Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide. From a point P between them on the road, the angle of elevation of the top of one pole is $60^{\circ}$ and the angle of depression from the top of the other pole at point P is $30^{\circ}$. Find the heights of the poles and the distance of point P from the poles.
AB, CD: Poles of height 'h'
AC: Road 80m wide
P: Point of observation
In the figure, $AP=x$ meter, $PC=80-x$ meter
In triangle PAB,
$\dfrac{BA}{AP}=\tan 60^{\circ} = \sqrt{3}$
i.e. $\dfrac{h}{x} = \sqrt{3}$ $\implies$ $x = \dfrac{h}{\sqrt{3}}$ $\cdots$ (1)
In triangle PCD,
$\dfrac{CD}{PC} = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$
i.e. $\dfrac{h}{80-x} = \dfrac{1}{\sqrt{3}}$ $\implies$ $80-x = h\sqrt{3}$ $\cdots$ (2)
Therefore, from equations (1) and (2) we have,
$80-\dfrac{h}{\sqrt{3}} = h\sqrt{3}$
i.e. $80=\left(\dfrac{1}{\sqrt{3}}+\sqrt{3}\right)h$
i.e. $\dfrac{4h}{\sqrt{3}} = 80$ $\implies$ $h = \dfrac{80\sqrt{3}}{4} = 20\sqrt{3} = 34.64$
$\therefore$ Height of poles = 34.64 m
$\therefore$ From equation (1), $x = \dfrac{20\sqrt{3}}{\sqrt{3}} = 20$
$\therefore$ Distance of P from AB $=20$ m
and distance of P from CD $= 80-20=60$ m