Solve Angle of Depression Word Problems � Distances, Heights, and Line-of-Sight (HSG-SRT.C.8)

1) A cliff is 85 m high. An observer at the top sees a ship with an angle of depression of 12°. Find the horizontal distance (nearest meter).

406 m

363 m

328 m

399.7 m

Use the tangent ratio:

tan 12° = 85 / x

Solve for x:

x = 85 / tan 12° ≈ 85 / 0.2126 ≈ 399.7

Explanation

Use the tangent ratio:

tan 12° = 85 / x

Solve for x:

x = 85 / tan 12° ≈ 85 / 0.2126 ≈ 399.7

2) From the top of a 60 m lighthouse, the angle of depression to a boat is 25°. What is the horizontal distance from the base to the boat (nearest meter)?

140 m

128.7 m

112 m

224 m

Use the tangent ratio, with 60 m as the vertical side and x as the horizontal distance:

tan 25° = 60 / x

Solve for x by dividing 60 by tan 25°:

x = 60 / tan 25° ≈ 60 / 0.4663 ≈ 128.7

Explanation

Use the tangent ratio, with 60 m as the vertical side and x as the horizontal distance:

tan 25° = 60 / x

Solve for x by dividing 60 by tan 25°:

x = 60 / tan 25° ≈ 60 / 0.4663 ≈ 128.7

3) A drone is 150 ft above the ground. The angle of depression to a target is 35°. Find the horizontal distance to the target (nearest foot).

210 ft

107 ft

257 ft

214 ft

Use the tangent ratio, with 150 ft as the vertical side and x as the horizontal distance:

tan 35° = 150 / x

Solve for x:

x = 150 / tan 35° ≈ 150 / 0.7002 ≈ 214.2

Rounded to the nearest foot,

x ≈ 214 ft.

Explanation

Use the tangent ratio, with 150 ft as the vertical side and x as the horizontal distance:

tan 35° = 150 / x

Solve for x:

x = 150 / tan 35° ≈ 150 / 0.7002 ≈ 214.2

Rounded to the nearest foot,

x ≈ 214 ft.

4) A fire tower is 100 ft tall. The angle of depression to a campsite is 18°. Which equation correctly finds the horizontal distance d?

Tan(18°) = d / 100

Tan(18°) = 100 / d

Sin(18°) = d / 100

Cos(18°) = 100 / d

In the right triangle, the vertical side (opposite) is 100 ft and the horizontal side (adjacent) is d.

The tangent ratio is:

tan 18° = opposite / adjacent = 100 / d

So the correct equation is:

tan(18°) = 100 / d

Explanation

In the right triangle, the vertical side (opposite) is 100 ft and the horizontal side (adjacent) is d.

The tangent ratio is:

tan 18° = opposite / adjacent = 100 / d

So the correct equation is:

tan(18°) = 100 / d

5) From the top of a 30 m sea wall, the angle of depression to a buoy is 28°. Which expression gives the horizontal distance x?

X = 30 sin(28°)

X = 30 cos(28°)

X = 30 / tan(28°)

X = 30 tan(28°)

From the top of the wall (30 m high), the angle of depression is 28°, so we use tangent:

tan 28° = 30 / x

Solve for x:

x = 30 / tan 28°

Explanation

From the top of the wall (30 m high), the angle of depression is 28°, so we use tangent:

tan 28° = 30 / x

Solve for x:

x = 30 / tan 28°

6) From a building 45 m tall, the angle of depression to a car is 40°. What is the straight-line distance from the observer to the car (nearest meter)?

59 m

70 m

47 m

35 m

Let s be the straight-line (slant) distance from the observer to the car.

The vertical side is 45 m, angle at the observer is 40°.

Use sine:

sin 40° = 45 / s

Solve for s:

s = 45 / sin 40° ≈ 45 / 0.6428 ≈ 70.0

Hence, the line-of-sight distance is approximately 70 m.

Explanation

Let s be the straight-line (slant) distance from the observer to the car.

The vertical side is 45 m, angle at the observer is 40°.

Use sine:

sin 40° = 45 / s

Solve for s:

s = 45 / sin 40° ≈ 45 / 0.6428 ≈ 70.0

Hence, the line-of-sight distance is approximately 70 m.

7) A helicopter hovers at 900 m altitude. The angle of depression to a landing zone is 10°. What is the horizontal distance (nearest meter)?

5107 m

915 m

5286 m

5061 m

Use the tangent ratio, with 900 m vertical and x horizontal:

tan 10° = 900 / x

Solve for x:

x = 900 / tan 10° ≈ 900 / 0.1763 ≈ 5107

Compare with the choices.

The closest option is 5133 m.

Hence, the horizontal distance is approximately 5133 m.

Explanation

Use the tangent ratio, with 900 m vertical and x horizontal:

tan 10° = 900 / x

Solve for x:

x = 900 / tan 10° ≈ 900 / 0.1763 ≈ 5107

Compare with the choices.

The closest option is 5133 m.

Hence, the horizontal distance is approximately 5133 m.

8) A lifeguard stands on a 6 m tower. The angle of depression to a swimmer is 15°. What is the horizontal distance (nearest tenth of a meter)?

22.4 m

1.6 m

17.9 m

23.2 m

Use the tangent ratio: tan 15° = 6 / x

Solve for x: x = 6 / tan 15° ≈ 6 / 0.2679 ≈ 22.4

Hence, the swimmer is approximately 22.4 m from the tower.

Explanation

Use the tangent ratio: tan 15° = 6 / x

Solve for x: x = 6 / tan 15° ≈ 6 / 0.2679 ≈ 22.4

Hence, the swimmer is approximately 22.4 m from the tower.

9) A plane at 2800 ft sees a runway with an angle of depression of 6°. What is the straight-line distance to the runway (nearest foot)?

26,791 ft

26,847 ft

26,915 ft

26,056 ft

Let s be the straight-line distance to the runway. The vertical side is 2800 ft and the angle at the plane is 6°.

Use sine: sin 6° = 2800 / s

Solve for s:

s = 2800 / sin 6° ≈ 2800 / 0.10453 ≈ 26,791

Explanation

Let s be the straight-line distance to the runway. The vertical side is 2800 ft and the angle at the plane is 6°.

Use sine: sin 6° = 2800 / s

Solve for s:

s = 2800 / sin 6° ≈ 2800 / 0.10453 ≈ 26,791

10) A mountain lookout is 1250 m above the valley floor. The angle of depression to a trail marker is 4°. Find the horizontal distance (nearest meter).

17,872 m

17,897 m

17,859 m

17,916 m

Use tangent:

tan 4° = 1250 / x

Solve for x:

x = 1250 / tan 4° ≈ 1250 / 0.06993 ≈ 17,872

Explanation

Use tangent:

tan 4° = 1250 / x

Solve for x:

x = 1250 / tan 4° ≈ 1250 / 0.06993 ≈ 17,872

11) A camera 18 ft high views a point on the ground at a 32° angle of depression. What is the slant distance to the point (nearest foot)?

38 ft

19 ft

34 ft

45 ft

Let s be the slant distance from the camera to the ground point.

The vertical side is 18 ft, angle at the camera is 32°.

Use sine: sin 32° = 18 / s

Solve for s: s = 18 / sin 32° ≈ 18 / 0.5299 ≈ 34.0

Hence, the slant distance is approximately 34 ft.

Explanation

Let s be the slant distance from the camera to the ground point.

The vertical side is 18 ft, angle at the camera is 32°.

Use sine: sin 32° = 18 / s

Solve for s: s = 18 / sin 32° ≈ 18 / 0.5299 ≈ 34.0

Hence, the slant distance is approximately 34 ft.

12) A rescue team on a cliff 210 m high spots a boat at an angle of depression of 27°. What is the horizontal distance (nearest meter)?

412 m

451 m

399 m

427 m

The vertical drop is 210 m and the angle of depression is 27°.

Use tangent:

tan 27° = 210 / x

Solve for x:

x = 210 / tan 27° ≈ 210 / 0.5095 ≈ 412.3

Hence, the horizontal distance is approximately 412 m.

Explanation

The vertical drop is 210 m and the angle of depression is 27°.

Use tangent:

tan 27° = 210 / x

Solve for x:

x = 210 / tan 27° ≈ 210 / 0.5095 ≈ 412.3

Hence, the horizontal distance is approximately 412 m.

13) A kite is 120 ft above the ground. The string makes a 20° angle below the horizontal at the kite (angle of depression to the person). Find the string length (nearest foot).

315 ft

128 ft

321 ft

351 ft

Let L be the length of the kite string.

The vertical side is 120 ft, and the angle between the string and the horizontal is 20°.

Use sine:

sin 20° = 120 / L

Solve for L:

L = 120 / sin 20° ≈ 120 / 0.3420 ≈ 350.9

Rounded to the nearest foot:

L ≈ 351 ft.

Explanation

Let L be the length of the kite string.

The vertical side is 120 ft, and the angle between the string and the horizontal is 20°.

Use sine:

sin 20° = 120 / L

Solve for L:

L = 120 / sin 20° ≈ 120 / 0.3420 ≈ 350.9

Rounded to the nearest foot:

L ≈ 351 ft.

14) From a 50 m crane, the operator looks down at an angle of depression of 37° to a crate. What is the horizontal distance (nearest meter)?

40 m

67 m

66 m

53 m

The crane is 50 m tall, and the angle of depression is 37°.

Use tangent:

tan 37° = 50 / x

Solve for x:

x = 50 / tan 37° ≈ 50 / 0.7536 ≈ 66.4

Hence, the horizontal distance is approximately 66 m.

Explanation

The crane is 50 m tall, and the angle of depression is 37°.

Use tangent:

tan 37° = 50 / x

Solve for x:

x = 50 / tan 37° ≈ 50 / 0.7536 ≈ 66.4

Hence, the horizontal distance is approximately 66 m.

15) A lighthouse 75 ft tall has two ships A and B on the same side. Angle of depression to A is 12°, to B is 7°. How much farther (horizontally) is B than A?

384 ft

319 ft

307 ft

258 ft

Let d_A be the horizontal distance to ship A and d_B to ship B.

For ship A (angle 12°):

d_A = 75 / tan 12° ≈ 75 / 0.2126 ≈ 352.9

For ship B (angle 7°):

d_B = 75 / tan 7° ≈ 75 / 0.1228 ≈ 610.7

Difference in distances:

Δd = d_B − d_A ≈ 610.7 − 352.9 ≈ 257.8

Rounded to the nearest foot,

Δd ≈ 258 ft farther.

Explanation

Let d_A be the horizontal distance to ship A and d_B to ship B.

For ship A (angle 12°):

d_A = 75 / tan 12° ≈ 75 / 0.2126 ≈ 352.9

For ship B (angle 7°):

d_B = 75 / tan 7° ≈ 75 / 0.1228 ≈ 610.7

Difference in distances:

Δd = d_B − d_A ≈ 610.7 − 352.9 ≈ 257.8

Rounded to the nearest foot,

Δd ≈ 258 ft farther.

16) An observer is standing on a 20 m platform and looks down at an angle of depression of 30° to a point on the ground. Which formula correctly gives the line-of-sight distance s?

S = 20 / cos(30°)

S = 20 / sin(30°)

S = 20 tan(30°)

S = 20 sin(30°)

Let s be the line-of-sight distance.

The platform is 20 m high and the angle of depression is 30°.

Use sine at the observer:

sin 30° = opposite / hypotenuse = 20 / s

Solve for s:

s = 20 / sin 30° = 20 / 0.5 = 40

So the correct formula is:

s = 20 / sin(30°)

Explanation

Let s be the line-of-sight distance.

The platform is 20 m high and the angle of depression is 30°.

Use sine at the observer:

sin 30° = opposite / hypotenuse = 20 / s

Solve for s:

s = 20 / sin 30° = 20 / 0.5 = 40

So the correct formula is:

s = 20 / sin(30°)

17) A bridge deck is 14 m above water. A boat is seen at a 10° angle of depression. What is the horizontal distance (nearest meter)?

80 m

79 m

81 m

82 m

Use the tangent ratio:

tan 10° = 14 / x

Solve for x:

x = 14 / tan 10° ≈ 14 / 0.1763 ≈ 79.4

Hence, the horizontal distance is approximately 79 m.

Explanation

Use the tangent ratio:

tan 10° = 14 / x

Solve for x:

x = 14 / tan 10° ≈ 14 / 0.1763 ≈ 79.4

Hence, the horizontal distance is approximately 79 m.

18) An airplane descends along a 3° glide path. If it is 1.5 miles horizontally from the runway threshold, what is its altitude drop (nearest foot)?

415 ft

277 ft

414 ft

138 ft

First convert the horizontal distance to feet:

1.5 miles = 1.5 × 5280 = 7920 ft

Use tangent for the glide path:

Drop = 7920 × tan 3°

Compute:

tan 3° ≈ 0.05241

Drop ≈ 7920 × 0.05241 ≈ 414.9

Rounded to the nearest foot:

Drop ≈ 415 ft.

Explanation

First convert the horizontal distance to feet:

1.5 miles = 1.5 × 5280 = 7920 ft

Use tangent for the glide path:

Drop = 7920 × tan 3°

Compute:

tan 3° ≈ 0.05241

Drop ≈ 7920 × 0.05241 ≈ 414.9

Rounded to the nearest foot:

Drop ≈ 415 ft.

19) A hiker stands on a 90 ft cliff. The angle of depression to a raft is 33°. What is the straight-line distance to the raft (nearest foot)?

170 ft

164 ft

165 ft

168 ft

Let s be the slant distance to the raft.

The vertical side is 90 ft, angle of depression is 33°.

Use sine:

sin 33° = 90 / s

Solve for s:

s = 90 / sin 33° ≈ 90 / 0.5446 ≈ 165.3

Hence, the straight-line distance is approximately 165 ft.

Explanation

Let s be the slant distance to the raft.

The vertical side is 90 ft, angle of depression is 33°.

Use sine:

sin 33° = 90 / s

Solve for s:

s = 90 / sin 33° ≈ 90 / 0.5446 ≈ 165.3

Hence, the straight-line distance is approximately 165 ft.

20) A tower casts a shadow on level ground. From the top, the angle of depression to the tip of the shadow is 41°. If the shadow length is 28 m, how tall is the tower (nearest meter)?

25 m

24 m

27 m

26 m

The shadow on the ground is 28 m, and the angle of depression from the top gives the angle at the top between vertical and line of sight.

But for height vs shadow, we use tangent at the ground angle (same acute angle).

Use tangent: tan 41° = height / 28

Solve for height: height = 28 × tan 41° ≈ 28 × 0.8693 ≈ 24.3

Rounded to the nearest meter:

height ≈ 24 m.

Explanation

The shadow on the ground is 28 m, and the angle of depression from the top gives the angle at the top between vertical and line of sight.

But for height vs shadow, we use tangent at the ground angle (same acute angle).

Use tangent: tan 41° = height / 28

Solve for height: height = 28 × tan 41° ≈ 28 × 0.8693 ≈ 24.3

Rounded to the nearest meter:

height ≈ 24 m.

Angle of Depression Problems: Calculate Horizontal, Vertical & Slant Distances

1) A cliff is 85 m high. An observer at the top sees a ship with an angle of depression of 12°. Find the horizontal distance (nearest meter).

2)

What first name or nickname would you like us to use?

2) From the top of a 60 m lighthouse, the angle of depression to a boat is 25°. What is the horizontal distance from the base to the boat (nearest meter)?

3) A drone is 150 ft above the ground. The angle of depression to a target is 35°. Find the horizontal distance to the target (nearest foot).

4) A fire tower is 100 ft tall. The angle of depression to a campsite is 18°. Which equation correctly finds the horizontal distance d?

5) From the top of a 30 m sea wall, the angle of depression to a buoy is 28°. Which expression gives the horizontal distance x?

6) From a building 45 m tall, the angle of depression to a car is 40°. What is the straight-line distance from the observer to the car (nearest meter)?

7) A helicopter hovers at 900 m altitude. The angle of depression to a landing zone is 10°. What is the horizontal distance (nearest meter)?

8) A lifeguard stands on a 6 m tower. The angle of depression to a swimmer is 15°. What is the horizontal distance (nearest tenth of a meter)?

9) A plane at 2800 ft sees a runway with an angle of depression of 6°. What is the straight-line distance to the runway (nearest foot)?

10) A mountain lookout is 1250 m above the valley floor. The angle of depression to a trail marker is 4°. Find the horizontal distance (nearest meter).

11) A camera 18 ft high views a point on the ground at a 32° angle of depression. What is the slant distance to the point (nearest foot)?

12) A rescue team on a cliff 210 m high spots a boat at an angle of depression of 27°. What is the horizontal distance (nearest meter)?

13) A kite is 120 ft above the ground. The string makes a 20° angle below the horizontal at the kite (angle of depression to the person). Find the string length (nearest foot).

14) From a 50 m crane, the operator looks down at an angle of depression of 37° to a crate. What is the horizontal distance (nearest meter)?

15) A lighthouse 75 ft tall has two ships A and B on the same side. Angle of depression to A is 12°, to B is 7°. How much farther (horizontally) is B than A?

16) An observer is standing on a 20 m platform and looks down at an angle of depression of 30° to a point on the ground. Which formula correctly gives the line-of-sight distance s?

17) A bridge deck is 14 m above water. A boat is seen at a 10° angle of depression. What is the horizontal distance (nearest meter)?

18) An airplane descends along a 3° glide path. If it is 1.5 miles horizontally from the runway threshold, what is its altitude drop (nearest foot)?

19) A hiker stands on a 90 ft cliff. The angle of depression to a raft is 33°. What is the straight-line distance to the raft (nearest foot)?

20) A tower casts a shadow on level ground. From the top, the angle of depression to the tip of the shadow is 41°. If the shadow length is 28 m, how tall is the tower (nearest meter)?