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


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