In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 40°, find ∠AQB and ∠AMB.

Join OA and OB.

We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

∠PAO = ∠PBO = 90°

In quadrilateral AOPB, By angle sum property of quadrilateral,

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

⇒ 90° + 40° + 90° + ∠AOB = 360°

⇒ 220° + ∠AOB = 360°

⇒ ∠AOB = 360° - 220°

⇒ ∠AOB = 140°.

Arc AB subtends ∠AOB at center and ∠AQB on the remaining part of the circle.

⇒ ∠AQB = $\dfrac{1}{2}$ ∠AOB

⇒ ∠AQB = $\dfrac{140°}{2}$

⇒ ∠AQB = 70°.

Sum of opposite angles in cyclic quadrilateral is 180°.

⇒ ∠AQB + ∠AMB = 180°

⇒ 70° + ∠AMB = 180°

⇒ ∠AMB = 180° - 70°

⇒ ∠AMB = 110°.

Hence, ∠AMB = 110° and ∠AQB = 70°.