Question

Question 4
The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower is ${30}^{\circ }$. Find the height of the tower.

Answer 1

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
As per question,

In right $\Delta$ABC,
tan ${30}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{30}$
$\Rightarrow AB=10\sqrt{3}$
Thus, the height of the tower is $10\sqrt{3}$ m.