The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m.

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.

The total height of the tree is $x+y$.

In $\triangle A B C$

$\angle C=30^{\circ}$ and $\angle B=90^{\circ} .$

$\therefore \angle A=60^{\circ} .$

So, on using sine rule, we get:

$\frac{A B}{\sin 30^{\circ}}=\frac{B C}{\sin 60^{\circ}}=\frac{A C}{\sin 90^{\circ}}$

$\Rightarrow \frac{x}{\sin 30^{\circ}}=\frac{15}{\sin 60^{\circ}}=\frac{y}{\sin 90^{\circ}}$

So, $\frac{x}{\sin 30^{\circ}}=\frac{15}{\sin 60^{\circ}}$

$\Rightarrow \frac{x}{\frac{1}{2}}=\frac{15}{\frac{\sqrt{3}}{2}}$

$\Rightarrow x=\frac{15}{=}=5 \sqrt{3}$

Also,

$\frac{15}{\sin 60^{\circ}}=\frac{y}{\sin 90^{\circ}}$

$\Rightarrow \frac{15}{\frac{\sqrt{3}}{2}}=y$

$\Rightarrow y=\frac{30}{\sqrt{3}}=10 \sqrt{3}$

So, the height of the tree is $x+y=5 \sqrt{3}+10 \sqrt{3} \mathrm{~m}$

$=15 \sqrt{3} \mathrm{~m}$