[tex]m\angle BCA=180^o-5x[/tex]
[tex]\angle CAB=\angle ACD\\\\m\angle CAB=33^o[/tex]
In the triangle ABC:
[tex]m\angle ABC+m\angle BCA+m\angle CAB=180^o\\\\2x+180^o-5x+33^o=180^o\\\\-3x+213^o=180^o\ \ \ |-213^o\\\\-3x=-33^o\ \ \ |:(-3)\\\\x=11^o[/tex]
