Q23 of 107 Page 12

The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was 60°. Find the height of the tower.

Let the height of tower = h (m)


In ∆ABC,


tan 30° =


=


=


√3h = 40 ---------------(1)


In ∆ABD,



tan 60° =


√3 =


h =√3 -------(2)


on substituting value of h from eqn. (2) in eqn. (1)


h = √3× (√3h-40)


h = 3h-40√3


2h = 40√3


h = 20√3 m.


Therefore height of tower is 20√3 m.


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