Not enough information? Do some construction!

Image from GeoGebra .

Motivation for construction: $AM=BC$ means that we would want to "move" a shape with base $AM$ (probably a triangle) to $BC$ .

Hence we have point $D$ such that $\triangle BCD \equiv \triangle AMC$ .

By construction, $BD=AC=AB$ , thus $\triangle ABD$ is an isosceles triangle.

Also, $\angle MAC = 180^\circ - 2*80^\circ = 20^\circ \Rightarrow \angle ABD = 80^\circ - 20^\circ\ = 60^\circ$

Hence $\triangle ABD$ is an equilateral triangle.

Furthermore, note that $AD=BD=AC$ , so $\triangle ACD$ is isosceles.

$\angle CAD = 60^\circ - 20^\circ = 40^\circ$ $\Rightarrow \angle ADC = \frac {180^\circ-40^\circ}{2} = 70^\circ$ $\Rightarrow \angle ACM = \angle BDC = 70^\circ-60^\circ=10^\circ$

Finally, $\angle MCB = 80^\circ-10^\circ = \boxed{70^\circ}$

$\text{Without loss of generality let BC=1.}\\ \text{Applying Sin Law to triangle ABC we get } \dfrac{BC}{Sin20}= \dfrac{AB}{Sin80}.\\ AB= 2.87938. ~~ \therefore ~MB=1.87938.\\ In ~\Delta ~MBC,\\ Applying~ Cos ~Law,~~MC^2=1.87938^2+1^2 - 2*1.87938*1*Cos80, \therefore~ MC=1.9696.\\ Applying ~ Sin ~ Law, ~~~~~~~~\dfrac{SinMCB}{1.87938}=\dfrac{Sin80}{1.9696}.\\ \therefore ~\angle MCB=\Large \color{#D61F06}{70^o}.$

I hope it is clear enough, it is only 2 equations, but a calculator is needed to get the final number.