In the acute triangle $\triangle A B C$, $A B=c, A C=1$, $\angle A=60^{\circ}$, and the circumradius $R \leqslant 1$ of $\triangle A B C$, then ).
(A) $\frac{1}{2}<c \leqslant 1$.
(B) $1<c \leqslant \sqrt{3}$.
(C) $\sqrt{3}<c \leqslant 2$.
( D) ) $c=2$.
Solve As shown in Figure 5, construct $C D \perp A B \exists^{\circ} D$, $B E \perp A C$ at $E$. Since $A B C$ is an acute triangle, $D$ and $E$ are located on $A B$ and $A C$ respectively. Therefore,
$$
\begin{array}{l}
c>A D=A C \cos 60^{\circ} \\
=\frac{1}{2}, \\
c=2 A E<2 A C=2 .
\end{array}
$$

Therefore, the answer is (A).