Question

The angle of elevation of the top of a vertical tower from two points. $30\text{m}$ apart, and on the same straight line passing through the base of tower, are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. The height of the tower is

• A. $10\text{m}$
• B. $15\text{m}$
• C. $15\sqrt{3}\text{m}$
• D. $30\text{m}$

Answer 1

The correct option is C $15\sqrt{3}\text{m}$

Let $AB$ is the tower of height $h$, and $BC$ equal to $x.$

In $△ABC,$

$\tan {60}^{\circ }=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$

$\Rightarrow x=\frac{h}{\sqrt{3}}\dots \left(i\right)$

In $△ABD,$

$\tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{(x+30)}$

$\Rightarrow x+30=\sqrt{3}h$

$\Rightarrow \sqrt{3}h-30\dots \left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right),$

$\begin{array}{cc}\frac{h}{\sqrt{3}}& =\sqrt{3}h-30\\ h& =3h-30\sqrt{3}\\ 2h& =30\sqrt{3}\\ h& =15\sqrt{3}\text{m}\end{array}$

So, height of the tower is $15\sqrt{3}\text{m}.$