In the $\Delta AOB,\,\,\angle AOB = {60^ \circ },$ and $\angle OBA = \angle OAB$ (since $OA=OB=AB$ radius of same circle). $\therefore$ $\Delta AOB$ is a equilateral triangle. Let the height of tower is $h$ Given distance between two points $A$ & $B$ lie on boundary of circular park, subtends an angle of ${60^ \circ }$ 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 ${30^ \circ }$ . In $\Delta OAC\,\,\tan {30^ \circ } = {h \over a}$ $\Rightarrow {1 \over {\sqrt 3 }} = {h \over a} \Rightarrow h = {a \over {\sqrt 3 }}$