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

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

According to question,

⇒ α + β = 90°

⇒ β = 90° - α

From figure,

In △ABD,

$\Rightarrow \text{tan α} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan α} = \dfrac{AB}{BD} \\[1em] \Rightarrow \text{tan α} = \dfrac{h}{b} ………..(1)$

In △ABC,

$\Rightarrow \text{tan β} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan β} = \dfrac{AB}{BC} \\[1em] \Rightarrow \text{tan β} = \dfrac{h}{a} \\[1em] \Rightarrow \text{tan (90° - α)} = \dfrac{h}{a} \\[1em] \Rightarrow \text{cot α} = \dfrac{h}{a} ……….(2)$

Multiplying (1) by (2) we get,

⇒ tan α cot α = $\dfrac{h}{a} \times \dfrac{h}{b}$

⇒ $\dfrac{\text{sin α}}{\text{cos α}} \times \dfrac{\text{cos α}}{\text{sin α}} = \dfrac{h^2}{ab}$

⇒ 1 = $\dfrac{h^2}{ab}$

⇒ h² = ab

⇒ h = $\sqrt{ab}$ meters.

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