Question

A tree breaks due to storm and the broken part bends so that the top of tree touches the ground, making an angle of 30° with the ground. If the distance between the foot of the tree to the point, where the top touches the ground is 8 m, show that the height of the tree is $8\sqrt{3}\mathrm{m}.$

Answer 1

Let $AB$ be the tree broken at point $C$ so that $CB$ takes the position $CD$ such that ∠$CDA={30}^{\mathrm{o}}$ and $AD=8$ m.
Let:
$AC=x$ m and $CB=CD=y$ m

From right ∆$CAD$, we have:
$\frac{AC}{AD}=\tan {30}^{\mathrm{o}}=\frac{1}{\sqrt{3}}$

⇒ $\frac{x}{8}=\frac{1}{\sqrt{3}}$

⇒ $x=\frac{8}{\sqrt{3}}$

Again, we have:
$\cos {30}^{\mathrm{o}}=\frac{AD}{CD}=\frac{8}{y}$

⇒ $\frac{8}{y}=\frac{\sqrt{3}}{2}$

⇒ $y=\frac{16}{\sqrt{3}}$
Height of the tree = $AB=(x+y)=(\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}})=\frac{24}{\sqrt{3}}=\frac{24}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=8\sqrt{3}$ m