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

The relationship is:

tan 35° = height / 18

So, first write the tangent ratio:

tan 35° = height / 18

Then, solve for height by multiplying both sides by 18:

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

Hence, the flagpole’s height is approximately 12.6 m.

Explanation

The relationship is:

tan 35° = height / 18

So, first write the tangent ratio:

tan 35° = height / 18

Then, solve for height by multiplying both sides by 18:

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

Hence, the flagpole’s height is approximately 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

The relationship is:

sin 68° = 14 / L

So, first write the sine ratio:

sin 68° = 14 / L

Then, solve for L by multiplying both sides by L and dividing by sin 68°:

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

Hence, the ladder’s length is approximately 15.1 ft.

Explanation

The relationship is:

sin 68° = 14 / L

So, first write the sine ratio:

sin 68° = 14 / L

Then, solve for L by multiplying both sides by L and dividing by sin 68°:

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

Hence, the ladder’s length is approximately 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

The vertical component is found with sine:

vertical = 250 · sin 22°

So, first substitute into the formula:

vertical = 250 · sin 22°

Then, estimate the value:

vertical ≈ 250 · 0.3746 ≈ 93.6 m

Hence, the vertical gain is approximately 93.6 m.

Explanation

The vertical component is found with sine:

vertical = 250 · sin 22°

So, first substitute into the formula:

vertical = 250 · sin 22°

Then, estimate the value:

vertical ≈ 250 · 0.3746 ≈ 93.6 m

Hence, the vertical gain is approximately 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

The tangent ratio gives:

tan 27° = 48 / d

So, first write the equation:

tan 27° = 48 / d

Then, solve for d by multiplying both sides by d and dividing by tan 27°:

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

Hence, the horizontal distance is approximately 94.3 m.

Explanation

The tangent ratio gives:

tan 27° = 48 / d

So, first write the equation:

tan 27° = 48 / d

Then, solve for d by multiplying both sides by d and dividing by tan 27°:

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

Hence, the horizontal distance is approximately 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

The maximum slope is given as rise : run = 1 : 12.

So, for a rise of 0.9 m, use the ratio:

run = 12 × rise = 12 × 0.9 = 10.8 m

Hence, the minimum horizontal run is 10.8 m.

Explanation

The maximum slope is given as rise : run = 1 : 12.

So, for a rise of 0.9 m, use the ratio:

run = 12 × rise = 12 × 0.9 = 10.8 m

Hence, the minimum horizontal run is 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

The relationship is:

sin 55° = 32 / L

So, first write the sine ratio:

sin 55° = 32 / L

Then, solve for L:

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

Hence, the guy-wire length is approximately 39.1 ft.

Explanation

The relationship is:

sin 55° = 32 / L

So, first write the sine ratio:

sin 55° = 32 / L

Then, solve for L:

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

Hence, the guy-wire length is approximately 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°

Use the tangent ratio for the stair angle:

tan θ = rise / run

So, first substitute the values:

tan θ = 7.75 / 10.5 ≈ 0.7381

Then, find the angle whose tangent is 0.7381:

θ = arctan(0.7381) ≈ 36.5°

Hence, the stair angle is approximately 36.5°.

Explanation

Use the tangent ratio for the stair angle:

tan θ = rise / run

So, first substitute the values:

tan θ = 7.75 / 10.5 ≈ 0.7381

Then, find the angle whose tangent is 0.7381:

θ = arctan(0.7381) ≈ 36.5°

Hence, the stair angle is approximately 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

The horizontal distance is found with cosine:

horizontal = 80 · cos 50°

So, first substitute:

horizontal = 80 · cos 50°

Then, estimate:

horizontal ≈ 80 · 0.6428 ≈ 51.4 m

Hence, the horizontal distance is approximately 51.4 m.

Explanation

The horizontal distance is found with cosine:

horizontal = 80 · cos 50°

So, first substitute:

horizontal = 80 · cos 50°

Then, estimate:

horizontal ≈ 80 · 0.6428 ≈ 51.4 m

Hence, the horizontal distance is approximately 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

Use the tangent ratio:

tan 4° = 62 / d

So, first write the equation:

tan 4° = 62 / d

Then, solve for d:

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

Hence, the horizontal distance is approximately 887 m.

Explanation

Use the tangent ratio:

tan 4° = 62 / d

So, first write the equation:

tan 4° = 62 / d

Then, solve for d:

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

Hence, the horizontal distance is approximately 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

Let the distance from the observer to the billboard be 30 m.

Height to the top: 30 · tan 25°

Height to the bottom: 30 · tan 7°

So, the billboard’s height is:

height = 30 · (tan 25° − tan 7°)

Then, estimate:

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

Hence, the billboard’s height is approximately 10.3 m.

Explanation

Let the distance from the observer to the billboard be 30 m.

Height to the top: 30 · tan 25°

Height to the bottom: 30 · tan 7°

So, the billboard’s height is:

height = 30 · (tan 25° − tan 7°)

Then, estimate:

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

Hence, the billboard’s height is approximately 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°

Use the sine ratio with slope length as the hypotenuse:

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

So, first write:

sin θ = 1/3

Then, find the angle:

θ = arcsin(1/3) ≈ 19.5°

Hence, the slope angle is approximately 19.5°.

Explanation

Use the sine ratio with slope length as the hypotenuse:

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

So, first write:

sin θ = 1/3

Then, find the angle:

θ = arcsin(1/3) ≈ 19.5°

Hence, the slope angle is approximately 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

Use the angle of depression and tangent:

tan 35° = 6 / d

So, first write:

tan 35° = 6 / d

Then, solve for d:

d = 6 / tan 35° ≈ 6 / 0.7002 ≈ 8.6 m

Hence, the horizontal distance is approximately 8.6 m.

Explanation

Use the angle of depression and tangent:

tan 35° = 6 / d

So, first write:

tan 35° = 6 / d

Then, solve for d:

d = 6 / tan 35° ≈ 6 / 0.7002 ≈ 8.6 m

Hence, the horizontal distance is approximately 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

Use cosine to find the horizontal component:

horizontal = 1.20 · cos 12°

So, first substitute:

horizontal ≈ 1.20 · 0.9781 ≈ 1.17 km

Hence, the horizontal distance traveled is approximately 1.17 km.

Explanation

Use cosine to find the horizontal component:

horizontal = 1.20 · cos 12°

So, first substitute:

horizontal ≈ 1.20 · 0.9781 ≈ 1.17 km

Hence, the horizontal distance traveled is approximately 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

Use the tangent ratio with height and shadow:

tan 28° = 80 / shadow

So, first write:

tan 28° = 80 / shadow

Then, solve for shadow:

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

Hence, the shadow length is approximately 150.4 m.

Explanation

Use the tangent ratio with height and shadow:

tan 28° = 80 / shadow

So, first write:

tan 28° = 80 / shadow

Then, solve for shadow:

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

Hence, the shadow length is approximately 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°

Use the tangent ratio for the angle:

tan θ = opposite / adjacent = 45 / 120

So, first compute the ratio:

tan θ = 45 / 120 = 0.375

Then, find the angle:

θ = arctan(0.375) ≈ 20.6°

Hence, the cable’s angle with the ground is approximately 20.6°.

Explanation

Use the tangent ratio for the angle:

tan θ = opposite / adjacent = 45 / 120

So, first compute the ratio:

tan θ = 45 / 120 = 0.375

Then, find the angle:

θ = arctan(0.375) ≈ 20.6°

Hence, the cable’s angle with the ground is approximately 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°

Use the sine ratio with the ladder as the hypotenuse:

sin θ = opposite / hypotenuse = 17 / 18

So, first write:

sin θ = 17 / 18 ≈ 0.9444

Then, find the angle:

θ = arcsin(17/18) ≈ 70.5°

Rounded to the nearest tenth, θ ≈ 70.7°.

Hence, the ladder’s angle with the ground is approximately 70.7°.

Explanation

Use the sine ratio with the ladder as the hypotenuse:

sin θ = opposite / hypotenuse = 17 / 18

So, first write:

sin θ = 17 / 18 ≈ 0.9444

Then, find the angle:

θ = arcsin(17/18) ≈ 70.5°

Rounded to the nearest tenth, θ ≈ 70.7°.

Hence, the ladder’s angle with the ground is approximately 70.7°.

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

First find the vertical difference between eye level and the tank top:

vertical difference = 25.0 − 1.6 = 23.4 m

Use the tangent ratio:

tan 33° = 23.4 / d

So, write the equation:

tan 33° = 23.4 / d

Then, solve for d:

d = 23.4 / tan 33° ≈ 23.4 / 0.6494 ≈ 36.0 m

Hence, the horizontal distance is approximately 36.0 m.

Explanation

First find the vertical difference between eye level and the tank top:

vertical difference = 25.0 − 1.6 = 23.4 m

Use the tangent ratio:

tan 33° = 23.4 / d

So, write the equation:

tan 33° = 23.4 / d

Then, solve for d:

d = 23.4 / tan 33° ≈ 23.4 / 0.6494 ≈ 36.0 m

Hence, the horizontal distance is approximately 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

Use the Pythagorean theorem since the zipline is the hypotenuse:

Length² = horizontal² + vertical²

So, first substitute:

Length² = 95² + 18² = 9025 + 324 = 9349

Then, take the square root:

Length = √9349 ≈ 96.7 m

Hence, the zipline length is approximately 96.7 m.

Explanation

Use the Pythagorean theorem since the zipline is the hypotenuse:

Length² = horizontal² + vertical²

So, first substitute:

Length² = 95² + 18² = 9025 + 324 = 9349

Then, take the square root:

Length = √9349 ≈ 96.7 m

Hence, the zipline length is approximately 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

Use the sine ratio for vertical height:

height = 40 · sin 72°

So, first substitute:

height ≈ 40 · 0.9511 ≈ 38.0 m

Hence, the vertical height of the boom’s tip is approximately 38.0 m.

Explanation

Use the sine ratio for vertical height:

height = 40 · sin 72°

So, first substitute:

height ≈ 40 · 0.9511 ≈ 38.0 m

Hence, the vertical height of the boom’s tip is approximately 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

Use the tangent ratio with the mast height as opposite:

height = 120 · tan 14°

So, first substitute:

height ≈ 120 · 0.2493 ≈ 29.9 m

Rounded to the nearest tenth, height ≈ 30.0 m.

Hence, the mast’s height is approximately 30.0 m.

Explanation

Use the tangent ratio with the mast height as opposite:

height = 120 · tan 14°

So, first substitute:

height ≈ 120 · 0.2493 ≈ 29.9 m

Rounded to the nearest tenth, height ≈ 30.0 m.

Hence, the mast’s height is approximately 30.0 m.

Model Real Situations with SOHCAHTOA

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

2)

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

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.

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

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.