Ta có: $\widehat{AMB}=\frac{180-\widehat{B}}{2}\implies \widehat{MAC}=\frac{180-\widehat{B}}{2}-\widehat{C}$.Tương tự ta có: $\widehat{NAB}=\frac{180-\widehat{C}}{2}-\widehat{B}$.
$\implies \widehat{NAM}=90^0-(\widehat{AMB}-\widehat{NAB})=90^0-[180^0-\frac{3}{2}*90^0]=45^0$