A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.

Let the tower be represented by CD, the initial position of the car by A and the second position of the car by B.
Let the distance travelled from A to B be d1 in time t1 and distance between B to C be d2 in time t2.
In ΔDCB, tan45° = CD/BC. ∴ BC = CD = d2
In ΔACD, tan30° = CD/AC ∴ 1/√3 = d2/(d1 + d2) ∴ d1 + d2 = √3d2
Since the speed of the car is uniform throughout its journey, we can say t1/d1 = t2/d2
Given t1 = 12 minutes
d1 = √3d2 - d2 = d2(√3 - 1)
12 / [d2(√3 - 1)] = t2 / d2
12 = t2(√3 - 1)
t2 = 12 / (√3 - 1) ≈ 12 / (1.732 - 1) ≈ 12 / 0.732 ≈ 16.39 minutes.