# 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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