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
⇒ 15√3-√3h = 2h
⇒ h(2+√3) = 15√3
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.