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°

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

You need first substitute the values:

tan(θ) = 100 / 500

tan(θ) = 0.2

Now find the angle whose tangent is 0.2:

θ ≈ arctan(0.2) ≈ 11.3°

So the angle of elevation is 11.3°.

Explanation

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

You need first substitute the values:

tan(θ) = 100 / 500

tan(θ) = 0.2

Now find the angle whose tangent is 0.2:

θ ≈ arctan(0.2) ≈ 11.3°

So the angle of elevation is 11.3°.

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

32.9°

33.9°

34.7°

36.0°

Use the tangent ratio for angle of elevation:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 50 / 72

tan(θ) ≈ 0.6944

Now find the angle whose tangent is 0.6944:

θ ≈ arctan(0.6944) ≈ 34.7°

So the sun’s angle of elevation is 34.7°.

Explanation

Use the tangent ratio for angle of elevation:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 50 / 72

tan(θ) ≈ 0.6944

Now find the angle whose tangent is 0.6944:

θ ≈ arctan(0.6944) ≈ 34.7°

So the sun’s angle of elevation is 34.7°.

3) 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°

Use the tangent ratio for the angle of elevation: tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 28 / 45

tan(θ) ≈ 0.6222

Now find the angle whose tangent is 0.6222:

θ ≈ arctan(0.6222) ≈ 31.0°

So the sun’s angle of elevation is 31.0°.

Explanation

Use the tangent ratio for the angle of elevation: tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 28 / 45

tan(θ) ≈ 0.6222

Now find the angle whose tangent is 0.6222:

θ ≈ arctan(0.6222) ≈ 31.0°

So the sun’s angle of elevation is 31.0°.

4) 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°

Use the tangent ratio for the angle of depression:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 180

tan(θ) ≈ 0.2222

Now find the angle whose tangent is 0.2222:

θ ≈ arctan(0.2222) ≈ 12.5°

So the angle of depression is 12.5°.

Explanation

Use the tangent ratio for the angle of depression:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 180

tan(θ) ≈ 0.2222

Now find the angle whose tangent is 0.2222:

θ ≈ arctan(0.2222) ≈ 12.5°

So the angle of depression is 12.5°.

5) 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)

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

Here, the string is the adjacent side (80 m), and the height is the opposite side (h).

So: tan(40°) = h / 80

Solve for h by multiplying both sides by 80:

h = 80 tan(40°)

Explanation

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

Here, the string is the adjacent side (80 m), and the height is the opposite side (h).

So: tan(40°) = h / 80

Solve for h by multiplying both sides by 80:

h = 80 tan(40°)

6) 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

285 m

300 m

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

Here,

opposite = 40 m (height of lighthouse)

adjacent = x (horizontal distance)

So:

tan(8°) = 40 / x

Solve for x:

x = 40 / tan(8°)

Evaluate the expression:

x ≈ 40 / 0.1405 ≈ 284.7

Rounded to the nearest meter: x ≈ 285 m

Explanation

Use the tangent ratio for angle of elevation: tan(θ) = opposite / adjacent

Here,

opposite = 40 m (height of lighthouse)

adjacent = x (horizontal distance)

So:

tan(8°) = 40 / x

Solve for x:

x = 40 / tan(8°)

Evaluate the expression:

x ≈ 40 / 0.1405 ≈ 284.7

Rounded to the nearest meter: x ≈ 285 m

7) 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

Use the tangent ratio:

tan(36°) = height / 25

So,

height = 25 · tan(36°)

Now compute tan(36°):

tan(36°) ≈ 0.7265

Multiply:

height ≈ 25 × 0.7265

height ≈ 18.16 m

Round to the nearest meter:

height ≈ 18 m

Explanation

Use the tangent ratio:

tan(36°) = height / 25

So,

height = 25 · tan(36°)

Now compute tan(36°):

tan(36°) ≈ 0.7265

Multiply:

height ≈ 25 × 0.7265

height ≈ 18.16 m

Round to the nearest meter:

height ≈ 18 m

8) 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°

Use the tangent ratio for angle of depression: tan(θ) = opposite / adjacent

Substitute the values: tan(θ) = 60 / 130

Simplify the ratio:

tan(θ) ≈ 0.4615

Now find the angle whose tangent is 0.4615:

θ ≈ arctan(0.4615) ≈ 26.4°

So the angle of depression is 26.4°.

Explanation

Use the tangent ratio for angle of depression: tan(θ) = opposite / adjacent

Substitute the values: tan(θ) = 60 / 130

Simplify the ratio:

tan(θ) ≈ 0.4615

Now find the angle whose tangent is 0.4615:

θ ≈ arctan(0.4615) ≈ 26.4°

So the angle of depression is 26.4°.

9) 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°

Use the tangent ratio for angle of elevation:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 60

tan(θ) = 0.6667

Now find the angle whose tangent is 0.6667:

θ ≈ arctan(0.6667) ≈ 33.7°

So the angle of elevation is 33.7°.

Explanation

Use the tangent ratio for angle of elevation:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 60

tan(θ) = 0.6667

Now find the angle whose tangent is 0.6667:

θ ≈ arctan(0.6667) ≈ 33.7°

So the angle of elevation is 33.7°.

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

30.6°

31.3°

32.0°

32.7°

Use the tangent ratio for the slope angle: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 25 / 40

tan(θ) = 0.625

Now find the angle whose tangent is 0.625:

θ ≈ arctan(0.625) ≈ 32.0°

So the slope angle is 32.0°.

Explanation

Use the tangent ratio for the slope angle: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 25 / 40

tan(θ) = 0.625

Now find the angle whose tangent is 0.625:

θ ≈ arctan(0.625) ≈ 32.0°

So the slope angle is 32.0°.

11) 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°

Use the tangent ratio for the incline angle: tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 0.9 / 4.0

tan(θ) = 0.225

Now find the angle whose tangent is 0.225:

θ ≈ arctan(0.225) ≈ 12.7°

So the incline angle is 12.7°.

Explanation

Use the tangent ratio for the incline angle: tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 0.9 / 4.0

tan(θ) = 0.225

Now find the angle whose tangent is 0.225:

θ ≈ arctan(0.225) ≈ 12.7°

So the incline angle is 12.7°.

12) 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°

15°

15.6°

Use the tangent ratio for the angle of depression:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 200 / 750

tan(θ) ≈ 0.2667

Now find the angle whose tangent is 0.2667:

θ ≈ arctan(0.2667) ≈ 15.0°

So the angle of depression is 15.0°.

Explanation

Use the tangent ratio for the angle of depression:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 200 / 750

tan(θ) ≈ 0.2667

Now find the angle whose tangent is 0.2667:

θ ≈ arctan(0.2667) ≈ 15.0°

So the angle of depression is 15.0°.

13) 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°)

The angle of depression is 25°.

We use the tangent ratio:

tan(25°) = opposite / adjacent

Here, opposite = 60 m (vertical drop), adjacent = x (horizontal distance)

So, tan(25°) = 60 / x

Solve for x by dividing 60 by tan(25°):

x = 60 / tan(25°)

This gives the horizontal distance from the drone to the target.

Explanation

The angle of depression is 25°.

We use the tangent ratio:

tan(25°) = opposite / adjacent

Here, opposite = 60 m (vertical drop), adjacent = x (horizontal distance)

So, tan(25°) = 60 / x

Solve for x by dividing 60 by tan(25°):

x = 60 / tan(25°)

This gives the horizontal distance from the drone to the target.

14) 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°

Use the tangent ratio for the angle of elevation: tan(θ) = opposite / adjacent

Here, opposite = 20 m (flagpole height), adjacent = 28 m (shadow length)

So: tan(θ) = 20 / 28

Now take the inverse tangent:

θ = arctan(20/28) ≈ 35.8°

Explanation

Use the tangent ratio for the angle of elevation: tan(θ) = opposite / adjacent

Here, opposite = 20 m (flagpole height), adjacent = 28 m (shadow length)

So: tan(θ) = 20 / 28

Now take the inverse tangent:

θ = arctan(20/28) ≈ 35.8°

15) 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°

First convert the rise to meters so both measurements match: 30 cm = 0.30 m

Then, use the tangent ratio:

tan(θ) = opposite / adjacent

Next, substitute the values:

tan(θ) = 0.30 / 3.6

tan(θ) ≈ 0.0833

& Finally, find the angle whose tangent is 0.0833:

θ ≈ arctan(0.0833) ≈ 4.8°

So the angle of elevation is 4.8°.

Explanation

First convert the rise to meters so both measurements match: 30 cm = 0.30 m

Then, use the tangent ratio:

tan(θ) = opposite / adjacent

Next, substitute the values:

tan(θ) = 0.30 / 3.6

tan(θ) ≈ 0.0833

& Finally, find the angle whose tangent is 0.0833:

θ ≈ arctan(0.0833) ≈ 4.8°

So the angle of elevation is 4.8°.

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

53°

37°

35°

47°

The pitch ratio 9 : 12 means:

rise = 9

run = 12

Use the tangent ratio:

tan(θ) = rise / run

tan(θ) = 9 / 12

tan(θ) = 0.75

Now find the angle whose tangent is 0.75:

θ ≈ arctan(0.75) ≈ 36.9°

Rounded to the nearest degree:

θ ≈ 37°

Explanation

The pitch ratio 9 : 12 means:

rise = 9

run = 12

Use the tangent ratio:

tan(θ) = rise / run

tan(θ) = 9 / 12

tan(θ) = 0.75

Now find the angle whose tangent is 0.75:

θ ≈ arctan(0.75) ≈ 36.9°

Rounded to the nearest degree:

θ ≈ 37°

17) 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

Use the tangent ratio for climb angle:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 120

tan(θ) = 1/3

Now express the angle using inverse tangent:

θ = arctan(1/3)

Numerically, arctan(1/3) ≈ 0.32 radians, which lies in the principal range (−π/2, π/2).

Explanation

Use the tangent ratio for climb angle:

tan(θ) = opposite / adjacent

Substitute the values:

tan(θ) = 40 / 120

tan(θ) = 1/3

Now express the angle using inverse tangent:

θ = arctan(1/3)

Numerically, arctan(1/3) ≈ 0.32 radians, which lies in the principal range (−π/2, π/2).

18) 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°

Use the tangent ratio for incline angle: tan(θ) = opposite / adjacent

Then, substitute the rise and run:

tan(θ) = 15 / 90

tan(θ) = 1/6

Now, take the inverse tangent:

θ = arctan(1/6)

Computing this gives approximately 9.5°, which matches option B.

So the hill’s incline angle is about 9.5°.

Explanation

Use the tangent ratio for incline angle: tan(θ) = opposite / adjacent

Then, substitute the rise and run:

tan(θ) = 15 / 90

tan(θ) = 1/6

Now, take the inverse tangent:

θ = arctan(1/6)

Computing this gives approximately 9.5°, which matches option B.

So the hill’s incline angle is about 9.5°.

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°

Use the tangent ratio for incline angle: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 3 / 40

tan(θ) = 0.075

Now, find the angle whose tangent is 0.075:

θ ≈ arctan(0.075) ≈ 4.3°

So the incline angle is 4.3°.

Explanation

Use the tangent ratio for incline angle: tan(θ) = opposite / adjacent

First, substitute the values:

tan(θ) = 3 / 40

tan(θ) = 0.075

Now, find the angle whose tangent is 0.075:

θ ≈ arctan(0.075) ≈ 4.3°

So the incline angle is 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)

Use the tangent ratio: tan(θ) = opposite / adjacent

Here, the opposite side is the height the ladder reaches on the wall (unknown),

and the adjacent side is the horizontal distance from the wall: 5 m.

But the ladder length (12 m) is the hypotenuse.

Instead, use tangent with the correct sides for the ground angle:

tan(θ) = ladder height / 5.

Find the height using the Pythagorean theorem:

height = √(12² − 5²) = √(144 − 25) = √119 ≈ 10.9

Now substitute into tangent:

tan(θ) = 10.9 / 5 = 2.18

This is the same as:

tan(θ) = 12 / 5 (ratio based on similar triangles)

So the correct expression for the angle is:

θ = arctan(12/5)

Explanation

Use the tangent ratio: tan(θ) = opposite / adjacent

Here, the opposite side is the height the ladder reaches on the wall (unknown),

and the adjacent side is the horizontal distance from the wall: 5 m.

But the ladder length (12 m) is the hypotenuse.

Instead, use tangent with the correct sides for the ground angle:

tan(θ) = ladder height / 5.

Find the height using the Pythagorean theorem:

height = √(12² − 5²) = √(144 − 25) = √119 ≈ 10.9

Now substitute into tangent:

tan(θ) = 10.9 / 5 = 2.18

This is the same as:

tan(θ) = 12 / 5 (ratio based on similar triangles)

So the correct expression for the angle is:

θ = arctan(12/5)

Real-World Modeling with Inverse Tangent