The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt{ab}$ metres.

Given,

AB is the tower of height h meters, BC = a meters and BD = b meters.

From figure,

In △ABD,

$\Rightarrow \tan (90° - \theta) = \dfrac{AB}{BD} \\[1em] \Rightarrow \tan (90° - \theta) = \dfrac{h}{b} \\[1em] \Rightarrow \cot \theta = \dfrac{h}{b}….(1) \\[1em] \text{ In △ABC,} \\[1em] \Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan \theta = \dfrac{AB}{BC} \\[1em] \Rightarrow \tan \theta = \dfrac{h}{a}…..(2)$

Multiplying (1) by (2) we get,

$\Rightarrow \tan \theta \cot \theta = \dfrac{h}{a} \times \dfrac{h}{b} \\[1em] \Rightarrow \dfrac{\sin \theta}{\cos \theta} \times \dfrac{\cos \theta}{\sin \theta} = \dfrac{h^2}{ab} \\[1em] \Rightarrow 1 = \dfrac{h^2}{ab} \\[1em] \Rightarrow h^2 = ab \\[1em] \Rightarrow h = \sqrt{ab}.$

Hence, proved that h = $\sqrt{ab}$ meters.