The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower
In the fig PQ be the height of the tower.
In ∆QRT
tan 45° =
1 =
h = x ……….(1)
In ∆QRT
tan 60° =
√3 =
√3x = h+40 ………….(2)
On substituting the value of x from eqn (1) in eqn (2)
√3h = h+40
h (√3-1) = 40
h = ⇒ 54.64m
Therefore height of the tower is = h+40 = 54.64+40 ⇒ 94.64m