Given that tangent PA and PB are inclined to each other at angle of 80°.
\(\therefore\) \(\angle\) APB = 80°.
= \(\angle\) OPA = \(\frac {\angle APB}2 = \frac {80°}2 = 40°\)(\(\because\) \(\triangle\) OAP \(\cong \, \triangle\) OBP by SSS congruence criteria)
\(\because\) OA \(\perp\) AP
= \(\angle\) OAP = 90°.
Now, \(\angle\) POA + \(\angle\) OAP + \(\angle\) OPA = 180°(Sum of angles in a triangle)
= \(\angle\) POA + 90° + 40° = 180°
= \(\angle\) POA = 180°-130° = 50°
