Question

A man standing at a certain distance from a building, observes the angle of elevation of its top as ${60}^{\circ }$. He walks 30 meters away from the building. Now, the angle of elevation of the building's top is ${30}^{\circ }$. How high is the building?

• A. 15 m
• B. 153 m
• C. 15(3-1) m
• D. 30 m
• E. 53 m

Answer 1

The correct option is B

153 m

Let the height of the building be `h', and let `d' be the original distance between the man and the building. The following figure depicts the given situation:

We have,

tan ${60}^{\circ }$ = $\sqrt{3}=\frac{h}{d}$

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

tan ${30}^{\circ }$ = $\frac{1}{\sqrt{3}}=\frac{h}{d+30}$

$\Rightarrow \sqrt{3}h=d+30=\frac{h}{\sqrt{3}}+30$

$\Rightarrow h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=30$

$\Rightarrow h=15\sqrt{3}m$