Question

A man observes the angle of elevation of the top of a building to be ${30}^{\circ }$. He walks towards it in horizontal line through its base. On covering 60 m the angle of elevation changes to ${60}^{\circ }$. Find the height of the building correct to the nearest metre.

• A. 50 m
• B. 51 m
• C. 52 m
• D. 55 m

Answer 1

The correct option is C

52 m

Let 'D' be the first point of observation and 'B' be the top of the building.

$\angle ADB={30}^{\circ }$.

Given, on walking 60m towards the building the angle of elevation becomes ${60}^{\circ }$.

$\therefore$ DC = 60m and $\angle ACB={60}^{\circ }$.

Let the height of the building = h and CA = x.

In $\Delta ACB$

$tan{60}^{\circ }=\frac{h}{x}$

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

$\therefore h=x\sqrt{3}$ .......(i)

In $\Delta ADB$

$tan{30}^{\circ }=\frac{h}{60+x}$

$\frac{1}{\sqrt{3}}=\frac{x\sqrt{3}}{60+x}$ .... from (i)

$60+x=3x$

$\therefore 2x=60$

$x=30m$

$\therefore$ height of the building (h) $=x\sqrt{3}$

$=30\left(1.732\right)$

$=51.96m$

$=52m$