Question

Two pillars of equal height and stand on either side of a roadway which is $150$ m wide. At a point in the roadway between the pillars, the angle of elevation of the top of pillars are ${60}^0$and ${30}^0$. then find height of pillars-

Answer 1

Let us consider the $CD$ and $AB$ are the poles and $O$ is the point where elevation angle is made.

In $\Delta ABO$

$\begin{array}{c}\frac{AB}{BO}=\tan {60}^{\circ }\\ \frac{AB}{BO}=\sqrt{3}\\ BO=\frac{AB}{\sqrt{3}}\end{array}$

In $\Delta CDO$

$\begin{array}{c}\frac{CD}{DO}=\tan {30}^{\circ }\\ \frac{CD}{150-BO}=\frac{1}{\sqrt{3}}\\ CD\sqrt{3}=150-\frac{AB}{\sqrt{3}}\\ CD\sqrt{3}+\frac{AB}{\sqrt{3}}=150\end{array}$

Since the poles are of equal heights

therefore $AB=CD$

So,

$\begin{array}{c}CD\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)=150\\ CD\left(\frac{3+1}{\sqrt{3}}\right)=150\\ CD\left(\frac{4}{\sqrt{3}}\right)=150\\ CD=37.5\sqrt{3}\end{array}$

$\begin{array}{c}BO=\frac{AB}{\sqrt{3}}\\ =\frac{CD}{\sqrt{3}}\\ =\frac{37.5\sqrt{3}}{\sqrt{3}}\\ =37.5\\ DO=BD-BO\\ =150-37.5\\ =112.5\end{array}$

Hence the height of poles are $37.5\sqrt{3}$ and the distance of point $O$ from either of the poles is $37.5m$ and $112.5m$