In the given figure, if ∠AOC=120∘, then ∠ABC = _______.
Given that ∠AOC=120∘.
Consider the arc ABC. Take a point on the circumference of the circle outisde this arc. Join AP and PC.
Arc ABC subtends ∠AOC at centre and ∠APC on the remaining part of the circle at a point P.
Since the angle subtended by an arc at the centre is double the angle subtended by it on any remaining part of the circle,
∠APC=12∠AOC=12×120∘=60∘
Consider the cyclic quadrilateral APCB. Since opposite angles are supplementary,
∠ABC+∠APC=180∘.
⟹∠ABC=180∘−∠APC=180∘−60∘=120∘