Question

A straight tree breaks due to storm and the broken part bends so that the top of the trees touches the ground making an angle of 30° with the ground. the distance from the foot of the tree to the point, where the top touches the ground is 10 m. The height of the tree is

(a) $10\sqrt{3}\mathrm{m}$
(b) $\frac{10\sqrt{3}}{3}\mathrm{m}$
(c) $10(\sqrt{3}+1)\mathrm{m}$
(d) $10(\sqrt{3}-1)\mathrm{m}$

Answer 1

(a) $10\sqrt{3}\mathrm{m}$
Let $AB$be the tree broken at point $C$ such that $CB$ takes the position $CD$ so that ∠$ADC={30}^o$ and $AD=10$m.


Let:
$AC=x$ m and $CB=CD=y$ m. So, the height of the tree is $AB=(x+y)$ m.
Now, in ∆$ADC$, we have:
$\frac{AC}{AD}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{x}{10}=\frac{1}{\sqrt{3}}$
⇒ $x=\frac{10}{\sqrt{3}}$ m
Again,
$\frac{AD}{CD}=\cos {30}^o=\frac{\sqrt{3}}{2}$

⇒ $\frac{10}{y}=\frac{\sqrt{3}}{2}$
⇒ $y=\frac{20}{\sqrt{3}}$ m
∴ Height of the tree = $(x+y)=(\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}})=\frac{30\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}$ m