Question

In the given figure, PA and PB are two tangents to the circle with centre O. If $\angle$ APB = ${60}^o$ then find the measure of $\angle$ OAB.

Answer 1

Join OB

We know that the radius and tangent are perpendicular at their point of contact

$\angle OBP+\angle OBP=\angle OAP={90}^{\circ }$

In quadrilateral $AOBP$

$\angle AOB+\angle OBP+\angle APB+\angle OAP={360}^{\circ }$ [Angle sum property of a quadrilateral]

$\to \angle AOB+{90}^{\circ }+{60}^{\circ }+{90}^{\circ }={360}^{\circ }$

$\to {240}^{\circ }+\angle AOB={360}^{\circ }$

$\to \angle AOB={120}^{\circ }$

Now, In an isosceles $△AOB$

$\to \angle AOB+\angle OAB+\angle OBA={180}^{\circ }$
[Angle sum property of a triangle]

$\to 120°+2\angle OAB={180}^{\circ }$

$\to \angle OAB={30}^{\circ }$