Five. As shown in Figure 2, it is known that
$ABCD$ is a cyclic quadrilateral,
\[
\begin{array}{r}
\angle D=60^{\circ}, \quad AB=3 \text{ cm}, \\
BC=2 \text{ cm}, \text{ find the length of } AC. \\
\text{ (10 points) }
\end{array}
\]
Five. Solution (as shown in Figure 2)
\[
\because A B C D \text { is a cyclic quadrilateral,}
\]
\[
\begin{aligned}
\text { and } \angle \mathrm{D} & =60^{\circ}, \\
\therefore \angle \mathrm{B} & =120^{\circ} \\
\because \mathrm{AB} & =3 \mathrm{~cm}, \mathrm{BC}=2 \mathrm{~cm}, \\
\therefore & \mathrm{AC}^{2}=3^{2}+2^{2}-2 \times 3 \times 2 \cos 120^{\circ}=19 .
\end{aligned}
\]

Therefore, $\mathrm{AC}=\sqrt{19}$.