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

(N/A) Let $AB$ be the original height of the tree. Let the tree break at point $B$,and the top $A$ touches the ground at point $A^{\prime}$.
In the right-angled triangle $\triangle BCA^{\prime}$,where $\angle BCA^{\prime} = 90^{\circ}$ and $\angle BA^{\prime}C = 30^{\circ}$:
Given $CA^{\prime} = 8\, m$.
Using $\tan 30^{\circ} = \frac{BC}{CA^{\prime}}$:
$\frac{BC}{8} = \frac{1}{\sqrt{3}} \implies BC = \frac{8}{\sqrt{3}}\, m$.
Using $\cos 30^{\circ} = \frac{CA^{\prime}}{BA^{\prime}}$:
$\frac{8}{BA^{\prime}} = \frac{\sqrt{3}}{2} \implies BA^{\prime} = \frac{16}{\sqrt{3}}\, m$.
The total height of the tree is $AB = BC + BA^{\prime}$ (since $BA^{\prime} = BA$):
$AB = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}\, m$.
Rationalizing the denominator:
$AB = \frac{24 \sqrt{3}}{3} = 8\sqrt{3}\, m$.
Thus,the height of the tree is $8\sqrt{3}\, m$.