A 1.8 m tall man is standing at some distance from a building. The angle of elevation from his eyes to the top of the building increases from 30∘ to 60∘ as he walks 20 m towards the building. Find the height of the building (in m).
(10√3 + 1.8) m
The situation can be represented by the figure above.
tan(∠ACB)=ABBC⇒BC=ABtan60∘.....(1)tan(∠ADB)=ABBD⇒tan30∘=ABBC+CD⇒BC+CD=ABtan30∘.....(2) on subtracting equation(1) from (2)⇒CD=ABtan30∘−ABtan60∘⇒AB=10√3
Hence, the height of the building is
= AB + BG = (10√3+1.8)m.