wsk-PreCourse.qxd

CHAPTER 10 Chapter Opener Chapter Readiness Quiz (p. 536) 1. B; 20 2 21 2 29 2 2. G; 6090 400 441 841 30 841 841 3. B; 5 2 z 2 9 2 25 z 2 81 z 2 56 z 56 Lesson 10.1 7.5 10.1 Checkpoint (p. 538) 1. 27 5.2; This is reasonable because 27 is between the perfect squares 25 and 36. So, 27 should be between 25 and 36, or 5 and 6. The answer 5.2 is between 5 and 6. 2. 46 6.8; This is reasonable because 46 is between the perfect squares 36 and 49. So, 46 should be between 36 and 49, or 6 and 7. The answer 6.8 is between 6 and 7. 3. 8 2.8; This is reasonable because 8 is between the perfect squares 4 and 9. So, 8 should be between 4 and 9, or 2 and 3. The answer 2.8 is between 2 and 3. 4. 97 9.8; This is reasonable because 97 is between the perfect squares 81 and 100. So, 97 should be between 81 and 100, or 9 and 10. The answer 9.8 is between 9 and 10. 5. 3 5 3 5 6. 11 6 11 6 15 66 7. 3 27 3 27 8. 53 3 53 3 81 59 9 5 3 15 9. 20 4 5 10. 8 4 2 4 5 25 4 2 22 11. 75 25 3 12. 112 16 7 25 3 53 16 7 47 10.1 Guided Practice (p. 539) 1. The radicand in the epression 25 is 25. 2. D; 36 6 3. A; 3 2 3 2 6 4. B; 3 6 3 6 18 9 2 9 2 32 5. C; 32 16 2 16 2 42 6. a 2 b 2 c 2 2 2 4 2 c 2 4 16 c 2 20 c 2 20 c 4 5 c 4 5 c 25 c 7. c 25 4.5 8. 49 7 9. 28 4 7 10. 72 36 2 4 7 27 11. 54 9 6 9 6 36 36 2 62 10.1 Practice and Applications (pp. 539 541) 12. 13 3.6; This is reasonable because 13 is between the perfect squares 9 and 16. So, 13 should be between 9 and 16, or 3 and 4. The answer 3.6 is between 3 and 4. 13. 6 2.4; This is reasonable because 6 is between the perfect squares 4 and 9. So, 6 should be between 4 and 9, or 2 and 3. The answer 2.4 is between 2 and 3. 14. 91 9.5; This is reasonable because 91 is between the perfect squares 81 and 100. So, 91 should be between 81 and 100, or 9 and 10. The answer 9.5 is between 9 and 10. 15. 34 5.8; This is reasonable because 34 is between the perfect squares 25 and 36. So, 34 should be between 25 and 36, or 5 and 6. The answer 5.8 is between 5 and 6. 16. 106 10.3; This is reasonable because 106 is between the perfect squares 100 and 121. So, 106 should be between 100 and 121, or 10 and 11. The answer 10.3 is between 10 and 11. Geometry, Concepts and Skills 163

17. 148 12.2; This is reasonable because 148 is between the perfect squares 144 and 169. So, 148 should be between 144 and 169, or 12 and 13. The answer 12.2 is between 12 and 13. 18. 62 7.9; This is reasonable because 62 is between the perfect squares 49 and 64. So, 62 should be between 49 and 64, or 7 and 8. The answer 7.9 is between 7 and 8. 19. 186 13.6; This is reasonable because 186 is between the perfect squares 169 and 196. So, 186 should be between 169 and 196, or 13 and 14. The answer 13.6 is between 13 and 14. 20. a 2 b 2 c 2 2 2 5 2 c 2 2 5 c 2 7 c 2 7 c 21. a 2 b 2 c 2 13 2 6 2 c 2 13 36 c 2 49 c 2 49 c 7 c 22. a 2 b 2 c 2 23. 4 2 1 2 2 19 2 10 2 c 2 16 1 2 19 10 c 2 17 2 29 c 2 17 29 c 4.1 24. 2 11 2 6 2 25. 2 5 2 13 2 2 11 36 2 5 13 2 25 2 8 25 8 26. 7 2 7 2 5 2.8 14 27. 5 5 5 5 28. 3 11 3 11 25 5 33 29. 25 7 25 7 30. 10 43 43 10 235 430 31. 11 22 11 22 32. 65 2 65 65 242 6 6 5 5 121 2 36 5 121 2 180 112 33. 53 2 53 53 34. 72 2 72 72 5 5 3 3 7 7 2 2 25 3 49 2 75 98 35. 18 9 2 36. 50 25 2 9 2 25 2 32 52 37. 48 16 3 38. 60 4 15 16 3 4 15 43 215 39. 56 4 14 40. 125 25 5 4 14 25 5 214 55 41. 200 100 2 42. 162 81 2 100 2 81 2 102 92 43. 44 4 11 4 11 211 44. Yes, the epression can be simplified further. 80 4 20 4 20 220 24 5 24 5 2 2 5 45 45. No, the epression cannot be simplified further. 46. A lw 47. A lw 14 10 83 25 14 10 8 2 3 5 140 163 5 4 35 1615 4 35 62.0 235 11.8 48. A lw 49. A lw 92 46 43 43 9 4 2 6 4 4 3 3 362 6 16 3 3612 48 364 3 364 3 723 124.7 164 Geometry, Concepts and Skills

50. A lw 51. A lw 83 46 32 14 8 4 3 6 323 6 3218 329 2 329 2 32 3 2 32 14 328 34 7 34 7 3 2 7 67 962 15.9 135.8 52. A 1 4 s2 3 1 4 (30)2 3 1 4 (900)3 2253 389.7 ft 2 53. (1 m) 2 (100 cm) 2 10,000 cm; A 1 4 s2 3 10,000 1 4 s2 3 40,000 s 2 3 23,094 s 2 23,094 s 152 cm s 10.1 Standardized Test Practice (p. 541) 54. C; 169 13 55. H; 220 14.8, 14 14.8 15 56. D; a 2 b 2 c 2 210 2 8 2 c 2 40 64 c 2 104 c 2 104 c 4 26 c 4 26 c 226 c 10.2 c 10.1 Mied Review (p. 541) 57. 45ma1 90 ma1 45 58. 1806151ma1 180112ma1 68ma1 59. 1808725ma1 180112ma1 68ma1 60. 9 4 61. 644 62. 2 3 13 16 3 10.1 Algebra Skills (p. 541) 63. ( 5) 2 5 64. 4(2 1) 8 4 65. (3 4) 3 2 4 66. 5( 2) 5 2 10 67. 3(1 ) 3 3 68. 2 ( 6) 2 6 6 Lesson 10.2 10.2 Geo-Activity (p. 542) 2. 45, 45, 90 3. Answers may vary. 4. Answers may vary, but will be 2 1.4 times the answer in Step 3. 5. The answer should be the same as in Step 4. 10.2 Checkpoint (pp. 543 544) 1. 2 2. 2 42 52 3. 2 4. 2 32 2 62 2 3 2 2 2 6 2 2 2 2 2 3 6 5. The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, 90180. So, 290, and 45. So, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 8 2 8 2 5.7. 6. The triangle is an isosceles right triangle. By the Converse of the Base Angles Theorem its sides opposite the congruent angles are congruent. By the Triangle Sum Theorem, 90180. So, 290, and 45. So, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 12 2 12 2 8.5. Geometry, Concepts and Skills 165

10.2 Guided Practice (p. 545) 1. An isosceles right triangle has 2 congruent sides. 2. An isosceles right triangle has 2 congruent angles. The measures of the angles are 45, 45, and 90. 3. 2 62 4. 2 22 2 2 5. 2 6 2 12 4 3 23 10.2 Practice and Applications (pp. 545 547) 6. 2 7. 2 22 8. 2 82 72 9. 2 10. 2 5 2 10 3 2 6 11. 2 12. 2 10 2 20 4 5 25 102 2 10 13. 2 14. 2 42 2 52 2 4 5 15. 2 16. 2 2 2 82 2 1 8 17. 2 18. 2 142 2 (1.4) 2 14 2 cm 19. No; you cannot determine the measures of the other two angles. 20. Yes; the triangle has congruent acute angles, so 290 and 45. So, the triangle is a 45 45 90 triangle. 21. No; the triangle is isosceles, but there is no information about the angle measures. 22. The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, 90180. So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 4 2 4 2 2.8. 23. The right triangle has congruent acute angles. By the Triangle Sum Theorem, 90180. So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 9 2 9 2 6.4. 24. By the Triangle Sum Theorem, 4545180. So, 90180, and 90. Therefore the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 32 2 32 2 22.6. 25. The right triangle has congruent acute angles. By the Triangle Sum Theorem, 90180. So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 8 y2 8 y 2 5.7 y. 26. The triangle is an isosceles right triangle. By the Base Angles Theorem, its acute angles are congruent. By the Triangle Sum Theorem, 90180. So, 290, and 45. Since the measure of each acute angle is 45, the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 20 2 20 2 14.1. 166 Geometry, Concepts and Skills

27. By the Triangle Sum Theorem, 4590180. So, 135180, and 45. Therefore the triangle is a 45 45 90 triangle. By the 45 45 90 Triangle Theorem, 2 35 2 35 2 24.7. 28. T ADB, T ACD, and T BCD; since D is on the perpendicular bisecr of **** AB, **** AD DB **** by the Perpendicular Bisecr Theorem. mdb 90, so T ADB is an isosceles right triangle. By the Base Angles Theorem, ab. By the Triangle Sum Theorem, 90180. 2 90 45. Since m mab 45, T ADB is a 45 45 90 triangle. mcd 90, so by the Triangle Sum Theorem, mdc 180mCD m 1809045 45. So, T ACD is a 45 45 90 triangle. mabcd 90, so by the Triangle Sum Theorem, mabdc 180maBCD mab 1809045 45. So, T BCD is a 45 45 90 triangle. 29. Lengths may vary, but AC, CB, and CD are all equal. Since T ACD and T BCD are 45 45 90 triangles, they are isosceles triangles by the Converse of the Base Angles Theorem. Therefore, AC CD and CD BC. 30. Lengths may vary, but AD DB and each is equal 2 1.4 times the length in Eercise 29. 31. If the has length 52, the s should have length 5 2, or 5. If the s have length 5, then the 2 has length 5 2, or 10. 32. r 2 1 2 1 2 1 1 2 r 2; s 2 2 2 1 2 2 1 3 s 3; t 2 3 2 1 2 3 1 4 t 4 2; u 2 2 2 1 2 4 1 5 u 5; v 2 5 2 1 2 5 1 6 v 6; w 2 6 2 1 2 6 1 7 w 7 33. The right triangle with s of length 1 and of length r 2 is a 45 45 90 triangle. 10.2 Standardized Test Practice (p. 547) 34. a. 363180 12180 15; m 3(3 15)45; mab 3(3 15)45; mac 6(6 15)90 b. b 12; c 122 c. Check using the Pythagorean Theorem. 12 2 12 2 122 2 144 144 144 2 288 288 10.2 Mied Review (p. 547) 35. a 2 b 2 c 2 36. a 2 b 2 c 2 24 2 10 2 26 2 9 2 14 2 16 2 676 676 277 256 The triangle is right. The triangle is acute. 37. a 2 b 2 c 2 9 2 40 2 41 2 1681 1681 The triangle is right. 38. 24 4 6 39. 63 9 7 4 6 9 7 26 37 40. 52 4 13 41. 64 8 4 13 213 42. 80 16 5 43. 196 14 16 5 45 44. 250 25 10 45. 117 9 13 25 10 510 9 13 313 10.2 Algebra Skills (p. 547) 9 46. 1 0 0.9 47. 3 0.6 5 48. 2 3 0.6 0.67 49. 33 0.33 100 50. 4 9 0.4 0.44 51. 3 0.15 20 52. 4 7 50 53. 1 0.16 0.17 6 Geometry, Concepts and Skills 167

Lesson 10.3 10.3 Activity (p. 548) 1. 2. 3. AC AC AD CD A D 4. Lengths will vary, but AC 2 AD and CD 3 AD 1.73 AD. AC 5. The ratio is 2:1 and A D the ratio C D is approimately 1.73:1. AD 10.3 Checkpoint (pp. 550 551) 1. 2 shorter 2 7 14 2. longer shorter 3 33 3. longer shorter 3 103 4. longer shorter 3 6 3 6 3 3.5 5. 2 shorter 42 2 21 ; longer shorter 3 y 213 36.4 10.3 Guided Practice (p. 552) 1. Two special right triangles are 45 45 90 triangle and 30 60 90 triangle. 2. true 3. false 4. false 5. true 6. true 7. true 8. 2 shorter 2 5 10; longer shorter 3 y 53 CD A D 10 5 8.7 2 1.74 20 10 17.3 2 1.73 50 25 43.3 2 1.732 9. 2 shorter 8 2 h 4 h 10. 2 shorter 4 2 a 2 a; longer shorter 3 b 23 10.3 Practice and Applications (pp. 552 554) 11. 2 shorter 2 3 6 12. 2 shorter 2 8 16 13. longer shorter 3 113 shorter 3 11 shorter 2 shorter 2 11 22 14. 2 shorter 2 9 18 15. 2 shorter 2 1 2 16. longer shorter 3 123 shorter 3 12 shorter 2 shorter 2 12 24 17. longer shorter 3 43 18. longer shorter 3 63 19. longer shorter 3 133 20. longer shorter 3 203 168 Geometry, Concepts and Skills

21. longer shorter 3 163 22. longer shorter 3 23 3 29 2 3 6 23. longer shorter 3 4 3 4 3 2.3 24. longer shorter 3 7 3 7 3 4.0 25. longer shorter 3 18 3 18 3 10.4 26. longer shorter 3 10 3 10 3 5.8 27. longer shorter 3 143 3 14 28. longer shorter 3 12 3 12 3 6.9 29. 2 shorter 60 2 h 30 ft h 30. 2 shorter 10 2 5 ; longer shorter 3 y 53 31. 2 shorter 8 2 4 ; longer shorter 3 y 43 32. 2 shorter 12 2 6 ; longer shorter 3 y 63 33. 2 shorter 9 2 9 ; 2 longer shorter 3 y 9 2 3 34. 2 shorter 80 2 40 ; longer shorter 3 y 403 35. 2 shorter 15 2 1 5 ; 2 longer shorter 3 y 1 5 3 2 36. 2 shorter 18 2 h 9 in. h Your shoulders should be lifted a height of 9 inches. 37. Divide the triangle in two 30 60 90 triangles. The length of the shorter of each triangle is 15 feet. The length of the longer of each triangle is 153 feet, by the 30 60 90 Triangle Theorem. Area 1 2 bh 1 30 153 390 ft2 2 38. Divide the triangle in two 30 60 90 triangles. The length of the shorter of each triangle is 9 inches. The length of the longer of each triangle is 93 inches, by the 30 60 90 Triangle Theorem. Area 1 2 bh 1 18 93 140 in.2 2 Geometry, Concepts and Skills 169

39. Divide the triangle in two 30 60 90 triangles. The length of the shorter of each triangle is 3.5 centimeters. The length of the longer of each triangle is 3.53 centimeters, by the 30 60 90 Triangle Theorem. Area 1 2 bh 1 7 3.53 21 cm2 2 40. The area of the triangle is given by Area 1 2 bh. If an equilateral triangle with sides of length s is divided in two 30 60 90 triangles, then the length of the s base is s and the height is 3. Then 2 Area 1 2 bh 1 2 s s 1 3 2 4 s2 3. 41. 30 60 2 shorter 2 15 15 cm 30 cm; longer shorter 3 153 cm 42. Divide one of the si equilateral triangles in two 30 60 90 triangles. The length of the shorter of each triangle is 1 cm. The length of the longer of 2 each triangle is 1 2 3. 2 longer 2 1 2 3 3 1.73 cm 10.3 Standardized Test Practice (pp. 554 555) 43. C 44. F; 2 shorter 12 2 shorter 6 shorter longer shorter 3 63 perimeter 12 6 63 28.4 cm 10.3 Mied Review (p. 555) wins 45. 1 0 5 l osses 6 3 46. l osses wins 1 60 3 5 wins 10 47. 1 0 numbe r of games 10 6 16 5 8 losses 6 48. numbe r of games 10 6 1 66 3 8 49. Yes; by the AA Similarity Postulate, T ABC T FHG. 50. Yes; by the SAS Similarity Theorem, T PQR T STU. 10.3 Algebra Skills (p. 555) 51. 5 4 5(4) 4 20 4 16 52. 10 1 10(4) 1 40 1 41 53. 2 7 (4) 2 7 16 7 9 54. ( 3)( 3) (4 3)(4 3) (1)(7) 7 55. 2 2 1 2(4) 2 (4) 1 32 4 1 37 56. 5 2 2 3 5(4) 2 2(4) 3 80 8 3 69 Quiz 1 (p. 555) 1. 8 3 8 3 24 46 4 6 26 2. 2 15 215 30 3. 8 18 818 14 4 12 4. 80 5 80 5 40 0 20 5. 27 93 9 3 33 6. 17 6 16 11 16 11 411 7. 52 413 4 13 213 8. 18 0 36 5 36 5 65 9. longer shorter 3 2 3 2 3 10. 2 3 2 6 11. 2 shorter 28 2 14 ; longer shorter 3 Lesson 10.4 y 143 10.4 Activity (p. 556) 1. Triangle longer shorter 2. Each ratio of the lengths is 0.84. s horter longer A 5 cm 4.2 cm 0.84 B 10 cm 8.4 cm 0.84 C 15 cm 12.6 cm 0.84 D 20 cm 16.8 cm 0.84 3. Answers will vary, but the ratios in the fourth column will always be equal; the ratio of lengths depends on the measures of its angles, not the size of the right triangle. 10.4 Checkpoint (pp. 558 560) opposite as 1. tan S 6 le g adjacent as 8 3 0.75; 4 opposite ar tan R 8 le g adjacent ar 6 4 1.3333 3 170 Geometry, Concepts and Skills

opposite as 2. tan S 1 0 le g adjacent as 24 5 0.4167; 12 opposite ar tan R 2 4 le g adjacent ar 10 1 2 2.4 5 3. tan 35 0.7002 4. tan 85 11.4301 5. tan 10 0.1763 6. 904446 tan 44 8 and tan 46 8 7. 903753 tan 37 4 and tan 53 4 8. 905931 tan 59 5 and tan 31 5 9. tan 34 o pposite 10. tan 55 o pposite adjacent adjacent tan 34 7 tan 551 8 tan 347 tan 5518 7 18 tan 34 tan 55 10.4 12.6 11. tan 60 o pposite adjacent tan 60 2 0 20 tan 60 34.6 10.4 Guided Practice (p. 560) 1. The acute angles in T DEF are ad and ae. 2. The opposite ad is EF ****, and the adjacent ad is DF ****. 3. tan A o pposi adjace te nt 1 2 4 9 1.3333 3 4. tan A o pposi adjace te nt 5 5 1 5. tan A o pposite 8 1 0.5774 adjacent 8 3 3 6. tan 25 0.4663 7. tan 62 1.8807 8. tan 80 5.6713 9. tan 43 0.9325 10.4 Practice and Applications (pp. 560 562) 10. tan A o pposite adjacent 1 85 11. tan A o pposi adjace te nt 6 3 6 3 12. tan A o pposite 2 adjacent 3 13. tan P o pposite 7 0.2917; adjacent 2 4 tan R o pposi adjace te nt 2 4 3.4286 7 14. tan P o pposi adjace te nt 1 2 3 0.3429; 5 tan R o pposi adjace te nt 3 5 1 2.9167 2 15. tan P o pposi adjace te nt 2 0 1 5 4 1.3333; 3 tan R o pposi adjace te nt 1 5 2 0 3 0.75 4 16. tan 28 0.5317 17. tan 54 1.3764 18. tan 5 0.0875 19. tan 89 57.2900 20. tan 67 2.3559 21. tan 40 0.8391 22. tan 12 0.2126 23. tan 83 8.1443 24. The measure of the other acute angle is 905634. tan 56 9 tan 569 9 tan 56 6.1, tan 34 9 9 tan 34 6.1 25. The measure of the other acute angle is 903951. tan 39 3 3 tan 3933 33 tan 39 40.8, tan 51 3 3 33 tan 51 40.8 26. The measure of the other acute angle is 903555. tan 35 7 0 tan 3570 70 tan 35 100.0, tan 55 7 0 70 tan 55 100.0 Geometry, Concepts and Skills 171

27. tan 41 o pposite 28. tan 29 o pposite adjacent adjacent tan 41 5 2 12 tan 29 tan 4152 tan 2912 52 12 tan 41 tan 29 59.8 21.6 29. tan 53 o pposite adjacent tan 53 2 0 tan 5320 20 tan 53 15.1 30. The tangent ratio can only be used for any acute angle of a right triangle. This triangle is not a right triangle. 31. tan 13 o pposite adjacent h tan 13 58.2 58.2(tan 13) h 13.4 h The height h of the slide is about 13.4 meters. 32. tan 70 o pposite adjacent tan 70 3 3 tan 70 8.2 33. tan 49 o pposite adjacent tan 49 1 4 tan 4914 14 tan 49 12.2 34. tan 34 o pposite adjacent tan 34 1 0 10(tan 34) 6.7 35. tan 40 o pposite adjacent tan 40 3 5 35(tan 40) 29.4 36. tan 35 o pposite adjacent tan 35 2 0 20(tan 35) 14.0 37. tan 50 o pposite adjacent tan 50 oppo site 10 10(tan 50) opposite 11.9 opposite tan 45 o pposite adjacent tan 45 11.9 (tan 45) 11.9 11.9 ta n 45 11.9 38. tan 42 o pposite adjacent d tan 42 4 0 40(tan 42) d 36 d The distance d is about 36 meters. 10.4 Standardized Test Practice (p. 562) 39. C; tan 38 1 0 tan 3810 10 tan 38 40. F; tan 50 6 y 6 tan 50y 7.2 y 10.4 Mied Review (p. 562) 41. V πr 2 h 42. V lwh π(6) 2 (9) 5 4 8 π 36 9 160 ft 3 1018 m 3 43. V 1 3 πr2 h 1 3 π (7)2 10 513 in. 3 172 Geometry, Concepts and Skills

10.4 Algebra Skills (p. 562) 44. 8 10 3 45. 4( 3) 32 5 10 0 4 12 32 5 10 4 20 2 5 46. 3 7 11 47. 6 5 3 4 2 7 11 3 5 4 2 18 3 9 9 3 48. 2 4 22 49. 5 18 2 21 2 5 22 3 18 21 5 20 3 39 4 13 Lesson 10.5 10.5 Checkpoint (pp. 563 565) 1. sin A opposite 1 5 17 cos A a djacent 1 87 2. sin A oppos ite 4 2 5 ; cos A a djacent 7 2 5 3. sin A opposite 1 80 4 5 ; cos A a djacent 6 1 0 3 5 4. sin A oppos ite 0 4 0.9756; 1 cos A a djacent 9 0.2195 4 1 5. sin A oppos ite 2 2 0.7071; cos A a djace hypot nt e 2 2 0.7071 6. sin A oppos ite 39 0.7806; 8 cos A a djace hypot nt e 0.625 7. sin 43 0.6820 8. cos 43 0.7314 9. sin 15 0.2588 10. cos 15 0.9659 11. cos 72 0.3090 12. sin 72 0.9511 13. cos 900 14. sin 901 15. sin A opposite sin 34 a 7 7(sin 34) a 3.9 a; cos A a djacent cos 34 b 7 7(cos 34) b 5.8 b 16. sin A opposite a sin 65 1 2 12(sin 65) a 10.9 a; cos A a djacent b cos 65 1 2 12(cos 65) b 5.1 b 17. sin 43 opposite sin 43 a 5 5(sin 43) a 3.4 a; cos 43 a djacent cos 43 b 5 5(cos 43) b 3.7 b 10.5 Guided Practice (p. 566) 1. B; cos D a djacen t ad DE D F 2. C; sin D o pposite ad EF D F 3. A; tan D o pposite EF adjacent D E 4. The value of sin 37 is constant, so sin D sin A. The sine of an acute angle of a right triangle depends on the measure of the angle, not on the size of the triangle. 5. sin A oppos ite 4 5 0.8 6. cos A a djacent a A 3 0.6 5 Geometry, Concepts and Skills 173

7. tan A o pposi adjace te nt 4 3 1.3333 8. sin B oppos ite 3 5 0.6 9. cos B a djace hypot nt ab e 4 5 0.8 10. tan B o pposi adjace te nt 3 4 0.75 10.5 Practice and Applications (pp. 566 568) 11. sin A oppos ite 1 1 6 1 ; cos A a djace hypot nt e 6 0 6 1 12. sin A oppos ite 5 3 1 3 ; 0 2 cos A a djacent 1 50 1 2 13. sin A oppos ite 3 6 3 9 1 2 13 ; cos A a djace hypot nt e 1 5 3 9 5 13 14. sin P oppos ite ap 1 2 3 0.3243; 7 cos P a djace hypot nt ap e 3 5 3 0.9459 7 15. sin P oppos ite ap 6 2 1 0.7714; 1 cos P a djace hypot nt ap e 71 0.6364 1 16. sin P oppos ite ap 3 9 1 3 0.3333; cos P a djacent ap 6 2 2 2 0.9428 9 3 17. sin 40 0.6428 18. cos 23 0.9205 19. sin 80 0.9848 20. cos 5 0.9962 21. sin 59 0.8572 22. cos 61 0.4848 23. sin 901 24. cos 77 0.2250 25. sin 46 o pposite sin 46 8 8(sin 46) 5.8 ; cos 46 a djacent y cos 46 8 26. sin 54 o pposite sin 54 1 y4 14(sin 54) y 11.3 y; cos 54 a djacent cos 54 1 4 14(cos 54) 8.2 27. sin 29 o pposite sin 29 1 1 11(sin 29) 5.3 ; cos 29 a djacent cos 29 1 y1 11(cos 29) y 9.6 y 28. sin 24 o pposite sin 24 1 y6 16(sin 24) y 6.5 y; cos 24 a djacent cos 24 1 6 16(cos 24) 14.6 29. sin 37 o pposite sin 37 1 5 15(sin 37) 9.0 ; cos 37 a djacent cos 37 1 y5 15(cos 37) y 12.0 y 8(cos 46) y 5.6 y 174 Geometry, Concepts and Skills

30. sin 68 o pposite sin 68 2 6 26(sin 68) 24.1 ; cos 68 a djacent cos 68 2 y6 26(cos 68) y 9.7 y 31. sin 70 o pposite 15 h sin 70 1 h5 70 15(sin 70) h 14 h The ladder reaches about 14 feet up the wall. 32. 8 ft 1 2 in. 96 in. 1 ft sin 22 o pposite y sin 22 9 6 96(sin 22) y 36 in. y; cos 22 a djacent cos 22 9 6 96(cos 22) 89 in. 33. Answers may vary. 34. (sin A) 2 (cos A) 2 1 35. When you drag point C, (sin A) 2 (cos A) 2 1. 37. sin 42 3 r4 34(sin 42) r 22.8 r, cos 48 3 r4 34(cos 48) r 22.8 r Both students get the correct answer. 10.5 Standardized Test Practice (p. 568) 38. C; sin 25 8 (sin 25) 8 8 sin 25 39. H; the sine of an angle cannot be greater than 1. 10.5 Mied Review (p. 568) 40. longer shorter 3 133 41. 2 shorter y 2 7 14; longer shorter 3 73 42. 2 shorter 16 2 8 ; longer shorter 3 y 83 43. tan 32 0.6249 44. tan 88 28.6363 45. tan 56 1.4826 46. tan 24 0.4452 47. tan 17 0.3057 48. tan 49 1.1504 36. Sample answer: A c b C a B 10.5 Algebra Skills (p. 568) 49. 18, 8, 1.8, 0.8, 0, 0.08, 1.8 50. 2146, 2164, 2416, 2461, 2614, 2641 51. 0.61, 0.6, 0.56, 0.5, 0.47 In T ABC shown, sin A a c, cos A b c, and tan A a b. sin A Then c os A a c b c a c c ac b a tan A. bc b Lesson 10.6 10.6 Checkpoint (pp. 570 572) 1. m tan 1 3.5 74.1 2. m tan 1 2 63.4 3. m tan 1 0.4402 23.8 Geometry, Concepts and Skills 175

4. Since tan A 5 0.5556, 9 m tan 1 0.5556 29.1 5. Since tan A 1 7 0.85, 20 m tan 1 0.85 40.4 6. Since tan A 1 6 1.6, 10 m tan 1 1.6 58.0 7. m sin 1 0.5 30 8. m cos 1 0.92 23.1 9. m sin 1 0.1149 6.6 10. m cos 1 0.5 60 11. m sin 1 0.25 14.5 12. m cos 1 0.45 63.3 13. 3 2 4 2 2 25 2 25 5 ; tan A 3 0.75 4 m tan 1 0.75 36.9; mab 90m 9036.9 53.1 14. 4 2 y 2 6 2 16 y 2 36 y 2 20 y 20 4.5; cos E 4 0.6667 6 mae cos 1 0.6667 48.2; mad 90maE 9048.2 41.8 15. z 2 5 2 7 2 z 2 25 49 z 2 24 z 24 4.9; sin H 5 0.7143 7 mah sin 1 0.7143 45.6; mag 90maH 9045.6 44.4 10.6 Guided Practice (p. 572) 1. Solving a right triangle means finding the measures of both acute angles and the lengths of all three sides. 2. true 3. false 4. 1990 71 5. 4 2 5 2 2 6. 9 2 2 13 2 16 25 2 81 2 169 41 2 2 88 41 88 6.4 9.4 7. m tan 1 5.4472 79.6 8. m sin 1 0.8988 64.0 9. m cos 1 0.3846 67.4 10. max 906030; 2 shortest 4 2 2 ; longest shortest 3 y 23 3.5 11. d 2 12 2 14 2 d 2 144 196 d 2 52 d 52 7.2; cos D 1 2 0.8571 14 mad cos 1 0.8571 31.0; mae 90maD 9031 59.0 10.6 Practice and Applications (pp. 573 575) 12. m tan 1 0.5 26.6 13. m tan 1 1.0 45 14. m tan 1 2.5 68.2 15. m tan 1 0.2311 13.0 16. m tan 1 1.509 56.5 17. m tan 1 4.125 76.4 176 Geometry, Concepts and Skills

18. Use the Pythagorean Theorem; 48 2 55 2 (QS) 2 5329 (QS) 2 5329 QS 73 QS 19. Sample answer: Use the inverse tangent function; tan Q 5 5 1.1458 48 maq tan 1 1.1458 48.9 20. Sample answer: Use the fact that aq and as are complements; mas 90maQ 9048.9 41.1 21. 7 2 7 2 2 22. 6 2 2 2 2 49 49 2 36 4 2 98 2 40 2 98 40 72 210 9.9 ; 6.3 ; tan A 7 1 7 tan m tan 1 1 45 23. 20 2 21 2 2 400 441 2 841 2 841 29 ; tan A 2 0 0.9524 21 m tan 1 0.9524 43.6 24. m sin 1 0.75 48.6 25. m cos 1 0.1518 81.3 26. m sin 1 0.6 36.9 27. m cos 1 0.45 63.3 28. m cos 1 0.1123 83.6 29. m sin 1 0.6364 39.5 30. (2.5) 2 (horizontal distance) 2 20 2 A 2 6 1 3 6.25 (horizontal distance) 2 400 (horizontal distance) 2 393.75 horizontal distance 393.75 m tan 1 1 18.4 3 horizontal distance 19.8 ft sin(ramp angle) 2. 5 0.125 20 sin 1 0.125 7.2, so the measure of the ramp angle is about 7.2. This ramp meets the standards. 31. Sample answer: 28 ft (Any ramp greater than or equal 27.6 ft will meet the standards) 32. Answers may vary. altitude 33. tan(glide angle) distan ce runway tan(glide angle) 1 5.7 59 tan(glide angle) 0.2661 tan 1 0.2661 14.9, so the measure of the glide angle is about 14.9. 6 34. sin A 1 2 1 0.5 2 m sin 1 (0.5) 30 8 35. cos A 1 0 4 0.8 5 m cos 1 0.8 36.9 36. cos A 6 8 3 0.75 4 m cos 1 0.75 41.4 37. (LM) 2 16 2 17 2 (LM) 2 256 289 (LM) 2 33 LM 33 5.7; cosak 1 6 0.9412 17 mak cos 1 0.9412 19.7; mal 90maK 9019.7 70.3 38. (NQ) 2 4 2 14 2 (NQ) 2 16 196 (NQ) 2 180 NQ 180 13.4; 4 sin N 0.2857 1 4 man sin 1 0.2857 16.6; map 90maN 9016.6 73.4 Geometry, Concepts and Skills 177

39. (ST) 2 6 2 15 2 (ST) 2 36 225 (ST) 2 189 ST 189 13.7; 6 sin S 0.4 1 5 mas sin 1 0.4 23.6; mar 90maS 9023.6 66.4 40. tan 1 BC ; the diagram gives the lengths AB and BC, A B but does not give the length AC. 10.6 Standardized Test Practice (p. 574) 41. B 42. G 10.6 Mied Review (p. 575) 43. C 2πr 2π(8) 16π 50 cm; A πr 2 π(8) 2 64π 201 cm 2 44. C 2πr 2π(15) 30π 94 in.; A πr 2 π(15) 2 225π 707 in. 2 45. C πd π(34) 107 yd; A πr 2 π(17) 2 289π 908 yd 2 46. V 4 3 πr3 47. V 4 3 πr3 4 3 π(10)3 4 3 π(14)3 4189 ft 3 11,494 cm 3 4 3 πr3 48. V 2 4 3 π(5)3 2 25 0 π 3 262 in. 3 10.6 Algebra Skills (p. 575) 49. 0.36 0.194 0.554 50. $8.42 $2.95 $5.47 51. 7 4.65 32.55 52. 55.40 0.04 1385 53. 700 0.35 2000 54. $22.50 0.08 $1.80 Quiz 2 (p. 575) 1. tan 42 6 tan 426 6 tan 42 6.7 2. tan 63 2 2(tan 63) 3.9 y 3. sin 40 1 2 4. sin 21 19 12(sin 40) y 19(sin 21) 7.7 y; 6.8 ; tan 40 7.7 6.8 tan 21 y tan 407.7 y (tan 21)6.8 7. 7 6. 8 y tan 40 tan 21 9.2 y 17.7 5. tan 29 1 4 14(tan 29) 7.8 6. sin 35 8 8(sin 35) 4.6 ; tan 35 4.6 y y tan 354.6 4. 6 y tan 35 y 6.6 7. tan 72 3.0777 8. sin 52 0.7880 9. cos 36 0.8090 10. man 904050; m tan 40 1 6 16 tan 40m 13.4 m; 16 2 (13.4) 2 q 2 435.56 q 2 435.56 q 20.9 q 178 Geometry, Concepts and Skills

11. q 2 7 2 8 2 q 2 49 64 q 2 15 q 15 3.9; cos Q 7 0.875 8 maq cos 1 0.875 29.0; map 90maQ 9029.0 61.0 12. k 2 3 2 12.4 2 k 2 9 153.76 k 2 144.76 k 144.76 12.0; 3 sin L 0.2419 12.4 mal sin 1 0.2419 14.0; mak 90maL 9014.0 76.0 Chapter 10 Summary and Review (pp. 576 579) 1. A radical is an epression written with a radical symbol. 2. The number or epression inside the radical symbol is the radicand. 3. A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. 4. A right triangle with side lengths 9, 93, and 18 is a 30 60 90 triangle. 5. To solve a right triangle means determine that lengths of all three sides of the triangle and the measures of both acute angles. 6. A right triangle with side lengths 4, 4, and 42 is a 45 45 90 triangle. 7. If af is an acute angle of a right triangle, then opposite af tangent of af. le g adjacent af 8. If af is an acute angle of a right triangle, then cosine of af a djacent af. 9. If af is an acute angle of a right triangle, then sine of af opposite af. 10. 13 13 13 13 169 13 11. 2 72 2 72 144 12 12. 7 10 7 10 70 13. 47 2 47 47 4 4 7 7 16 7 112 14. 3 19 3 19 57 15. 5 5 5 5 25 5 16. 6 18 6 18 108 36 3 36 3 63 17. 311 2 311 311 3 3 11 11 9 11 99 18. 27 9 3 9 3 33 19. 72 36 2 36 2 62 20. 150 25 6 25 6 56 21. 68 4 17 4 17 217 22. 108 36 3 36 3 63 23. 80 16 5 16 5 45 24. 7500 2500 3 2500 3 503 25. 507 169 3 169 3 133 26. A lw 5225 5 2 2 5 1010 31.6 27. A lw 4633 4 3 6 3 1218 50.9 28. A lw 2210 22 10 220 8.9 29. 2 152 30. 2 31. 2 52 2 7 2 5 2 7 2 10 14 32. 2 33. 2 192 2 32 2 19 3 34. 2 10 2 10 2 7.1 35. 2 shorter 2 25 50; longer shorter 3 y 253 43.3 36. 2 shorter 2 19 38; longer shorter 3 y 193 32.9 Geometry, Concepts and Skills 179

37. longer shorter 3 943 3 94 ; 2 shorter y 2 94 y 188 38. longer shorter 3 45 3 45 3 26.0 ; 2 shorter y 2 45 3 90 y 3 52.0 opposite 39. tan A 1 6 le g adjacent 30 8 0.5333; 15 opposite ab tan B 3 0 le g adjacent ab 16 1 5 1.875 8 opposite 20 1 40. tan A 0.5774; le g adjacent 20 3 3 opposite ab tan B 20 3 3 1.7321 le g adjacent ab 20 opposite 41. tan A 1 2 3 1.5; le g adjacent 8 2 opposite ab tan B le g adjacent ab 1 82 2 0.6667 3 42. tan 17 0.3057 43. tan 81 6.3138 44. tan 36 0.7265 45. tan 24 0.4452 46. sin A oppos ite 1 5 2 5 3 0.6; 5 cos A a djace hypot nt e 2 0 2 5 4 0.8 5 47. sin A oppos ite 5 9 0.5556; cos A a djace hypot nt e 2 14 9 0.8315 48. sin A oppos ite 30 2 6 2 0.7071; 0 2 cos A a djace hypot nt e 30 2 6 0 2 2 0.7071 49. sin 57 0.8387 50. sin 12 0.2079 51. cos 31 0.8572 52. cos 75 0.2588 53. sin 31 1 0 10(sin 31) 5.2 ; cos 31 1 y0 10(cos 31) y 8.6 y 54. m tan 1 3.2145 72.7 55. m sin 1 0.0888 5.1 56. m cos 1 0.2243 77.0 57. m tan 1 1.2067 50.4 58. 3 2 6 2 c 2 45 c 2 45 c 6.7 c; tan A 3 0.5 6 m tan 1 0.5 26.6; mab 90m 9026.6 63.4 59. 14 2 23 2 q 2 725 q 2 725 q 26.9 q; tan R 1 4 0.6087 23 mar tan 1 0.6087 31.3; map 90maR 9031.3 58.7 60. maz 90maX 9055 35; sin 55 3 3 sin 55 3(0.8192) 2.5 ; cos 55 3 z 3 cos 55 z 3(0.5736) z 1.7 z 180 Geometry, Concepts and Skills

Chapter 10 Test (p. 580) 1. 15 7 15 7 105 2. 311 2 311 311 3 3 11 11 9 11 99 3. 23 2 25 2 2 4 3 4 5 2 12 20 2 32 2 32 16 2 42 4. D 5. A 6. B 7. C 8. 2 47 2 47 2 33.2 9. tan 38 3 10. y tan 70 8 3(tan 38) 8(tan 70) y 2.3 ; 22.0 y; cos 38 3 y cos 70 8 y cos 383 cos 708 3 8 y cos 38 cos 70 y 3.8 23.4 y 11. sin 35 1 0 12. sin 50 15 10(sin 35) y 15(sin 50) 5.7 y; 11.5 ; cos 35 1 0 cos 50 y 15 10(cos 35) 15 cos 50y 8.2 9.6 y 13. tan 70 2.7475 14. cos 14 0.9703 15. tan 31 0.6009 16. sin 26 0.4384 17. cos 30 0.8660 18. tan 451 19. sin 5 0.0872 20. tan 10 0.1763 21. m tan 1 5.2 79.1 22. m tan 1 7 81.9 23. m sin 1 0.3091 18.0 24. m sin 1 0.5318 32.1 25. m cos 1 0.6264 51.2 26. m cos 1 0.3751 68.0 27. mak 903060; 2 shortest 9 2 KL 9 KL 2 4.5 KL; longest shortest 3 JL 4 5 3 JL 7.8 28. maf 902565; 12 tan 25 D E DE(tan 25) 12 12 DE tan 25 DE 25.7; 12 sin 25 D F DF(sin 25) 12, 12 DF sin 25 DF 28.4 29. 4 2 QR 2 6 2 16 QR 2 36 QR 2 20 QR 20 QR 4.5; cos P 4 6 2 0.6667 3 map cos 1 0.6667 48.2; mar 90maP 9048.241.8 30. tan X 1 80 0.5625 320 max tan 1 0.5625 29.4 Chapter 10 Standardized Test (p. 581) 1. A; 124 4 31 4 31 231 2. J; 2 7 2 14 3. B; tan J o pposite 15 adjacent 1 12 4. J; sin 1 0.8170 54.8 5. C; longer shorter 3 PS 483 Geometry, Concepts and Skills 181

6 6. G; sin A 0.5455 1 1 m sin 1 0.5455 33.1 6 7. B; sin 32 P R PR sin 326 6 PR sin 32 11.3 8. J; tan 72 3 h0 30(tan72) h 92.3 h 9. D; tan 67 8 8(tan 67) 18.8, cos 67 8 y y cos 678 8 y cos 67 y 20.5 Chapter 10 Algebra Review (p. 583) 1. y 1 4 5y 7 Solve for y in Equation 1. y 1 y 1 Substitute for y in Equation 2. 4 5y 7 4 5(1 ) 7 4 5 5 7 5 7 2 2 Substitute for in the revised Equation 1. y 1 1 (2) 1 2 3 Check that (2, 3) is a solution. y 1 (2) 3 1 1 1 4 5y 7 4(2) 5(3) 7 8 15 7 7 7 The solution is (2, 3) 182 Geometry, Concepts and Skills 2. 2y 9 3 y 1 Solve for y in Equation 2. 3 y 1 3 1 y Substitute for y in Equation 1. 2(3 1) 9 6 2 9 7 7 1 Substitute for in revised Equation 2. y 3 1 3(1) 1 4 Check that (1, 4) is a solution. 1 2(4) 9 1 8 9 9 9 3(1) 4 1 3 4 1 1 1 The solution is (1, 4). 3. 3 y 3 7 2y 1 Solve for y in Equation 1. 3 y 3 y 3 3 Substitute for y in Equation 2 7 2(3 3) 1 7 6 6 1 6 1 5 Substitute for in revised Equation 1. y 3 3 3 3(5) 3 15 18 Check that (5, 18) is a solution. 3(5) 18 3 15 18 3 3 3 7(5) 2(18) 1 35 36 1 1 1 The solution is (5, 18).

4. y 4 y 16 Solve for y in Equation 2. y 16 y 16 Substitute for y in Equation 1. (16 ) 4 16 4 2 12 6 Substitute for in revised Equation 2. y 16 16 6 10 Check that (6, 10) is a solution. y 4 6 10 4 4 4 y 16 6 10 16 16 16 The solution is (6, 10). 5. y 1 2 y 4 Solve for y in Equation 1. y 1 y 1 Substitute for y in Equation 2. 2 y 4 2 (1 ) 4 2 1 4 3 3 1 Substitute for in revised Equation 1. y 1 1 1 2 Check that (1, 2) is a solution. y 1 1 2 1 1 1 2 y 4 2(1) 2 4 2 2 4 4 4 The solution is (1, 2). 6. 6 y 2 4 3y 6 Solve for y in Equation 1. 6 y 2 6 2 y Substitute for y in Equation 2. 4 3y 6 4 3(6 2) 6 4 18 6 6 22 0 0 Substitute for in revised Equation 1. y 6 2 6(0) 2 2 Check that (0, 2) is a solution. 6 y 2 6(0) (2) 2 0 2 2 2 2 4 3y 6 4(0) 3(2) 6 0 6 6 6 6 The solution is (0, 2). 7. 2 3y 5 4y 3 Solve for in Equation 2. 4y 3 3 4y Substitute for in Equation 1. 2 3y 5 2(3 4y) 3y 5 6 8y 3y 5 11y 11 y 1 Substitute for y in revised Equation 2. 3 4y 3 4(1) 1 Check that (1, 1) is a solution. 2 3y 5 2(1) 3(1) 5 2 3 5 5 5 4y 3 (1) 4(1) 3 1 4 3 3 3 The solution is (1, 1). Geometry, Concepts and Skills 183

8. 3 2y 5 3y 9 Solve for in Equation 2. 3y 9 3y 9 Substitute for in Equation 1. 3(3y 9) 2y 5 9y 27 2y 5 11y 22 y 2 Substitute for y in revised Equation 2. 3y 9 3(2) 9 6 9 3 Check that (3, 2) is a solution. 3 2y 5 3(3) 2(2) 5 9 4 5 5 5 3y 9 (3) 3(2) 9 3 6 9 9 9 The solution is (3, 2). 9. 5 2y 7 2 4y 22 Solve for in Equation 2. 2 4y 22 2 22 4y 11 2y Substitute for in Equation 1. 5 2y 7 5(11 2y) 2y 7 55 10y 2y 7 12y 48 y 4 Substitute for y in revised Equation 2. 11 2y 11 2(4) 11 8 3 Check that (3, 4) is a solution. 5 2y 7 5(3) 2(4) 7 15 8 7 7 7 2 4y 22 2(3) 4(4) 22 6 16 22 22 22 The solution is (3, 4). 184 Geometry, Concepts and Skills