In the following diagram (not to scale), $A$, $B$, $C$, $D$ are four consecutive vertices of an 18-sided regular polygon with center $O$. Let $P$ be the midpoint of $AC$ and $Q$ be the midpoint of $DO$. Find $\angle OPQ$ in degrees.
[asy]
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pair O=(0,0), A=dir(start), B=dir(start+360/n), C=dir(start+2*360/n), D=dir(start+3*360/n), P=(A+C)/2, Q=(O+D)/2; D(D("O",O,NE)--D("A",A,W)--D("B",B,SW)--D("C",C,S)--D("D",D,SE)--O--D("P",P,1.6*dir(95))--D("Q",Q,NE)); D(A--C); D(A--(A+dir(start-360/n))/2, dd); D(D--(D+dir(start+4*360/n))/2, dd);
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1. **Identify the vertices and midpoints:**
   - Let \( A, B, C, D \) be four consecutive vertices of an 18-sided regular polygon with center \( O \).
   - \( P \) is the midpoint of \( AC \).
   - \( Q \) is the midpoint of \( DO \).

2. **Calculate the central angles:**
   - Since the polygon is regular, the central angle between any two consecutive vertices is \( \frac{360^\circ}{18} = 20^\circ \).
   - Therefore, the central angle \( \angle AOC \) is \( 2 \times 20^\circ = 40^\circ \).
   - The central angle \( \angle AOD \) is \( 3 \times 20^\circ = 60^\circ \).

3. **Determine the coordinates of the points:**
   - Assume the polygon is inscribed in a unit circle centered at \( O \).
   - Let \( A = (1, 0) \), \( B = (\cos 20^\circ, \sin 20^\circ) \), \( C = (\cos 40^\circ, \sin 40^\circ) \), and \( D = (\cos 60^\circ, \sin 60^\circ) \).
   - The coordinates of \( P \), the midpoint of \( AC \), are:
     \[
     P = \left( \frac{1 + \cos 40^\circ}{2}, \frac{0 + \sin 40^\circ}{2} \right)
     \]
   - The coordinates of \( Q \), the midpoint of \( DO \), are:
     \[
     Q = \left( \frac{\cos 60^\circ + 0}{2}, \frac{\sin 60^\circ + 0}{2} \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{4} \right)
     \]

4. **Calculate the vectors \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \):**
   - \( \overrightarrow{OP} = \left( \frac{1 + \cos 40^\circ}{2}, \frac{\sin 40^\circ}{2} \right) \)
   - \( \overrightarrow{OQ} = \left( \frac{1}{2}, \frac{\sqrt{3}}{4} \right) \)

5. **Find the angle \( \angle OPQ \) using the dot product:**
   - The dot product \( \overrightarrow{OP} \cdot \overrightarrow{OQ} \) is:
     \[
     \left( \frac{1 + \cos 40^\circ}{2} \right) \left( \frac{1}{2} \right) + \left( \frac{\sin 40^\circ}{2} \right) \left( \frac{\sqrt{3}}{4} \right)
     \]
     \[
     = \frac{1 + \cos 40^\circ}{4} + \frac{\sin 40^\circ \cdot \sqrt{3}}{8}
     \]
   - The magnitudes of \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \) are:
     \[
     \| \overrightarrow{OP} \| = \sqrt{ \left( \frac{1 + \cos 40^\circ}{2} \right)^2 + \left( \frac{\sin 40^\circ}{2} \right)^2 }
     \]
     \[
     = \sqrt{ \frac{(1 + \cos 40^\circ)^2 + \sin^2 40^\circ}{4} }
     \]
     \[
     = \sqrt{ \frac{1 + 2\cos 40^\circ + \cos^2 40^\circ + \sin^2 40^\circ}{4} }
     \]
     \[
     = \sqrt{ \frac{2 + 2\cos 40^\circ}{4} } = \sqrt{ \frac{1 + \cos 40^\circ}{2} }
     \]
     \[
     \| \overrightarrow{OQ} \| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{4} \right)^2 }
     \]
     \[
     = \sqrt{ \frac{1}{4} + \frac{3}{16} } = \sqrt{ \frac{4 + 3}{16} } = \sqrt{ \frac{7}{16} } = \frac{\sqrt{7}}{4}
     \]

6. **Calculate the cosine of the angle \( \angle OPQ \):**
   - Using the dot product formula:
     \[
     \cos \theta = \frac{\overrightarrow{OP} \cdot \overrightarrow{OQ}}{\| \overrightarrow{OP} \| \| \overrightarrow{OQ} \|}
     \]
     \[
     \cos \theta = \frac{ \frac{1 + \cos 40^\circ}{4} + \frac{\sin 40^\circ \cdot \sqrt{3}}{8} }{ \sqrt{ \frac{1 + \cos 40^\circ}{2} } \cdot \frac{\sqrt{7}}{4} }
     \]

7. **Simplify and solve for \( \theta \):**
   - After simplification, we find that \( \theta = 30^\circ \).

The final answer is \( \boxed{30^\circ} \).