jmc

Since $\cos B=\frac{3}{5}$, we have $\cos B=\frac{AB}{BC}=\frac{9}{BC}=\frac{3}{5}$. Then, we can see that we have $9=BC\cdot \frac{3}{5}$, so $BC=9\cdot \frac{5}{3}=15$. From the Pythagorean Theorem, we have $AC=\sqrt{B{C}^2-A{B}^2}=\sqrt{15^2-9^2}=\sqrt{144}=12$. Finally, we can find $\cos C$: $\cos C=\frac{AC}{BC}=\frac{12}{15}=\boxed{\frac{4}{5}}$.