Question

The angle of elevation of the a tower as seen by an observer is ${30}^{\circ }$. The observer is at a distance of $30\sqrt{3}$m from the tower. If the eye level of the observer is $1.5$ m above the ground level, then find the height of the tower.

Answer 1

Let $BD$ be the height of the tower and $AE$ be the distance of the eye level of the observer from the ground level.

Draw $EC$ parallel to $AB$ such that $AB=EC$.

Given $AB=EC=30\sqrt{3}\text{m}$ and $AE=BC=1.5\text{m}$

In right angled $△DEC$,

$\tan {30}^{\circ }=\frac{CD}{EC}$

$\Rightarrow CD=EC\tan {30}^{\circ }=\frac{30\sqrt{3}}{\sqrt{3}}$

$\therefore CD=30\text{m}$

Thus, the height of the tower, $BD=BC+CD$

$=1.5+30=31.5\text{m}$.