\(a,\)Vì \(Ax//Dy\) nên \(\widehat{xAD}=180-\widehat{CDB}\left(trong.cùng.phía\right)\)
Ta có \(\widehat{ACB}=\widehat{CDB}+\widehat{CBD}\left(góc.ngoài\right);\widehat{CBy}=\widehat{BCD}+\widehat{BDC}\left(góc.ngoài\right)\)
\(\Rightarrow\widehat{xAD}+\widehat{ACB}+\widehat{CBy}=180-\widehat{CDB}+\widehat{CDB}+\widehat{CBD}+\widehat{BCD}+\widehat{CDB}=180+\widehat{CDB}+\widehat{CBD}+\widehat{BCD}=180+180=360\)
\(b,\widehat{yBC}-\widehat{ACB}=30\Leftrightarrow\widehat{yBC}=30+\widehat{ACB}\\ \Leftrightarrow\widehat{xAC}+\widehat{ACB}+30+\widehat{ACB}=360\left(câu.a\right)\\ \Leftrightarrow2\widehat{ACB}+110=330\\ \Leftrightarrow\widehat{ACB}=110\\ \Leftrightarrow\widehat{BCD}=180-\widehat{ACB}=70\left(kề.bù\right)\\ \widehat{xAD}+\widehat{CDB}=180\left(chứng.minh.trên\right)\\ \Leftrightarrow\widehat{CDB}=180-\widehat{xAD}=70\\ \Leftrightarrow\widehat{CBD}=180-\widehat{BCD}-\widehat{CDB}=40\)
