Heights and Distances

357. The Car's Countdown · From 30° to the tower's base

The time from the 45° position to the tower is always ~16.39 minutes.

ABCD12 mint = ?∠APB = 45°∠APB = 45°Car
A moving object's distance to a tower follows from angles of elevation taken at two moments, using tan. The difference of the two distances is the distance travelled, and dividing by the speed gives the time between observations.

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Selina ICSE: Heights and Distances

What this lesson covers

Try to break it

Drag the car along the ground. The angle of depression from the tower top grows as the car approaches — from 30° to 45° to 60°. At constant speed, time and distance covered are directly proportional. At 30°, the car is h·√3 away; at 45°, h away; at 60°, h/√3 away. The time between two of those positions divides the corresponding distance to give the speed.

The proof, step by step

Prove that the car takes about 16.39 minutes to reach the tower.

  • In ΔABD, tan 45° = AB/BD ⇒ BD = AB.
  • In ΔABC, tan 30° = AB/BC ⇒ BC = AB√3.
  • BC = BD + CD ⇒ AB√3 = AB + CD ⇒ CD = AB(√3 - 1).
  • Time is proportional to distance. t_DB / 12 = BD / CD = 1 / (√3 - 1).
  • t_DB = 12 / (√3 - 1) ≈ 16.39 minutes.

Worked example

A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

Let AB be the tower. In ΔABD, BD = AB. In ΔABC, BC = AB√3. CD = BC - BD = AB(√3 - 1). Since speed is constant, time ∝ distance. Time(DB) = 12 × BD/CD = 12/(√3 - 1) ≈ 16.39 min.

  • 12 minutes
  • 14.5 minutes
  • 16.39 minutes — correct
  • 18 minutes
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