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(C) Let the tree be $AB$ and it breaks at point $C$. The upper part $AC$ touches the ground at point $D$.Given: $BD = 10 	ext{ m}$ and $\angle BDC =

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$10(2+\sqrt{3})$

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(C) Let the tree be $AB$ and it breaks at point $C$. The upper part $AC$ touches the ground at point $D$.Given: $BD = 10 	ext{ m}$ and $\angle BDC = 60^{\circ}$.In $	riangle BCD$,$	an 60^{\circ} = \frac{BC}{BD} \implies \sqrt{3} = \frac{BC}{10} \implies BC = 10\sqrt{3} 	ext{ m}$.Also,$\cos 60^{\circ} = \frac{BD}{CD} \implies \frac{1}{2} = \frac{10}{CD} \implies CD = 20 	ext{ m}$.Since $AC = CD$,the original height of the tree is $AB = BC + AC = BC + CD = 10\sqrt{3} + 20 = 10(2 + \sqrt{3}) 	ext{ m}$.

$A$ tree breaks at a certain height and the upper part touches the ground,making an angle of $60^{\circ}$ with the ground at a distance of $10 \text{ m}$ from its foot. The original height of the tree was (in $\text{m}$):

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