Question

The height of a tree is 10 m. It is bent by the wind in such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom did the tree get bent?

Answer 1

Let $AB$be the tree broken at point $C$ such that $CB$ takes the position $CD$ and ∠$ADC={60}^o$.
If $CB=CD=y$ m, then $AC=10-y$ m.

In ∆$ADC$, we have:
$$\begin{gathered} \frac{AC}{CD}=\sin 60° \\ \Rightarrow \frac{10-y}{y}=\sin {60}^{°}=\frac{\sqrt{3}}{2} \end{gathered}$$
⇒ $20-2y=\sqrt{3}y$
⇒ $y(\sqrt{3}+2)=20$
⇒ $y=\frac{20}{(\sqrt{3}+2)}$m
Now,
$AC=10-y=10-\frac{20}{(\sqrt{3}+2)}=\frac{(10\sqrt{3}+20)-20}{(\sqrt{3}+2)}=\frac{10\sqrt{3}}{(\sqrt{3}+2)}$

$=\frac{10\times 1.732}{(1.732+2)}=\frac{17.32}{3.73}=4.64$ m

Hence, the height from the bottom from where the tree got bent is 4.64 m.