Question

An observer $1.6m$ tall is $20\sqrt{3}m$ away from a tower. The angle of elevation from his eye to the top of the tower is ${30}^{\circ }$. The heights of the tower is:

• A. $21.6m$
• B. $23.2m$
• C. $24.72m$
• D. None of these

Answer 1

Answer: (A)

$21.6m$

Let $AB$ be the observer and $CD$ be the tower.
Draw $BE\perp CD$.
Then, $CE=AB=1.6m$,
$BE=AC=20\sqrt{3}m$.
$\frac{DE}{BE}=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
$\Rightarrow DE=\frac{20\sqrt{3}}{\sqrt{3}}m=20m$.
$\therefore CD=CE+DE=(1.6+20)m=21.6m$.