Four, Given that the area of $\triangle \mathrm{ABC}$ is $5 \mathrm{~cm}^{2}, a=4 \mathrm{~cm}$, $b=5 \mathrm{~cm}$, find the value of $\cos C$. (10 points)
$$
\begin{array}{l}
\because \mathrm{S}_{\triangle \mathrm{ABC}}=\frac{1}{2} \mathrm{ab} \sin C, \\
6=\frac{1}{2} \times 4 \times 5 \sin C, \\
\sin C=\frac{3}{5} . \\
\cos C= \pm \sqrt{1-\sin ^{2} C}= \pm \sqrt{1-\left(\frac{3}{5}\right)^{2}}= \pm \frac{4}{5} .
\end{array}
$$

〈or set $\mathrm{y}=3 \mathrm{k}, \mathrm{r}=5 \mathrm{k}$, then $\mathrm{x}= \pm 4 \mathrm{k}$,
$$
\left.\cos C= \pm \frac{4 k}{5 k}= \pm \frac{4}{5}\right\rangle
$$