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.

Solution:

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 $x+y=5 \sqrt{3}+10 \sqrt{3} \mathrm{~m}$

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