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A tree $12 m$ high is broken by the wind in such a way that its top touches the ground and makes an angle $60^{\circ}$ with the ground. At what height from the bottom the tree is broken by the wind?

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5.569 m

1.732 m

5.59 m

2.456 m

answer is A.

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Detailed Solution

Given,

Height of the tree =

AC + BC = 12 m

,
Considering

AC = x m

,
We have the formula,

tanθ = \frac{height}{base}

We have to find the height from the bottom,
In

ΔABC

\sin 60^{\circ} = \frac{x}{12 - x}

\Rightarrow

\frac{\sqrt{3}}{2} = \frac{x}{12 - x}

\Rightarrow

\sqrt{3} (12 - x) = 2 x

\Rightarrow 12 \sqrt{3} - \sqrt{3} x = 2 x

\Rightarrow 12 \sqrt{3} = 2 x + \sqrt{3} x

\Rightarrow 12 \sqrt{3} = x (2 + \sqrt{3})

\Rightarrow

x = \frac{12 \sqrt{3}}{2 + \sqrt{3}}

\Rightarrow

x = \frac{(12 \sqrt{3}) (2 - \sqrt{3})}{(2 + \sqrt{3}) (2 - \sqrt{3})}

\Rightarrow x = \frac{(12 \sqrt{3}) (2 - \sqrt{3})}{4 - 3}

\Rightarrow

x = 12 \sqrt{3} (2) - 12 \times 3

\Rightarrow

x = 24 \sqrt{3} - 36

\Rightarrow x = 5.569 m

(Taking

\sqrt{3} = 1.732

)
Thus, the value of x is 5.569m.
Therefore the correct option is 1.

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