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The upper part of a tree broken by wind makes an angle $30^{\circ}$ with the ground and the distance from the foot to the point where the top of the tree meets the ground is $8 m$ . What was the height of the tree?

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8 \sqrt{3} m

5 \sqrt{3} m

10 \sqrt{3} m

13 \sqrt{3} m

answer is A.

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Detailed Solution

Let

l

be the broken part of the tree and

h

be the unbroken part of the tree. Question Image

So, according to the diagram,

\Rightarrow \tan 30^{\circ} = \frac{AB}{BC}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{8}

\Rightarrow h = \frac{8}{\sqrt{3}}

[equation I]
Now,

\Rightarrow \cos 30^{\circ} = \frac{BC}{AC}

\Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{l}

\Rightarrow l = \frac{8 \times 2}{\sqrt{3}}

\Rightarrow l = \frac{16}{\sqrt{3}}

[equation II]
And, we know,

\Rightarrow

Height of the tree

= l + h

= \frac{16}{\sqrt{3}} + \frac{8}{\sqrt{3}}

[From equation I and II]

= \frac{16 + 8}{\sqrt{3}}

= \frac{24}{\sqrt{3}}

= \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

[Rationalize numerator and denominator by

\sqrt{3}

]

= \frac{24 \sqrt{3}}{3}

= 8 \sqrt{3} m

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