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 complimentary. Prove that the height of the tower is $\sqrt{ab}$.

Answer 1

Let $AB$ be the tower and $C,D$ are two points on the same side of the tower such that $BD=b$ and $BC=a$
In $△ABC$
$\frac{h}{b}=\tan (90-\alpha )=\cot \alpha ...\left(1\right)$
In $△ABD$,
$\frac{h}{a}=\tan \alpha ....\left(2\right)$
Multiplying $\left(1\right)$ and $\left(2\right)$ we get:
$\Rightarrow \tan \alpha .\cot \alpha =1$
$\Rightarrow \frac{h}{a}\times \frac{h}{b}=1$
$\Rightarrow h^2=ab$
$\Rightarrow h=\sqrt{ab}$