In the given figure, O is the centre of a circle, ∠OAB = 30° and ∠OCB = 55°. Find ∠BOC and ∠AOC.
Given: and
Here,
In ΔAOB
OA = OB (radius)
∠OAB = ∠OBA (angles opposite to equal sides are equal)
∴ ∠ OBA = 30°
Now, by angle sum property
∠AOB + ∠OBA + ∠OAB = 180°
∠AOB + 30° + 30° = 180°
∠AOB = 180° – 30° – 30°
∠AOB = 120°
Now, Consider Δ BOC
OC = OB (radius)
∠OCB = ∠OBC (angles opposite to equal sides are equal)
∴ ∠ OBA = 55°
Now, by angle sum property
∠BOC + ∠OBC + ∠OCB = 180°
∠BOC + 55° + 55° = 180°
∠BOC = 180° – 55° – 55° = 70°
∴ ∠BOC = 70°
Here,
∠AOB = ∠AOC + ∠BOC
120° = ∠AOC + 70°
∠AOC = 120° – 70°
∠AOC = 50°
∴ ∠AOC = 50°
∴