The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 meters towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres (Use ).
Let the height of the tower = h (m)
In ∆ABC,
tan 30° =
=
=
√3h = 150+
-------- (1)
In ∆ABD,
tan 60° =
√3 = ⇒ h = √3
-------(2)
on substituting value of x from eqn.(2)i eqn.(1)
h-3h = -150
2h = 150√3
h = ⇒ 75
h= 129.9 m.
Hence height of tower is 129.9 m.