[tex]z-x=160^\circ-x-z\implies2z=160^\circ\implies z=80^\circ[/tex]
By the law of sines, in triangle ABD we have
[tex]\dfrac{\sin x}b=\dfrac{\sin(160^\circ-x)}a\implies\dfrac ab=\dfrac{\sin(160^\circ-x)}{\sin x}[/tex]
and in triangle ABC,
[tex]\dfrac{\sin80^\circ}a=\dfrac{\sin20^\circ}b\implies\dfrac ab=\dfrac{\sin80^\circ}{\sin20^\circ}[/tex]
By substitution we end up with
[tex]\dfrac{\sin(160^\circ-x)}{\sin x}=\dfrac{\sin80^\circ}{\sin20^\circ}[/tex]
[tex]\dfrac{\sin160^\circ\cos x-\cos160^\circ\sin x}{\sin x}=\dfrac{\sin80^\circ}{\sin^\circ}[/tex]
[tex]\sin160^\circ\cot x-\cos160^\circ=\dfrac{\sin80^\circ}{\sin20^\circ}[/tex]
[tex]\cot x=\dfrac{\sin80^\circ+\sin20^\circ\cos160^\circ}{\sin20^\circ\sin160^\circ}[/tex]
[tex]x=\cot^{-1}\left(\dfrac{\sin80^\circ+\sin20^\circ\cos160^\circ}{\sin20^\circ\sin160^\circ}\right)=10^\circ[/tex]
There's probably some sequence of identities you can apply to the right side to somehow end up with [tex]\cot x=\cot10^\circ[/tex] but I'm not seeing it right away... I just used a calculator at this point.
