Question : The upper part of a tree broken at a certain height makes an angle of 60° with the ground at a distance of 10 metres from its foot. The original height of the tree was:
Option 1: $20\sqrt{3}$ metres
Option 2: $10{\sqrt3}$ metres
Option 3: $10\left (2+{\sqrt3} \right)$ metres
Option 4: $10\left (2-{\sqrt3}\right)$ metres
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Correct Answer: $10\left (2+{\sqrt3} \right)$ metres
Solution : Let CD be the height of the tree and it breaks from the point B which touches the ground at A. As AB = BD, the total height of the tree CD = BC + AB In $\triangle ABC$, $\tan 60°=\frac{BC}{10}$ ⇒ $\sqrt{3}=\frac{BC}{10}$ ⇒ $BC=10\sqrt{3}$ metres Again in $\triangle ABC$, $(AB)^{2}=(BC)^{2}+(AC)^{2}$ ⇒ $(AB)^{2}=(10\sqrt{3})^{2}+(10)^{2}$ ⇒ $(AB)^{2}=300+100$ ⇒ $(AB)^{2}=400$ ⇒ $AB=20$ metres Original Height of the tree $=AB+BC$ $=20+10\sqrt{3}$ $=10(2+\sqrt{3})$ metres Hence, the correct answer is $10(2+\sqrt{3})$ metres.
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