Ans: A
For some positive constant $k$,
For some positive constant $k$,
$\begin{array}{rcl}
AB^2 + BC^2 & = & (8k)^2 + (15k)^2 \\
& = & 289k^2
\end{array}$
and
$\begin{array}{rcl}
AC^2 & = & (17k)^2 \\
& = & 289k^2
\end{array}$
$\because AB^2+BC^2 = AC^2 = 289k^2$,
$\therefore$ by the converse of Pythagorus Theorem, $\Delta ABC$ is a right-angled triangle with $\angle B = 90^\circ$.
Hence, we have
$\begin{array}{rcl}
\cos A : \cos C & = & \dfrac{AB}{AC} : \dfrac{BC}{AC} \\
& = & AB : BC \\
& = & 8 : 15
\end{array}$
