2007-II-23

Ans: B
In $\Delta ABC$,

$\begin{array}{rcl}
\angle ABC & = & 180^\circ – 90^\circ – 30^\circ \\
& = & 60^\circ
\end{array}$

Since $BD$ is the angle bisector of $\angle ABC$, then $\angle CBD= 30^\circ$.

In $\Delta ABC$,

$\begin{array}{rcl}
\sin 30^\circ & = & \dfrac{BC}{AB} \\
\dfrac{1}{2} & = & \dfrac{BC}{c} \\
BC & = & \dfrac{c}{2}
\end{array}$

In $\Delta BCD$,

$\begin{array}{rcl}
\tan \angle CBD & = & \dfrac{CD}{BC} \\
CD & = & BC \times \tan \angle CBD \\
CD & = & \dfrac{c}{2} \times \tan 30^\circ \\
CD & = & \dfrac{c}{2} \times \dfrac{1}{\sqrt{3}} \\
CD & = & \dfrac{c}{2\sqrt{3}}
\end{array}$