Do \(\widehat{A}=\widehat{D}=60^{0}(so le trong)\)
\(\Rightarrow\)\(AB//CD\)
\(\Rightarrow\)\(x+\widehat{C}=180^0(trong cùng phía) \)
\(\Rightarrow\)\(x+80^0=180^0\)
\(x=100^0\)
+)Vì \(\widehat{A1}\)=\(\widehat{B1}\)(gt)
Mà 2 góc này ở vị trí so le trong
⇒AB//CD
+Vì AB//CD(cm trên)
⇒\(\widehat{B1}\)+\(\widehat{C1}\)=1800
⇒\(\widehat{B1}\)=1800-\(\widehat{C1}\)
⇒\(\widehat{B1}\)=1800-800
⇒\(\widehat{B1}\)=1000
⇒x=1000
Ta có: AB//CD
nên \(x+80^0=180^0\)
hay \(x=180^0-80^0=100^0\)