In figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.

1.	In figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.
Answer» 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°.

Answer»

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°.

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