Model Real Situations with SOHCAHTOA

  • 10th Grade
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| Attempts: 11 | Questions: 20 | Updated: Dec 11, 2025
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1) A flagpole casts an 18 m shadow when the sun's elevation is 35°. Find the flagpole's height to the nearest tenth.

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.

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About This Quiz
Model Real Situations With Sohcahtoa - Quiz

Are you ready to connect trigonometry to real-life measurement and design problems? This quiz presents applied contexts such as determining heights of buildings, distances across water, ladder positions, and ramp inclines. You will model situations using right triangles, select appropriate trigonometric ratios, and solve for missing quantities to the nearest... see moreunit or tenth. By combining geometry and trigonometric reasoning, this quiz strengthens the ability to interpret and solve practical problems involving right triangles in everyday settings.
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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.

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.

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

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.

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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.

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.

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

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.

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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.

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.

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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.

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°.

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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.

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.

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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.

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.

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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.

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.

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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.

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°.

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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.

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.

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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.

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.

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

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.

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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.

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°.

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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.

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°.

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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.

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.

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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.

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.

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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.

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.

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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.

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.

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A flagpole casts an 18 m shadow when the sun's elevation is 35°....
A ladder reaches 14 ft up a wall at a 68° angle with the ground....
A drone flies 250 m along a straight line at a 22° angle above...
A building is 48 m tall. From a point on level ground, the angle of...
A ramp has rise 0.9 m and must meet a maximum slope of 1:12. Find the...
A guy-wire is attached to the top of a 32 ft pole and makes a 55°...
A staircase has rise 7.75 in and run 10.5 in per step. Find the stair...
A kite has 80 m of taut string at a 50° angle with the ground....
A boat measures a 4° angle of elevation to the top of a 62 m...
From 30 m away, the angle of elevation to the top of a billboard is...
A ski slope drops 300 m over a slope length of 900 m. Find its angle...
A security camera is mounted 6.0 m above level ground. The line of...
A paraglider flies 1.20 km along a straight path at a descent angle of...
A wind turbine is 80 m tall. Find the shadow length to the nearest...
A bridge cable runs from a ground anchor to the top of a 45 m tower...
A roof access ladder of length 18 ft reaches a vertical height of 17...
A water tank's top is 25.0 m above ground. An observer with eye level...
Two platforms are connected by a zipline. The lower platform is 18 m...
A crane boom 40 m long is raised to 72° above the ground. Find the...
From 120 m offshore, the angle of elevation to the top of a sailboat's...
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