Question

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

Answer 1

Let $AB$ be the tree of height $12$ metres. Suppose the tree is broken by the wind at point $C$ and the part $CB$ assumes the position $CO$ and meets the ground at $O.$ Let $AC=x.$ Then, $CO=CB=12-x.$

Total height of tree, $AB=12m$

$AC=xm$

$OC=CB=12-x$

It is given that $\angle AOC={60}^{\circ }$

In $\Delta OAC$, we have

$\sin {60}^{\circ }=\frac{AC}{OC}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{x}{12-x}$

$\Rightarrow 12\sqrt{3}-\sqrt{3}x=2x$

$\Rightarrow 12\sqrt{3}=x(2+\sqrt{3})$

$\Rightarrow x=\frac{12\sqrt{3}}{2+\sqrt{3}}$

$\Rightarrow x=\frac{12\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$\Rightarrow x=12\sqrt{3}(2-\sqrt{3})$

$\Rightarrow x=24\sqrt{3}-36=5.569$ metres

Hence, the tree is broken at a height of $5.569$ metres from the ground.