Question

A tower stands vertically on the ground. From a point on the ground which is $20m$ away from the foot of the tower is found to be $60°$. The height of the tower is:

Answer 1

Let , $AB$ is the height of a tower and $CB=20m$ is a point away from the foot of the tower.

In $∆ABC$

$$\begin{gathered} \Rightarrow \tan C=\frac{AB}{CB} \\ \Rightarrow \tan 60°=\frac{AB}{20} \\ \Rightarrow \sqrt{3}=\frac{AB}{20} \\ \Rightarrow AB=20\sqrt{3}m \end{gathered}$$

Hence , the height of tower is $20\sqrt{3}m$