Question

The shadow of a tower standing on a plane ground is found to be $40m$ longer when the sun's altitude reduces to ${30}^{\circ }$ from ${60}^{\circ }$. Find the height of the tower.

Answer 1

The given scenario is shown in the attached diagram.

Given:

$\angle ADB={30}^o$

$\angle ACB={60}^o$

$CD=40m$

To find $AB$

Solution:

$\tan \angle ADB=\frac{AB}{DB}$

$\tan \left({30}^o\right)=\frac{AB}{BC+CD}$

$\frac{1}{\sqrt{3}}=\frac{AB}{BC+40}..............\left(i\right)$

$\tan \angle ACB=\frac{AB}{BC}$

$\tan \left({60}^o\right)=\frac{AB}{BC}$

$\sqrt{3}=\frac{AB}{BC}...............\left(ii\right)$

Dividing (ii) by (i), we get

$\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\frac{\frac{AB}{BC}}{\frac{AB}{BC+40}}$

Solving, we get $BC=20m$

Substituting $BC$ in (ii), we get $AB=20\sqrt{3}m$