Cho hình vẽ sau: Biết \( \widehat {{M_3}} = \widehat {{N_2}} = {140^0}\) Hãy tính \(\widehat {{M_4}} + \widehat {{N_2}};\widehat {{M_3}} + \widehat {{N_1}}\)

Ta có: \( \widehat {{M_3}} + \widehat {{M_4}} = {180^0}\) (kề bù)

\(\begin{array}{*{20}{l}} { \Rightarrow \widehat {{M_4}} = {{180}^0} - \widehat {{M_3}} = {{180}^0} - {{140}^0} = {{40}^0}}\\ { \Rightarrow \widehat {{M_4}} + {\mkern 1mu} \widehat {{N_2}} = {{40}^0} + {{140}^0} = {{180}^0}} \end{array}\)

Ta có: \( \widehat {{N_2}} + \widehat {{N_1}} = {180^0}\) (kề bù)

\(\begin{array}{*{20}{l}} { \Rightarrow \widehat {{N_1}} = {{180}^0} - \widehat {{N_2}} = {{180}^0} - {{140}^0} = {{40}^0}}\\ { \Rightarrow \widehat {{M_3}} + \widehat {{N_1}} = {{140}^0} + {{40}^0} = {{180}^0}} \end{array}\)