A man on the top of a vertical 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°, then how soon, after this, will the car reach the tower?

As per the given question,

Given, time taken for the angle of depression to change from $$30^{\circ}$$ to $$40^{\circ}$$ is 12 minutes, so BC = 12s and CD = st

where, s = speed and t = time 

In triangle ABC,

tan 45$$^{\circ} = \frac{AB}{12s}$$ (or) AB = st

In triangle ABD,

tan 30$$^{\circ} = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{AB}{s(12 + t)}$$

$$\frac{1}{\sqrt{3}} = \frac{st}{s(12 + t)}$$   ($$\because$$ AB = st)

$$12 + t - \sqrt{3}t = 0$$ (or) $$ 12 - 0.73t = 0$$ (or) t = 16.23 (Approx)

Hence, option B is the correct answer.