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 ${60}^{\circ }$ with the ground. At what height from the bottom the tree is broken by the wind? [3 MARKS]

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.



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

In $\Delta OAC$, we have

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

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

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

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

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

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

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

$sin60=\frac{12-x}{x}$ [$\frac{1}{2}$ MARK]

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

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

$12\sqrt{3}=x(2+\sqrt{3})$ [1 MARK]

$x=\frac{12\sqrt{3}}{(2+\sqrt{3})}\frac{(2-\sqrt{3})}{(2-\sqrt{3})}$ [$\frac{1}{2}$ MARK]

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