Tangents and radii are perpendicular at the point of contact.In quadrilateral CAPB, we have:$\angle APB = 55^{\circ}, \angle CAP = \angle CBP = 90^{\circ}$By angle sum property of quadrilateral:$\Rightarrow \angle ACB = 125^{\circ}$Now, we know that angle subtended at the center is twice the angle subtended at any point on the arc, so we have$\angle AQB = \frac{1}{2} \angle ACB = 62\frac{1}{2}^{\circ}$