Correct option is: (B) 70°
Given the PA & PB are tangents to circle whose centre is O.
We know that tangent is perpendicular to radius drawn through the point of contact.
\(\therefore\) AP \(\perp\) OP & AQ \(\perp\) OQ.
= \(\angle\) OPA = 90° & \(\angle\)OQA = 90°
Given that \(\angle\)POQ = 110°
\(\because\) Sum of angles in a quadrilateral is 360°.
\(\therefore\) \(\angle\)OPA + \(\angle\)POQ + \(\angle\)OQA + \(\angle\)PAQ = 360°.
= 90° +110° + 90° + \(\angle\)PAQ = 360°.
= \(\angle\)PAQ = 360°- 290° = 70°
