12. In $\triangle A B C$, $\angle B A C=30^{\circ}, \angle A B C=70^{\circ}, M$ is a point inside the triangle, $\angle M A B=\angle M C A=20^{\circ}$, find the degree measure of $\angle M B A$.
12. Take a point $E$ on the extension of $BC$ such that $\angle EAC = 10^{\circ}$. Draw a perpendicular from $B$ to $AC$, intersecting $AE$ and $AM$ at $D$ and $F$ respectively. Connect $DM$, $DC$, and $CF$. Given $\angle MAC = 10^{\circ}$ and $\angle ACB = 80^{\circ}$, we know $AF \perp BC$, so $F$ is the orthocenter of $\triangle ABC$. Also, $\angle CAB = 30^{\circ}$, thus $\angle FCA = 60^{\circ}$. Since $\angle MCA = 20^{\circ}$, we have $\angle MCF = 40^{\circ}$. It is easy to see that $\angle ADB = 80^{\circ} = \angle ACB$, hence $A, B, C, D$ are concyclic. Therefore, $\angle DCA = \angle DBA = 60^{\circ}$, and $\angle MCD = 80^{\circ}$.

In $\triangle AFD$, it is easy to see that $\angle MFD = 80^{\circ} = \angle MCD$, so $D, M, F, C$ are concyclic. Thus, $\angle MDF = \angle MCF = 40^{\circ}$, which means $DM$ bisects $\angle ADB$. Since $AM$ bisects $\angle DAB$, $M$ is the incenter of $\triangle ABD$, and $\angle MBA = \frac{1}{2} \angle ABD = 30^{\circ}$.