Let $PQRS$ be a square with side length 12. Point $A$ lies on segment $\overline{QR}$ with $\angle QPA = 30^\circ$, and point $B$ lies on segment $\overline{PQ}$ with $\angle SRB = 60^\circ$. What is $AB$?
1. Given that $PQRS$ is a square with side length 12, we know that $PQ = QR = RS = SP = 12$.
2. Point $A$ lies on segment $\overline{QR}$ such that $\angle QPA = 30^\circ$. Since $PQ = 12$, we can use trigonometry to find $QA$.
   - In $\triangle QPA$, $\angle QPA = 30^\circ$ and $PQ = 12$.
   - Using the sine function, $\sin(30^\circ) = \frac{QA}{PQ} = \frac{QA}{12}$.
   - Since $\sin(30^\circ) = \frac{1}{2}$, we have $\frac{1}{2} = \frac{QA}{12}$.
   - Solving for $QA$, we get $QA = 12 \cdot \frac{1}{2} = 6$.
3. Point $B$ lies on segment $\overline{PQ}$ such that $\angle SRB = 60^\circ$. Since $PQ = 12$, we can use trigonometry to find $QB$.
   - In $\triangle SRB$, $\angle SRB = 60^\circ$ and $RS = 12$.
   - Using the cosine function, $\cos(60^\circ) = \frac{QB}{RS} = \frac{QB}{12}$.
   - Since $\cos(60^\circ) = \frac{1}{2}$, we have $\frac{1}{2} = \frac{QB}{12}$.
   - Solving for $QB$, we get $QB = 12 \cdot \frac{1}{2} = 6$.
4. Now, we have $QA = 6$ and $QB = 6$. Since $\angle BQA = 90^\circ$, $\triangle BQA$ is a right triangle with legs $QA$ and $QB$.
5. To find the length of $AB$, we use the Pythagorean theorem in $\triangle BQA$:
   \[
   AB = \sqrt{QA^2 + QB^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}
   \]

The final answer is $\boxed{6\sqrt{2}}$.