Question

How do I work this out?

Answer 1

We can find the length of BC:

$(BC)^2+6^2=12^2\implies BC=6\sqrt3$

Then by the law of sines,

$\dfrac{\sin55^\circ}{BC}=\dfrac{\sin90^\circ}{CD}\iff CD=\dfrac{6\sqrt3}{\sin55^\circ}\approx12.7\,\mathrm{cm}$