A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $ 30^{\circ} $ with the ground. The distance between the foot of the tree to the point where the top touches the ground is $ 8 \mathrm{~m} $. Find the height of the tree.
Given:
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of \( 30^{\circ} \) with the ground.
The distance between the foot of the tree to the point where the top touches the ground is \( 8 \mathrm{~m} \).
To do:
We have to find the height of the tree.
Solution:
Let $AB$ be the original height of the tree and $CD$ be the broken part of the tree that touches the ground at $D$.
Let point $D$ be the point where the top of the broken tree part touches the ground.
From the figure,
$\mathrm{AD}=8 \mathrm{~m}, \angle \mathrm{CDA}=30^{\circ}$
Let the height of the tree from the ground where the tree broke be $\mathrm{CA}=x \mathrm{~m}$ and the height of the broken part be $\mathrm{DC}=y \mathrm{~m}$.
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { CA }}{DA}$
$\Rightarrow \tan 30^{\circ}=\frac{x}{8}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{x}{8}$
$\Rightarrow x=\frac{8}{\sqrt3} \mathrm{~m}$........(i)
Similarly,
$\cos \theta=\frac{\text { Adjacent }}{\text { Hypotenuse }}$
$=\frac{\text { DA }}{CD}$
$\Rightarrow \cos 30^{\circ}=\frac{8}{y}$
$\Rightarrow \frac{\sqrt3}{2}=\frac{8}{y}$
$\Rightarrow y=\frac{8(2)}{\sqrt3}=\frac{16}{\sqrt3} \mathrm{~m}$
Therefore,
$x+y=\frac{8}{\sqrt3}+\frac{16}{\sqrt3}$
$=\frac{8+16}{\sqrt3}$
$=\frac{24}{\sqrt3}$
$=\frac{24\sqrt3}{\sqrt3 \times \sqrt3}$
$=\frac{24\sqrt3}{3}$
$=8\sqrt3 \mathrm{~m}$
Therefore, the height of the tree is $8\sqrt3 \mathrm{~m}$.
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