Question

Answer 1

Do AD//BE \(\Rightarrow\widehat{EAD}=180^o-\widehat{AEB}=180^o-50^o=130^o\)

Có: \(\widehat{FAD}=180^o-\widehat{AFC}=180^o-40^o=140^o\)

\(\Rightarrow\widehat{EAF}=360^o-\widehat{EAD}-\widehat{FAD}=360^o-140^o-130^o=90^o\)

Answer 2

có : EB//DA

`=> góc BEA + góc DAF = 180^0`( trong cùng phía bù nhau)

`=> 50^0 + DAF = 180^0`

`=>            góc DAF = 130^0`

có : AD//CF

`=> DAF + CFA = 180^0`( TCP bù nhau)

`=> DAF  + 40^0 = 180^0`

`=>  DAF             = 140^0`

Lại có :

`EAD + DAF + EAF + 360^0`

`=> 130^0 + 140^0 + EAF = 360^0`

`=>  270^0 + EAF = 360^0`

`=> EAF                = 90^0`