Question 51, Given $\triangle A B C$ satisfies $A B=1, A C=2, \cos B+\sin C=1$, find the length of side $B C$.

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Question 51, Solution: According to the conditions, $\frac{\sin B}{\sin C}=\frac{A C}{A B}=2 \Rightarrow \sin B=2 \sin C$, so $1=\cos B+\sin C=$ $\cos B+\frac{\sin B}{2}=\cos B+\frac{\sqrt{1-\cos ^{2} B}}{2}$, solving this gives $\cos B=\frac{3}{5}$. Consequently, $\sin C=\frac{2}{5}$, hence $\cos C=\frac{\sqrt{21}}{5}$ or $-\frac{\sqrt{21}}{5}$. According to the projection theorem, $B C=A B \cdot \cos B+A C \cdot \cos C$.
If $\cos C=-\frac{\sqrt{21}}{5}$, then $B C=A B \cdot \cos B+A C \cdot \cos C=1 \times \frac{3}{5}-2 \times \frac{\sqrt{21}}{5}<0$, which does not satisfy the conditions;
If $\cos C=\frac{\sqrt{21}}{5}$, then $B C=A B \cdot \cos B+A C \cdot \cos C=1 \times \frac{3}{5}+2 \times \frac{\sqrt{21}}{5}=\frac{3+2 \sqrt{21}}{5}$.
In conclusion, the length of $B C$ is $\frac{3+2 \sqrt{21}}{5}$.