2008-II-22

Ans: C
Consider $\Delta ABD$.

$\begin{array}{rcl}
\tan 30^\circ & = & \dfrac{BD}{AD} \\
AD & = & \dfrac{BD}{\tan 30^\circ} \\
& = & \dfrac{BD}{1/\sqrt{3}} \\
& = & \sqrt{3} BD
\end{array}$

Consider $\Delta CBD$.

$\begin{array}{rcl}
\tan 45^\circ & = & \dfrac{BD}{DC} \\
DC & = & \dfrac{BD}{\tan 45^\circ} \\
& = & BD
\end{array}$

Hence, we have

$\begin{array}{rcl}
AD : DC & = & \sqrt{3} BD : BD \\
& = & \sqrt{3} : 1
\end{array}$