Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$ 
In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$.

What I Tried: Here is a picture to keep track of the angle-chasing .

The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.
After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?
Thank You.
 A: 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, $\triangle BMO \cong \triangle BMC$.
$\therefore 180° - \angle BMC = \frac{\angle OBC}{2} + \angle BCO - 20° = 70°$
A: 
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.
A: Let $\angle MBA = \theta$, $\angle MBC = 80 - \theta$
Using trig form of Ceva's theorem,
$$ \sin MAC \sin MCB \sin MBA = \sin MCA \sin MBC \sin MAB$$
$$ \sin 10 \sin 40 \sin \theta = \sin 30 \sin (80-\theta) \sin 20 $$
$$ 4\sin 10 \cos 20 =\dfrac{\sin (80-\theta)}{\sin \theta} $$
$$ 2(\sin 30 - \sin 10) =\sin 80 \cot \theta - \cos 80 $$
$$ 1 - \sin 10 =\sin 80 \cot \theta $$
$$ \cot \theta = \dfrac{1 - \sin 10}{\cos 10} $$
$$ \cot \theta = \dfrac{(\cos 5 - \sin 5)^2}{\cos^2 5 - \sin^2 5} $$
$$ \cot \theta = \dfrac{\cos 5 - \sin 5}{\cos 5 + \sin 5} $$
$$ \cot \theta = \dfrac{1 - \tan 5}{1 + \tan 5} = \tan 40$$
$$\therefore \angle MBA = 50$$
$$\Rightarrow \boxed{180 - \angle BMC = 70}$$
