In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50°, find :

(i) ∠AOB

(ii) ∠OAB

(iii) ∠ACB

(i) We know that,

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

∠OBP = ∠OAP = 90°

In quadrilateral AOPB, By angle sum property of quadrilateral,

⇒ ∠OBP + ∠OAP + ∠AOB + ∠APB = 360°

⇒ 90° + 90° + ∠AOB + 50° = 360°

⇒ 230° + ∠AOB = 360°

⇒ ∠AOB = 360° - 230°

⇒ ∠AOB = 130°.

Hence, ∠AOB = 130°.

(ii) From figure,

OA = OB (Radii)

In triangle OAB,

∠OBA = ∠OAB [Angles opposite to equal sides]

In triangle OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 130° = 180°

⇒ 2∠OAB = 180° - 130°

⇒ 2∠OAB = 50°

⇒ ∠OAB = 25°.

Hence, ∠OAB = 25°.

(iii) Arc AB subtends ∠AOB at center and ∠ACB on the remaining part of the circle.

⇒ ∠ACB = $\dfrac{1}{2}$ ∠AOB

⇒ ∠ACB = $\dfrac{130°}{2}$

⇒ ∠ACB = 65°.

Hence, ∠ACB = 65°.