Question

In the given figure, seg $AB$ is a diameter of the circle, $m\left(arcAKC\right)=$${30}^0$. Find the value of $m\left(arcBMC\right).$.

Answer 1

We are given $\angle ABC={30}^o$, now we are to find $\angle BAC$ as this is the measure of $arcBMC$.

Now in the given figure if we join $A$ and $C$ then $\Delta$ ABC is right-angled triangle.[ As semi-circular angle is right-angle].

Then from $\Delta$ ABC we get $\angle ACB={90}^o$ and $\angle ABC={30}^o$.

Then the remaining angle $\angle CAB$ will be of ${60}^o$.

So $m\left(arcBMC\right)={60}^o$.