Question

Question 10
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 ${60}^{\circ }$ and ${30}^{\circ }$, respectively. Find the height of the poles and the distances of the point from the poles.

Answer 1

Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
As per question,
$AB=CD$,
$OB+OD=80m$
Now,

In right $\Delta CDO$,
tan ${30}^{\circ }=CD/OD$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{CD}{OD}$
$\Rightarrow CD=\frac{OD}{\sqrt{3}}...\left(i\right)$
Also,
In right $\Delta ABO$,
tan ${60}^{\circ }=\frac{AB}{OB}$
$\Rightarrow \sqrt{3}=\frac{AB}{(80-OD)}$
$\Rightarrow AB=\sqrt{3}(80-OD)$
$AB=CD$
$\Rightarrow \sqrt{3}(80-OD)=\frac{OD}{\sqrt{3}}$
$\Rightarrow 3(80-OD)=OD$
$\Rightarrow 240-3OD=OD$
$\Rightarrow 4OD=240$
$\Rightarrow OD=60m$
Putting the value of OD in equation (i)
$CD=\frac{OD}{\sqrt{3}}\Rightarrow CD=\frac{60}{\sqrt{3}}\Rightarrow CD=20\sqrt{3}m$
Also,
$OB+OD=80m\Rightarrow OB=(80-60)m=20m$
Thus, the height of the poles are $20\sqrt{3}$ m and distance from the point of elevation are 20 m and 60 m respectively.