(A) Let $O$ be the centre of the circular park and $P$ be the top of the tower. Thus,$OP$ is the height of the tower.Since $O$ is the centre,$OA = OB

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(A) Let $O$ be the centre of the circular park and $P$ be the top of the tower. Thus,$OP$ is the height of the tower.Since $O$ is the centre,$OA = OB = R$ (radius of the park).Given that $\angle AOB = 60^{\circ}$ and $OA = OB$,$\Delta AOB$ is an equilateral triangle.Therefore,$OA = OB = AB = a$.In the right-angled triangle $\Delta AOP$,where $\angle OAP = 30^{\circ}$ is the angle of elevation:$\tan 30^{\circ} = \frac{OP}{OA}$$\frac{1}{\sqrt{3}} = \frac{OP}{a}$$OP = \frac{a}{\sqrt{3}}$Thus,the height of the tower is $\frac{a}{\sqrt{3}}$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

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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 $AB = a$ 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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$\frac{a}{\sqrt{3}}$
B
$a\sqrt{3}$
C
$\frac{2a}{\sqrt{3}}$
D
$2a\sqrt{3}$

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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 $AB = a$ 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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