Question

The angle of elevation of the top of a building from the foot of the tower is ${30}^{\circ }$ and the angle of elevation of the top of the tower from the foot of the building is ${60}^{\circ }$. If the tower is 50 m high, find the height of the building.

Answer 1

Let CD be the height of the tower equal to 50 m.
Let AB be the height of the building.
BC be the distance between the foot of the building and the tower.
Elevations are ${30}^{\circ }$ and ${60}^{\circ }$ from the tower and the building respectively.
As per question,
In right $\Delta BCD$,
tan ${60}^{\circ }=\frac{CD}{BC}$
$\Rightarrow \sqrt{3}=\frac{50}{BC}$
$\Rightarrow BC=\frac{50}{\sqrt{3}}m$
Also,
In right $\Delta ABC$,
tan ${30}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BC}$
$\Rightarrow AB=\frac{50}{3}m$
Thus, the height of the building is $\frac{50}{3}m$.