Two radii OA and OB of a circle c with midpoint O are perpendicular. Another circle touches c in point Q and the radii in points C and D, respectively. Determine $ \angle{AQC}$.
1. **Label the centers and points:**
   - Let \( O \) be the center of the circle \( c \).
   - Let \( P \) be the center of the smaller circle that touches \( c \) at point \( Q \) and the radii \( OA \) and \( OB \) at points \( C \) and \( D \), respectively.
   - Since \( OA \) and \( OB \) are perpendicular, \( \angle AOB = 90^\circ \).

2. **Identify the geometric configuration:**
   - The smaller circle touches \( c \) at \( Q \) and the radii \( OA \) and \( OB \) at points \( C \) and \( D \), respectively.
   - This implies that \( OCPD \) forms a square because \( OC \) and \( OD \) are radii of the larger circle, and \( CP \) and \( DP \) are radii of the smaller circle, all of which are equal in length.

3. **Calculate \( \angle CQD \):**
   - Since \( OCPD \) is a square, \( \angle CPD = 90^\circ \).
   - The angle \( \angle CQD \) is half of \( \angle CPD \) because \( Q \) is the point of tangency and lies on the line segment \( CPD \).
   \[
   \angle CQD = \frac{1}{2} \angle CPD = \frac{1}{2} \times 90^\circ = 45^\circ
   \]

4. **Symmetry considerations:**
   - By symmetry, \( \angle AQC = \angle BQD \).

5. **Calculate \( \angle AQB \):**
   - Since \( \angle AOB = 90^\circ \), the angle subtended by the arc \( AB \) at the circumference is:
   \[
   \angle AQB = 180^\circ - \frac{1}{2} \angle AOB = 180^\circ - \frac{1}{2} \times 90^\circ = 180^\circ - 45^\circ = 135^\circ
   \]

6. **Determine \( \angle AQC \):**
   - If \( C \) lies on \( OA \):
     \[
     \angle AQC = \frac{1}{2} (\angle AQB - \angle CQD) = \frac{1}{2} (135^\circ - 45^\circ) = \frac{1}{2} \times 90^\circ = 45^\circ
     \]
   - If \( C \) lies on \( OB \):
     \[
     \angle AQC = \angle CQD + \angle CQD = 45^\circ + 45^\circ = 90^\circ
     \]

The final answer is \( \boxed{ 45^\circ } \) (if \( C \) lies on \( OA \)) or \( 90^\circ \) (if \( C \) lies on \( OB \)).