Question

The shadow of a tower standing on the level ground to be $60\text{m}$ shorter when the sun's altitude changes from ${30}^{\circ }$ to ${60}^{\circ }$, find the height of the tower.

• A. $90\sqrt{30}\text{m}$
• B. $30\sqrt{3}\text{m}$
• C. $90\text{m}$
• D. None of the above

Answer 1

The correct option is B $30\sqrt{3}\text{m}$

In $△ABC,$

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

$\Rightarrow \sqrt{3}=\frac{AB}{CB}$

$\Rightarrow CB=\frac{AB}{\sqrt{3}}\dots \left(i\right)$

In $△ABD,$

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{DB}$

$\Rightarrow DB=\sqrt{3}AB$

$\Rightarrow DC+CB=\sqrt{3}AB$

$\Rightarrow 60+CB=\sqrt{3}AB$

$\Rightarrow CB=\sqrt{3}AB-60\dots \left(ii\right)$

On comparing $\left(i\right)$ & $\left(ii\right),$

$\frac{AB}{\sqrt{3}}=\sqrt{3}AB-60$

$\Rightarrow \frac{AB}{\sqrt{3}}-\sqrt{3}AB=-60$

$\Rightarrow \frac{AB-3AB}{\sqrt{3}}=-60$

$\Rightarrow 2AB=60\sqrt{3}$

$\Rightarrow AB=30\sqrt{3}$

So, height of the tower is equal to $30\sqrt{3}\text{m}$.