a) Ta có: a⊥c,b⊥c
=> a//b
b) Ta có: a//b
\(\Rightarrow\widehat{B_1}+\widehat{A_1}=180^0\)(trong cùng phía)
\(\Rightarrow\widehat{B_1}=180^0-120^0=60^0\)
c) Ta có: \(\widehat{A_1}=\widehat{B_2}\)(2 góc so le trong và a//b)
\(\Rightarrow\widehat{A_2}+\widehat{B_2}=\widehat{A_2}+\widehat{A_1}=180^0\)(kề bù)