Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of :

(i) ∠OCA,

(ii) ∠OAC.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOB = 2∠ACB

⇒ 70° = 2∠ACB

⇒ ∠ACB = $\dfrac{70}{2}$ = 35°.

(i) From figure,

∠OCA = ∠ACB = 35°.

Hence, ∠OCA = 35°.

(ii) Since, BC is a straight line,

∴ ∠AOB + ∠AOC = 180°

⇒ 70° + ∠AOC = 180°

⇒ ∠AOC = 180° - 70° = 110°.

In △OAC,

⇒ ∠OCA + ∠OAC + ∠AOC = 180°

⇒ ∠OAC + 35° + 110° = 180°

⇒ ∠OAC + 145° = 180°

⇒ ∠OAC = 180° - 145° = 35°.

Hence, ∠OAC = 35°.