Question

The angle of elevation of the top of a tower from two. points at distance a and b meters from the base and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt{ab}$ meter.

Answer 1

$TD\to AB=\sqrt{ab}$

In $\Delta$ABP In $\Delta$ABQ

$\tan \theta =\frac{P}{B}$ $\tan 90-\theta =\cot \theta =\frac{P}{B}$

$=\frac{AB}{a}$ ……$\left(1\right)$ $\tan \theta =\frac{B}{P}\Rightarrow \frac{b}{AB}$ ………….$\left(2\right)$

By equating equation $\left(1\right)$ & $\left(2\right)$, we get

$\frac{AB}{a}=\frac{b}{AB}$

$A{B}^2=ab$

$AB=\sqrt{ab}$

Hence proved.