Given, PA and PB from a point P to a circle with center O are inclined to each other at an angle of 80°

Since tangents to a circle is always perpendicular to its radius,

So, ∠OAP = ∠OBP = 90

Now, in quadrilateral OABP,

∠OAP + ∠APB + ∠PBO + ∠BOA = 360

=> 90 + 90 + 80 + ∠BOA = 360

=> 260 + ∠BOA = 360

=> ∠BOA = 360 - 260

=> ∠BOA = 100

Now, in traingle OPB and OPA,

AP = BP   {tangents froma point}

OA = OB   {Radii of the circle}

OP = OP   {common side}

By SSS congruent criterian,

 ΔOPB ≅ ΔOPA

Hence, ∠POB = ∠POA  {by CPCT}

Now, ∠POA = ∠AOB/2 = 100/2 = 50