Ta có:
\(4x+2x+2x+x=360^o\)
\(\left(4+2+2+1\right)x=360^o\)
\(9x=360^o\)
\(x=40^o\)
=>\(\widehat{D}=40^o\)
\(\widehat{B}=\widehat{C}=2\cdot40^o=80^o\)
\(\widehat{A}=40^o\cdot4=160^o\)
Xét tứ giác ABCD có:
\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}+\widehat{ADC}=360\) độ
\(\Rightarrow4x+2x+2x+x=360\) độ
\(\Rightarrow9x=360\) độ \(\Rightarrow x=40\) độ
\(\Rightarrow\left\{{}\begin{matrix}\widehat{BAC}=4x=160\\\widehat{ABC}=2x=80\\\widehat{ACB}=2x=80\\\widehat{ADC}=x=40\end{matrix}\right.\)( đơn vị là độ)
Ta có: \(4x+2x+2x+x=360^0\)
\(\Leftrightarrow9x=360^0\)
\(\Leftrightarrow x=40^0\)
\(\Leftrightarrow\widehat{A}=160^0;\widehat{B}=80^0;\widehat{C}=80^0;\widehat{D}=40^0\)
