Ta có : $\widehat{A_{1}}$ = `120^o`
=> $\widehat{A_{3}}$ = `120^o`
Ta có : $\widehat{A_{3}}$ + $\widehat{A_{4}}$ = `180^o` ( kề bù )
=> `120^o` + $\widehat{A_{4}}$ = `180^o`
=> $\widehat{A_{4}}$ = `180^o - 120^o = 60^o`
Ta có : a // b
=> $\widehat{B_{1}}$ = $\widehat{A_{3}}$ = `120^o` ( SLT )
=> $\widehat{B_{2}}$ = $\widehat{A_{4}}$ = `60^o` ( SLT )
Vậy $\widehat{A_{3}}$ = `120^o`
$\widehat{A_{4}}$ = `60^o`
$\widehat{B_{1}}$ = `120^o`
$\widehat{B_{2}}$ = `60^o`
