Let O be the centre of a circle. PA and PB are tangents to the circle from a point P outside the circle and A and B are points on the circle. If angle APB = 50°, then angle OAB is equal to:

PA and PB are tangents from external point P, so PA = PB and angle APO = angle BPO. Given angle APB = 50°, each half is 25°. OA is perpendicular to PA (tangent perpendicular to radius at point of contact), so angle OAP = 90°. In triangle OAP: 90° + 25° + angle AOP = 180°, so angle AOP = 65°. In triangle OAB: OA = OB (radii), so it's isosceles with angle OAB = angle OBA. Central angle AOB = 2 × angle APB = 100°. So angle OAB + angle OBA = 80°, giving angle OAB = 40°. Wait, let me recalculate: angle OAP = 90°, angle APO = 25°, so angle AOP = 180° - 90° - 25° = 65°. Triangle OAB has angle AOB = 2 × 65° = 130°. Then angle OAB = angle OBA = (180° - 130°)/2 = 25°.