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A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B(=a)$ subtends an angle of $60^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $30^{\circ}$. The height of the tower is
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$\frac{a}{\sqrt{3}}$
$\frac{a}{\sqrt{3}}$
$\triangle \mathrm{O A B}$ is equilateral
$\therefore \mathrm{OA}=\mathrm{OB}=\mathrm{AB}=\mathrm{a}$
Now $\tan 30^{\circ}=\frac{h}{a}$
$\therefore h=\frac{a}{\sqrt{3}} \text {. }$
$\therefore \mathrm{OA}=\mathrm{OB}=\mathrm{AB}=\mathrm{a}$
Now $\tan 30^{\circ}=\frac{h}{a}$
$\therefore h=\frac{a}{\sqrt{3}} \text {. }$
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