Model Real Scenarios with Angles of Depression � Construction, Navigation, Safety (HSG-SRT.C.8)

1) From the roof of a 42 m building, the angle of depression to a truck is 19°. What is the horizontal distance to the truck (nearest meter)?

116 m

128 m

122 m

132 m

x = 42 / tan 19° ≈ 121.99 ⇒ ≈ 122 m.

Explanation

x = 42 / tan 19° ≈ 121.99 ⇒ ≈ 122 m.

2) A drone flies at 320 ft altitude. The angle of depression to a landing pad is 14°. What is the line-of-sight distance to the pad (nearest foot)?

1,254 ft

1,323 ft

1,198 ft

1,406 ft

s = 320 / sin 14° ≈ 1322.2 ⇒ ≈ 1,323 ft.

Explanation

s = 320 / sin 14° ≈ 1322.2 ⇒ ≈ 1,323 ft.

3) A lookout on a 72 m cliff sees a boat at an angle of depression of 11°. Find the horizontal distance to the boat (nearest meter).

346 m

359 m

370 m

388 m

x = 72 / tan 11° ≈ 370.5 ⇒ ≈ 370 m.

Explanation

x = 72 / tan 11° ≈ 370.5 ⇒ ≈ 370 m.

4) A surveyor is 90 m from a tower. The angle of elevation to the top is 31°. Estimate the tower’s height (nearest meter).

49 m

54 m

57 m

60 m

height = 90 × tan 31° ≈ 54.1 ⇒ ≈ 54 m.

Explanation

height = 90 × tan 31° ≈ 54.1 ⇒ ≈ 54 m.

5) A hiker stands on a ledge 27 m above a lake. The angle of depression to a canoe is 22°. What is the line-of-sight distance to the canoe (nearest meter)?

67 m

70 m

72 m

75 m

s = 27 / sin 22° ≈ 72.1 ⇒ ≈ 72 m.

Explanation

s = 27 / sin 22° ≈ 72.1 ⇒ ≈ 72 m.

6) A sign that is 18 m tall casts a 40 m shadow. What is the sun’s angle of elevation (nearest degree)?

12°

20°

24°

26°

tan θ = 18 / 40 = 0.45 ⇒ θ ≈ 24°.

Explanation

tan θ = 18 / 40 = 0.45 ⇒ θ ≈ 24°.

7) A walkway rises 1.2 m over a horizontal run of 9.0 m. What is the angle of elevation of the walkway (nearest degree)?

3°

8°

11°

17°

tan θ = 1.2 / 9 = 0.1333 ⇒ θ ≈ 7.6° ⇒ 8°.

Explanation

tan θ = 1.2 / 9 = 0.1333 ⇒ θ ≈ 7.6° ⇒ 8°.

8) A person stands on a pier 7 m above the water. The angle of depression to a buoy is 29°. What is the horizontal distance to the buoy (nearest tenth of a meter)?

11.3 m

12.6 m

13.2 m

14.5 m

x = 7 / tan 29° ≈ 12.63 ⇒ ≈ 12.6 m.

Explanation

x = 7 / tan 29° ≈ 12.63 ⇒ ≈ 12.6 m.

9) A balloon is 150 m above a field. A marker on the field is 520 m away horizontally. What is the angle of depression from the balloon to the marker (nearest degree)?

14°

16°

18°

20°

tan θ = 150 / 520 ≈ 0.2885 ⇒ θ ≈ 16°.

Explanation

tan θ = 150 / 520 ≈ 0.2885 ⇒ θ ≈ 16°.

10) A lighthouse is 38 m tall. The angle of depression to a ship is 16°. How far is the ship from the base horizontally (nearest meter)?

125 m

133 m

141 m

149 m

x = 38 / tan 16° ≈ 132.5 ⇒ ≈ 133 m.

Explanation

x = 38 / tan 16° ≈ 132.5 ⇒ ≈ 133 m.

11) A camera mounted 12 ft high views a point on the ground at a 27° angle of depression. What is the line-of-sight distance to the point (nearest foot)?

24 ft

25 ft

26 ft

28 ft

s = 12 / sin 27° ≈ 26.4 ⇒ ≈ 26 ft.

Explanation

s = 12 / sin 27° ≈ 26.4 ⇒ ≈ 26 ft.

12) A crane operator is 24 m above ground level. The angle of depression to a pallet is 34°. Find the horizontal distance to the pallet (nearest meter).

33 m

36 m

38 m

41 m

x = 24 / tan 34° ≈ 35.6 ⇒ ≈ 36 m.

Explanation

x = 24 / tan 34° ≈ 35.6 ⇒ ≈ 36 m.

13) A paraglider is at an altitude of 410 m. The line-of-sight distance to a landing zone is 900 m. What is the angle of depression to the landing zone (nearest degree)?

15°

20°

22°

27°

sin θ = 410 / 900 ≈ 0.4556 ⇒ θ ≈ 27°.

Explanation

sin θ = 410 / 900 ≈ 0.4556 ⇒ θ ≈ 27°.

14) When the sun’s angle of elevation is 41°, a tree casts an 18 m shadow. How tall is the tree (nearest meter)?

14 m

16 m

18 m

20 m

height = 18 × tan 41° ≈ 15.6 ⇒ ≈ 16 m.

Explanation

height = 18 × tan 41° ≈ 15.6 ⇒ ≈ 16 m.

15) A security post is 14 ft high. The angle of depression to a package on the ground is 33°. What is the horizontal distance to the package (nearest foot)?

17 ft

19 ft

22 ft

25 ft

x = 14 / tan 33° ≈ 21.6 ⇒ ≈ 22 ft.

Explanation

x = 14 / tan 33° ≈ 21.6 ⇒ ≈ 22 ft.

16) A bridge is 32 m above the river. A boat is 220 m away horizontally. What is the angle of depression from the bridge to the boat (nearest degree)?

7°

8°

9°

10°

tan θ = 32 / 220 ≈ 0.1455 ⇒ θ ≈ 8.3° ⇒ 8°.

Explanation

tan θ = 32 / 220 ≈ 0.1455 ⇒ θ ≈ 8.3° ⇒ 8°.

17) A quadcopter is 95 m above the ground. The angle of depression to a target is 26°. What is the line-of-sight distance to the target (nearest meter)?

205 m

217 m

228 m

239 m

s = 95 / sin 26° ≈ 216.8 ⇒ ≈ 217 m.

Explanation

s = 95 / sin 26° ≈ 216.8 ⇒ ≈ 217 m.

18) A road descends at a constant 8° over a horizontal distance of 600 m. What is the change in elevation (nearest meter)?

79 m

82 m

84 m

88 m

drop = 600 × tan 8° ≈ 84.3 ⇒ ≈ 84 m.

Explanation

drop = 600 × tan 8° ≈ 84.3 ⇒ ≈ 84 m.

19) An observer on a dune 11 m high looks down at a flag with an angle of depression of 17°. What is the line-of-sight distance to the flag (nearest meter)?

35 m

38 m

40 m

42 m

s = 11 / sin 17° ≈ 37.6 ⇒ ≈ 38 m.

Explanation

s = 11 / sin 17° ≈ 37.6 ⇒ ≈ 38 m.

20) An observation deck is 55 m high. The angle of depression to a van in the parking lot is 13°. Find the horizontal distance to the van (nearest meter).

228 m

232 m

238 m

244 m

x = 55 / tan 13° ≈ 238.3 ⇒ ≈ 238 m.

Explanation

x = 55 / tan 13° ≈ 238.3 ⇒ ≈ 238 m.

Real-World Modeling with Angles of Depression/Elevation