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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