Question

A tree is broken by the wind. The top of that tree struck the ground at an angle of ${30}^o$ and at a distance of $30$m from the root. Find the height of the whole tree.

Answer 1

Segment $AB$ represents the height of the tree.

The tree breaks at point $D$.

$\therefore$ Segment $AD$ is the broken part of tree which then takes the position of $DC$.

$\therefore AD=DC$

Given $m\angle DCB={30}^{\circ },BC=30m$

In right angled $\Delta DBC$,

$tan{30}^{\circ }=\frac{BD}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{BD}{30}$

$\Rightarrow BD=\frac{30}{\sqrt{3}}$

$\therefore BD=10\sqrt{3}m$

Also $cos{30}^{\circ }=\frac{BC}{DC}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{30}{DC}$

$\Rightarrow DC=\frac{60}{\sqrt{3}}$

$\therefore DC=20\sqrt{3}m$

Hence $AD=DC=20\sqrt{3}m$

From the figure, $AB=AD+DB$

$\therefore AB=20\sqrt{3}+10\sqrt{3}$

$\Rightarrow AB=30\sqrt{3}m$

$\therefore AB=30\times 1.73$

$\therefore AB=51.9m$

Hence, the height of tree is $51.9m$.