Question

Answer 1

Ta có: \(\widehat{A_1}\) kề bù với \(\widehat{A_2}\)

\(\Rightarrow\)\(\widehat{A_1}+\)\(\widehat{A_2}\)\(=180^0\)

\(\Rightarrow120^0+\)\(\widehat{A_2}\)\(=180^0\)

\(\Rightarrow\widehat{A_2}=180^0-120^0=60^0\)

Ta lại có: \(\widehat{B_1}\) so le trong với \(\widehat{A_2}\)

\(\)⇒a//b