We want to calculate angles BMC and AMC, as you can observe, these angles are supplementary.
Thus using triangle relations, we start by calculating angle BMC.
We can agree that;
[tex]\begin{gathered} \angle CBM+\angle MCB+\angle BMC=180 \\ \angle CBM=37 \\ \angle MCB=\frac{86}{2}=43 \\ \angle BMC=180-43-37=100^o \end{gathered}[/tex]And;
[tex]\begin{gathered} \angle BMC+\angle AMC=180 \\ 100+\angle AMC=180 \\ \angle AMC=80^o \end{gathered}[/tex]Thus;
[tex]\begin{gathered} \angle AMC=80^0 \\ \angle BMC=100^o \end{gathered}[/tex]