Let $P$ and $Q$ be the midpoints of sides $AB$ and $BC,$ respectively, of $\triangle ABC.$ Suppose $\angle A = 30^{\circ}$ and $\angle PQC = 110^{\circ}.$ Find $\angle B$ in degrees.
1. Given that $P$ and $Q$ are the midpoints of sides $AB$ and $BC$ respectively in $\triangle ABC$, we can use the Midpoint Theorem. The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. Therefore, $PQ \parallel AC$.

2. Since $PQ \parallel AC$, the corresponding angles are equal. Thus, $\angle BPQ = \angle A = 30^\circ$.

3. We are given that $\angle PQC = 110^\circ$. Since $PQ \parallel AC$, $\angle PQC$ is an exterior angle to $\triangle BPQ$. The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. Therefore, $\angle PQC = \angle BPQ + \angle B$.

4. Substituting the known values, we get:
   \[
   \angle PQC = \angle BPQ + \angle B
   \]
   \[
   110^\circ = 30^\circ + \angle B
   \]

5. Solving for $\angle B$, we subtract $30^\circ$ from both sides:
   \[
   \angle B = 110^\circ - 30^\circ = 80^\circ
   \]

The final answer is $\boxed{80^\circ}$