In the given figure, is the center of the circle and

Calculate the values of (i) (ii)

(i) ∠AOC + ∠AOB = 180°[Because BC is a straight line]

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC = 110°

OA = OC [Radius]

∴ ∠OAC = ∠OCA ________________ (i)

In triangle AOC,

∠OAC + ∠OCA + ∠AOC = 180°[Sum of angles of triangle]

⇒ 2 ∠OCA + 110° = 180°[From equation (i)]

⇒ 2 ∠OCA = 70°

⇒ 2 ∠OCA = 70°

⇒ ∠OCA = 35°

(ii)

∠AOC + ∠AOB = 180°[Because BC is a straight line]

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC = 110°

OA = OC [Radius]

∴ ∠OAC = ∠OCA ________________ (i)

In triangle AOC,

∠OAC + ∠OCA + ∠AOC = 180°[Sum of angles of triangle]

⇒ 2 ∠OAC + 110° = 180°[From equation (i)]

⇒ 2 ∠OAC = 70°

⇒ 2 ∠OAC = 70°

⇒ ∠OAC = 35°