A man on the top of a rock rising on a seashore observed a boat coming towards it. If it takes 10 minutes for the angle of depression to change from 30° to 60°, how soon will the boat reach the shore?

Let AB be the rock of height ’h’ metres.

Let C and D be the two positions of boat such

that ∠ACB=∠XBC=30° and ∠ADB=∠XBD = 60°

Let CD = x metre and AD = y metre

In $\Delta \mathrm{ABD}, \tan 60^{\circ}=\frac{h}{y}$

$\Rightarrow \sqrt{3}=\frac{h}{y}$

In $\Delta \mathrm{ABC}, \tan 30^{\circ}=\frac{h}{x+y}=\frac{1}{\sqrt{3}}$

∴ From equations (i) and (ii)

$\frac{\sqrt{3} y}{x+y}=\frac{1}{\sqrt{3}}$

$\Rightarrow 3 y=x+y$

$\Rightarrow x=2 y$

$\Rightarrow y=\frac{x}{2}$

∴ Time taken in covering x metres = 10 minutes

$\therefore$ Required time $=\frac{10}{2}=5$ minutes