Question

hính đúng k sai

Answer 1

EF//BC

=>\(\widehat{BEF}=130^0\)

Ta có: \(\widehat{BEF}+\widehat{AEF}=180^0\)(hai góc kề bù)

=>\(\widehat{AEF}+130^0=180^0\)

=>\(\widehat{AEF}=50^0\)

Ta có: \(\widehat{AFE}+\widehat{EFC}=180^0\)(hai góc kề bù)

=>\(\widehat{AFE}+140^0=180^0\)

=>\(\widehat{AFE}=40^0\)

Xét ΔAEF có \(\widehat{AEF}+\widehat{AFE}+\widehat{FAE}=180^0\)

=>\(x+50^0+40^0=180^0\)

=>\(x=90^0\)