Bài 2:
Ta có: AB//CD
nên \(\widehat{A}+\widehat{D}=180^0\)
mà \(\widehat{A}-\widehat{D}=40^0\)
nên \(\left\{{}\begin{matrix}\widehat{A}=\dfrac{180^0+40^0}{2}=110^0\\\widehat{D}=70^0\end{matrix}\right.\)
Ta có: \(\widehat{C}-2\widehat{D}=30^0\)
nên \(\widehat{C}=30^0+140^0=170^0\)
=>\(\widehat{B}=10^0\)