 60, C = 80º c 2 = a 2 + b 2 − 2ab cos C The length of the highway is 2.92 + 3.07 + 0.23 = 6.22 miles. = 502 + 602 − 2 ⋅ 50 ⋅ 60 cos 80º = 6100 − 6000 cos 80º c = 6100 − 6000 cos80º ≈ 71.12 The houses are approximately 71.12 feet apart. 848 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Review Exercises 60. Find angle AMB and subtract from 80˚ to obtain θ . 58. Find the lengths of the two unknown sides of the middle triangle: x 2 = 1002 + 1252 − 2 (100 )(125 ) cos 50º 80º M Α 10 = 25, 625 − 25, 000cos 50º θ x = 25, 625 − 25, 000 cos 50º ≈ 97.75 feet 40 y 2 = 702 + 502 − 2 ( 70 )( 50 ) cos100º B = 7400 − 7000 cos100º 40 =4 10 ∠AMB = tan −1 4 ≈ 76.0º θ ≈ 80º − 76.0º ≈ 4.0º The bearing is S4.0°E . tan ∠AMB = y = 7400 − 7000 cos100º ≈ 92.82 feet Find the areas of the three triangles: 1 K1 = (100)(125) sin 50º ≈ 4787.78 ft 2 2 1 K 2 = (50)(70) sin100º ≈ 1723.41 ft 2 2 1 s = (50 + 97.75 + 92.82) = 120.285 2 K3 = (120.285 )( 70.285 )( 22.535)( 27.465 ) 61. Extend the tangent line until it meets a line extended through the centers of the pulleys. Label these extensions x and y. The distance between the points of tangency is z. Two similar triangles are formed. ≈ 2287.47 ft 2 The approximate area of the lake is 4787.78 + 1723.41 + 2287.47 = 8798.67 ft 2 . z βα 6.5 59. Construct a diagonal. Find the area of the first triangle and the length of the diagonal: d y 2.5 24 + y 6.5 = where 24 + y is the y 2.5 hypotenuse of the larger triangle and y is the hypotenuse of the smaller triangle. Solve for y: 6.5 y = 2.5(24 + y ) 6.5 y = 60 + 2.5 y 4 y = 60 y = 15 B 40º 100 ft K1 = x Thus, 20 ft A 100º 50 ft 15 α 1 ⋅ 50 ⋅100 ⋅ sin 40º ≈ 1606.969 ft 2 2 d 2 = 502 + 1002 − 2 ⋅ 50 ⋅100 cos 40º = 12,500 − 10, 000 cos 40º Use the Pythagorean Theorem to find x: x 2 + 2.52 = 152 x 2 = 225 − 6.25 = 218.75 d = 12,500 − 10, 000 cos 40º ≈ 69.566914 feet x = 218.75 ≈ 14.79 Use the Law of Sines to find B : sin B sin100º = 20 69.566914 20sin100º sin B = 69.566914 20sin100º B = sin −1 ≈ 16.447º 69.566914 A ≈ 180º − 100º − 16.447º = 63.553º Use the Pythagorean Theorem to find z: ( z + 14.79)2 + 6.52 = ( 24 + 15 ) 2 ( z + 14.79)2 = 1521 − 42.25 = 1478.75 z + 14.79 = 1478.75 ≈ 38.45 z ≈ 23.66 2.5 ≈ 0.1667 15 2.5 α = cos −1 ≈ 1.4033 radians 15 β ≈ π − 1.4033 ≈ 1.7383 radians Find α : cos α = Find the area of the second triangle: 1 K 2 = ⋅ 20 ⋅ 69.566914 ⋅ sin 63.553º ≈ 622.865 ft 2 2 The cost of the parcel is approximately 100(1606.969 + 622.865) = $222,983.40 . 849 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions π 63. d = − 3cos t 2 The arc length on the top of the larger pulley is about: 6.5(1.7383) = 11.30 inches. The arc length on the top of the smaller pulley is about: 2.5(1.4033) = 3.51 inches. π 64. d = − 5cos t 3 The distance between the points of tangency is about 23.66 inches. 65. d = 6sin(2 t ) a. Simple harmonic b. 6 feet c. π seconds The length of the belt is about: 2(11.30 + 3.51 + 23.66) = 76.94 inches. 62. Let x = the hypotenuse of the larger right triangle in the figure below. Then 24 − x is the hypotenuse of the smaller triangle. The two triangles are similar. 24 − x 6.5 βα z x y d. 1 oscillation/second π 66. d = 2 cos(4 t ) a. Simple harmonic b. 2 feet α β 2.5 6.5 x = 2.5 24 − x 2.5 x = 6.5(24 − x) 2.5 x = 156 − 6.5 x 9 x = 156 52 x= 3 20 24 − x = 3 c. π seconds 2 d. 2 oscillation/second π 67. d = − 2 cos(π t ) a. Simple harmonic b. 2 feet c. 2 seconds d. 6.5 3 = 52 8 3 3 α = cos−1 ≈ 67.98º 8 cos α = 1 oscillation/second 2 π 68. d = − 3sin t 2 a. Simple harmonic b. 3 feet c. 4 seconds β = 180º − 67.98º = 112.02º z = 6.5 tan 67.98º ≈ 16.07 in y = 2.5 tan 67.98º ≈ 6.18 in d. The arc length on the top of the larger pulley is: π 6.5 112.02 ⋅ ≈ 12.71 inches. 180 69. a. The arc length on the top of the smaller pulley is π 2.5 112.02 ⋅ ≈ 4.89 inches. 180 1 oscillation/second 4 2π 2 (0.75) 2 d = −15e−0.75 t / 2(40) cos − 5 4(40) 2 4π2 0.5625 d = −15e−0.75 t / 80 cos t − 25 6400 b. The distance between the points of tangency is z + y ≈ 16.07 + 6.18 = 22.25 inches. 15 0 25 The length of the belt is about: 2(12.71 + 4.89 + 22.25) = 79.7 inches. –15 850 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall t Chapter 8 Review Exercises 70. a. π 2 (0.65) 2 d = −13e−0.65 t / 2(25) cos − 2 4(25) 2 π 0.4225 d = −13e−0.65 t / 50 cos − 4 2500 2 t 0 t 15 –20 13 b. 20 c. d. The displacement of the bob at the start of the second oscillation is about 19.51 meters. e. It approaches zero, since e −0.5 t / 60 → 0 as t→∞. 20 0 –13 71. a. b. c. Damped motion with a bob of mass 20 kg and a damping factor of 0.6 kg/sec. 15 meters leftward 73. y = 2sin x + cos ( 2 x ) , 0 ≤ x ≤ 2π x 74. y = 2 cos ( 2 x ) + sin , 2 0 ≤ x ≤ 2π 15 25 0 –15 d. e. 72. a. b. The displacement of the bob at the start of the second oscillation is about 13.92 meters. It approaches zero, since e t→∞. −0.6 t / 40 → 0 as Damped motion with a bob of mass 30 kg and a damping factor of 0.5 kg/sec. 20 meters leftward 851 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions Chapter 8 Test 4. Use the Law of Sines to find b : a b = sin A sin B b 12 = sin 41° sin 22° 12 ⋅ sin 22° b= ≈ 6.85 sin 41° C = 180° − A − B = 180° − 41° − 22° = 117° Use the Law of Sines to find c: a c = sin A sin C c 12 = sin 41° sin117° 12 ⋅ sin117° c= ≈ 16.30 sin 41° 1. opposite = 3; adjacent = 6; hypotenuse = ? (hypotenuse)2 = 32 + 62 = 45 hypotenuse = 45 = 3 5 opp 3 = = hyp 3 5 adj 6 cos θ = = = hyp 3 5 opp 3 1 tan θ = = = adj 6 2 sin θ = csc θ = 1 5 2 5 ⋅ ⋅ 5 5 5 5 = 5 5 = 2 5 5 hyp 3 5 = = 5 opp 3 hyp 3 5 5 = = adj 6 2 adj 6 cot θ = = =2 opp 3 sec θ = 5. Use the law of cosines to find A . a 2 = b2 + c 2 − 2bc cos A 82 = (5)2 + (10)2 − 2(5)(10) cos A 64 = 25 + 100 − 100 cos A 100cos α = 61 cos A = 0.61 2. sin 40° − cos 50° = sin 40° − sin ( 90° − 50° ) = sin 40° − sin 40° =0 A = cos −1 (0.61) ≈ 52.41° 3. Use the law of cosines to find a: a 2 = b 2 + c 2 − 2bc cos A Use the law of sines to find B . a b = sin A sin B 8 5 = sin 52.41° sin B 5 sin B = ( sin 52.41° ) 8 sin B ≈ 0.495 Since b is not the longest side of the triangle, we have that B < 90° . Therefore = (17)2 + (19) 2 − 2(17)(19) cos 52° ≈ 289 + 361 − 646(0.616) = 252.064 a = 252.064 ≈ 15.88 Use the law of sines to find B . a b = sin A sin B 15.88 17 = sin 52° sin B 17 sin B = (sin 52°) ≈ 0.8436 15.88 Since b is not the longest side of the triangle, we know that B < 90 . Therefore, B = sin −1 (0.8436) ≈ 57.5° B = sin −1 (0.495) ≈ 29.67° C = 180° − A − B = 180° − 52.41° − 29.67° = 97.92° 6. A = 55º , C = 20º , a = 4 B = 180º − A − C = 180º − 55º − 20º = 105º Use the law of sines to find b. C = 180° − A − B = 180° − 52° − 57.5° = 70.5° 852 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Test 10. a = 8, b = 5, c = 10 sin A sin B = a b sin 55º sin105º = 4 b 4 sin105º b= ≈ 4.72 sin 55º 1 1 a + b + c ) = ( 8 + 5 + 10 ) = 11.5 ( 2 2 K = s ( s − a)( s − b)( s − c) s= = 11.5(11.5 − 8)(11.5 − 5)(11.5 − 10) = 11.5(3.5)(6.5)(1.5) Use the law of sines to find c. sin C sin A = c a sin 20º sin 55º = 4 c 4sin 20º c= ≈ 1.67 sin 55º = 392.4375 ≈ 19.81 square units 11. Let α = the angle formed by the ground and the ladder. 7. a = 3, b = 7, A = 40º Use the law of sines to find B sin B sin A = b a sin B sin 40º = 7 3 7 sin 40º sin B = ≈ 1.4998 3 There is no angle B for which sin B > 1 . Therefore, there is no triangle with the given measurements. 12 ft A Then sin A = 10.5 12 10.5 A = sin −1 ≈ 61.0° 12 The angle formed by the ladder and ground is about 61.0° . 12. Let A = the angle of depression from the balloon to the airport. 8. a = 8, b = 4, C = 70º 5 mi A c 2 = a 2 + b 2 − 2ab cos C c 2 = 82 + 42 − 2 ⋅ 8 ⋅ 4 cos 70º = 80 − 64 cos 70º MSFC c = 80 − 64 cos 70º ≈ 7.62 600 ft AP Note that 5 miles = 26,400 feet. 600 1 tan A = = 26400 44 1 A = tan −1 = 1.3° 44 The angle of depression from the balloon to the airport is about 1.3° . a 2 = b 2 + c 2 − 2bc cos A cos A = 10.5 ft b 2 + c 2 − a 2 42 + 7.622 − 82 10.0644 = = 2bc 2(4)(7.62) 60.96 10.0644 A = cos −1 ≈ 80.5º 60.96 B = 180º − A − C ≈ 180º −80.5º −70º = 29.5º 13. We can find the area of the shaded region by subtracting the area of the triangle from the area of the semicircle. Since triangle ABC is a right triangle, we can use the Pythagorean theorem to find the length of the third side. a2 + b2 = c2 9. a = 8, b = 4, C = 70º 1 ab sin C 2 1 = (8)(4)sin 70° ≈ 15.04 square units 2 K= a 2 + 6 2 = 82 a 2 = 64 − 36 = 28 a = 28 = 2 7 853 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions sin A sin 40° = OB AB sin 70° sin 40° = 5 AB 5sin 40° AB = ≈ 3.420 sin 70° Now, AB is the diameter of the semicircle, so the 3.420 = 1.710 . radius is 2 1 1 2 ASemicircle = π r 2 = π (1.710 ) ≈ 4.593 sq. units 2 2 The area of the triangle is 1 1 A = bh = 2 7 ( 6 ) = 6 7 square cm . 2 2 The area of the semicircle is 2 1 1 A = π r 2 = π ( 4 ) = 8π square cm . 2 2 Therefore, the area of the shaded region is 8π − 6 7 ≈ 9.26 square centimeters . ( ) 14. Begin by adding a diagonal to the diagram. 5 72° 11 1 ab sin(O ) 2 1 = (5)(5)(sin 40°) ≈ 8.035 sq. units 2 ATriangle = d 7 8 1 (5)(11)(sin 72°) ≈ 26.15 sq. units 2 By the law of cosines, d 2 = (5)2 + (11)2 − 2(5)(11)(cos 72°) Aupper = ATotal = ASemicircle + ATriangle ≈ 4.593 + 8.035 ≈ 12.63 sq. units 17. Using Heron’s formula: 5 x + 6 x + 7 x 18 x s= = = 9x 2 2 K = 9 x(9 x − 5 x )(9 x − 6 x)(9 x − 7 x) = 25 + 121 − 110(0.309) = 112.008 d = 112.008 ≈ 10.58 = 9 x ⋅ 4 x ⋅ 3x ⋅ 2 x ( Using Heron’s formula for the lower triangle, 7 + 8 + 10.58 s= = 12.79 2 Alower = 12.79(5.79)(4.79)(2.21) ) = 216 x 4 = 6 6 x 2 Thus, (6 6) x 2 = 54 6 x2 = 9 x=3 The sides are 15, 18, and 21. = 783.9293 ≈ 28.00 sq. units Total Area = 26.15 + 28.00 = 54.15 sq. units 18. Since we ignore all resistive forces, this is simple harmonic motion. Since the rest position ( t = 0 ) 15. Use the law of cosines to find c: c 2 = a 2 + b 2 − 2ab cos C is the vertical position ( d = 0 ) , the equation will = (4.2) 2 + (3.5) 2 − 2(4.2)(3.5) cos 32° have the form d = a sin(ωt ) . ≈ 17.64 + 12.25 − 29.4(0.848) = 4.9588 42° c = 4.9588 ≈ 2.23 Madison will have to swim about 2.23 miles. 5 feet a 16. Since ΔOAB is isosceles, we know that 180° − 40° ∠A = ∠B = = 70° 2 Then, 854 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Cumulative Review g ( x ) = x 2 − 3x − 4 ≥ 0 . Now, the period is 6 seconds, so 2π 6= ω= ω= x 2 − 3x − 4 ≥ 0 ( x − 4 )( x + 1) ≥ 0 ω 2π x = 4, x = −1 are the zeros. 6 π Interval Test Number −∞ < x < −1 −2 −1 < x < 4 0 4< x<∞ 5 radians/sec 3 From the diagram we see a = sin 42° 5 a = 5 ( sin 42° ) The domain of f ( x ) = x 2 − 3x − 4 is πt Thus, d = 5 ( sin 42° ) sin or 3 πt d ≈ 3.346 ⋅ sin . 3 {x x ≤ −1 or x ≥ 4 } . 4. y = 3sin (π x ) Amplitude: Chapter 8 Cumulative Review A = 3 =3 2π Period: T= Phase Shift: φ 0 = =0 ω π 3x2 + 1 = 4 x 1. g ( x) Pos./Neg. 6 Positive −4 Negative 6 Positive π =2 3x2 − 4 x + 1 = 0 ( 3x − 1)( x − 1) = 0 x= 1 or x = 1 3 1 The solution set is ,1 . 3 2. Center (−5, 1) ; Radius 3 π 5. y = −2cos ( 2 x − π ) = −2 cos 2 x − 2 Amplitude: A = − 2 = 2 ( x − h )2 + ( y − k )2 = r 2 ( x − (−5) )2 + ( y − 1)2 = 32 ( x + 5 )2 + ( y − 1)2 = 9 3. 2π =π 2 Period: T= Phase Shift: φ π = ω 2 f ( x ) = x 2 − 3x − 4 f will be defined provided 855 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions 6. tan θ = −2, 3π < θ < 2π , so θ lies in quadrant IV. 2 f. 1 θ 2 5 a. b. c. 3π < θ < 2π , so sin θ < 0 . 2 2 5 2 5 ⋅ =− sin θ = − 5 5 5 1 + cos θ 1 =− cos θ = − 2 2 =− 7. a. sin(2θ ) = 2 sin θ cos θ 0 cos(2θ ) = cos θ − sin θ 2 2 5 2 5 = − − 5 5 5 20 = − 25 25 15 3 =− =− 25 5 e. y = ex , 0 ≤ x ≤ 4 60 2 5 5 = 2 − 5 5 20 4 =− =− 25 5 d. b. 2 4 y = sin x , 0 ≤ x ≤ 4 0 4 −1.5 c. y = e x sin x , 0 ≤ x ≤ 4 10 4 0 −50 5 1− 1 − cos θ 1 5 = sin θ = 2 2 2 = 0 1.5 3π < θ < 2π 2 3π 1 < θ <π 4 2 1 1 Since θ lies in Quadrant II, sin θ > 0 . 2 2 = 5 5 1+ 2 5+ 5 5 =− 2 3π < θ < 2π , so cos θ > 0 2 1 5 5 ⋅ = cos θ = 5 5 5 2 3π < θ < 2π 2 3π 1 < θ <π 4 2 1 1 Since θ lies in Quadrant II, cos θ < 0 . 2 2 d. y = 2 x + sin x , 0 ≤ x ≤ 4 8 5− 5 5 2 0 5− 5 10 0 4 856 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall 5+ 5 10 Chapter 8 Cumulative Review y=x e. y = ex b. y = x2 f. y = ln x c. y= x g. y = sin x h. y = cos x 8. a. d. y = x3 857 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions i. y = tan x 10. 3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0 Let f ( x ) = 3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0 f ( x) has at most 5 real zeros. Possible rational zeros: p = ±1, ±2, ±4, ±8; q = ±1, ± 2, ±3; p 1 1 2 4 8 = ±1, ± , ± , ±2, ± , ±4, ± , ±8, ± q 2 3 3 3 3 9. a = 20, c = 15, C = 40o sin C sin A = c a sin 40o sin A = 15 20 20sin 40o sin A = 15 20sin 40o A = sin −1 15 Using the Bounds on Zeros Theorem: 10 8 f ( x) = 3 x5 − x 4 + 7 x3 − 14 x 2 + 12 x − 3 3 10 8 a4 = − , a3 = 7, a2 = −14, a1 = 12, a0 = − 3 3 8 Max 1, − + 12 + 3 = Max {1, 39} = 39 1,833,832 sin C sin B1 = b1 c −15 sin 40o sin 81.0o = b1 15 sin 40 o 15 −1 3 −10 From the graph it appears that there are x1 intercepts at ,1, and 2. 3 sin C sin B2 = b2 c Using synthetic division with 1: 1 3 − 10 21 − 42 36 − 8 3 −7 14 − 28 8 sin 40o sin19.0o = b2 15 b2 ≈ 3 −2,865,248 ≈ 23.05 B2 = 180o − A2 − C ≈ 180o − 40o − 121.0o = 19.0o 15sin19.0o 3 The smaller of the two numbers is 15. Thus, every zero of f lies between –15 and 15. Graphing using the bounds: (Second graph has a better window.) B1 = 180o − A1 − C ≈ 180o − 40o − 59.0o = 81.0o 15sin 81.0o 10 8 10 1 + Max − , 12 , − 14 , 7 , − 3 3 = 1 + 14 = 15 A1 ≈ 59.0o or A2 ≈ 121.0o For both values, A + C < 180º . Therefore, there are two triangles. b1 ≈ − 14 + 7 + − 3 ≈ 7.60 − 7 14 − 28 8 0 Two triangles: A1 ≈ 59.0o , B1 ≈ 81.0o , b1 ≈ 23.05 Since the remainder is 0, x − 1 is a factor. The other factor is the quotient: 3x 4 − 7 x3 + 14 x 2 − 28 x + 8 . A2 ≈ 121.0o , B2 ≈ 19.0o , b2 ≈ 7.60 . Using synthetic division with 2 on the quotient: sin 40o or 858 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Cumulative Review 2 3 − 7 14 − 28 8 6 −2 24 − 8 3 −1 −4 12 11. R( x) = Since the remainder is 0, x − 2 is a factor. The other factor is the quotient: 3x3 − x 2 + 12 x − 4 . Domain: ( x ≠ − 5, x ≠ 3} 2x +1 = 0 x−4 = 0 or 1 2 The y-intercept is x=− R(0) = ) 3x + 12 = 3 x + 4 = 3 ( x + 2i )( x − 2i ) . x=4 2 ⋅ 02 − 7 ⋅ 0 − 4 R(− x) = Factoring, 0 + 2 ⋅ 0 − 15 2 = −4 4 = . − 15 15 2(− x)2 − 7(− x) − 4 = 2x2 + 7 x − 4 (− x) + 2(− x) − 15 x 2 − 2 x − 15 is neither R ( x ) nor − R ( x) , so there is no symmetry. 1 f ( x) = 3( x − 1) ( x − 2 ) x − ( x + 2i )( x − 2i ) 3 The real zeros are 1,2, and {x 2 x2 − 7 x − 4 = 0 ( 2 x + 1)( x − 4 ) = 0 0 12 0 2 (2 x + 1)( x − 4) ( x − 3)( x + 5) The x-intercepts are the zeros of p( x) : 1 Since the remainder is 0, x − is a factor. The 3 other factor is the quotient: 2 = R is in lowest terms. 1 on the quotient: 3 1 3 − 1 12 4 3 1 0 4 3 x + 2 x − 15 2 p( x) = 2 x 2 − 7 x − 4; q ( x ) = x 2 + 2 x − 15; n = 2; m = 2 0 Using synthetic division with 2 x2 − 7 x − 4 1 . The imaginary 3 2 which The vertical asymptotes are the zeros of q ( x) : x 2 + 2 x − 15 = 0 ( x + 5)( x − 3) = 0 zeros are −2i and 2i. Therefore, over the complex numbers, the equation 3x5 − 10 x 4 + 21x3 − 42 x 2 + 36 x − 8 = 0 has solution 1 set −2i, 2i, , 1, 2 . 3 x+5 = 0 or x − 3 = 0 x = −5 x=3 Since n = m , the line y = 2 is the horizontal asymptote. 26 R ( x ) intersects y = 2 at , 2 , since: 11 2 x2 − 7 x − 4 =2 x 2 + 2 x − 15 2 x 2 − 7 x − 4 = 2 x 2 + 2 x − 15 ( ) 2 x − 7 x − 4 = 2 x + 4 x − 30 −11x = −26 26 x= 11 2 2 Graphing utility: 8 −10 8 −4 859 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions Graph by hand: 14. Test Value number of f Interval −6 ( −5, −0.5 ) −1 ( −0.5,3) 0 ≈ 0.27 ( 3, 4 ) 3.5 ≈ −0.94 ( 4, ∞ ) 5 0.55 a. Location Point Above x-axis Below −0.3125 x-axis ( −∞, −5) ≈ 12.22 Above x-axis Below x-axis Above x-axis f ( x) = 4 x + 5 ; g ( x) = x 2 + 5 x − 24 f ( x) = 0 4x + 5 = 0 4 x = −5 5 x=− 4 ( −6,12.22 ) ( −1, −0.3125) 5 The solution set is − . 4 ( 0, 0.27 ) ( 3.5, −0.94 ) b. f ( x ) = 13 4 x + 5 = 13 4x = 8 x=2 The solution set is {2} . ( 5, 0.55 ) c. f ( x) = g ( x) 4 x + 4 = x 2 + 5 x − 24 0 = x 2 + x − 28 x= = 12. The solution set is ( ) = ln (12) x x ln ( 3) = ln (12 ) x= ln (12 ) ln ( 3) −1 ± 117 −1 ± 3 13 = 2 2 −1 − 3 13 −1 + 3 13 , . 2 2 3x = 12 ln 3 −1 ± 12 − 4(1)(−29) 2(1) d. ≈ 2.26 The solution set is {2.26}. 13. log3 ( x + 8 ) + log3 x = 2 f ( x) > 0 4x + 5 > 0 4 x > −5 5 x>− 4 5 5 The solution set is x x > − or − , ∞ . 4 4 log3 ( x + 8 )( x ) = 2 ( x + 8 )( x ) = 32 g ( x) ≤ 0 e. x2 + 8x = 9 x + 5 x − 24 ≤ 0 ( x + 8 )( x − 3) ≤ 0 2 x2 + 8x − 9 = 0 ( x + 9 )( x − 1) = 0 x = −9 or x = 1 x = −9 is extraneous because it makes the original logarithms undefined. The solution set is {1} . x = −8, x = 3 are the zeros. Interval Test number −9 ( −∞, −8 ) 0 ( −8,3) 4 ( 3, ∞ ) The solution set is {x g ( x) Pos./Neg. 12 Positive −24 Negative 12 Positive − 8 ≤ x ≤ 3 } or [ −8,3] . 860 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Projects f. y = f ( x) = 4 x + 5 The graph of f is a line with slope 4 and yintercept 5. b y = f − = f (−2.5) 2a = (−2.5)2 + 5(−2.5) − 24 = −30.25 The vertex is ( −2.5, −30.25) . g. y = g ( x) = x 2 + 5 x − 24 The graph of g is a parabola with y-intercept −24 and x-intercepts −8 and 3. The xcoordinate of the vertex is b 5 5 x=− =− = − = −2.5 . 2a 2 (1) 2 The y-coordinate of the vertex is Chapter 8 Projects C Project I C 1. a a b O B c Q O a Q sin a = b c 2. A CQ , OC 2 + CQ 2 = OQ 2 OQ For ΔOPQ , we have that ( PQ )2 = (OP )2 + (OQ )2 − 2(OQ)(OP) cos c For ΔCPQ , we have that P ( PQ )2 = (CQ )2 + (CP )2 − 2(CQ)(CP) cos C Triangles ΔOCP and ΔOCQ are plane right triangles. 3. C 0 = (OP )2 + (OQ )2 − 2(OQ)(OP) cos c − (CQ) 2 + (CP) 2 − 2(CQ )(CP) cos C 2(OQ )(OP ) cos c = (OQ )2 − (CQ) 2 + (OP) 2 − (CQ )2 + 2(CQ )(CP ) cos C b O sin b = P 4. CP , OC 2 + CP 2 = OP 2 OP From part a: (OC ) 2 = (OQ )2 − (CQ) 2 (OC ) 2 = (OP) 2 − (CP)2 861 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions Thus, 2(OQ )(OP ) cos c = OQ 2 − CQ 2 + OP 2 2. N Lewiston/Clarkton 43.5° − CQ 2 + 2(CQ)(CP) cos C 2(OQ )(OP ) cos c = OC 2 + OC 2 + 2(CQ)(CP) cos C 45° 46.5° N 45° N c Lemhi 43.5° 2OC 2 2(CQ)(CP) cos C cos c = + 2(OQ )(OP ) 2(OQ)(OP) 45° 117.0°W 113.5°W 2 (CQ)(CP) cos C OC cos c = + (OQ)(OP) (OQ )(OP ) O = 117.0° − 113.5° = 3.5° cos c = cos 45° cos 43.5° + sin 45° sin 43.5° cos 3.5° cos c = 0.99875 c = 2.87° OC 2 (CQ)(CP) cos C + (OQ )(OP ) (OQ)(OP) OC OC CQ CP = ⋅ + ⋅ ⋅ cos C OQ OP OQ OP = cos a cos b + sin a sin b cos C cos c = 5. π s = rc = 3960 2.87° ⋅ ≈ 198.4 ° 180 It is about 198.4 miles from Lemhi to Lewiston and Clarkston. Project II 1. Putting this on a circle of radius 1 in order to apply the Law of Cosines from A: N 3. Great Falls 42.5° 45° 4. 47.5° N 45° N c They traveled 202.5 + 198.4 miles just to go from Great Falls to Lewiston and Clarkston. Lewiston and Clarkston 282 miles 42° 48.5° 45° 42.5° 0° Great Falls x y Lemhi 0° α α = 180° − 48.5° − 42° = 89.5° 111.5°W 113.5°W O = 113.5° − 111.3° = 2.2° cos c = cos 45° cos 42.5° + sin 45° sin 42.5° cos 2.2° cos c = 0.99870 c = 2.93° π s = rc = 3960 2.93° ⋅ ≈ 202.5 180° x 282 = sin 48.5° sin 89.5° 282sin 48.5° x= ≈ 211.2 sin 89.5° y 282 = sin 42° sin 89.5° 282sin 42° y= ≈ 188.8 sin 89.5° Using a plane triangle, they traveled 211.2 + 188.7 = 399.9 miles. It is 202.5 miles from Great Falls to Lemhi. The mileage by using spherical triangles and that by using a plane triangle are relatively close. The total mileage is basically the same in each case. This is because compared to the surface of the Earth, these three towns are very close to each other and the surface can be approximated very closely by a plane. 862 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8 Projects Project III 2. f1 = sin(π t ) 1 f3 = sin(3π t ) 3 1 f5 = sin(5π t ) 5 1 f 7 = sin(7π t ) 7 1 f9 = sin(9π t ) 9 a. 184 .5 ft 6° 84° 13 ft b. Let h = the height of the tower with a lean of 6° . ft 6° 184.5 1. Project IV f1 = sin(π t ) h 84° 1 f1 + f3 = sin(π t ) + sin(3π t ) 3 1 1 f1 + f3 + f 5 = sin(π t ) + sin(3π t ) + sin(5π t ) 3 5 f1 + f3 + f5 + f 7 1 1 1 = sin(π t ) + sin(3π t ) + sin(5π t ) + sin(7π t ) 3 5 7 f1 + f3 + f5 + f 7 + f9 1 1 = sin(π t ) + sin(3π t ) + sin(5π t ) 3 5 1 1 + sin(7π t ) + sin(9π t ) 7 9 h = sin 84° 184.5 h = 184.5sin 84° = 183.5 feet 13 ft 2 ft 3 1 84.5 ft c. 13 ft − 16 in = 11 ft 8 in = 11 11 _23 ft 11.67 184.5 11.67 α = sin −1 ≈ 4° 184.5 d. sin α = 3. If one graphs each of these functions, one observes that with each iteration, the function becomes more square. e. 1 4. f1 + f13 + f3 = sin(π t ) + cos(2π t ) + sin(3π t ) 2 By adding in the cosine term, the curve does not become as flat. The waves at the “tops” and the “bottoms” become deeper. h = sin 86° 184.5 h = 184.5sin 86° ≈ 184.1 ft f. The angles are relatively small and for part (d), where 4° was acquired, it was arrived at by rounding. g. Answers will vary. 863 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 8: Applications of Trigonometric Functions Project V Project VI highest point a. a. Answers will vary. Mountain b. b. 50 yd d x d 410 = sin147.5° sin 22.5° 410sin147.5° d= sin 22.5° d = 575.7 feet 70 yd 40° Palm Tree sin α sin 40° = 70 50 70sin 40° sin α = ≈ 0.8999 50 70sin 40° α = sin −1 ≈ 64° 50 nc e ista d l α im a β− min l d= α 180−β β x L c. y = height 147.5° 32.5° 410’ α l 410 = sin10° sin 22.5° 410sin10° l= sin 22.5° l = 186.0 feet y = height d L = sin(180° − β ) sin( β − α ) L sin(180° − β ) d= sin( β − α ) c. α = 180° − 64° = 116° Mountain 50 yd nc e ista d l a 22.5° in im l =m 10° Rock l L = sin α sin( β − α ) L sin α l= sin( β − α ) Rock β 116° x fence home plate α 70 yd d. 40° Palm Tree The third angle (i.e., at the rock) is 24° . x 70 = sin 24° sin116° 70sin 24° x= ≈ 32 sin116° The treasure is about 32 yards away from the palm tree. d L = sin(180° − β ) sin( β − α ) L sin(180° − β ) d= sin( β − α ) L sin(180°) cos(− β ) + cos180° sin(− β ) d= sin( β − α ) L (0) cos( β ) + (−1) ( − sin( β ) ) sin( β − α ) L sin( β ) d= sin( β − α ) d= y l y = l sin β sin β = y= L sin α sin β sin( β − α ) 864 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall