Question

A vertical tower stands on a plane and is surmounted by a  pole of height 5 m. At a point A on the plane, the angle of elevation of the top of pole is 60° and the angle of depression of the point A from the top of tower is 45° Find the height of the tower.

Answer 1

BC is a tower of height h m. A pole CD is surmounted on this tower with height 5 m. 

Point A is x m away from the foot of the tower.

Now,

∠DAB = 60° and ∠XCA = 45° = ∠CAB

From right angled ∆ABC

tan 45° = BC/AB

⇒ t = h/x

⇒ h = x

From right angled ∆ABD,

tan 60° = BD/AB

Hence, height of the tower = 6.83 m