9.39 In the acute triangle $\triangle ABC$, $AC=1, \quad AB$ $=c, \angle A=60^{\circ},$ and the circumradius $R \leqslant 1$ of $\triangle ABC$. Then
(A) $\frac{1}{2}<c \leqslant 1$.
(B) $1<c \leqslant \sqrt{3}$.
(C) $\sqrt{3}<c \leqslant 2$.
(D) $c=2$.
[Solution 1] Draw $CD \perp AB$. Since $\triangle ABC$ is an acute triangle, $D$ is on $AB$,
thus $c=AB>AD=AC \cos A=\cos A=\frac{1}{2}$.
By the Law of Sines, we have
$$
c=AB=2R \sin C<2R \leqslant 2.
$$

Combining (1) and (2), we get $\frac{1}{2}<c<2$.
Therefore, the answer is $(A)$.
[Solution 2] Consider an equilateral triangle that satisfies the given conditions, then $c=1$.
Therefore, the answer is $(A)$.
[Solution 3] As shown in the figure, draw $CB' \perp AB$, then
$$
AB' = AC \cdot \cos A = \cos A = \frac{1}{2}.
$$

Also, draw $CB'' \perp AC$, then
$$
AB'' = \frac{AC}{\cos A} = \frac{1}{\cos A} = 2.
$$

Since $\triangle ABC$ is an acute triangle, $B$ should lie between $B'$ and $B''$ (excluding the endpoints).
$$
\text{Thus, } \frac{1}{2}<c<2.
$$

Therefore, the answer is $(A)$.
Note: From Solution 3, we find that the condition “$\triangle ABC$ has an circumradius $R \leqslant 1$” is redundant.
In fact, given $\angle A=60^{\circ}$, and the triangle is acute, we have $30^{\circ}<B<90^{\circ}$.
Thus, $\frac{1}{2}<\sin B<1$. Also, $R=\frac{AC}{2 \sin B}=\frac{1}{2 \sin B}<1$.
That is, from the given conditions, we can deduce $R<1$. Moreover, when $R=1$, $\triangle ABC$ is a right triangle, which contradicts the given conditions!