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 equal to:

In quadrilateral OAPB, the angles at A and B are $90^{\circ}$ because the radius is perpendicular to the tangent at the point of contact. The sum of angles in quadrilateral OAPB is $360^{\circ}$. Therefore, $\angle AOB = 180^{\circ} - 80^{\circ} = 100^{\circ}$. Since $\triangle OPA \cong \triangle OPB$, $\angle POA = \frac{1}{2} \angle AOB = \frac{100^{\circ}}{2} = 50^{\circ}$.