Question

Trigonometry. I can't figure it out. ​

Answer 1

You also seem to be given that $m\angle ADC=50^\circ$, according to the diagram.

In triangle ABC, we have

$\cos\angle CAB=\dfrac39=\dfrac13\implies m\angle CAB=m\angle CAD=\cos^{-1}\dfrac13$

By the law of sines,

$\dfrac{\sin\angle ADC}{AC}=\dfrac{\sin\angle CAD}{CD}$

$\implies CD=\dfrac{(9\,\mathrm{cm})\sin\left(\cos^{-1}\frac13\right)}{\sin50^\circ}\approx11.1\,\mathrm{cm}$