Question

From the top of a tower of height $60$m, the angles of depression of the top and the bottom of a building are observed to be ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the building.

• A. $20$m
• B. $60$m
• C. $80$m
• D. $40$m

Answer 1

Answer: (D)

$40$m

Let $BE=h,AB=60-h,BC=ED$

Distance between tower and building be $ED=x$

In $\Delta AED,\tan {60}^{\circ }=\frac{AE}{ED}=\frac{60}{x}$

$\Rightarrow \sqrt{3}=\frac{60}{x}$

$\Rightarrow x=20\sqrt{3}m$

In $\Delta ABC,\tan {30}^{\circ }=\frac{AB}{BC}=\frac{60-h}{20\sqrt{3}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-h}{20\sqrt{3}}$

$\Rightarrow h=40m$