Vì AB//CD nên \(\left\{{}\begin{matrix}\widehat{A}+\widehat{D}=180^0\\\widehat{B}+\widehat{C}=180^0\end{matrix}\right.\left(trong.cùng.phía\right)\)
Mà \(\widehat{A}-\widehat{D}=30^0;\widehat{B}=2\widehat{C}\Rightarrow\left\{{}\begin{matrix}\widehat{A}=\left(180^0+30^0\right):2=105^0\\\widehat{D}=180^0-105^0=75^0\\3\widehat{C}=180^0\end{matrix}\right.\)
\(\Rightarrow\widehat{C}=60^0\Rightarrow\widehat{B}=120^0\)