Given: ∠OAB = 30°, ∠OCB = 57°
In ΔOAB, AO = BO (Both are radii of the circle).
Thus ∠OAB = ∠OBA = 30° (angles opposite to equal sides are equal)
In ΔAOB, sum of all angles of a triangle is 180°.
∴ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 30° + 30° + ∠AOB = 180°
⇒ ∠AOB = 180° - 30° - 30°
⇒ ∠AOB = 120° …..…… (1)
Now, in triangle OBC, OC and OB are radius of the circle and are thus equal.
∴ ∠OBC = ∠OCB = 57° (angles opposite to equal sides are equal)
In ΔAOB, sum of all angles of a triangle is 180°.
∴ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 57° + 57° + ∠BOC = 180°
⇒ ∠BOC = 180° - 57° - 57°
⇒ ∠BOC = 66° …………… (2)
Now, from equation (1), we have:
∠AOB = 120°
⇒ ∠AOC + ∠COB = 120°
⇒ ∠AOC + 66° = 120°
⇒ ∠AOC = 120° - 66°
⇒ ∠AOC = 54°
Therefore, ∠AOC = 54° and ∠BOC = 66°.
