If the angles of elevation of a tower from two points distant a and b (a > b) from its foot and in the same straight line from it and on the same side, are 30° and 60°, then the height of the tower is :

1. $\sqrt{a + b}$

2. $\sqrt{ab}$

3. $\sqrt{a - b}$

4. $\sqrt{\Big(\dfrac{a}{b}\Big)}$

Let height of the tower (AB) = h.

Let BC = b and BD = a.

In △ABC,

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

In △ABD,

$\Rightarrow \tan 30^\circ = \dfrac{AB}{BD} \\[1em] \Rightarrow \tan 30^\circ = \dfrac{h}{a} \\[1em] \Rightarrow \tan (90^{\circ} - 60^\circ) = \dfrac{h}{a} \\[1em] \Rightarrow \cot 60^\circ = \dfrac{h}{a} \\[1em] \Rightarrow \dfrac{1}{\tan 60^\circ} = \dfrac{h}{a} \\[1em] \Rightarrow \dfrac{1}{\dfrac{h}{b}} = \dfrac{h}{a} \text{ [From (i)]} \\[1em] \Rightarrow \dfrac{b}{h} = \dfrac{h}{a} \\[1em] \Rightarrow h^2 = ab \\[1em] \Rightarrow h = \sqrt{ab}$

Hence, option 2 is the correct option.