Ta có: Aa//Bb
\(\Rightarrow\widehat{ABb}+\widehat{BAa}=180^0\)( trong cùng phía)
\(\Rightarrow\widehat{ABb}=180^0-\widehat{BAa}=180^0-60^0=120^0\)
Ta có: Cc//Bb
\(\Rightarrow\widehat{CBb}+\widehat{BCc}=180^0\)(trong cùng phía)
\(\Rightarrow\widehat{CBb}=180^0-\widehat{BCc}=180^0-70^0=110^0\)
Ta có: \(\widehat{ABC}=360^0-\widehat{ABb}-\widehat{BCc}=360^0-120^0-110^0=130^0\)
