Question

A tree 12 m high, is broken by the wind in such a way that its top touches the ground and makes an angle 60^o with the ground. At what height from the bottom the tree is broken by the wind?

Answer 1

Let AC be the tree that got broken from B and that met the ground at D.
Thus, we have:

$$\begin{gathered} \mathrm{In}△BCD, \\ \sin 60°=\frac{BC}{BD} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{12-h} \\ \Rightarrow 12\sqrt{3}-h\sqrt{3}=2h \\ \Rightarrow 12\sqrt{3}=2h+h\sqrt{3} \\ \Rightarrow h=\frac{12\sqrt{3}}{2+\sqrt{3}} \\ \Rightarrow h=12\sqrt{3}\left(2-\sqrt{3}\right)=5.52\mathrm{m} \end{gathered}$$

Thus, the tree was broken at a height of 5.52 m from the bottom.