Question

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

Answer 1

Let height of the tower be ${}^{\prime }{h}^{\prime }$,
From figure,
$tan\theta =\frac{h}{b}$___________(1)
$tan(\frac{\pi }{2}-\theta )=\frac{h}{a}$__________(2)
=> $cot\theta =\frac{h}{a}$
On multiplying (1) & (2), we get
$cot\theta .tan\theta =\frac{h^2}{ab}$
=> $\frac{h^2}{ab}=1$
Therefore,
$h=\sqrt{ab}$ meters
Hence proved.