# Two poles of equal heights are standing opposite each other on either side of the road, which is $80\ m$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^{o}$ and $30^{o}$ respectively. Find the height of the poles and the distances of the point from the poles.

Given: Two poles $AB$ and $CD$ of equal heights, Let $AB=CD=h$, Distance between the poles, $BD=80\ m$

To do: To  find the height of the poles and the distances of the point from the poles.

Solution:

Let the given point be $E$ such that $BE=x$ and $ED=80−x$

In $\vartriangle ABE$

$tan30^{o}=\frac{AB}{BE}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}$

$\Rightarrow x=\sqrt{3}h$    .........$( 1)$

In $\vartriangle CED$

$tan60^{o}=\frac{CD}{ED}$

$\sqrt{3}=\frac{h}{80-x}$

$\Rightarrow h=( 80-x)\sqrt{3}$            .........$( 2)$

Put value of x in equation $( 2)$, we get

$h=( 80-\sqrt{3}h)\sqrt{3}$

$\Rightarrow h=80\sqrt{3}-3h$

$\Rightarrow h+3h=80\sqrt{3}$

$\Rightarrow 4h =80\sqrt{3}$

$\Rightarrow h=\frac{80\sqrt{3}}{4}$

$\Rightarrow h=20\sqrt{3}\ m$

And from $( 1)$, we know

$x=h\sqrt{3}$

$\Rightarrow x=20\sqrt{3}\times\sqrt{3}$

$\Rightarrow x=60\ m$

$\therefore DE=80-x=80-60=20\ m$

Hence, the heights of each pole is $20\sqrt{3}\ m$ and distance of the point from the poles are $60\ m$ and $20\ m$.

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Updated on: 10-Oct-2022

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