A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that

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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^ at the foot of the tower, and the angle of elevation of the top of the tower from A or B is 30^ . The height of the tower is:

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![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/d52a7157-2a45-49e0-a872-3e77070e408b.png) Let the foot of the tower be O and the height be h .Let the radius of the circular park be r . So OA = OB = r .In OAB , AOB = 60^ and OA = OB .This means OAB is an equilateral triangle.Therefore, AB = OA = OB = a . So r = a .The angle of elevation of the top of the tower from A is 30^ .In the right-angled triangle formed by the tower and point A : 30^= h/OA = h/a 1/√(3) = h/a h = a/√(3)

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 $6 0^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $3 0^{\circ}$ . The height of the tower is:

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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∘ at the foot of the tower, and the angle of elevation of the top of the tower from A or B is 30∘. The height of the tower is: