\(a,AB//CD\Rightarrow\widehat{A}+\widehat{D}=180^0\left(2.góc.TCP\right)\\ \Rightarrow\widehat{A}=180^0-\widehat{D}=180^0-60^0=120^0\)
\(b,AB//CD\Rightarrow\widehat{B}+\widehat{C}=180^0\left(1\right)\\ 5\widehat{B}=4\widehat{C}\Rightarrow\widehat{B}=\dfrac{4}{5}\widehat{C}\left(2\right)\)
Thay (2) vào (1) \(\Rightarrow\dfrac{4}{5}\widehat{C}+\widehat{C}=180^0\Rightarrow\dfrac{9}{5}\widehat{C}=180^0\Rightarrow\widehat{C}=100^0\)
\(\Rightarrow\widehat{B}=100^0\cdot\dfrac{4}{5}=80^0\)
a: \(\widehat{A}=180^0-60^0=120^0\)
b: Ta có: \(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=180^0\\5\cdot\widehat{B}-4\widehat{C}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\cdot\widehat{B}+5\cdot\widehat{C}=900^0\\5\cdot\widehat{B}-4\cdot\widehat{C}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{C}=100^0\\\widehat{B}=80^0\end{matrix}\right.\)