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Trigonometry Identities

Determine whether each equation is true for all real numbers x . Explain your reasoning. 1. 2 x + 3 x = 5 x 2. –(4 x – 10) = 10 – 4 x 3. = 4 x 4. = x + 1. x 2 + 1 x – 1. 4 x 2 x. Trigonometry Identities. ALGEBRA 2 LESSON 14-1.

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Trigonometry Identities

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  1. Determine whether each equation is true for all real numbers x. Explain your reasoning. 1. 2x + 3x = 5x2. –(4x – 10) = 10 – 4x 3. = 4x4. = x + 1 x2 + 1 x – 1 4x2 x Trigonometry Identities ALGEBRA 2 LESSON 14-1 (For help, go to Lessons 1-1, 1-2, and 9-4.) 14-1

  2. x2 + 1 x – 1 4x2 x Trigonometry Identities ALGEBRA 2 LESSON 14-1 Solutions 1. 2x + 3x = (2 + 3)x Distributive Property = 5x Addition So 2x + 3x = 5x is true for all real numbers x. 2. –(4x – 10) = –4x + 10 Distributive Property = 10 – 4x Commutative Property So –(4x – 10) = –4x + 10 is true for all real numbers x. 3. = 4x is not defined for x = 0, and therefore not true for all real numbers x. 4. = x + 1 is not defined for x = 1, and therefore not true for all real numbers x. 14-1

  3. 1 + 2 sin cos = (sin + cos )2Rewrite the equation. = sin2 + 2 sin cos + cos2Multiply. = (sin2 + cos2 ) + 2 sin cos Rewrite the expression. = 1 + 2 sin cos Pythagorean identity Trigonometry Identities ALGEBRA 2 LESSON 14-1 Verify the identity (sin + cos )2 = 1 + 2sin cos . 14-1

  4. sin tan cos cot cos sin + = sin + cos Tangent identity sin cos cos sin sin • cos sin cos • sin cos + = Simplify. cos + sin = Simplify. sin + cos = Rewrite the expression. Trigonometry Identities ALGEBRA 2 LESSON 14-1 Verify the identity + = sin + cos . 14-1

  5. (1 + cot2 )(sec2 – 1) = csc2 • tan2Pythagorean identities sin2 cos2 1 sin2 = •    Reciprocal and Tangent identities 1 cos2 = Simplify. = sec2Reciprocal identity Trigonometry Identities ALGEBRA 2 LESSON 14-1 Simplify the trigonometric expression (1 + cot2 )(sec2 – 1). 14-1

  6. pages 766–768  Exercises 1. cos cot = cos = = – sin 2. sin cot = sin = cos 3. cos tan = cos = sin 4. sin sec = sin = = tan 5. cos sec = cos = 1 6. tan cot = = 1 7. sin csc = sin = = 1 8. cot = = • = ÷ = 9. 1 10. sin2 11. tan2 12. –cot2 13. csc 14. sin 15. cos 16. 1 17. sin 18. 1 19. 1 20. 1 cos sin 1 – sin2 sin 1 sin cos sin sin cos 1 cos sin cos 1 cos sin cos cos sin 1 sin sin sin cos sin 1 sin cos 1 1 sin 1 cos csc sec Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  7. 45. sin2 tan2 = sin2 = (1 – cos2 ) = = – = tan2 – sin2 46. sec – sin tan = – sin = – = = = cos 21. sec 22. 1 23. sec2 24. sec2 25. cot 26. csc 27. –tan2 28. tan 29. sec 30. csc 31. sin2 32. sin2 33. sin 34. sec 35. sec csc2 36. 1 37. 1 38. 1 39. ± 1 – cos2 40. 41. 42. ± 1 + cot2 43. ± csc2 – 1 44. ± 1 + tan2 sin2 cos2 sin2 cos2 sin2 – sin2 cos2 cos2 sin2 cos2 sin2 cos2 cos2 sin cos 1 cos ± 1 – cos2 cos 1 cos sin2 cos 1 – sin2 cos ± 1 – sin2 sin cos2 cos Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  8. 47. sin cos (tan + cot ) = sin cos = + = sin2 + cos2 = 1 48. = • = = = = 49. = • = = = sin 50. (cot + 1)2 = cot2 + 2 cot + 1 = cot2 + 1 + 2 cot = csc2 + 2 cot 51. 52. 1 – sin 53. Check students’ work. 54. When checking a root of an equation, you substitute the solution into the equation to see if it is true. When verifying an identity, you substitute equivalent expressions until both sides are the same. cos sin sin cos + 1 – sin2 sin2 sin2 cos cos cos2 sin sin 1 – sin cos 1 – sin cos cos cos (1 – sin )cos 1 – sin2 (1 – sin )cos cos2 (1 – sin )cos (1 – sin )(1 – sin ) cos 1 + sin 1 cos sec cot + tan sin cos sin cos cos sin sin cos + sin cos2 + sin2 sin 1 Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  9. 55. 1 + sec = 1 + = + 1 = 56. = + = cot + 1 57. + = + = = cos2 + sin2 = 1 58. sin2 tan2 + cos2 tan2 = tan2 (sin2 + cos2 ) 58.(continued) = tan2 (1) = tan2 = sec2 – 1 59. 1 60. csc2 61. If n2 > n1, then 1 > 2; if n2 < n1, then 1 < 2; if n2 = n1, then 1 = 2. 62. C 63. H 64. C 65. F 66. [2] (sec + 1)(sec – 1) = sec2 + sec – sec – 1 = sec2 – 1 = tan2 [1] does not show work 1 cos 1 cos 1 + cos cos 1 tan 1 + tan tan tan tan cotsin sec tancos csc sin cos cos sin cos sin 1 sin 1 cos cos 1 cos sin 1 sin + Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  10. 67.[2] Replace tan and sec with their equivalents. This results in . Combine the terms in the denominator by multiplying the first term by , so the terms have a common denominator. This leaves . Invert and multiply: • . Simplify and substitute – sin2 for cos2 – 1, which yields – . The final answer is – csc . [1] omits explanation OR includes minor error sin cos 1 cos cos – cos cos sin cos cos cos2 – 1 sin cos cos2 – 1 cos 1 sin Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  11. 70. 71. 72. 68.[4] Using the Pythagorean identities, 1 – sin2x is replaced by cos2x. This leaves = sec x. The next step is to cancel cos x, leaving = sec x, and since = sec x, then sec x = sec x, which proves the identity. [3] appropriate methods with minor error [2] correct steps without explanation [1] steps with a minor error and without explanation 69. cos x cos2x 1 cosx 1 cos x Trigonometry Identities ALGEBRA 2 LESSON 14-1 14-1

  12. Trigonometry Identities ALGEBRA 2 LESSON 14-1 78. 211° 79. 267° 80. 79° 81. 133° 82. 111° 83. 84. 73. 74. 75. 35° 76. 45° 77. 135° 14-1

  13. Answers may vary. Sample: = + = tan2 + 1 = 1 + tan2 = sec2 cot2 cot2 1 cot2 1 + cot2 cot2 – tan2 1 2 cos2 – 1 Trigonometry Identities ALGEBRA 2 LESSON 14-1 1 + cot2 cot2 1. Verify the identity = sec2 . 2. Simplify (1 – sec )(1 + sec ). 3. Express in terms of cos . 1 + tan2 1 – tan2 14-1

  14. For each function ƒ, find ƒ–1. 1. ƒ(x) = x + 1 2. ƒ(x) = 2x – 3 3. ƒ(x) = x2+ 4 Find each value. 4. sin 30° 5. cos 6. cos 135° 7. tan – 8. tan 315° 9. sin – 4 7 3 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 (For help, go to Lessons 7-7, 13-2, and 13-6.) 14-2

  15. Solutions 2. ƒ(x) = 2x – 3; rewrite as y = 2x – 3 Inverse: x = 2y – 3 x + 3 = 2y = y y = + ƒ–1(x) = + 1. ƒ(x) = x + 1; rewrite as y = x + 1 Inverse: x = y + 1 x – 1 = y ƒ–1(x) = x – 1 3. ƒ(x) = x2 + 4; rewrite as y = x2 + 4 Inverse: x = y2 + 4 x – 4 = y2 ± x – 4 = y ƒ–1(x) = ± x – 4, x 4 4. sin 30° = 0.5 5. cos = = 6. cos 135° = – 7. tan – = 0 8. tan 315° = –1 9. sin – = – > – 4 x + 3 2 3 2 2 2 2 2 3 2 x 2 x 2 3 2 7 3 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  16. Refer to the graph of the inverse of y = cos in Example 1. 2 2 2 2 2 3 2 The line x = 0 intersects the graph at (0, ) and (0, – ). 5 2 So the measures of two angles whose cosine is 0 are and – . Other points of intersection are (0, ), (0, ), and so on. The measures of all the angles whose cosine is 0 can be written as + n, where n is an integer. b. Find the radian measures of the angles whose cosine is –1.5. The line x = –1.5 does not intersect the graph. –1.5 is not in the domain of the inverse of y = cos . Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 a. Find the radian measures of the angles whose cosine is 0. There is no angle whose cosine is –1.5. 14-2

  17. Draw a unit circle and mark the points on the circle that have x-coordinates of . 2 2 2 2 2 2 2 2 2 2 These points and the origin form 45°–45°–90° triangles. 458 and 3158 are the measures of two angles whose cosine is . All their coterminal angles also have a cosine . The measure of all the angles whose cosine is can be written as 45° + n • 360° and 315° + n • 360°. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 Use a unit circle to find the degree measure of the angles whose cosine is . 14-2

  18. sin–1 (0.98) 1.37 Use a calculator. The angle is in Quadrant I. The sine function is also positive in Quadrant II, as shown in the figure at the right. So – 1.37 ≈ 1.77 is another solution. The radian measures of all the angles whose sine is 0.98 can be written as 1.37 + 2 n and 1.77 + 2 n. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 Use a calculator and an inverse function to find the radian measures of all the angles whose sine is 0.98. 14-2

  19. tan–1 1.34 0.93 Use a calculator. The tangent function is also positive in Quadrant III, as shown in the figure at the right. So 0.93 + is another solution. The radian measures of all angles whose tangent is 0.93 can be written as 0.93 = n. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 Use a calculator and an inverse function to find the measures in radians of all the angles whose tangent is 1.34. 14-2

  20. 7 cos – 3 = 2 cos 5 cos = 3 Add 3 – 2 cos to each side. 3 5 cos = Divide each side by 5. 3 5 cos–1 0.93 Use the inverse function to find one value for . The cosine function is also positive in Quadrant IV. So another value of is 2 – 0.93 5.36. The two solutions between 0 and 2 are approximately 0.93 and 5.36. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 Solve 7 cos – 3 = 2 cos for 0 < < 2 . 14-2

  21. 2 sin cos – cos = 0 cos (2 sin – 1) = 0 Factor. cos = 0 or 2 sin – 1 = 0 Use the Zero-Product Property. 1 2 cos = 0 sin = Solve for sin . 6 5 6 2 3 2 = and = and Use the unit circle. 6 2 5 6 3 2 The four values of are , , , and . Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 Solve 2 sin cos – cos = 0 for 0 < < 2 . 14-2

  22. d = 40 sin + 40 Substitute 60 for d. 60 = 40 sin + 40 20 = 40 sin (t – 6) 12 (t – 6) 12 (t – 6) 12 (t – 6) 12 (t – 6) 12 Subtract 40 from each side. = sin 1 2 Divide each side by 40 and simplify. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 The noise level d in decibels close to an electronic alarm device is modeled by d = 40 sin + 40, where t is the number of seconds after the device is activated. How many seconds does it take for the sound to reach a noise level of 60decibels? 14-2

  23. Use the inverse of the sine to solve for t. sin–1 = 1 2 0.5236 = Evaluate the inverse. 12 12 12 0.5236 = t – 6 Multiply each side by . (t – 6) 12 (t – 6) 12 0.5236 + 6 = tAdd 6 to each side. Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 (continued) 8 = tSimplify. The device will reach 60 decibels in 8 seconds. 14-2

  24. pages 773–776  Exercises 1. – + 2 n 2. 0 + n 3. + 2 n 4. 30° + n • 360° and 150° + n • 360° 5. 30° + n • 360° and 210° + n • 360°, or 30° + n • 180° 6. 210° + n • 360° and 330° + n • 360° 7. 120° + n • 360° and 300° + n • 360°, or 120° + n • 180° 8. 0.79 + 2 n and 3.93 + 2 n 9. 0.38 + 2 n and 2.76 + 2 n10. –0.89 + 2 n and 4.04 + 2 n 11. 1.89 + 2 n and 5.03 + 2 n, or 1.89 + n 12. 2.67 + 2 n and 3.62 + 2 n 13. no solution 14. 1.37 + 2 n and 4.51 + 2 n 15. 0.95 + 2 n and 5.33 + 2 n 16. , 17. , 18. , 19. , 20. 0.84, 5.44 21. 0.46, 3.61 11 6 5 6 5 4 3 4 2 6 6 4 4 2 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  25. 22. 0 23. 2.11, 5.25 24. no solution 25. , 26. , , 27. , 28. 0, , , 29. , , , 30. 0, , , 31. 0, 32. 0, 33. , 34.a.t = • cos–1 b. 1.16 s, 1.33 s, 1.54 s c. 3.16 s, 3.33 s, 3.54 s 35. 30° + n • 360° and 150° + n • 360° 36. 60° + n • 360° and 300° + n • 360° 37. 210° + n • 360° and 330° + n • 360° 38. , 39. 40. 0.10, 3.24 41. 0.34, 2.80 42. 3.04, 6.18 2 h –4 11 6 4 3 5 3 3 2 3 2 3 4 5 3 7 4 5 4 7 4 5 4 7 6 3 4 3 2 4 2 2 3 4 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  26. < < < < – – – – 43. 0 44. 0.0028 s, 0.019 s 45. a. 30° + n • 360° x 150° + n • 360° b. 150° + n • 360° x 390° + n • 360° c. Find the values of x where the graphs intersect, and then choose the appropriate interval. 46. 0 + 2 n, + 2 n, + 2 n 47. + 2 n, + 2 n 48. + 2 n, + 2 n, + 2 n 49. + n 50. + 2 n, + 2 n, + n 51. 0 + 2 n, + 2 n, 1.25 + 2 n, 4.39 + 2 n 52. + 2 n, + 2 n, + 2 n 53. + n, + n 54. + 2 n, + 2 n 55. 1.107 + n, + n 56. The student misinterpreted the meaning of cos–1 0.5 as being equal to . 57. The student divided both sides of the equation by sin , which in the given interval can be equal to zero. Since division by zero is not possible, this is where the error was. 2 3 4 3 1 cos 0.5 11 6 7 6 3 2 3 2 5 6 5 6 5 6 5 6 3 4 4 2 6 2 6 6 2 2 2 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  27. 58. 2.09 + 2 n, 4.19 + 2 n 59. 0.79 + n, 2.36 + n 60. 0 + n 61. 0.79 + n, 2.36 + n 62. 0 + n, 0.79 + n, 2.36 + n 63. 2.09 + 2 n, 4.19 + 2 n 64. Answers may vary. Sample: In the first equation, you isolate the variable x to get the solution. In the trigonometric equation, you first isolate the trig. part, sin , but then you must continue to solve for . 65. + 2 n 66.a. Answers may vary. Sample: cos = –1, 2 cos = –2, 3 cos = –3 b. Start with cos = –1, and then multiply both sides of the equation by any nonzero number. 67. sin–1 68. cos–1(y) 69. sin–1 – 2 70. cos–1 – 71. cos–1(y – 1) y 2 y 3 y 4 1 2 3 2 1 2 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  28. 80. [2] 2 cos = 2 cos = cos–1 = = and [1] finds solution in Quadrant I only 81.[4] 2 sin2 = –sin 2 sin2 + sin = 0 sin (2 sin + 1) = 0 sin = 0 2 sin + 1 = 0 sin = – sin–1 0 = 0, sin–1 – = , [3] appropriate methods, with minor error [2] answer only, without work shown [1] finds only one solution for each instance where it might equal 0 72. cos–1 73.a.K = 1250( – sin ) ft2 b. 2.08 74.a. 1:55 A.M., 11:05 A.M., and 2:55 P.M. b. 12:00 midnight to 1:55 A.M., 11:05 A.M. to 2:55 P.M. 75. D 76. H 77. C 78. G 79. D 1 y – 1 2 4 2 2 2 2 1 2 11 6 1 2 7 4 7 6 4 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  29. 82. cot 83. tan2 84. 1 85. 1 86. sin 87. tan 88.y = 4 cos 89.y = 3 cos 90. y = 3 cos 2 91. y = cos 92. 4 4 2 3 91 100 Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 14-2

  30. < – 3 2 0.72 + 2 n, 5.56 + 2 n 4 2 7 4 5 4 3 2 5 4 7 4 3 4 , , , , , , Solving Trigonometric Equations Using Inverses ALGEBRA 2 LESSON 14-2 1. Use a unit circle and 30°–60°–90° triangles to find the degree measures of the angles that have a sine of – . 2. Use a calculator and inverse functions to find the radian measures of the angles whose cosine is 0.75. Solve each equation for 0 < 2 . 3. 2 cos2 – 1 = 0 4. 2 sin cos + cos = 0 240° + n • 360°, 300° + n • 360° 14-2

  31. ABC is similar to RST. Complete the following proportions. 1. = 2. = 3. = 4. = a b a c c b c s t r r t Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 (For help, go to Skills Handbook page 844.) 14-3

  32. Solutions a b a c 1. = 2. = 3. = 4. = c b c a t s r s t r r t Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  33. A tourist visiting Washington D.C. is seated on the grass at point A and is looking up at the top of the Washington Monument. The angle of her line of sight with the ground is 27°. Given that sin 27° 0.45, cos 27° 0.89, and tan 27° 0.51, find her approximate distance AC from the base of the monument. height of monument distance from monument tan 27° = Definition of tan 555 AC tan 27° = Substitute. 555 AC 0.51 = Use a calculator in degree mode. 555 0.51 AC = 1088 Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 The distance of the visitor from the monument is approximately 1088 feet. 14-3

  34. 7 25 In PQR, R is a right angle and cos P = . Find sin P, tan P, and cos Q in fraction and in decimal form. Step 1: Draw a diagram. p r p r p q 24 25 length of leg opposite P length of the hypotenuse sin P = = = = 0.96 length of leg opposite P length of leg adjacent P 24 7 tan P = = = 3.43 24 25 length of leg adjacent Q length of the hypotenuse cos Q = = = = 0.96 Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 Step 2:  Use the Pythagorean Theorem to find p. r2 = p2 + q2 252 = p2 + 72 625 = p2 + 49 576 = p2 24 = p Step 3: Calculate the ratios. 14-3

  35. x 50 tan 42° = Definition of tan x = 50 tan 42° Solve for x. 45 Use a calculator in degree mode. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 A man 6 feet tall is standing 50 feet from a tree. When he looks at the top of the tree, the angle of elevation is 42°. Find the height of the tree to the nearest foot. In the right triangle, the length of the leg adjacent to the 42° angle is 50 ft. The length of the leg opposite the 42° angle is unknown. You need to find the length of the leg opposite the 42° angle. Use the tangent ratio. The height of the tree is approximately 45 + 6 or 51 ft. 14-3

  36. In KMN, N is a right angle, m = 7, and n = 25. Find m K to the nearest tenth of a degree. Step 2:  Use a cosine ratio. cos K = = 0.28 m K = cos–1 0.28 73.74° Use a calculator. Step 1: Draw a diagram. 7 25 Since K is acute, the other solutions of cos–1 0.28 do not apply. To the nearest tenth of a degree, m K is 73.7°. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  37. Let = the measure of the angle of the slope of the road. 800 6515 tan = 800 6515 = tan–1Use the inverse of the tangent function. 7.0 Use a calculator. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 A straight road that goes up a hill is 800 feet higher at the top than at the bottom. The horizontal distance covered is 6515 feet. To the nearest degree, what angle does the road make with level ground? You know the length of the leg opposite of the angle you need to find. You know the length of the hypotenuse. So, use the tangent ratio. The angle the road makes with level ground is approximately 7°. 14-3

  38. pages 782–785  Exercises 1.a. 8333 ft b. 8824 ft 2.a. 0.88 b. 2.13 c. 0.53 d. 2.13 e. 1.13 f. 0.53 3. a. 0.22 b. 0.98 c. 4.44 d. 4.56 e. 0.22 f. not defined 4. sin P = 0.92, cos P = 0.38, tan P = = 2.4, csc P = 1.08, sec P = = 2.6 15 17 12 13 17 8 9 41 5 13 8 15 40 41 12 5 17 8 40 9 13 12 17 15 41 9 13 5 8 15 9 41 Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  39. 9. 45.0° 10. 18.4° 11. 48.6° 12. 60.0° 13. 19.6° 14. 7.3° 15. 74.3° 16. 3.0° 17. 68.0° 18. a 8.7, m A 60.0°, m B 30.0° 19. c 7.8, m A 39.8°, m B 50.2° 20.a = 9, m A 36.9°, m B 53.1° 5. 41.8 6. 25.2 7. 10.6 8. a. 300 ft b. 445 ft c. Answers may vary. Sample: The flagpole must be straight, the ground must be flat, and the flagpole and the ground must be perpendicular. You assume these things so that the flagpole and the ground form a right triangle. By having a right triangle you can use its properties to find the missing parts. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  40. 26. sin = , tan = , csc = , sec = , cot = 27. sin = , tan = 2 6, csc = , sec = 5, cot = 21. c 10.2, m A 52.6°, m B 37.4° 22.a 8.0, m A 61.8°, m B 28.2° 23.b 14.0, m A 50.6°, m B 39.4° 24. a.m A = cos–1 b. 37° c. 53° 25. cos = , tan = , csc = , sec = , cot = 20 39 117 1200 d 20 7 55 3 55 8 2 6 5 5 6 12 8 55 55 7 39 117 3 39 20 3 39 7 3 55 55 8 3 6 12 Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  41. 30. sin = , cos = , tan = , csc = , cot = 31. sin = , cos = , tan = , csc = , sec = 28. sin = , cos = , csc = , sec = , cot = 29. sin = , cos = , tan = , sec = , cot = 24 25 32. cos = 0.937, tan = 0.374, csc = 2.857, sec = 1.068, cot = 2.676 33. sin = 0.192, cos = 0.981, tan = 0.196, sec = 1.019, cot = 5.103 34. a.d = b. 115.5 ft, 130.5 ft 35.a = 15, m A 61.9°, m B 28.1° 36.c 12.2, m A 35.0°, m B 55.0° 37.a 7.9, b 6.2, m B = 38° 38.a 3.9, c 6.9, m B = 55.8° 7 25 9 16 25 24 25 7 16 7 35 100 sin 7 24 41 4 33 7 41 4 33 4 5 7 16 5 7 9 9 7 35 4 7 4 33 33 7 33 33 4 41 41 5 41 41 4 5 Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  42. 39.a 26.8, c 28.1, m A = 72.8° 40.a 19.8, b 2.9, m A = 81.7° 41. 35.5° 42. 33.4 ft 43. 20.3 m2 44.a. 46.6 ft b. 136.2 ft 45.a. 12 b. Answers may vary. Sample: cos E = sin E = 0.385 mE = sin–1 0.385 22.6° EF = 13 cos 22.6° 12 46. Using inverse sine, you can find that = 30°. Since sine is positive in the first and second quadrants, another solution is 150°. All the solutions would be 30° + n • 360° and 150° + n • 360°. 47. Since APQ and ABC are similar triangles, = . So, cos = AQ = = = = cos A. EF 13 AQ AP AC AB 5 13 AQ 1 AQ AP AC AB Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  43. 51. cos2A + sin2A 1 + 1 b2 + a2c2 c2 = c2 52.a. 5.9 units b. 31.9 units2 c. 6.2 units2 53.a. 72° b. 19.0 cm 54. A 55. I 56. A 57. G 58. C 59.[2] tan A = m A tan–1 1.421 m A 54.9° tan B m B tan–1 0.7037 m B 35.1° [1] answer only, without work shown 60. 180° + n • 360° 61. no solution 62. 45° + n • 180° 63. 0° + n • 360° and 180° + n • 360° 48.a. 67 ft b. 3.4° 49. sec A = 50. tan A = 135 95 a c b c 1 cos A 1 b c c b 95 135 c b c b sin A cos A a c b c a b 2 2 a b a b Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 14-3

  44. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 64. 65. 66. 67. 68.P(0) 0.167772 P(1) 0.335544 P(2) 0.293601 P(3) 0.146801 P(4) 0.045875 P(5) 0.009175 P(6) 0.001147 P(7) 0.000082 P(8) 0.000003 14-3

  45. Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 69.P(0) 0.016796 P(1) 0.089580 P(2) 0.209019 P(3) 0.278692 P(4) 0.232243 P(5) 0.123863 P(6) 0.041288 P(7) 0.007864 P(8) 0.000655 70.P(0) 0.000003 P(1) 0.000082 P(2) 0.001147 P(3) 0.009175 P(4) 0.045875 P(5) 0.146801 P(6) 0.293601 P(7) 0.335544 P(8) 0.167772 14-3

  46. 4 5 4 3 4 5 , , 2 3 13 2 13 2 , , Right Triangles and Trigonometric Ratios ALGEBRA 2 LESSON 14-3 1. Find sin A, tan A, and cos B for the triangle shown to the right. 2. Find csc P, sec Q, and cot Q for the triangle shown to the right. 3. In KST, S is a right angle, s = 17, and t = 15. Find m T and m K to the nearest tenth of a degree. 4. A flagpole that is 35 feet tall casts a shadow 22 feet long at a certain time of the morning. What is the angle of elevation of the sun to the nearest degree? 61.9°, 28.1° 58° 14-3

  47. Simplify each expression. 1. 2. 3. 4. 5. 6. Find the area of a triangle with the given base b and height h. 7.b = 3 cm, h = 4 cm 8.b = 6 in., h = 15 in. 9.b = 5.2 mm, h = 12.6 mm 10. b = 6.17 ft, h = 3.25 ft sin 30° 6 sin 45° 4 sin 60° 10 sin 30° 12 sin 45° 8 sin 60° 9 Area and the Law of Sines ALGEBRA 2 LESSON 14-4 (For help, go to Lesson 13-4.) 14-4

  48. Solutions sin 30° 6 0.5 6 6 50 1 12 1 4 sin 45° 4 4 3 2 2 2 2 2 3 2 3 2 2 2 3 2 2 2 2 16 2 2 3 20 3 2 2 2 3 2 3 18 2 8 1 10 sin 60° 10 10 sin 30° 12 0.5 12 5 120 1 24 1 8 sin 45° 8 8 1 9 sin 60° 9 9 Area and the Law of Sines ALGEBRA 2 LESSON 14-4 1. = = = 2. = = ÷ 4 = • = 3. = = ÷ 10 = • = 4. = = = 5. = = ÷ 8 = • = 6. = = ÷ 9 = • = 14-4

  49. Solutions (continued) 1 2 1 2 7. A = bh = (3 cm)(4 cm) = 6 cm2 8.A = bh = (6 in.)(15 in.) = 45 in.2 9.A = bh = (5.2 mm)(12.6 mm) = 32.76 mm2 10. A = bh = (6.17 ft)(3.25 ft) 10.03 ft2 1 2 1 2 1 2 1 2 1 2 1 2 Area and the Law of Sines ALGEBRA 2 LESSON 14-4 14-4

  50. 1 2 The area K = bc sin A. h 40 b = 23, and since sin 25° = , h = 40 sin 25°. 1 2 So K = (23)(40 sin 25°) 194.4 cm2. Area and the Law of Sines ALGEBRA 2 LESSON 14-4 Find the area of the triangle shown below. 14-4

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