Heights and Distances

359. Climbing the Tangent · finding tower height from walking distances

The tangent ratios lock the tower height to 120 m when angles match 3/5 and 4/5.

ABQhx50 mα = 31°α = 31°β = 39°β = 39°P
Observing a tower from two distances gives two tangent equations for the same height. Solving them together eliminates the unknown distance and pins down the tower's height.

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Selina ICSE: Heights and Distances

What this lesson covers

Try to break it

Drag P along the ground. With tan α = 3/5 at P and tan β = 4/5 at Q (50 m closer to the tower), the tower height comes out as 120 m — fixed by the two ratios and the 50 m gap. If the tower were shorter or taller, the two tangent ratios at exactly 50 m apart wouldn't both fit; only h = 120 m makes both conditions true.

The proof, step by step

Prove that the tower height is 120 m from the given tangent ratios.

  • Let the height of the tower AB = h m and BQ = x m.
  • In right Δ AQB, tan β = AB/QB = h/x = 4/5 ⇒ 5h = 4x.
  • In right Δ APB, tan α = AB/PB = h/(x+50) = 3/5 ⇒ 5h = 3x + 150.
  • Equating (2) and (3): 4x = 3x + 150 ⇒ x = 150.
  • Substitute x = 150 into (2): 5h = 600 ⇒ h = 120 m.

Worked example

From a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is 3/5. On walking 50 m towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be 4/5. Find the height of the tower.

Let height = h, BQ = x. tan β = h/x = 4/5 ⇒ x = 1.25h. tan α = h/(x+50) = 3/5. Substituting x: h/(1.25h+50) = 3/5 ⇒ 5h = 3.75h + 150 ⇒ 1.25h = 150 ⇒ h = 120 m.

  • 100 m
  • 120 m — correct
  • 150 m
  • 180 m
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