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.


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