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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