[Geometry]
Problem: In triangle $A B C, A B=5, A C=6$, and $\tan \angle B A C=-\frac{4}{3}$. Compute the area of $\triangle A B C$.

Solution: Let $s=\sin \angle B A C$. Then $s>0$ and $\frac{s}{-\sqrt{1-s^{2}}}=-\frac{4}{3}$, which gives $s=\frac{4}{5}$. The area of triangle $A B C$ is therefore $\frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C=\frac{1}{2} \cdot 5 \cdot 6 \cdot \frac{4}{5}=\mathbf{1 2}$.

Final answer: 12