A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 60° with the man's eye when at a distance of 60 metres from the tower. After 9 seconds, the angle of depression becomes 30°. The speed of the boat, assuming that it is running in still water, will be:

48 kmph

The correct answer is Option (3) → 48 kmph

Step-by-Step Calculation:

1. Determine the height of the tower ($h$):

Let the height of the tower be $h$. Initially, the boat is at a distance of 60 m from the base of the tower, and the angle of depression is 60°.

In a right-angled triangle formed by the tower and the initial position of the boat:

$\tan(60^\circ) = \frac{h}{60}$

$\sqrt{3} = \frac{h}{60} ⇒h = 60\sqrt{3} \text{ m}$

2. Determine the distance after 9 seconds ($d_2$):

After 9 seconds, the boat has moved further away, and the angle of depression becomes 30°. Let the new distance from the tower be $d_2$.

$\tan(30^\circ) = \frac{h}{d_2}$

$\frac{1}{\sqrt{3}} = \frac{60\sqrt{3}}{d_2}$

$d_2 = 60\sqrt{3} \times \sqrt{3} = 60 \times 3 = 180 \text{ m}$

3. Calculate the distance traveled by the boat:

The boat traveled from 60 m to 180 m away from the tower.

$\text{Distance traveled} = 180 \text{ m} - 60 \text{ m} = 120 \text{ m}$

4. Calculate the speed of the boat:

The boat covers this distance in 9 seconds.

$\text{Speed (m/s)} = \frac{\text{Distance}}{\text{Time}} = \frac{120}{9} = \frac{40}{3} \text{ m/s}$

5. Convert the speed to kmph:

To convert from m/s to km/h, we multiply by 3.6 (or $\frac{18}{5}$):

$\text{Speed (km/h)} = \frac{40}{3} \times \frac{18}{5}$

$\text{Speed (km/h)} = 8 \times 6 = 48 \text{ kmph}$

Conclusion:

The speed of the boat is 48 kmph.