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.
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