Question

Question 13
The angle of elevation of the top of a tower 30m high from the foot of another tower in the same plane is ${60}^{\circ }$ and the angle of elevation of the top of the second tower from the foot of the first tower is ${30}^{\circ }$. Find the distance between the two towers and also the height of the tower.

Answer 1

Let distance between the two towers = AB = x m

And height of the other tower = PA = h m

Given that, height of the tower = QB = 30 m and $\angle QAB={60}^{\circ },\angle PBA={30}^{\circ }$

Now, in $\Delta QAB$, $tan{60}^{\circ }=\frac{QB}{AB}=\frac{30}{x}$

$\Rightarrow \sqrt{3}=\frac{30}{x}$

$\therefore x=\frac{30}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}m$

And in $\Delta PBA$,

$tan{30}^{\circ }=\frac{PA}{AB}=\frac{h}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{10\sqrt{3}}$ $[\because x=10\sqrt{3}m]$

$\Rightarrow h=10m$

Hence, the required distance and height are $10\sqrt{3}m$ and 10 m, respectively.