A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from 30° to 45°; then after this, the time taken (in min) by the car to reach the foot of the tower, is?

Given, ∠HOA = 30° and ∠HOB = 45°. Let OH = h, then tan(∠HOA) = HA/h or HA = h × tan30° tan(∠HOB) = HB/h or HB = h × tan45° It takes 18 mins to travel distance AB, hence speed = (HB - HA)/18 = (h - (h/√3))/18 Time taken to travel HA = HA/speed = (h/√3)/((h - h/√3)/18) = 18 × √3/(√3 - 1) = 18 × √3(√3 + 1)/(√3 - 1)(√3 + 1) = 18 × √3(√3 + 1)/2 = 9(√3 + 1) Option A is correct.