Question

In the following figure, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.

Answer 1

Given, ∠OAB = 30° and ∠OCB = 57°

In ΔAOB, AO = OB   ...[Both are the radius of a circle]

⇒ ∠OBA = ∠BAO = 30°  ...[Angles opposite to equal sides are equal]

In ΔAOB,

⇒ ∠AOB + ∠OBA + ∠BAO = 180°  ...[By angle sum property of a triangle]

∴ ∠AOB + 30° + 30° = 180° 

∴ ∠AOB = 180° – 2(30°)

= 180° – 60°

= 120°  ...(i)

Now, in ΔAOB,

OC = OB   ...[Both are the radius of a circle]

⇒ ∠OBC = ∠OCB = 57°  ...[Angles opposite to equal sides are equal]

In ΔOCB,

∠COB + ∠OCB + ∠CBO = 180°  ...[By angle sum property of triangle]

∴ ∠COB = 180° – (∠OCB + ∠OBC)

= 180° – (57° + 57°)

= 180° – 114°

= 66°  ...(ii)

From equation (i), ∠AOB = 120°

⇒ ∠AOC + ∠COB = 120° 

⇒ ∠AOC + 66° = 120°   ...[From equation (ii)]

∴ ∠AOC = 120° – 66° = 54°