In the given figure, TP and TQ are two tangents to the circle with centre O, touching at A and C respectively. If ∠BCQ = 55° and ∠BAP = 60°, find :

(i) ∠OBA and ∠OBC

(ii) ∠AOC

(iii) ∠ATC

(i) ∠OAP = 90° [As TP is tangent at point A]

From figure,

⇒ ∠OAB = ∠OAP - ∠BAP = 90° - 60° = 30°.

⇒ OA = OB (Radius of same circle)

⇒ ∠OBA = ∠OAB = 30° (Angles opposite to equal sides are equal)

∠OCQ = 90° [As TQ is tangent at point C]

From figure,

∠OCB = ∠OCQ - ∠BCQ = 90° - 55° = 35°.

⇒ OB = OC (Radius of same circle)

⇒ ∠OBC = ∠OCB = 35° (Angles opposite to equal sides are equal)

Hence, ∠OBA = 30° and ∠OBC = 35°.

(ii) We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

⇒ ∠AOC = 2∠ABC

⇒ ∠AOC = 2(∠OBA + ∠OBC)

⇒ ∠AOC = 2(30° + 35°)

⇒ ∠AOC = 2 × 65° = 130°.

Hence, ∠AOC = 130°.

(iii) In quadrilateral OATC,

⇒ ∠AOC + ∠OAT + ∠OCT + ∠ATC = 360°

⇒ 130° + 90° + 90° + ∠ATC = 360°

⇒ 310° + ∠ATC = 360°

⇒ ∠ATC = 360° - 310° = 50°.

Hence, ∠ATC = 50°.