From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are 45° and 60° respectively. Find the height of the transmission tower.
In ∆OPQ,
tan 45° =
1 =
OP = 20 -------(1)
Now in ∆OPR
tan 60° =
=
=
20√3 = h+20
h = 20√3-20
h = 20(√3-1) m.
Therefore height of transmission tower is 20(√3-1) m.