a) AB//CD
\(=>\widehat{ABC}+\widehat{BCD}=180^o\) (cặp góc trong cùng phía)
\(=>\widehat{BCD}=180^o-\widehat{ABC}\\ =>\widehat{BCD}=x=180^o-140^o=40^o\)
b) \(\widehat{MNP}=180^o-70^o=110^o\)
MN//PQ
\(=>\left\{{}\begin{matrix}\widehat{MQP}+\widehat{QMN}=180^o\\\widehat{MNP}+\widehat{QPN}=180^o\end{matrix}\right.\\ =>\left\{{}\begin{matrix}60^o+x=180^o\\110^o+y=180^o\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=180^o-60^o=120^o\\y=180^o-110^o=70^o\end{matrix}\right.\)
c) Xét tứ giác HGKI ta có:
\(\widehat{H}+\widehat{G}+\widehat{K}+\widehat{I}=360^o\\ =>4x+3x+2x+x=360^o\\ =>10x=360^o\\ =>x=\dfrac{360^o}{10}\\ =>x=36^o\)
d) Xét tứ giác VUTS ta có:
\(\widehat{V}+\widehat{U}+\widehat{T}+\widehat{S}=360^o\\ =>2x+x+90^o+90^o=360^o\\ =>3x+180^o=360^o\\ =>3x=360^o-180^o=180^o\\ =>x=\dfrac{180^o}{3}=60^o\)
