Hình mk vẽ chỉ mang tính minh họa ko phải chính xác nên bn chú ý nhé
Do AB//CD \(\Rightarrow\widehat{A}+\widehat{D}=180^0\)
Ta có\(\left\{{}\begin{matrix}\widehat{A}+\widehat{D}=180^0\\\widehat{A}-\widehat{D}=20^0\end{matrix}\right.\)\(\Rightarrow\widehat{A}=100^0;\widehat{D}=20^0\)
Cx vì: AB//CD \(\Rightarrow\widehat{B}+\widehat{C}=180^0\)
Mà \(\widehat{B}=2\widehat{C}\Rightarrow2\stackrel\frown{C}+\widehat{C}=3\widehat{C}=180^0\Rightarrow C=60^0\)\(\Rightarrow\widehat{B}=120^0\)
Vậy \(\widehat{A}=100^0;\widehat{B}=120^0;\widehat{C}=60^0;\widehat{D}=80^0\)
a) A+D=180
A-D =20
=> A= 100; D= 80
b) B+C= 180
B=2C
=> B= 120; C =60
