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.


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