Question

Find the value of x in each case. PLEASE SHOW WORK!!!

Answer 1

$m\angle BCA=180^o-5x$

$\angle CAB=\angle ACD\\\\m\angle CAB=33^o$

In the triangle ABC:

$m\angle ABC+m\angle BCA+m\angle CAB=180^o\\\\2x+180^o-5x+33^o=180^o\\\\-3x+213^o=180^o\ \ \ |-213^o\\\\-3x=-33^o\ \ \ |:(-3)\\\\x=11^o$