Question

In the given figure, $O$ is the center of the circle and is $ABC$ subtends an angle ${130}^o$ at the center. If $AB$ extended to $P$, find $\angle PBC$.

Answer 1

Consider a point $D$ on the $arcCA$ and join $DC$ and $AD$

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment

So we get

$\angle 2=2\angle 1$

By substituting the values

${130}^o=2\angle 1$

So we get

$\angle 1={65}^o$

From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle

$\angle PBC=\angle 1$

So we get $\angle PBC={65}^o$

Therefore, $\angle PBC={65}^o$.