2010-II-21

Ans: B
In $\Delta ABD$,

$\begin{array}{rcl}
\tan \alpha & = & \dfrac{AD}{BD} \\
AD & = & BD \tan \alpha \ldots \unicode{x2460}
\end{array}$

In $\Delta ACD$,

$\begin{array}{rcl}
\sin \beta & = & \dfrac{AD}{AC} \\
AD & = & AC \sin \beta \ldots \unicode{x2461}
\end{array}$

By solving $\unicode{x2460}$ and $\unicode{x2461}$, we have

$\begin{array}{rcl}
BD \tan \alpha & = & AC \sin \beta \\
\dfrac{AC}{BD} & = & \dfrac{\tan\alpha}{\sin\beta}
\end{array}$