The angle of elevation of the top of a tower from a certain point is 60°. If the observer moves 10 m away from the tower, the angle of elevation of the top of the tower decreases by 15°. The height of the tower is:

23.65 m

The correct answer is Option (3) → 23.65 m

Given:

Initial angle of elevation = 60°
After moving 10 m away, angle = 45°

Let height of tower = $h$
Initial distance from tower = $x$

From first triangle:

$\tan 60^\circ = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$

From second triangle:

$\tan 45^\circ = \frac{h}{x + 10} \Rightarrow 1 = \frac{h}{x + 10} \Rightarrow h = x + 10$

Equating both expressions for $h$:

$x\sqrt{3} = x + 10$

$x(\sqrt{3} - 1) = 10$

$x = \frac{10}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{10(\sqrt{3} + 1)}{2}$

$x = 5(\sqrt{3} + 1)$

Now, $h = x\sqrt{3} = 5(\sqrt{3} + 1)\sqrt{3} = 5(\sqrt{9} + \sqrt{3}) = 5(3 + \sqrt{3})$