A triangle $ABC$ is inscribed in a circle with centre $O$. If $\angle AOB = 140^{\circ} $ and $\angle BOC = 120^{\circ}$, then $\angle A$ and $\angle B$ are -

Using the central angle theorem, angle AOC equals 360 minus 140 minus 120, which is 100 degrees. Angle A is an inscribed angle subtending arc BC, so angle A equals 120 divided by 2, which is 60 degrees. Similarly, angle B subtends arc AC, meaning angle B equals 100 divided by 2, which is 50 degrees.

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