Ans: A
Note that
$\begin{array}{rcl}
AC^2 + BC^2 & = & 12^2 + 5^2 \\
& = & 169 \\
& = & 13^2 \\
& = & AB^2
\end{array}$
Therefore, by the converse of the Pythagoras Theorem, $\Delta ABC$ is a right-angled triangle with $\angle ACB=90^\circ$. Hence, we have
$\begin{array}{rcl}
\tan \theta & = & \dfrac{BC}{AC} \\
& = & \dfrac{5}{12}
\end{array}$
