Ans: D
Since $AB:BC:AC = 3:4:5$, then let $AB=3k$, $BC= 4k$ and $AC = 5k$, where $k$ is a constant. Consider $\Delta ABC$,
Since $AB:BC:AC = 3:4:5$, then let $AB=3k$, $BC= 4k$ and $AC = 5k$, where $k$ is a constant. Consider $\Delta ABC$,
$\begin{array}{rcl}
AB^2 + BC^2 & = & (3k)^2 + (4k)^2 \\
& = & 9k^2 + 16k^2 \\
& = & 25k^2
\end{array}$
Also,
$\begin{array}{rcl}
AC^2 & = & (5k)^2 \\
& = & 25k^2 \\
& = & AB^2 + BC^2
\end{array}$
Therefore, by the converse of the Pythagoras Theorem, $\Delta ABC$ is a right-angled triangle with $\angle ABC = 90^\circ$. Hence, we have
$\begin{array}{rcl}
\tan A : \cos C & = & \dfrac{BC}{AB} : \dfrac{BC}{AC} \\
& = & \dfrac{1}{AB} : \dfrac{1}{AC} \\
& = & AC : AB \\
& = & 5 : 3
\end{array}$
