\(1,\\ a//b\Rightarrow\left\{{}\begin{matrix}\widehat{CAB}+\widehat{ACD}=180^0\left(trong.cùng.phía\right)\\\widehat{ABD}+\widehat{BDC}=180^0\left(trong.cùng.phía\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\widehat{ACD}=180^0-100^0=80^0\\y=\widehat{BDC}=180^0-120^0=60^0\end{matrix}\right.\)
\(2,\\ a//b\Rightarrow\widehat{A_2}+\widehat{B_2}=180^0\left(trong.cùng.phía\right)\\ \widehat{A_2}-\widehat{B_2}=30^0\\ \Rightarrow\widehat{A_2}=\left(180^0+30^0\right):2=110^0\\ \Rightarrow\widehat{B_2}=180^0-110^0=70^0\\ a//b\Rightarrow\left\{{}\begin{matrix}\widehat{A_2}=\widehat{B_1}=110^0\left(so.le.trong\right)\\\widehat{B_2}=\widehat{A_1}=70^0\left(so.le.trong\right)\end{matrix}\right.\)
Bài 1:
Ta có: a//b
\(\Rightarrow\widehat{C}+\widehat{A}=180^0\)(trong cùng phía)
\(\Rightarrow x+100^0=180^0\Rightarrow x=80^0\)
Ta có: a//b
\(\Rightarrow\widehat{B}+\widehat{D}=180^0\)(trong cùng phía)
\(\Rightarrow y+120^0=180^0\Rightarrow y=60^0\)
Bài 2:
Ta có: \(\left\{{}\begin{matrix}\widehat{A_2}+\widehat{B_2}=180^0\\\widehat{A_2}-\widehat{B_2}=30^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A_2}=\left(180^0+30^0\right):2=105^0\\\widehat{B_2}=180^0-105^0=75^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A_1}=\widehat{B_2}=75^0\\\widehat{A_2}=\widehat{B_1}=105^0\end{matrix}\right.\)( so le trong)
