[Geometry]
Problem: Let $T=3$. In triangle $A B C, A B=A C-2=T$, and $\mathrm{m} \angle A=60^{\circ}$. Compute $B C^{2}$.

Solution: By the Law of Cosines, $B C^{2}=A B^{2}+A C^{2}-2 \cdot A B \cdot A C \cdot \cos A=T^{2}+(T+2)^{2}-2 \cdot T \cdot(T+2) \cdot \frac{1}{2}=$ $T^{2}+2 T+4$. With $T=3$, the answer is 19 .

Final answer: 19