Question

The angle of elevation of the top of a tower from two distinct points which are at a distance of a meter and b meter away from its foot are complementary. Prove that the height of the tower is $\sqrt{ab}$ meter.

Answer 1

Based on the given information, we can draw the figure shown above.

Here, the angles of elevation are complementary.

So, if $\angle ACB=\alpha$

$\angle ADB={90}^o-\alpha$

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

Hence, in $△ABC,$

$\tan \alpha =\frac{h}{a}$ ___ (i)

Similarly, in $△ABD,$

$\tan ({90}^o-\alpha )=\frac{h}{b}$

$\tan ({90}^o-\theta )=\cot \theta$

$\therefore \cot \alpha =\frac{h}{b}$ ___ (ii)

Multiplying (i) and (ii), we get:

$\tan \alpha \times \cot \alpha =\frac{h}{a}.\frac{h}{b}$

$\Rightarrow \tan \alpha \times \frac{1}{\tan \alpha }=\frac{h^2}{ab}$

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

$\Rightarrow h^2=ab$

$\therefore h=\sqrt{ab}$

Hence, the height of the tower is $\sqrt{ab}m.\left[Henceproved\right]$