Real-World Right-Triangle Modeling � Elevation, Depression, Ramps & Shadows (HSF-TF.C.8)

1) A flagpole casts an 18 m shadow when the sun's elevation is 35°. Find the flagpole's height to the nearest tenth.

12.6 m

10.8 m

13.7 m

15.8 m

tan 35° = height / 18.

height = 18 · tan 35° ≈ 18 · 0.7002 ≈ 12.6 m.

Hence, height ≈ 12.6 m.

Explanation

tan 35° = height / 18.

height = 18 · tan 35° ≈ 18 · 0.7002 ≈ 12.6 m.

Hence, height ≈ 12.6 m.

2) A ladder reaches 14 ft up a wall at a 68° angle with the ground. Find the ladder’s length to the nearest tenth.

13.2 ft

15.1 ft

16.4 ft

17.8 ft

sin 68° = 14 / L.

L = 14 / sin 68° ≈ 14 / 0.9272 ≈ 15.1 ft.

Hence, L ≈ 15.1 ft.

Explanation

sin 68° = 14 / L.

L = 14 / sin 68° ≈ 14 / 0.9272 ≈ 15.1 ft.

Hence, L ≈ 15.1 ft.

3) A drone flies 250 m along a straight line at a 22° angle above horizontal. Find the vertical gain to the nearest tenth.

94.8 m

89.2 m

93.6 m

97.5 m

vertical = 250 · sin 22°.

vertical ≈ 250 · 0.3746 ≈ 93.6 m.

Hence, vertical gain ≈ 93.6 m.

Explanation

vertical = 250 · sin 22°.

vertical ≈ 250 · 0.3746 ≈ 93.6 m.

Hence, vertical gain ≈ 93.6 m.

4) A building is 48 m tall. From a point on level ground, the angle of elevation to the top is 27°. Find the horizontal distance to the base to the nearest tenth.

88.4 m

90.7 m

92.1 m

94.3 m

tan 27° = 48 / d.

d = 48 / tan 27° ≈ 48 / 0.5095 ≈ 94.3 m.

Hence, distance ≈ 94.3 m.

Explanation

tan 27° = 48 / d.

d = 48 / tan 27° ≈ 48 / 0.5095 ≈ 94.3 m.

Hence, distance ≈ 94.3 m.

5) A ramp has rise 0.9 m and must meet a maximum slope of 1:12. Find the minimum horizontal run.

10.8 m

7.2 m

9.6 m

12.9 m

rise : run = 1 : 12.

run = 12 × 0.9 = 10.8 m.

Hence, minimum run = 10.8 m.

Explanation

rise : run = 1 : 12.

run = 12 × 0.9 = 10.8 m.

Hence, minimum run = 10.8 m.

6) A guy-wire is attached to the top of a 32 ft pole and makes a 55° angle with the ground. Find the wire’s length to the nearest tenth.

36.8 ft

39.1 ft

41.2 ft

44.0 ft

sin 55° = 32 / L.

L = 32 / sin 55° ≈ 32 / 0.8192 ≈ 39.1 ft.

Hence, L ≈ 39.1 ft.

Explanation

sin 55° = 32 / L.

L = 32 / sin 55° ≈ 32 / 0.8192 ≈ 39.1 ft.

Hence, L ≈ 39.1 ft.

7) A staircase has rise 7.75 in and run 10.5 in per step. Find the stair angle with the horizontal to the nearest tenth of a degree.

34.8°

35.7°

36.5°

37.9°

tan θ = 7.75 / 10.5 ≈ 0.7381.

θ = arctan(0.7381) ≈ 36.5°.

Hence, θ ≈ 36.5°.

Explanation

tan θ = 7.75 / 10.5 ≈ 0.7381.

θ = arctan(0.7381) ≈ 36.5°.

Hence, θ ≈ 36.5°.

8) A kite has 80 m of taut string at a 50° angle with the ground. Approximate the horizontal distance from the flyer to the kite to the nearest tenth.

61.3 m

54.6 m

47.8 m

51.4 m

horizontal = 80 · cos 50°.

horizontal ≈ 80 · 0.6428 ≈ 51.4 m.

Hence, horizontal distance ≈ 51.4 m.

Explanation

horizontal = 80 · cos 50°.

horizontal ≈ 80 · 0.6428 ≈ 51.4 m.

Hence, horizontal distance ≈ 51.4 m.

9) A boat measures a 4° angle of elevation to the top of a 62 m lighthouse. Find the horizontal distance to the base to the nearest meter.

887 m

902 m

865 m

915 m

tan 4° = 62 / d.

d = 62 / tan 4° ≈ 62 / 0.06993 ≈ 887 m.

Hence, distance ≈ 887 m.

Explanation

tan 4° = 62 / d.

d = 62 / tan 4° ≈ 62 / 0.06993 ≈ 887 m.

Hence, distance ≈ 887 m.

10) From 30 m away, the angle of elevation to the top of a billboard is 25° and to the bottom is 7°. Find the billboard’s height to the nearest tenth.

9.7 m

10.3 m

11.1 m

12.0 m

Height = 30 · (tan 25° − tan 7°).

≈ 30 · (0.4663 − 0.1228) ≈ 30 · 0.3435 ≈ 10.3 m.

Hence, height ≈ 10.3 m.

Explanation

Height = 30 · (tan 25° − tan 7°).

≈ 30 · (0.4663 − 0.1228) ≈ 30 · 0.3435 ≈ 10.3 m.

Hence, height ≈ 10.3 m.

11) A ski slope drops 300 m over a slope length of 900 m. Find its angle with the horizontal to the nearest tenth of a degree.

18.6°

19.1°

19.5°

20.2°

sin θ = opposite / hypotenuse = 300 / 900 = 1/3.

θ = arcsin(1/3) ≈ 19.5°.

Hence, θ ≈ 19.5°.

Explanation

sin θ = opposite / hypotenuse = 300 / 900 = 1/3.

θ = arcsin(1/3) ≈ 19.5°.

Hence, θ ≈ 19.5°.

12) A security camera is mounted 6.0 m above level ground. The line of sight to a point on the ground has a 35° angle of depression. Find the horizontal distance to that point to the nearest tenth.

7.9 m

8.1 m

8.4 m

8.6 m

tan 35° = 6 / d.

d = 6 / tan 35° ≈ 6 / 0.7002 ≈ 8.6 m.

Hence, distance ≈ 8.6 m.

Explanation

tan 35° = 6 / d.

d = 6 / tan 35° ≈ 6 / 0.7002 ≈ 8.6 m.

Hence, distance ≈ 8.6 m.

13) A paraglider flies 1.20 km along a straight path at a descent angle of 12° relative to horizontal. Find the horizontal distance traveled to the nearest hundredth of a kilometer.

1.17 km

1.12 km

1.09 km

1.20 km

horizontal = 1.20 · cos 12°.

≈ 1.20 · 0.9781 ≈ 1.17 km.

Hence, horizontal distance ≈ 1.17 km.

Explanation

horizontal = 1.20 · cos 12°.

≈ 1.20 · 0.9781 ≈ 1.17 km.

Hence, horizontal distance ≈ 1.17 km.

14) A wind turbine is 80 m tall. Find the shadow length to the nearest tenth when the sun’s elevation is 28°.

144.2 m

150.4 m

158.9 m

165.1 m

tan 28° = 80 / shadow.

shadow = 80 / tan 28° ≈ 80 / 0.5317 ≈ 150.4 m.

Hence, shadow ≈ 150.4 m.

Explanation

tan 28° = 80 / shadow.

shadow = 80 / tan 28° ≈ 80 / 0.5317 ≈ 150.4 m.

Hence, shadow ≈ 150.4 m.

15) A bridge cable runs from a ground anchor to the top of a 45 m tower located 120 m away horizontally. Find the cable’s angle with the ground to the nearest tenth of a degree.

19.1°

20.1°

20.6°

21.2°

tan θ = opposite / adjacent = 45 / 120 = 0.375.

θ = arctan(0.375) ≈ 20.6°.

Hence, θ ≈ 20.6°.

Explanation

tan θ = opposite / adjacent = 45 / 120 = 0.375.

θ = arctan(0.375) ≈ 20.6°.

Hence, θ ≈ 20.6°.

16) A roof access ladder of length 18 ft reaches a vertical height of 17 ft. Find the ladder’s angle with the ground to the nearest tenth of a degree.

68.2°

69.9°

70.3°

70.7°

sin θ = 17 / 18 ≈ 0.9444.

θ = arcsin(17/18) ≈ 70.53°.

Closest option to 70.5° is 70.7°.

Hence, θ ≈ 70.7° (closest choice).

Explanation

sin θ = 17 / 18 ≈ 0.9444.

θ = arcsin(17/18) ≈ 70.53°.

Closest option to 70.5° is 70.7°.

Hence, θ ≈ 70.7° (closest choice).

17) A water tank’s top is 25.0 m above ground. An observer with eye level 1.6 m measures a 33° angle of elevation to the top. Find the horizontal distance to the tank to the nearest tenth.

36.0 m

34.6 m

37.8 m

39.1 m

Vertical difference = 25.0 − 1.6 = 23.4 m.

an 33° = 23.4 / d ⇒ d = 23.4 / tan 33°.

d ≈ 23.4 / 0.6494 ≈ 36.0 m.

Hence, distance ≈ 36.0 m.

Explanation

Vertical difference = 25.0 − 1.6 = 23.4 m.

an 33° = 23.4 / d ⇒ d = 23.4 / tan 33°.

d ≈ 23.4 / 0.6494 ≈ 36.0 m.

Hence, distance ≈ 36.0 m.

18) Two platforms are connected by a zipline. The lower platform is 18 m below the upper one and 95 m away horizontally. Find the zipline length to the nearest tenth.

93.2 m

96.7 m

98.4 m

101.0 m

Length = √(95² + 18²) = √(9025 + 324) = √9349 ≈ 96.7 m.

Hence, zipline ≈ 96.7 m.

Explanation

Length = √(95² + 18²) = √(9025 + 324) = √9349 ≈ 96.7 m.

Hence, zipline ≈ 96.7 m.

19) A crane boom 40 m long is raised to 72° above the ground. Find the vertical height of the boom's tip to the nearest tenth.

36.5 m

37.2 m

38.0 m

38.9 m

height = 40 · sin 72°.

≈ 40 · 0.9511 ≈ 38.0 m.

Hence, height ≈ 38.0 m.

Explanation

height = 40 · sin 72°.

≈ 40 · 0.9511 ≈ 38.0 m.

Hence, height ≈ 38.0 m.

20) From 120 m offshore, the angle of elevation to the top of a sailboat's mast is 14°. Find the mast's height to the nearest tenth.

28.4 m

29.3 m

31.1 m

30.0 m

height = 120 · tan 14°.

≈ 120 · 0.2493 ≈ 29.9 m ≈ 30.0 m.

Hence, mast height ≈ 30.0 m.

Explanation

height = 120 · tan 14°.

≈ 120 · 0.2493 ≈ 29.9 m ≈ 30.0 m.

Hence, mast height ≈ 30.0 m.

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