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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