Tứ giác ABCD có AB//CD nên :

\(\widehat{A}\)+ \(\widehat{D}\)=180\(^0\)(trong cùng phía bù nhau)

Ta có: \(\widehat{A}\) = ( 180\(^0\) + 120\(^0\) ) : 2 = 100\(^0\)

\(\widehat{D}\)= 180\(^0\) - 100\(^0\) = 80\(^0\)

Ta lại có \(\widehat{B}\) = 2\(\widehat{C}\)

\(\widehat{B}\) + \(\widehat{C}\) = 180\(^0\)( trong cùng phía bù nhau)

Nên 2\(\widehat{C}\)+\(\widehat{C}\)= 180\(^0\)

3\(\widehat{C}\)= 180\(^0\)

=> \(\widehat{C}\)=60\(^0\)

=> \(\widehat{B}\)=120\(^0\)