A tree is broken by the wind. The top struck the ground at an angle of ${30}^{\circ }$ and at a distance of $30$ metres from the foot of the tree. The height of the tree in metres is

The correct option is D: $30\sqrt{3}$

Let $AB$ be the tree, broken at point $D$ such that the broken part $DB$ takes the position $DO$ and touches the ground at $O,$ $OA=30m$, $\angle AOD={30}^{\circ }$.

Let $AD=x,$ and $BD=OD=y$

In $△AOD,$ we have

$\tan {30}^{\circ }=\frac{AD}{OA}$

$\frac{1}{\sqrt{3}}=\frac{x}{30}$

$x=\frac{30}{\sqrt{3}}=\frac{30\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=10\sqrt{3}$

Again, In \triangle AOD, we have

$\cos {30}^{\circ }=\frac{OA}{OD}$

$\frac{\sqrt{3}}{2}=\frac{30}{y}$

$y=\frac{60}{\sqrt{3}}=\frac{60\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=20\sqrt{3}$

Height of the tree $=x+y$

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

$=30\sqrt{3}$ metres