smc

We begin by drawing a diagram. We extend $CB$ and $DA$ to meet at $E.$ This gives us a couple right triangles in $CED$ and $BEA.$ We see that $\angle E=30^{\circ }$. Hence, $△BEA$ and $△DEC$ are 30-60-90 triangles. Using the side ratios of 30-60-90 triangles, we have $BE=2BA=6$. This tells us that $CE=BC+BE=4+6=10$. Also, $EA=3\sqrt{3}$. Because $△DEC\sim △BEA$, we have $$\frac{10}{3\sqrt{3}}=\frac{CD}{3}.$$ Solving the equation, we have $$\begin{array}{rl}\frac{CD}{3}& =\frac{10}{3\sqrt{3}}\\ CD& =3\cdot \frac{10}{3\sqrt{3}}\\ CD& =\frac{10}{\sqrt{3}}\end{array}$$ Hence, $CD=\boxed{\frac{10}{\sqrt{3}}}$.