Question

The angle of elevation of the top of a vertical tower from a point on the ground is ${60}^{\circ }$. From another point 10 m vertically above the first, its angle of elevation is ${30}^{\circ }$ . Find the height of the tower.

Answer 1

Let the height of the tower be h cm.

Now, In $△PAB$

$\tan {60}^o=\frac{AP}{AB}\Rightarrow \sqrt{3}=\frac{h}{AB}\Rightarrow AB=\frac{h}{\sqrt{3}}----\left(1\right)$

And, In $△PCD$

$\tan {30}^o=\frac{PD}{CD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{CD}\Rightarrow CD=\sqrt{3}(h-10)----\left(2\right)$

Since, AB = CD, So, equation (2) becomes,

$AB=\sqrt{3}(h-10)----\left(3\right)$

Equating equation (1) and (3), we get,

$\frac{h}{\sqrt{3}}=\sqrt{3}(h-10)h=3(h-10)2h=30=h=15m$

Hence, the height of the tower will be 15 m