Ta có $ \widehat{C}=180^{\circ}-\widehat{A}-\widehat{B}=180^{\circ}-70^{\circ}-50^{\circ}=60^{\circ}$
Ta có CM là tia phân giác của góc C, do đó $\widehat{C1}=\widehat{C2}=30^{\circ}$
$\widehat{AMC}=180^{\circ}-\widehat{A}-\widehat{C2}=180^{\circ}-50^{\circ}-30^{\circ}=100^{\circ}$
$\widehat{BMC}=180^{\circ}-\widehat{B}-\widehat{C1}=180^{\circ}-70^{\circ}-30^{\circ}=80^{\circ}$