$\angle CMB = 180^\circ - (\angle MBC + \angle BCM) = 25^\circ$. $BC$ externally bisects $\angle MBN$. Reflection of $AB \equiv MB$ in $CM$ cuts $BN$ at $N'$ $\Longrightarrow$ $C$ is M-excenter of $\triangle MBN'$ $\Longrightarrow$ $\angle BCN' = 90^\circ - \frac{_1}{^2}\angle N'MB = 90^\circ - \angle CMB = 65^\circ = \angle BCA$ $\Longrightarrow$ points $N' \equiv BN \cap AC \equiv N$ are identical $\Longrightarrow$ $\angle NMC= \angle CMB= 25^\circ$.

let the internal bisector of $\angle NBM$ intersect $CM$ and $AC$ at $S,X$ respectively. than $\angle XBM=\angle XCB=10$ so $XMBC$ is cyclic and now $\angle XMS=\angle XMC=\angle XBC=80+10=90$ and since $\angle SBN=\angle XBN=10=65-55=\angle SCN$ we have that $SBCN$ is cyclic and now $\angle XNS=180-\angle SNC=180-(180-\angle SBC)=180-(180-90)=90$ so $\angle XMS=\angle XNS=90$ and now $XMSN$ is cyclic so $\angle CMN=\angle SMN=\angle SXN=\angle BXC=180-(90+65)=25$

Let us denote $ BM=x, BN=y$. From the sinus theorem applied in the triangles $BNC$ and $BMC$ we get that $ x=\frac{a \sin{55}}{\sin{25}}, y=\frac{a \sin {65}}{\sin{35}} $. Now, by using the same theorem in the triangle $MNB$, we have$ \frac{\sin{55}}{\sin{25}\sin{\angle MNC}} = \frac {\sin{65}}{\sin{35}\sin(\angle MNC +20) }$. After some simple computations, we obtain $\angle NMB =110$, so $\angle NMB =50$.