Let $ABCDEF$ be a regular hexagon. Points $P$ and $Q$ on tangents to its circumcircle at $A$ and $D$ respectively are such that $PQ$ touches the minor arc $EF$ of this circle. Find the angle between $PB$ and $QC$.
1. **Define the problem and setup:**
   Let $ABCDEF$ be a regular hexagon inscribed in a circle with center $O$. Points $P$ and $Q$ are on the tangents to the circumcircle at $A$ and $D$ respectively, such that the line $PQ$ touches the minor arc $EF$ of the circle. We need to find the angle between the lines $PB$ and $QC$.

2. **Consider the duality transform:**
   We will use the duality transform with respect to the circumcircle. Let $X$ be any point on the minor arc $EF$. Let $XA$ meet the tangent to the circle at $B$ at point $R$, and $XD$ meet the tangent at $C$ at point $S$. We need to prove that $\angle ROS = 150^\circ$.

3. **Analyze the angles:**
   We start by analyzing the angles formed by the tangents and the lines from $X$:
   - $\angle (DC; RB) = \angle (CS; BA) = \angle (SD; AR) = \frac{\pi}{2}$.
   This means that there is a spiral similarity that maps $S$, $D$, $C$ onto $A$, $R$, $B$ respectively.

4. **Use properties of spiral similarity:**
   Since there is a spiral similarity, we have:
   \[
   |CS| \cdot |BR| = |AB| \cdot |CD| = |OC| \cdot |OB|
   \]
   This implies that $\triangle OCS \sim \triangle RBO$.

5. **Calculate the angles:**
   From the similarity, we have:
   \[
   \angle COS + \angle ROB = 90^\circ
   \]
   Since $O$ is the center of the circle, $\angle BOC = 60^\circ$ (as $B$ and $C$ are adjacent vertices of the regular hexagon).

6. **Combine the angles:**
   Therefore, we have:
   \[
   \angle ROS = 90^\circ + \angle BOC = 90^\circ + 60^\circ = 150^\circ
   \]

7. **Relate to the original problem:**
   The angle between $PB$ and $QC$ is the same as $\angle ROS$ because of the properties of the tangents and the duality transform.

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