Let C be a circle with centre O and P be an external point to
C. Let PA and PB be two tangents to C with A and B being the points of tangency, respectively. If PA and PB are inclined to each other at an angle of 60°, then find∠POA.

Let  C  be a circle with center  O  and  P  be an external point.   PA  and  PB  are tangents from  P  to the circle, with  A  and  B  being the points of tangency. The angle between the tangents PA & PB is given as 60°. The angle between the radii OA and OB (subtended by the tangents at the center) is 2 x the angle between the tangents, So, angle AOB = 2 x 60° = 120°. The angle POA (half of angle AOB) is 60°, as the tangents subtend equal angles at the center.