A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at 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 CD is highway and AB is tower. C is the initial position of the car and D is the position after 6 seconds.
In ΔABD,
AB/DB = tan 60° ⇒ AB/DB = √3 ⇒ BD = AB/√3
In ΔABC,
AB/BC = tan 30°
⇒ AB/(BD + DC) = 1/√3
⇒ AB√3 = BD + DC
⇒ AB = AB/√3 + DC [putting the value of BD]
⇒ AB√3 – AB/√3 = DC
⇒ DC = (3AB – AB)/√3 = 2AB/√3
The time taken to travelled CD (= 2AB/√3) distance = 6 second
Therefore, the speed of the car = distance/time = (2AB/√3)/6 = AB/3√3 m/s
The time taken to travelled BD (= AB/√3) distance = distance/speed = (AB/√3)/ (AB/3√3) = 3 second.
Hence, car will take 3 seconds to reach the foot of the car.