Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of ${30}^o$ with the ground. The distance between the foot of the tree to the point where the top touches the ground is $8$ m. Find the height of the tree.

Answer 1

$\Rightarrow$ Let $AC$ was the original tree. Due to storm, it was broken into two parts. The broken part is $A^{\prime }B$ is making ${30}^o$ with the ground.

In $△A^{\prime }BC$,

$\Rightarrow$ $\tan {30}^o=\frac{BC}{A^{\prime }C}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{BC}{8}$

$\therefore$ $BC=\frac{8}{\sqrt{3}}m$

$\Rightarrow$ Now, $\cos {30}^o=\frac{A^{\prime }C}{A^{\prime }B}$

$\Rightarrow$ $\frac{\sqrt{3}}{2}=\frac{8}{A^{\prime }B}$

$\therefore$ $A^{\prime }B=\frac{16}{\sqrt{3}}m$

$\therefore$ Height of tree $=A^{\prime }B+BC=(\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}})m=\frac{24}{\sqrt{3}}m=8\sqrt{3}m$

$\therefore$ Height of the tree is $8\sqrt{3}m.$