Question

Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?

Answer 1

Let AB and CD be the two building standing on the road.

Suppose the height of the second building be h m.



Here, AB = 12 m, BD = 15 m, ∠CAE = 30º and AE ⊥ CD.

CE = CD − ED = (h − 12) m (ED = AB)

AE = BD = 15 m

In right ∆AEC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{CE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h-12}{15} \\ \Rightarrow h-12=\frac{15}{\sqrt{3}}=5\sqrt{3} \\ \Rightarrow h=\left(12+5\sqrt{3}\right)\mathrm{m} \end{gathered}$$
Thus, the height of the second building is $\left(12+5\sqrt{3}\right)$ m.