In the adjoining figure, AP is a man of height 1.8 m and BQ is a building 13.8 m high. If the man sees the top of the building by focussing his binoculars at an angle of 30° to the horizontal, find the distance of the man from the building.

Let AB = d meters, then PC = d meters.

From right-angled △PCQ,

tan 30° = $\dfrac{CQ}{PC}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{BQ - BC}{d}$

From figure,

BC = AP = 1.8 m

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{13.8 - 1.8}{d}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{12}{d}$

⇒ d = $12\sqrt{3}$ meters.

Hence, distance of man from building is $12\sqrt{3}$ meters.