Question

The angle of elevation of the top of a tower from two-point at a distance $4m$ and $9m$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6m.$

Answer 1

In the above figure, let $AB$ is the tower of height $h$ and point $C$ and $D$ are $4m$ and $9m$ respectively.

In $△ABC,$

$\tan \theta =\frac{AB}{AC}$

$\Rightarrow \tan \theta =\frac{h}{4}\dots \left(1\right)$

Now, in $△ABD,$

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

$\Rightarrow \cot \theta =\frac{h}{9}\dots \left(2\right)$

Multiply equation $\left(1\right)$ and $\left(2\right),$

$\tan \theta \times \cot \theta =\frac{h}{4}\times \frac{h}{9}$

$\Rightarrow 1=\frac{h^2}{36}$

$\Rightarrow h^2=36$

$\Rightarrow h=6m$

Hence, proved height of the tower is equal to $6m.$