Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$

Let $O$ be the circumcenter of $\triangle ABC$. Consider $\triangle OAC$,
$\angle AOC = 2\angle ABC = 160°$
$\angle OAC = \angle OCA = \frac{180° - 160­°}{2} = 10°$
Hence, $O$ lies on $AM$.

$\angle MOC = 10° + 10° = \angle MCO \Rightarrow MO = MC$.
$\angle OCB = \angle OBC = \frac{180° - 60°}{2} = \angle COB \Rightarrow BC = OB$.
Hence, $\triangle BMO \cong \triangle BMC$.
$\therefore 180° - \angle BMC = \frac{\angle OBC}{2} + \angle BCO - 20° = 70°$

Ah, I think I am late. May this gives you more geometric insight.

You can take $\angle BAF$ right-angle and $AF$ meet $BE$ at $F$ then $ABCF$ share same circle since $\angle CAF=\angle CBF$, that makes $\angle BCF$ also a right-angle, and $\angle I=60$, which indicates similar proportional

$$\frac{AC}{BF}=\frac{AI}{FI}=\frac1{2}$$

thus we spot $H$ the middle point of $BF$ for $\angle BAF$ and $\angle BCF$ both right-angle, $H$ is the circumcenter of four point system $ABCF$, moreover $\triangle AHC$ is equilateral, in which $CD$ is one of its perpendicular line. And $AE$ is actually the perpendicular line of $\triangle ABC$.

And now you can solve any angle in the graph with almost no calculation.