The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of tower is metres.
Let the height of tower is AB = h (m.)
Now in ∆ABD
tan 45° =
1 =
h = ------(1)
Now in ∆ABC
tan 30° =
=
=
√3h = 2 -------(2)
On substituting value of y from eqn. (1) in eqn. (2)
√3h = 2
√3h = 2
√3h-h = 2
h (√3-1) = 2
h =
on rationalsing above fraction we get,
h =
h =
Therefore height of tower is m.