Given, PA and PB are two tangents to a circle with center O and ∠APB = 80°
\(\therefore \angle APO=\frac{1}{2}\angle APB=40^\circ\)
[Since they are equally inclined to the line segment joining the center to that point and ∠OAP = 90°]
[Since tangents drawn from an external point are perpendicular to the radius at the point of contact]
Now, in triangle AOP:
