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°
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