If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^\circ$, then $\angle POA$ is

Since tangents from an external point are equal, $PA = PB$, and also $OA = OB$. Hence triangles $OAP$ and $OBP$ are congruent, so $OP$ bisects $\angle APB$. Therefore $\angle APO = 40^\circ$. In right triangle $OAP$, $\angle OAP = 90^\circ$ because radius is perpendicular to tangent so $\angle POA = 180^\circ - 90^\circ - 40^\circ = 50^\circ$.