Question

The angles of depression of the top and the bottom of a building $50\text{m}$ high are observed from the top of a tower are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the tower and the horizontal distance between the building and the tower.

Answer 1

In $△BTP$, we have

$\tan {30}^{\circ }=\frac{TP}{BP}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{TP}{BP}$

$\Rightarrow BP=TP\sqrt{3}\dots \left(i\right)$

In $△GTR$, we have

$\tan {60}^{\circ }=\frac{TR}{GR}$

$\Rightarrow \sqrt{3}=\frac{TR}{GR}$

$\Rightarrow GR=\frac{TR}{\sqrt{3}}\dots \left(ii\right)$

As $BP=GR$ so, on comparing $\left(i\right)$ and $\left(ii\right),$

$TP\sqrt{3}=\frac{TR}{\sqrt{3}}$

$3TP=TP+PR$

$2TP=PR$

$2TP=BG$

$TP=\frac{50}{2}\text{m}$$=25\text{m}$

Now,

$TR=TP+PR$

$TR=(25+50)\text{m}$

Height of tower is equal to $TR$,

$TR=75\text{m}$

Distance between building and tower is equal to,

$GR=\frac{TR}{\sqrt{3}}$

$\Rightarrow GR=\frac{75}{\sqrt{3}}\text{m}$

$\Rightarrow GR=25\sqrt{3}\text{m}$