Question

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 -

• A. ${60}^{\circ },{50}^{\circ }$
• B. ${50}^{\circ },{40}^{\circ }$
• C. ${40}^{\circ },{50}^{\circ }$
• D. ${20}^{\circ },{60}^{\circ }$

Answer 1

The correct option is A ${60}^{\circ },{50}^{\circ }$

We know the measure of inscribed angle is exactly half the measure of its intercepted arc.

Therefore, $\angle A=\frac{{120}^{\circ }}{2}={60}^{\circ };\angle B=\frac{{100}^{\circ }}{2}={50}^{\circ }$