Ans: C
In $\Delta ABC$,
$\begin{array}{rcl}
AC & = & \sqrt{3^2+4^2} \\
& = & 5\mbox{ cm}
\end{array}$
Note that $\Delta ABC \sim \Delta CND$, then we have
$\begin{array}{rcl}
\dfrac{AC}{CD} & = & \dfrac{BC}{ND} \\
\dfrac{5}{4} & = & \dfrac{3}{ND} \\
ND & = & \dfrac{12}{5}
\end{array}$
In $\Delta DNE$,
$\begin{array}{rcl}
\tan \theta & = & \dfrac{DE}{DN} \\
& = & \dfrac{6}{\frac{12}{5}} \\
& = & \dfrac{5}{2}
\end{array}$
