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° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Let AB be the first pole and CD be the second pole of each height h meters and E be the point between the road.

Let BE = x meters and ED = (80 - x) meters.

In △ABE,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{x} \\[1em] \Rightarrow h = x\sqrt{3} \text{ …….(1)}$

In △CED,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{ED} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{80 - x} \\[1em] \Rightarrow h = \dfrac{80 - x}{\sqrt{3}} \text{ …….(2)}$

From (1) and (2), we get :

$\Rightarrow x\sqrt{3} = \dfrac{80 - x}{\sqrt{3}} \\[1em] \Rightarrow 3x = 80 - x \\[1em] \Rightarrow 3x + x = 80 \\[1em] \Rightarrow 4x = 80 \\[1em] \Rightarrow x = \dfrac{80}{4} = 20 \text{ m}.$

Substituting value of x in equation (1), we get :

⇒ h = $20\sqrt{3}$ meters.

⇒ 80 - x = 80 - 20 = 60 meters.

Hence, height of poles = $20\sqrt{3}$ meter and distance of point from first tower = 20 m and from second tower = 60 m.