a: AB//DC
=>\(\widehat{ADC}+\widehat{BAD}=180^0\)
=>\(\widehat{ADC}=180^0-120^0=60^0\)
Ta có: AB//DC
=>\(\widehat{CBA}+\widehat{BCD}=180^0\)
=>\(\widehat{CBA}+30^0=180^0\)
=>\(\widehat{CBA}=150^0\)
b: Xét ΔMDC có \(\widehat{DMB}\) là góc ngoài tại đỉnh M
nên \(\widehat{DMB}=\widehat{MDC}+\widehat{MCD}\)
=>\(\widehat{DMB}=\widehat{MDC}+\widehat{yBC}\)