a,Vì MB//CN\(\Rightarrow\widehat{ACN}=\widehat{CAx}\)(2 góc so le trong)
mà \(\widehat{ACN}=55^0\)
\(\Rightarrow\widehat{CAx}=55^0\)
b, Theo bài ra ta có :\(\widehat{BAC}\)=\(\widehat{CAx}+\widehat{BAx}\)
\(\Rightarrow108^0=55^0+\widehat{BAx}\Rightarrow\widehat{BAx}=53^0\)
mà\(\widehat{BAx}=\widehat{ABM}\)(2 góc so le trong)
\(\widehat{BAx}=53^0\Rightarrow\widehat{ABM}=53^0\)
Vậy\(\widehat{CAx}=55^0\)
\(\widehat{ABM}=53^0\)