Question

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

50m 

Answer 1

From the given figure,

Step 1- Finding the value of base BC

the height of the tower $\left(CD\right)=50m$

$$\begin{gathered} In△BCD \\ \tan 60°=\frac{CD}{BC} \\ \Rightarrow \sqrt{3}=\frac{50}{BC} \\ \Rightarrow BC=\frac{50}{\sqrt{3}}m-------\left(eq1\right) \end{gathered}$$

Step 2- Finding the height of building AB

$$\begin{gathered} In△ABC \\ \because \tan \left(30°\right)=\frac{AB}{BC} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{{\displaystyle \frac{50}{\sqrt{3}}}}(fromeq1) \\ \Rightarrow AB=\frac{1}{\sqrt{3}}\times \frac{50}{\sqrt{3}}m \\ \Rightarrow AB=\frac{50}{3}m \end{gathered}$$

Hence, the height of the building is $AB=\frac{50}{3}m$