Appendix A — Answer Key, Chapter 9: Right Triangle Trigonometry

SOL G.TR.4 (d, e, g) · Covers textbook Chapter 9 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 112 across the chapter.

Conventions used in every answer below. Name the angle first — opposite and adjacent mean nothing until you have. The hypotenuse is always opposite the right angle. Side lengths are given to the nearest hundredth and angle measures to the nearest degree, both computed from unrounded values. Angles are in degrees. A ratio is unitless.

Three ways to verify a ratio: against the calculator's value for that angle; against the complement, since $\sin \theta = \cos(90° - \theta)$; and against $\sin^2 \theta + \cos^2 \theta = 1$, which is the Pythagorean Theorem in ratio form.

Lesson 9.1 — Naming and Finding the Ratios

Guided practice

1. The side of length $5$, and no — the hypotenuse is opposite the right angle, which does not move when you change acute angles.
2. Relative to $\angle A$, the $3$ is opposite and the $4$ is adjacent.
3. Relative to $\angle B$ they swap: the $3$ is adjacent and the $4$ is opposite.
4. $\sin \theta = \tfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \tfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan \theta = \tfrac{\text{opposite}}{\text{adjacent}}$.
5. $\sin A = \tfrac{3}{5}$, $\cos A = \tfrac{4}{5}$, $\tan A = \tfrac{3}{4}$.
6. $\sin B = \tfrac{4}{5}$, $\cos B = \tfrac{3}{5}$, $\tan B = \tfrac{4}{3}$.

Independent practice

7. $\tfrac{8}{17}$.
8. $\tfrac{15}{17}$.
9. $\tfrac{8}{15}$.
10. $\tfrac{15}{17}$.
11. $\tfrac{8}{17}$.
12. $\tfrac{15}{8}$.
13. The hypotenuse is $13$, and the tangent of the angle opposite the $5$ is $\tfrac{5}{12}$.
14. The adjacent side is $\sqrt{25^2 - 7^2} = \sqrt{576} = 24$, so $\cos \theta = \tfrac{24}{25}$ and $\tan \theta = \tfrac{7}{24}$.
15. Opposite $3$ and adjacent $4$ give a hypotenuse of $5$, so $\sin \theta = \tfrac{3}{5}$ and $\cos \theta = \tfrac{4}{5}$.
16. Because it is a length divided by a length, and the units cancel. That is why the same $\sin 35°$ applies whether the triangle is measured in inches or in miles.
17. They wrote the cosine. $\sin A = \tfrac{\text{opposite}}{\text{hypotenuse}}$.
18. Because a ratio is a choice of two sides from three, and there are exactly three such choices. (Reversing a choice gives the reciprocal ratios — cosecant, secant, cotangent — which this standard does not include.)

Exit ticket 9.1

19. $\tfrac{9}{41}$.
20. $\tfrac{40}{41}$.
21. $\tfrac{9}{40}$.
22. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent — all named relative to the chosen acute angle.

Lesson 9.2 — Verifying the Ratios

Guided practice

23. The acute angle $\theta$, and a right angle.
24. AA. Two pairs of congruent angles make the triangles similar, similar triangles have proportional sides, so the ratio of any two corresponding sides is the same in all three.
25. $\tan \theta = 0.75$.
26. Reading two sides off the triangle, and taking the calculator's value for that angle.
27. That the sides were named against the wrong angle — most often opposite and adjacent swapped.
28. $\sin 37° \approx 0.6018$ and $\cos 53° \approx 0.6018$; they are equal.

Independent practice

29. $\sin A = \tfrac{6}{10} = \tfrac{3}{5}$. The triangle is twice a $3$-$4$-$5$, and scaling does not change a ratio.
30. $\tan A = \tfrac{9}{12} = \tfrac{3}{4}$.
31. $\cos 60° = 0.5$, because $\cos 60° = \sin(90° - 60°) = \sin 30°$.
32. $\sin 70° \approx 0.9397$.
33. $45°$.
34. $\theta = 45°$. From $\sin \theta = \cos(90° - \theta)$, equality forces $\theta = 90° - \theta$, so $2\theta = 90°$.
35. $\left(\tfrac{5}{13}\right)^2 + \left(\tfrac{12}{13}\right)^2 = \tfrac{25 + 144}{169} = \tfrac{169}{169} = 1$.
36. $\dfrac{\sin A}{\cos A} = \dfrac{3/5}{4/5} = \dfrac{3}{5} \cdot \dfrac{5}{4} = \dfrac{3}{4} = \tan A$.
37. Yes, to the nearest degree. $\sin^{-1}(0.6) \approx 36.87°$, which rounds to $37°$; and $\sin 37° \approx 0.6018$, which rounds to $0.6$. The two reports are the same measurement stated with different precision.
38. $\tan \theta = \tfrac{3}{12} = 0.25$, so $\theta = \tan^{-1}(0.25) \approx 14°$ (unrounded, $14.036°$).
39. False. $\sin 50° \approx 0.766$ and $\cos 50° \approx 0.643$, and their sum is about $1.409$. The identity they were reaching for squares each term: $\sin^2 \theta + \cos^2 \theta = 1$.
40. The two acute angles of a right triangle add to $90°$, so one is $\theta$ and the other is $90° - \theta$. The side opposite $\theta$ is the side adjacent to $90° - \theta$, and the hypotenuse is shared. So $\sin \theta$ and $\cos(90° - \theta)$ are the same fraction of the same two sides.

Exit ticket 9.2

41. $\cos 65° \approx 0.4226$.
42. $\tan A = \tfrac{16}{30} = \tfrac{8}{15}$. The triangle is twice an $8$-$15$-$17$.
43. $\sin \theta = 0.6$, since $\sin^2 \theta = 1 - 0.64 = 0.36$ and $\theta$ is acute. (It is a $3$-$4$-$5$ in disguise.)
44. Against the calculator's value for that angle, and against the complementary angle using $\sin \theta = \cos(90° - \theta)$. A third is $\sin^2 \theta + \cos^2 \theta = 1$.

Lesson 9.3 — Finding a Missing Side

Guided practice

45. Sine — it is the ratio that names the opposite side and the hypotenuse.
46. Cosine — it names the adjacent side and the hypotenuse.
47. $\sin 35° = \dfrac{x}{12}$.
48. $x = 12 \sin 35° \approx 6.88$.
49. $\cos 40° = \dfrac{9}{h}$.
50. $h = \dfrac{9}{\cos 40°} \approx 11.75$.

Independent practice

51. $20 \sin 28° \approx 9.39$.
52. $20 \cos 28° \approx 17.66$.
53. $14 \tan 52° \approx 17.92$.
54. $\dfrac{14}{\cos 52°} \approx 22.74$.
55. $\dfrac{30}{\sin 63°} \approx 33.67$.
56. $\dfrac{30}{\tan 63°} \approx 15.29$.
57. Opposite $= 8 \sin 15° \approx 2.07$; adjacent $= 8 \cos 15° \approx 7.73$.
58. $5 \tan 71° \approx 14.52$.
59. $24 \sin 11° \approx 4.58$ ft.
60. $\dfrac{40}{\sin 55°} \approx 48.83$ ft.
61. Sine names the opposite side and the hypotenuse, and neither of the two legs is the hypotenuse — so sine cannot relate them. With both legs involved the ratio is tangent.
62. Because the last step is whatever undoes the equation. If the unknown is in the numerator, as in $\sin \theta = \tfrac{x}{12}$, multiplying both sides by $12$ isolates it. If it is in the denominator, as in $\cos \theta = \tfrac{9}{h}$, the unknown must first be multiplied up and then divided out. The method is identical; only the algebra of the last step differs.
63. $120 \sin 48° \approx 89.18$ ft.
64. The student gets $12(0.57) = 6.84$; the correct value is $12 \sin 35° \approx 6.88$. The unrounded ratio must be carried into the multiplication and the answer rounded once, at the end.

Exit ticket 9.3

65. $15 \sin 33° \approx 8.17$.
66. $15 \tan 33° \approx 9.74$.
67. $\dfrac{15}{\sin 33°} \approx 27.54$.
68. Name the sides relative to the given angle; choose the ratio that mentions the side you have and the side you want; solve, rounding once at the end.

Lesson 9.4 — Finding a Missing Angle

Guided practice

69. The opposite side, $7$, and the adjacent side, $10$.
70. Tangent.
71. $\tan \theta = \dfrac{7}{10} = 0.7$.
72. $\theta = \tan^{-1}(0.7) \approx 35°$.
73. $34.992°$.
74. It takes a ratio back to the angle that has it. $\tan^{-1}(0.7)$ answers "which angle has a tangent of $0.7$?"

Independent practice

75. $\sin \theta = \tfrac{9}{15} = 0.6$, so $\theta \approx 37°$ (unrounded, $36.87°$).
76. $\cos \theta = \tfrac{8}{17}$, so $\theta \approx 62°$ (unrounded, $61.93°$).
77. $\tan \theta = \tfrac{5}{9}$, so $\theta \approx 29°$ (unrounded, $29.05°$).
78. $\sin \theta = \tfrac{11}{14}$, so $\theta \approx 52°$ (unrounded, $51.79°$).
79. $\tan \theta = \tfrac{21}{20} = 1.05$, so $\theta \approx 46°$ (unrounded, $46.40°$).
80. $\tan^{-1}\!\left(\tfrac{7}{24}\right) \approx 16°$ and $\tan^{-1}\!\left(\tfrac{24}{7}\right) \approx 74°$. The check: $16° + 74° = 90°$, as the two acute angles of a right triangle must.
81. $\tan \theta = \tfrac{2}{18}$, so $\theta \approx 6°$ (unrounded, $6.34°$).
82. $\sin \theta = \tfrac{23}{25} = 0.92$, so $\theta \approx 67°$ (unrounded, $66.93°$).
83. $\sin^{-1}$ expects a ratio of the opposite side to the hypotenuse, and two legs give neither. The two legs are the opposite and the adjacent, so the function is $\tan^{-1}$.
84. Because the three angles of a triangle add to $180°$ and one of them is the right angle, leaving $90°$ for the other two. Whichever ratio is used, the two computations are describing those same two angles.
85. $\tan \theta = \tfrac{40}{500} = 0.08$, so $\theta \approx 5°$ (unrounded, $4.574°$).
86. The calculator was in radian mode. $0.6435$ is the angle in radians; in degrees it is $36.87°$, which rounds to $37°$.

Exit ticket 9.4

87. $\sin \theta = 0.6$, so $\theta \approx 37°$.
88. $\tan \theta = 1$, so $\theta = 45°$.
89. $\cos \theta = \tfrac{12}{13}$, so $\theta \approx 23°$ (unrounded, $22.62°$).
90. Name the sides relative to the unknown angle; choose the ratio that names both known sides; apply the matching inverse function, and round to the nearest degree.

Lesson 9.5 — Angles of Elevation and Depression

Guided practice

91. Both are measured from a horizontal — elevation upward from it, depression downward from it.
92. The two horizontals are parallel and the line of sight is a transversal, so the two angles are alternate interior angles and therefore congruent (Chapter 2).
93. $\tan 62° = \tfrac{h}{50}$, so $h = 50 \tan 62° \approx 94.04$ ft.
94. The angle at the boat. It is the alternate interior angle to the $24°$ depression angle, and the two horizontals are parallel, so it also measures $24°$.
95. $\tan 24° = \tfrac{80}{d}$, so $d = \tfrac{80}{\tan 24°} \approx 179.68$ ft.
96. Because the definition says from a horizontal. Measuring from the cliff face would give the complement, $66°$, and every answer built on it would be wrong.

Independent practice

97. $80 \tan 55° \approx 114.25$ ft.
98. $120 \tan 38° \approx 93.75$ ft.
99. $\dfrac{60}{\tan 41°} \approx 69.02$ ft.
100. $\dfrac{150}{\tan 12°} \approx 705.69$ ft.
101. $\dfrac{200}{\tan 35°} \approx 285.63$ ft.
102. $\dfrac{30{,}000}{\tan 7°} \approx 244{,}330.39$ ft, which is about $46$ miles.
103. $200 \sin 55° \approx 163.83$ ft.
104. $45 \tan 62° \approx 84.63$ ft above eye level, and the eyes are $5$ ft up, so the tree is about $89.63$ ft.
105. $\tan 5° = \tfrac{3}{r}$, so $r = \tfrac{3}{\tan 5°} \approx 34.29$ ft.
106. $\tan 9° = \tfrac{120}{d}$, so $d = \tfrac{120}{\tan 9°} \approx 757.65$ ft.
107. The angle of depression is measured from the horizontal, not from the cliff face. Placed against the face it would be the complement, $66°$. Inside the triangle the $24°$ appears at the boat, as the alternate interior angle — and it is from there that the $80$ ft is opposite and $d$ is adjacent.
108. Because the angle of elevation from the lower point and the angle of depression from the upper point are the same angle — alternate interior angles across two parallel horizontals. The triangle is the same triangle either way, so the equation and the answer are the same.

Exit ticket 9.5

109. $100 \tan 40° \approx 83.91$ ft.
110. $\dfrac{90}{\tan 20°} \approx 247.27$ ft.
111. $\dfrac{50}{\tan 30°} \approx 86.60$ ft.
112. Both are measured from a horizontal — the angle of elevation upward from the horizontal at the observer, and the angle of depression downward from the horizontal at the higher point.

Chapter 9 Review — answers

Review 1 (G.TR.4d).

• $\sin A = \tfrac{20}{29}$, $\cos A = \tfrac{21}{29}$, $\tan A = \tfrac{20}{21}$.
• $\sin B = \tfrac{21}{29}$, $\cos B = \tfrac{20}{29}$, $\tan B = \tfrac{21}{20}$. So $\sin A = \cos B$ and $\cos A = \sin B$, and the two tangents are reciprocals.
• $\left(\tfrac{20}{29}\right)^2 + \left(\tfrac{21}{29}\right)^2 = \tfrac{400 + 441}{841} = \tfrac{841}{841} = 1$. The verification is the Pythagorean Theorem: the numerator is $20^2 + 21^2$ and the denominator is $29^2$, so the identity holds exactly when the triangle is right — and $20$, $21$, $29$ is a Pythagorean Triple.
• $m\angle A = \sin^{-1}\!\left(\tfrac{20}{29}\right) \approx 44°$ (unrounded, $43.60°$). Then $\sin 44° \approx 0.6947$ while $\tfrac{20}{29} \approx 0.6897$. They differ because the angle was rounded before the sine was taken: $44°$ is not the triangle's angle, it is that angle to the nearest degree. Taking the sine of the unrounded $43.60°$ returns $0.6897$ exactly.

Review 2 (G.TR.4e).

• Opposite the $37°$: $26 \sin 37° \approx 15.65$. Adjacent to it: $26 \cos 37° \approx 20.76$.
• $15.65^2 + 20.76^2 = 244.92 + 430.98 = 675.90$, against $26^2 = 676$. The check is off by about $0.1$ because both legs were rounded before being squared. Computing from unrounded values closes the gap exactly.
• The classmate gets $26(0.6) = 15.6$; the correct value is $15.65$. Yours is correct: the ratio is carried unrounded and the answer is rounded once, at the end.
• $\tan^{-1}\!\left(\tfrac{16}{21}\right) \approx 37°$ (unrounded, $37.30°$) and $\tan^{-1}\!\left(\tfrac{21}{16}\right) \approx 53°$ (unrounded, $52.70°$). The check: $37° + 53° = 90°$.

Review 3 (G.TR.4g).

• $\tan 47° = \tfrac{h}{140}$, so $h = 140 \tan 47° \approx 150.13$ ft.
• $47°$. The horizontal at the drone and the horizontal at the observation point are parallel, and the line of sight is a transversal, so the angle of depression and the angle of elevation are alternate interior angles — congruent.
• What does not change is the triangle: it still has a $140$ ft horizontal leg and a $47°$ angle, so its vertical leg is still about $150.13$ ft. What changes is what that leg measures — it is now the height above eye level, not above the ground. The drone is about $150.13 + 5 = 155.13$ ft above the ground.
• $140 \tan 60° \approx 242.49$ ft, so the climb is about $242.49 - 150.13 = 92.36$ ft. (Computed from unrounded values the climb is $92.36$ ft as well.)