In the given figure, PAT is tangent at A to the circle with centre O. If ∠ABC = 35°, find :

(i) ∠TAC

(ii) ∠PAB

(i) ∠TAC = ∠ABC = 35° [Angles in the alternate segment are equal]

Hence, ∠TAC = 35°.

(ii) Since BC is a diameter, ∠BAC is an angle in a semicircle.Therefore, ∠BAC = 90°

⇒ ∠ACB + ∠BAC + ∠ABC = 180°

⇒ ∠ACB + 90° + 35° = 180°

⇒ ∠ACB = 180° - 90° - 35°

⇒ ∠ACB = 55°

⇒ ∠PAB = ∠ACB [Angles in the alternate segment are equal]

⇒ ∠PAB = 55°

Hence, ∠PAB = 55°.