Ans: D
In $\Delta ABD$,
In $\Delta ABD$,
$\begin{array}{rcl}
\angle BDA & = & 40^\circ – 20^\circ \\
& = & 20^\circ
\end{array}$
$\because \angle DBA = \angle BAD = 20^\circ$,
$\therefore BD=AB=10\mbox{ cm}$.
$\therefore BD=CD=10\mbox{ cm}$.
$\therefore \angle DCB = \angle DBC =40^\circ$.
In $\Delta DBC$,
$\begin{array}{rcl}
\angle BDC & = & 180^\circ-40^\circ-40^\circ \\
& = & 100^\circ
\end{array}$
By cosine law, we have
$\begin{array}{rcl}
BC^2 & = & 10^2+10^2-2(10)(10)\cos 100^\circ \\
BC^2 & = & 234.729~635~5 \\
BC & = & 15.320~888~86 \\
BC & \approx & 15 \mbox{ cm}
\end{array}$
