Q11 of 25 Page 9

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.

Formula Used:



Explanation:


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)


Substituting value of x From eqn. (2) in eqn. (1)


h =


h - = 10


= 10


= 10√3


⇒


h= ⇒ h=


⇒ 15+5√3


⇒ 23.66 m.


Therefore, height of the tower is 23.66 m.


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