$17 \cdot 35$ As shown in the figure, in $\triangle A B C$, if it is known that $\angle A$ : $\angle B: \angle C=3: 5: 10$; and $\triangle A^{\prime} B^{\prime} C \cong \triangle A B C$, then $\angle B C A^{\prime}: \angle B C B^{\prime}$ equals
(A) $1: 2$.
(B) $1: 3$.
(C) $1: 4$.
(D) $2: 3$.
(China Sichuan Junior High School Mathematics Competition, 1989)
［Solution］From the given information, we have
$$
\angle B B^{\prime} C=\angle B^{\{\prime\}} B C=180^{\circ} \cdot \frac{5}{3+5+10}=50^{\circ} \text {, }
$$

In $\triangle B^{\prime} B C$, $B C=B^{\prime} C$, so $\triangle B^{\prime} B C$ is an isosceles triangle.
Therefore, $\angle B C A=180^{\circ}-\angle B C B^{\prime}=180^{\circ}-\left(180^{\circ}-2 \angle B B^{\prime} C\right)$
$$
\begin{array}{l}
=180^{\circ}-\left(180^{\circ}-2 \cdot 50^{\circ}\right) \\
=100^{\circ},
\end{array}
$$

Also, $\angle B^{\prime} C A^{\prime}+\angle B C A-\angle B C A^{\prime}=180^{\circ}$,
Thus, $\angle B C A^{\prime}=2 \angle B C A-180^{\circ}=200^{\circ}-180^{\circ}=20^{\circ}$.
Therefore, $\angle B C A^{\prime}: \angle B C B^{\prime}=1: 4$.
Hence, the answer is $(C)$.