4. If $\triangle A B C$ has exactly one solution when $\angle A B C=60^{\circ}, A C=12, B C=k$, then the range of $k$ is
A. $k=8 \sqrt{3}$
B. $0<k \leqslant 12$
C. $k \geqslant 12$
D. $0<k \leqslant 12$ or $k=8 \sqrt{3}$
Answer
As shown in the figure, when $k \sin 60^{\circ}=12 \Rightarrow k=8 \sqrt{3}$, $\triangle A B C$ has exactly one solution; when $12 \geqslant k>0$, $\triangle A B C$ also has exactly one solution. Therefore, the answer is $D$.