A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? Give your answer to the nearest second.

Let the height of the tower AB = h.

Let the speed of the car be x m/s.

Let the time taken to travel from D to A be t seconds.

CD (distance) = speed × time = x m/s × 600 (s) = 600x meters.

DA = (tx) meters.

In △ABD,

$\Rightarrow \tan 45° = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{AD} \\[1em] \Rightarrow 1 = \dfrac{h}{tx} \\[1em] \Rightarrow h = tx ….(1)$

In △ABC,

$\Rightarrow \tan 30° = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{CA} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{600x + tx} \\[1em] \Rightarrow h = \dfrac{600x + tx}{\sqrt3} ….(2)$

From (1) and (2), we get :

$\Rightarrow tx = \dfrac{600x + tx}{\sqrt3} \\[1em] \Rightarrow \sqrt3tx = 600x + tx \\[1em] \Rightarrow \sqrt3t = 600 + t \\[1em] \Rightarrow \sqrt3t - t = 600 \\[1em] \Rightarrow t(\sqrt3 - 1) = 600 \\[1em] \Rightarrow t = \dfrac{600}{(\sqrt3 - 1)} \\[1em] \Rightarrow t = \dfrac{600}{(\sqrt3 - 1)} \times \dfrac{(\sqrt3 + 1)}{(\sqrt3 + 1)} \\[1em] \Rightarrow t = \dfrac{600(\sqrt3 + 1)}{(3 - 1)} \\[1em] \Rightarrow t = \dfrac{600(\sqrt3 + 1)}{2} \\[1em] \Rightarrow t = 300(\sqrt3 + 1) = 819.6 \text{ sec.}$

Time taken to reach = $\dfrac{819.6}{60}$ = 13.66 mins i.e. 13 min 40 seconds.

Hence, time taken to reach tower is 13 min 40 sec.