Crazy Angles

Say $\angle ABD=\angle ADB=\alpha$ , $\angle CBD=\beta$ , then $\angle ACB=\alpha-\beta$ , given $\angle ABC-\angle ACB=30^\circ$ , $\alpha+\beta-\alpha+\beta=30^\circ$ $\beta=15^\circ$ So the supplementary angle of $\angle CBD=165^\circ$ , quite easily done.

As AB=AD,so ABD=ADB,let x be its measure,So in Triangle ABD x+x+A=180 or A=180-2x........(1)Given that ABC-ACB=30 or B-30=C........(2)Now in the whole triangle ABC,A+B+C=180 or 180-2x+B+B-30=180 or 2B-2x=180-180+30 or 2(B-x)=30 or B-x=15 or ABC-ABD=15 or CBD=15,Supplementary angle of CBD=180-15 so supplementary angle of CBD=165.