Question

If a chord AB subtends an angle of ${60}^o$ at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is

(a) ${30}^o$ (b) ${60}^o$ (c) ${90}^o$ (d) ${120}^o$

Answer 1

In the quadrilateral AOBP
∠AOB = 60º (given)
∠OAP =∠OBP =90° (tangent is always perpendicular to the radius)
∠AOB+∠OBP+∠OAP+∠APB = 360°(sum of the interior angles of a quadrilateral)⇒60°+90°+90°+∠APB = 360°⇒∠APB = 360°−(60°+90°+90°) =360°−240° =120