Answer:
∠POA= ∠POB = 50°
Explanation:
The diagram of the given question has been drawn in the attachment.
From the diagram
Δ PAO is congruent to Δ PBO.
as PO is common in Δ
∠ PAO = ∠PBO = 90°
PA =PB [ tangent from an external point]
⇒∠POA= ∠POB
Also, ∠APB+ ∠AOB = 180°
⇒ 80° + ∠AOB = 180°
⇒∠AOB = 100°
⇒ ∠POA= ∠POB = 50°