Between ΔAOP and ΔBOP
∠OAP = ∠OBP = 90° (∵ tangent and line from origin meet at 90° to each other)
OA=OB = radius of the circle = r (say) and OP is the common side
Hence, we can say ΔAOP = ΔBOP
Therefore, we can say
∠OPA = ∠OPB = \(\frac{1}{2}\)(∠APB)
= \(\frac{1}{2}\)(80°) = 40° (∵ given that tangents PA and PB are inclined to each other by 80°)
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°