Question

If the angle of elevation of the top of a building from a point 50 m away its base is ${60}^{\circ }$, the height of the building is:

• A. $100\sqrt{3}$ m
• B. $50(\sqrt{3}-1)$ m
• C. $\frac{500}{\sqrt{3}}$ m
• D. $50\sqrt{3}$ m

Answer 1

Answer: (C)

$\frac{500}{\sqrt{3}}$ m

In $△ABC,\tan {30}^{\circ }=\frac{AB}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{500}$

$\Rightarrow h=500\times \frac{1}{\sqrt{3}}=\frac{500}{\sqrt{3}}$

$\Rightarrow h=\frac{500}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow h=\frac{500\sqrt{3}}{3}$

$\therefore$ height of the building is $\frac{500\sqrt{3}}{3}=288.67$m