Now, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangents PB.

So, OA is perpendicular to PA and OB is perpendicular to PB i.e. OA ⊥ PA and OB ⊥ PB

So, ∠OBP = ∠OAP = 90°

Now, in the quadrilateral AOBP,

The sum of all the interior angles will be 360°

So, ∠AOB+∠OAP+∠OBP+∠APB = 360°

Putting their values, we get,

∠AOB + 260° = 360°

∠AOB = 100°

Now, consider the triangles △OPB and △OPA. Here,

AP = BP (Since the tangents from a point are always equal)

OA = OB (Which are the radii of the circle)

OP = OP (It is the common side)

Now, we can say that triangles OPB and OPA are similar using SSS congruency.

∴△OPB ≅ △OPA

So, ∠POB = ∠POA

∠AOB = ∠POA+∠POB

2 (∠POA) = ∠AOB

By putting the respective values, we get,

=>∠POA = 100°/2 = 50°

As angle ∠POA is 50°.