We can find the length of BC:
[tex](BC)^2+6^2=12^2\implies BC=6\sqrt3[/tex]
Then by the law of sines,
[tex]\dfrac{\sin55^\circ}{BC}=\dfrac{\sin90^\circ}{CD}\iff CD=\dfrac{6\sqrt3}{\sin55^\circ}\approx12.7\,\mathrm{cm}[/tex]
