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