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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