A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Let AB be the vertical tower. Suppose D and C be the positions of the car when the angle of depression from the top of the tower is 30° and 60° respectively.

Consider the uniform speed of the car be ν m/sec.

Time taken for the angle of depression to change from 30° to 45° = 10 sec (Given)

∠EAD = ∠ADB = 30° (Alternate angles)

∠EAC = ∠ACB = 60° (Alternate angles)

Suppose AB = *h* m and BC = *x* m.

CD = Distance covered by car in 10 sec = ν m/sec × 10 sec = 10 ν m

In triangle ABD

In ∆ACB

From (1) and (2), we get

now time taken by the car to reach the tower is from C = x / v

therefore time taken by the car to reach the foot of the tower = 5 sec

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