A vertically straight tree, 15m height, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break?

Total height of the tree is 15 m.


I.e AB= 15m.


Let height at which tree is broken is h (m)


Therefore BC = h (m)


CD = AB-BC


= 15-h.


In ∆DBC,


sin 60° =


=


On cross-multiplication


√3(15-h) = 2h


153-3h = 2h


h(2+3) = 153


h =



on multiplying and dividing by 2-√3


h =


h = 6.96 m.


Therefore the tree is broken at 6.96 m from the ground.


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