Question

A man on the top of a rock rising on a seashore observes a boat coming towards it. If it takes 10 minutes for the angle of depression to change from ${30}^{\circ }$ to ${60}^{\circ }$, how soon the boat reaches the shore?

• A. 5 min
• B. 4 min
• C. 3 min
• D. 10 min

Answer 1

The correct option is A

5 min

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




Let C and D be the two positions of boat such that
$\begin{array}{ccc}& \angle ACB=\angle XBC={30}^{\circ }& \\ and& \angle ADB=\angle XBD={60}^{\circ }& \\ Let& CD=xmandAD=ym& \\ Now,in\Delta ABD,& tan{60}^{\circ }=\frac{h}{y}& \\ & \sqrt{3}=\frac{h}{y}& \dots \left(1\right)\\ andin\Delta ABC& tan{30}^{\circ }=\frac{h}{x+y}=\frac{1}{\sqrt{3}}& \dots \left(2\right)\end{array}$
From equation (1) and (2), we get
$\frac{\sqrt{3}y}{x+y}=\frac{1}{\sqrt{3}}\Rightarrow x=2y\Rightarrow y=\frac{x}{2}$
Since the boat takes 10 minutes to cover x m,

hence it will take $\frac{10}{2}$ = 5 minutes to cover $y=\frac{x}{2}$ metres.

Thus the required time = 5 minutes.