PA and PB are the two tangent of a circle with centre O, From a point P outside the circle. A and B are the points on the circle, If ∠APB = 72o, then ∠OAB is equal to ?

Tangents from an external point to a circle are equal in length, so PA = PB. Also, the radius is perpendicular to the tangent at the point of contact, so ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, the sum of angles is 360°, so ∠AOB = 360° - 90° - 90° - 72° = 108°. Triangle OAB is isosceles (OA = OB as radii), so ∠OAB = ∠OBA = (180° - 108°)/2 = 36°.