Question

In the given figure, O is the centre of a circle, ∠OAB = 30° and ∠OCB = 55°. Find ∠BOC and ∠AOC.

Answer 1

We have:
OC = OB (Radii of a circle)
⇒ ∠OBC = ∠OCB = 55°

In ΔOCB, we have:
∠BOC + ∠OCB + ∠OBC = 180° (Angle sum property of a triangle)
⇒ ∠BOC + 55° + 55° = 180°
⇒ ∠BOC = (180° - 110°) = 70°
∴ ∠BOC = 70°

Again, we have:
OA = OB
∠OBA = ∠OAB = 30°

In ΔOAB, we have:
∠AOB + ∠OBA + ∠OAB = 180° (Angle sum property of a triangle)
⇒ ∠AOB + 30° + 30° = 180°
⇒ ∠AOB = (180° - 60°) = 120°
Thus, ∠AOB = 120°

Now, ∠AOC + ∠BOC = 120° [Since ∠AOB = ∠AOC + ∠BOC]
⇒ ∠AOC + 70° = 120°
⇒ ∠AOC = (120° - 70°) = 50°
∴ ∠AOC = 50°