Xét tam giác ABC có:
\(\begin{array}{l}
\widehat A + \widehat B + \widehat C = {180^o}\\
\Rightarrow \widehat A + \widehat B = {180^0} - \widehat C = {180^0} - {56^0} = {124^0}\\
Ma\,\,4\widehat A = 3\widehat B\\
\Rightarrow \frac{{\widehat A}}{3} = \frac{{\widehat B}}{4} = \frac{{\widehat A + \widehat B}}{{3 + 4}} = \frac{{{{124}^0}}}{7}\\
\Rightarrow \widehat A = \frac{{{{372}^0}}}{7} \approx {53^o}\\
\widehat B = \frac{{{{496}^0}}}{7} \approx {71^0}
\end{array}\)
