On the level ground, the angle of elevation of a tower is 30°. On moving 20 m nearer, the angle of elevation is 60°. The height of the tower is:

1. 10 m

2. $10\sqrt{3}$ m

3. 15 m

4. 20 m

Let AB be the tower of height h.

In △ABC,

$\Rightarrow \tan 60^\circ = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{BC} \\[1em] \Rightarrow BC = \dfrac{h}{\sqrt{3}} ….(1)$

In △ABD,

$\Rightarrow \tan 30^\circ = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{BC + 20} \\[1em] \Rightarrow BC + 20 = h\sqrt{3} ….(2)$

Substituting value of BC from equation (1) in (2), we get :

$\Rightarrow \dfrac{h}{\sqrt{3}} + 20 = h\sqrt{3} \\[1em] \Rightarrow \dfrac{h + 20\sqrt{3}}{\sqrt{3}} = h\sqrt{3} \\[1em] \Rightarrow h + 20\sqrt{3} = h(3) \\[1em] \\[1em] \Rightarrow 20\sqrt{3} = 2h \\[1em] \Rightarrow h = 10\sqrt{3} \text{ m}$

Hence, option 2 is the correct option.