[Geometry]
Problem: Let $T=3$. In $\triangle A B C, A C=T^{2}, \mathrm{~m} \angle A B C=45^{\circ}$, and $\sin \angle A C B=\frac{8}{9}$. Compute $A B$.

Solution: From the Law of Sines, $\frac{A B}{\sin \angle A C B}=\frac{A C}{\sin \angle A B C}$. Thus $A B=\frac{8}{9} \cdot \frac{T^{2}}{1 / \sqrt{2}}=\frac{8 \sqrt{2}}{9} \cdot T^{2}$. With $T=3, A B=\mathbf{8} \sqrt{\mathbf{2}}$.

Final answer: $8 \sqrt{2}$