Ta có : \(Ax//By\)
\(\Rightarrow\widehat{xAB}+\widehat{ABm}=180^0\) ( hai góc trong cùng phía )
\(\Rightarrow150^0+\widehat{ABm}=180^0\)
\(\Rightarrow\widehat{ABm}=180^0-150^0=30^0\)
Mặt khác :
\(\widehat{mBC}+\widehat{BCy}=180^0\) ( hai góc trong cùng phía )
\(\Rightarrow\widehat{mBC}+130^0=180^0\)
\(\Rightarrow\widehat{mBC}=180^0-130^0=50^0\)
Lại có :
\(\widehat{ABm}+\widehat{mBC}=\widehat{ABC}\)
\(\Rightarrow30^0+50^0=\widehat{ABC}\)
\(\Rightarrow\widehat{ABC}=80^0\)
Vậy : \(\widehat{ABC}=80^0\)
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