In the △AOB,∠AOB=60∘, and ∠OBA=∠OAB (since OA=OB=AB radius of same circle). ∴△AOB is a equilateral triangle. Let the height of tower is h
m. Given distance between two points A&B lie on boundary of circular park, subtends an angle of 60∘ at the foot of the tower is AB i.e. AB=a. A tower OC stands at the center of a circular park. Angle of elevation of the top of the tower from A and B is 30∘. In △OACtan30∘=