Two poles of equal heights are standing opposite each other on either sides of the roads, which is 80 m wide. From a point between them on the road, the angle 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 of equal heights, distance between the poles is 80 m. from a point angle of the top of the elevation is $60^{o}$ and $30^{o}$

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

Solution:

Given $AB=80$ m

Let $AP=a$ meter,

$\therefore$, $PB=( 80\ -\ x)$ m

In $\vartriangle APC.$

$tan30^{o}=\frac{AC}{AP}$

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

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

In$\vartriangle BPD$,

$tan60^{o} =\frac{BD}{BP}$

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

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

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

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

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

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

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

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

$\therefore \ a=h\sqrt{3}$

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

$=60$ m

Thus $AP=a=60$ m and $BP=80-a=80-60=20$ m

Therefore, The height of the poles is $20\sqrt{3}$ m and the distance of the point from first pole is $60$ m and from the second pole is $20$ m.
 

Updated on: 10-Oct-2022

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