Question

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

Answer 1

Given,

the angle of elevation of the top of the tower from two points P & Q is at a distance of a & b.

Also given, to prove that the tower

height$=\sqrt{ab}$

($\because$ complementary angle $=({90}^o-\theta )$)

From $\Delta$ABP

$\tan \theta =\frac{AB}{BP}=\frac{AB}{a}$ ……..$\left(1\right)$

From $\Delta$ABQ

$\tan (90-\theta )=\frac{AB}{BQ}$

$(\because \tan (90-\theta )=\cot \theta )$

$(\cot \theta =\frac{1}{\tan \theta })$

We get,

$\cot \theta =\frac{BQ}{AB}=\frac{b}{AB}$ ……..$\left(2\right)$

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

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

$\Rightarrow A{B}^2=ab\iff AB=\sqrt{ab}$

$\therefore AB=$height$=\sqrt{ab}$

Hence proved.