Ans: D
In $\Delta ACD$,
$\begin{array}{rcll}
\tan \theta & = & \dfrac{AC}{CD} \\
AC & = & CD \tan\theta & \ldots \unicode{x2460}
\end{array}$
In $\Delta ABC$,
$\begin{array}{rcll}
\sin \phi & = & \dfrac{AB}{AC} \\
AC & = & \dfrac{AB}{\sin \phi} & \ldots \unicode{x2461}
\end{array}$
Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
\dfrac{AB}{\sin \phi} & = & CD \tan\theta \\
\dfrac{AB}{CD} & = & \tan \theta \sin \phi
\end{array}$
