Real-World Modeling With Inverse Tangent

1) A ship is 500 m from a 100 m cliff. Find the angle of elevation to the top of the cliff (nearest tenth).

9.5°

10.8°

11.3°

12.0°

tan(θ) = opposite/adjacent = 100/500 = 0.2.

Now, θ = arctan(0.2).

Using a calculator, θ ≈ 11.31°, which rounds to 11.3°.

Explanation

tan(θ) = opposite/adjacent = 100/500 = 0.2.

Now, θ = arctan(0.2).

Using a calculator, θ ≈ 11.31°, which rounds to 11.3°.

2) A camera is 60 m above a target and 130 m horizontally away. Find the angle of depression (nearest tenth).

22.9°

24.8°

26.4°

28.1°

The angle of depression equals the angle below the horizontal line of sight.

tan(θ) = opposite/adjacent = 60/130 = 0.4615.

So θ = arctan(0.4615) ≈ 24.8°.

Explanation

The angle of depression equals the angle below the horizontal line of sight.

tan(θ) = opposite/adjacent = 60/130 = 0.4615.

So θ = arctan(0.4615) ≈ 24.8°.

3) A 50 m tower casts a 72 m shadow. Find the sun’s angle of elevation (nearest tenth).

32.9°

33.9°

34.8°

36.0°

tan(θ) = opposite/adjacent = 50/72 = 0.6944.

θ = arctan(0.6944) ≈ 34.8°.

That’s the sun’s angle of elevation.

Explanation

tan(θ) = opposite/adjacent = 50/72 = 0.6944.

θ = arctan(0.6944) ≈ 34.8°.

That’s the sun’s angle of elevation.

4) A 40 m building is viewed from 60 m away horizontally. Find the angle of elevation to its top (nearest tenth).

31.5°

33.7°

35.2°

36.9°

tan(θ) = 40/60 = 0.6667.

θ = arctan(0.6667) ≈ 33.7°.

This represents the angle of elevation to the top.

Explanation

tan(θ) = 40/60 = 0.6667.

θ = arctan(0.6667) ≈ 33.7°.

This represents the angle of elevation to the top.

5) A track rises 25 m over 40 m horizontally. Find the slope angle (nearest tenth).

30.6°

31.3°

32.0°

32.7°

tan(θ) = 25/40 = 0.625.

θ = arctan(0.625) ≈ 32.0°.

So the slope angle is about 32°.

Explanation

tan(θ) = 25/40 = 0.625.

θ = arctan(0.625) ≈ 32.0°.

So the slope angle is about 32°.

6) A skateboard ramp rises 0.9 m over 4.0 m horizontally. Find the incline angle (nearest tenth).

11.9°

12.7°

13.4°

14.1°

tan(θ) = 0.9/4.0 = 0.225.

θ = arctan(0.225) ≈ 12.7°.

This is the angle of incline.

Explanation

tan(θ) = 0.9/4.0 = 0.225.

θ = arctan(0.225) ≈ 12.7°.

This is the angle of incline.

7) A radio tower is 28 m tall and its shadow is 45 m long. Find the sun’s angle of elevation (nearest tenth).

30.4°

31.9°

33.1°

34.6°

tan(θ) = 28/45 = 0.6222.

θ = arctan(0.6222) ≈ 31.9°.

That’s the sun’s elevation angle.

Explanation

tan(θ) = 28/45 = 0.6222.

θ = arctan(0.6222) ≈ 31.9°.

That’s the sun’s elevation angle.

8) From a 200 m cliff, the line of sight to the shore is a horizontal distance of 750 m. Find the angle of depression (nearest tenth).

13.6°

14.3°

14.9°

15.6°

tan(θ) = 200/750 = 0.2667.

θ = arctan(0.2667) ≈ 14.9°.

This represents the downward viewing angle.

Explanation

tan(θ) = 200/750 = 0.2667.

θ = arctan(0.2667) ≈ 14.9°.

This represents the downward viewing angle.

9) A lighthouse light is 40 m above water; the point where the beam hits is 180 m from the base. Find the angle of depression (nearest tenth).

11.7°

12.5°

13.2°

14.0°

tan(θ) = 40/180 = 0.2222.

θ = arctan(0.2222) ≈ 12.5°.

So, the beam’s angle of depression is about 12.5°.

Explanation

tan(θ) = 40/180 = 0.2222.

θ = arctan(0.2222) ≈ 12.5°.

So, the beam’s angle of depression is about 12.5°.

10) A camera on a drone points downward at an angle of 25° toward a target 60 m below. How far horizontally is the drone from the target?

H = 60 tan(25°)

X = 60 / tan(25°)

X = tan(60° / 25°)

X = 25 / tan(60°)

tan(25°) = opposite/adjacent = 60/x.

Rearrange → x = 60 / tan(25°).

This gives the horizontal distance.

Explanation

tan(25°) = opposite/adjacent = 60/x.

Rearrange → x = 60 / tan(25°).

This gives the horizontal distance.

11) A kite is flying on 80 m of string held at ground level. If the angle of elevation is 40°, how high is the kite?

80 tan(45°)

80 tan(40°)

Tan(40°)/80

Tan(80/40)

tan(40°) = opposite/adjacent = height/80.

Multiply both sides by 80 → height = 80 tan(40°).

This gives the kite’s height above ground.

Explanation

tan(40°) = opposite/adjacent = height/80.

Multiply both sides by 80 → height = 80 tan(40°).

This gives the kite’s height above ground.

12) A 20 m flagpole casts a 28 m shadow. Find the sun’s angle of elevation in degrees.

θ = arctan(20/28) ≈ 35.8°

θ = arctan(28/20) ≈ 54.2°

θ = arctan(1.4) ≈ 54.9°

θ = arctan(0.7) ≈ 34.9°

tan(θ) = 20/28 = 0.7143.

θ = arctan(0.7143) ≈ 35.8°.

That’s the sun’s elevation angle.

Explanation

tan(θ) = 20/28 = 0.7143.

θ = arctan(0.7143) ≈ 35.8°.

That’s the sun’s elevation angle.

13) A wheelchair ramp rises 30 cm over a 3.6 m horizontal run. What is the angle of elevation to the nearest tenth of a degree?

4.5°

3.8°

4.8°

6.2°

Convert units (optional since both are in meters): tan(θ) = 0.3 / 3.6 = 0.0833.

θ = arctan(0.0833) ≈ 4.8°.

So, the ramp’s incline angle is 4.8°.

Explanation

Convert units (optional since both are in meters): tan(θ) = 0.3 / 3.6 = 0.0833.

θ = arctan(0.0833) ≈ 4.8°.

So, the ramp’s incline angle is 4.8°.

14) A roof has pitch = 9 : 12 (rise : run). Find the roof angle to the nearest degree.

53°

37°

35°

47°

tan(θ) = rise/run = 9/12 = 0.75.

θ = arctan(0.75) ≈ 36.9°.

Rounded to the nearest degree, θ ≈ 37°.

Explanation

tan(θ) = rise/run = 9/12 = 0.75.

θ = arctan(0.75) ≈ 36.9°.

Rounded to the nearest degree, θ ≈ 37°.

15) A drone climbs 40 m while traveling 120 m horizontally. Find its climb angle in radians (principal range).

θ = arctan(120/40)

θ = arctan(4)

θ = arctan(1/3)

θ = π/2

tan(θ) = rise/run = 40/120 = 1/3.

Therefore, θ = arctan(1/3).

This is within the principal range (−π/2, π/2).

Explanation

tan(θ) = rise/run = 40/120 = 1/3.

Therefore, θ = arctan(1/3).

This is within the principal range (−π/2, π/2).

16) A hillside has a vertical rise of 15 m and a horizontal run of 90 m. Find the hill’s angle of incline in degrees.

θ = arctan(90/15)

θ = arctan(15/90) ≈ 9.5°

θ = arctan(1/6) ≈ 45°

θ = arctan(6) ≈ 80°

tan(θ) = 15/90 = 1/6.

θ = arctan(1/6) ≈ 9.46°, rounded to 9.5°.

So, the slope of the hill is about 9.5°.

Explanation

tan(θ) = 15/90 = 1/6.

θ = arctan(1/6) ≈ 9.46°, rounded to 9.5°.

So, the slope of the hill is about 9.5°.

17) A ship observes the top of a lighthouse at an angle of elevation of 8°. If the lighthouse is 40 m tall, how far is the ship from its base (to the nearest meter)?

285 m

279 m

283 m

300 m

tan(8°) = 40 / x → x = 40 / tan(8°).

x ≈ 283.2 m.

Rounded to the nearest meter, x = 283 m.

Explanation

tan(8°) = 40 / x → x = 40 / tan(8°).

x ≈ 283.2 m.

Rounded to the nearest meter, x = 283 m.

18) A tree casts a 25 m shadow when the sun’s elevation is 36°. Find the tree’s height to the nearest meter.

15 m

20 m

18 m

22 m

tan(36°) = height / 25 → height = 25 × tan(36°).

height ≈ 18.2 m.

Rounded to the nearest meter, height ≈ 18 m.

Explanation

tan(36°) = height / 25 → height = 25 × tan(36°).

height ≈ 18.2 m.

Rounded to the nearest meter, height ≈ 18 m.

19) A road rises 3 m for every 40 m of horizontal distance. Find the incline angle to the nearest tenth of a degree.

θ = arctan(3/40) ≈ 4.3°

θ = arctan(40/3) ≈ 85.7°

θ = arctan(37) ≈ 87.5°

θ = arctan(1/3) ≈ 18.4°

tan(θ) = 3/40 = 0.075.

θ = arctan(0.075) ≈ 4.29°.

Rounded to the nearest tenth, θ ≈ 4.3°.

Explanation

tan(θ) = 3/40 = 0.075.

θ = arctan(0.075) ≈ 4.29°.

Rounded to the nearest tenth, θ ≈ 4.3°.

20) A 12 m ladder leans against a wall. The foot is 5 m from the wall. Find the angle θ the ladder makes with the ground.

θ = arctan(12/5)

θ = arctan(5/12)

θ = arctan(17)

θ = arctan(2.4)

tan(θ) = opposite/adjacent = 12/5 = 2.4.

θ = arctan(2.4) or equivalently θ = arctan(12/5).

This gives the angle between the ladder and the ground.

Explanation

tan(θ) = opposite/adjacent = 12/5 = 2.4.

θ = arctan(2.4) or equivalently θ = arctan(12/5).

This gives the angle between the ladder and the ground.

Real-World Modeling with Inverse Tangent