Q. 113.9( 29 Votes )

The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.

Answer :

Let the height of the tower = h (m)



Let the point of 60° elevation is (m) away from the foot of the tower.


In ∆ABC,


tan 45° =


1 =


1 =


h = 10+ ----(1)


In ∆ABD,


tan 60° =


√3 =


h = √3


= ---------(2)


From eqn. (2) in eqn. (1)


h =


h - = 10


= 10


= 10√3



h= h=


15+53 23.66 m.


Therefore height of the tower is 23.66 m.


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