Question

Two poles of equal heights are standing opposite each other on either side of the road, which is $80\mathrm{m}$ wide.

From a point between them on the road, the angles of elevation of the top of the poles are $60°$ and $30°$, respectively.

Find the height of the poles and the distances of the point from the poles.

Answer 1

Step 1: Form a system of linear equations.

Let $\mathrm{AB}$ and $\mathrm{CD}$ be the poles of equal height.

$\mathrm{O}$ is the point from where the angle of elevations is taken and $\mathrm{BD}$ is the distance between the poles.

From the diagram, $\mathrm{OB}+\mathrm{OD}=80\mathrm{m}$ $...\left(\mathrm{i}\right)$

Now, in $\mathrm{\Delta CDO},\tan \left(30°\right)=\frac{\mathrm{CD}}{\mathrm{OD}}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{\mathrm{CD}}{\mathrm{OD}}$

$\Rightarrow$ $\mathrm{OD}=\sqrt{3}\mathrm{CD}$ $...\left(\mathrm{ii}\right)$

And, in $\mathrm{\Delta ABO},\tan \left(60°\right)=\frac{\mathrm{AB}}{\mathrm{OB}}$

$\Rightarrow$ $\sqrt{3}=\frac{\mathrm{AB}}{\mathrm{OB}}$

$\Rightarrow$ $\mathrm{OB}=\frac{\mathrm{AB}}{\sqrt{3}}$ $...\left(\mathrm{iii}\right)$

Step 2: Calculate the required height of the poles.

Using $\left(\mathrm{ii}\right)$ and $\left(\mathrm{iii}\right)$ in $\left(\mathrm{i}\right)$, we get;

$\frac{\mathrm{AB}}{\sqrt{3}}+\sqrt{3}\mathrm{CD}=80$

$\Rightarrow$$\mathrm{AB}\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)=80$ $\left[\because \mathrm{AB}=\mathrm{CD}\right]$

$\Rightarrow$ $\mathrm{AB}\left(\frac{1+3}{\sqrt{3}}\right)=80$

$\Rightarrow$ $\mathrm{AB}\left(\frac{4}{\sqrt{3}}\right)=80$

$\Rightarrow$ $\mathrm{AB}=80\left(\frac{\sqrt{3}}{4}\right)$

$\Rightarrow$ $\mathrm{AB}=20\sqrt{3}\mathrm{m}$ $...\left(\mathrm{iv}\right)$

$\Rightarrow$ $\mathrm{CD}=20\sqrt{3}\mathrm{m}$ $...\left(\mathrm{v}\right)$

Step 3: Calculate the required distance of the point from the poles.

Using $\left(\mathrm{iv}\right)$ in $\left(\mathrm{iii}\right)$, we get;

$\mathrm{OB}=\frac{20\sqrt{3}}{\sqrt{3}}$

$\Rightarrow$$\mathrm{OB}=20\mathrm{m}$

Since, $\mathrm{OB}+\mathrm{OD}=80$

$\Rightarrow$ $\mathrm{OD}=80-20$

$\Rightarrow$ $\mathrm{OD}=60\mathrm{m}$

Hence, the height of the poles is $20\sqrt{3}\mathrm{m}$ and the distances of the point from the poles are $20\mathrm{m}$ and $60\mathrm{m}$ from the first and second pole, respectively.