Đáp án:
$\begin{array}{l}
Co:\widehat {{B_1}} = {80^0}\\
\Rightarrow \widehat {{B_3}} = {80^0}\left( {\text{đối}\,\text{đỉnh}} \right)\\
\Rightarrow \widehat {{B_2}} = \widehat {{B_4}} = {180^0} - \widehat {{B_1}} = {100^0}\\
Do:m//n\\
\Rightarrow \widehat {{A_1}} = \widehat {{A_3}} = \widehat {{B_1}} = \widehat {{B_3}} = {80^0}\left( {\text{đồng}\,\text{vị}} \right)\\
\Rightarrow \widehat {{A_2}} = \widehat {{A_4}} = {100^0}
\end{array}$