Heights and Distances

354. The Tower's Height · using tangent to measure the sky

The height AB is always BC × tan(30°).

ABBC = 120 mBC = 120 mAB = 69.3 mAB = 69.3 m∠C = 30°∠C = 30°C
In heights-and-distances problems, tan(angle of elevation) = height / horizontal distance. So the tower height AB = BC × tan(angle), for example BC × tan 30°.

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Selina ICSE: Heights and Distances

What this lesson covers

Try to break it

Drag C closer to the tower or farther from it. The distance BC stretches; ∠ACB shifts. With ∠ACB = 30° and BC = 120 m, the tower height AB = BC × tan(30°) = 120/√3 m. Move C closer and the angle grows; move it farther and the angle shrinks — the height of the tower stays the same, but the readouts let you compute it.

The proof, step by step

Prove that the tower height AB equals BC × tan 30°.

  • In right ΔABC, ∠B = 90° (Tower is vertical).
  • tan(∠C) = AB/BC (Definition of tangent).
  • tan(30°) = AB/120 (Substitute values).
  • AB = 120 × (1/√3) = 69.28 m (Solve for AB).

Worked example

A man observes the top of a tower from a point 120 m away from its foot. If the angle of elevation of the top of the tower is 30°, find the height of the tower.

Using tan(30°) = Height/Distance, we get Height = 120 × tan(30°) = 120 × (1/√3) ≈ 69.28 m.

  • 69.28 m — correct
  • 120 m
  • 207.84 m
  • 40 m
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