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.

(N/A) Let $AB$ and $CD$ be the poles of equal height $h$,and $O$ be the point on the road $BD$ such that $BD = 80\, m$.
Let $BO = x\, m$,then $OD = (80 - x)\, m$.
In $\triangle ABO$,$\tan 60^{\circ} = \frac{AB}{BO} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots(1)$.
In $\triangle CDO$,$\tan 30^{\circ} = \frac{CD}{OD} \implies \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}} \quad \dots(2)$.
Equating $(1)$ and $(2)$:
$x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$
$3x = 80 - x$
$4x = 80 \implies x = 20\, m$.
So,$BO = 20\, m$ and $OD = 80 - 20 = 60\, m$.
Substituting $x$ in $(1)$:
$h = 20\sqrt{3}\, m$.
Thus,the height of the poles is $20\sqrt{3}\, m$ and the distances of the point from the poles are $20\, m$ and $60\, m$.