If the angle of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is

Answer: Option B

Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $$\theta$$ and (90° – $$\theta$$)
Let height of tower = h

$$\begin{array}{rl}& \text{Then}\text{in}\text{right}\Delta APB,\\ & \tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{AB}{PB}\\ & =\frac{h}{a}............\left(\text{i}\right)\\ & \text{Similarly in right }\Delta AQB,\\ & \tan \left({90}^{\circ }-\theta \right)=\frac{AB}{QB}=\frac{h}{b}\\ & \Rightarrow \cot \theta =\frac{h}{b}...........\left(\text{ii}\right)\\ & \text{Multiplying }\left(\text{i}\right)\text{ and }\left(\text{ii}\right)\\ & \tan \theta \cot \theta =\frac{h}{a}\times \frac{h}{b}\\ & \Rightarrow 1=\frac{h^2}{ab}\\ & \Rightarrow h^2=ab\\ & \Rightarrow h=\sqrt{ab}\\ & \therefore \text{Height}\text{of}\text{tower}=\sqrt{ab}\end{array}$$