$ 45$(if $ C$ lies on $ OA$) or $ 90$ (if $ C$ lies on $ OB$) Label the center of the circle touching circle c and the two radii as P. Then it immediately follows that OCPD is a square. Thus $ \angle CQD = \frac{1}{2} \angle CPD = 45$. By symmetry we get that $ \angle AQC = \angle BQD$. Also $ \angle AQB = 180 - \frac{1}{2} \angle AOB = 135$. Thus $ \angle AQC = \frac{1}{2} (135 - 45) = 45$ if $ C$ lies on $ OA$ or $ \angle AQC = 45+45 = 90$ if $ C$ lies on $ OB$