Question

A man on the top of a tower, standing on the sea shore, finds a boat coming towards him takes 10 minutes for the angle of depression to change from ${30}^{\circ }$ to ${60}^{\circ }$. How soon will the boat reach the sea-shore?

• A. 20 min
• B. 10 min
• C. 5 min
• D. 2.5 min

Answer 1

The correct option is C

5 min

Let the height of the tower be 'h'.

Let C and D be two positions of the boat such that $\angle ACB={60}^{\circ }$ and $\angle ADB={30}^{\circ }$ Let AC = x and AD = y.

In $\Delta ABC$

$tan{60}^{\circ }=\frac{h}{x}$

$\sqrt{3}=\frac{h}{x}$

$x=\frac{h}{\sqrt{3}}$ ....... (i)

In $\Delta ABD$

$tan{30}^{\circ }=\frac{h}{y}$

$\frac{1}{\sqrt{3}}=\frac{h}{y}$

$y=h\sqrt{3}$ .......... (ii)

To travel (y - x) it took 10 minutes for the boat

i.e. to travel $(h\sqrt{3}-\frac{h}{\sqrt{3}})=\frac{2h}{\sqrt{3}}$ it took 10 minutes for the boat.

$\therefore$ to travel x i.e. $\frac{h}{\sqrt{3}}$ it will take

$\frac{\frac{h}{\sqrt{3}}}{\frac{2h}{\sqrt{3}}}\times 10=5min$.
So, the boat will reach the sea-shore in 5 min.