find the measure of $AMC$ if $M$ is the midpoint of $BC$ then find the measure of $AMC$.

I tried to use the angles to find $AMC$ but I don't know how to use that $M$ is the midpoint of $BC$.


Continue $CA$, draw the ray $\vec{CA}$. Then draw $BN$, in which it is perpendicular to $\vec{CA}$. 
$\triangle BCN$ is a 30-60-90 triangle. as we know about this triangle. $BN=\frac12 BC$. So $BN=BM$. This tells us $\triangle ABN$, is isosceles, so $\angle BMN=\angle BNM=\frac{180^\circ-\angle MBN}2=\frac{120^\circ}{2}=60^\circ$. 


*

*Now we know $\triangle BMN$ is equilateral.


Therefor $$MN=BN=BM\qquad(1)\\
\text{and }\angle MBN=60^\circ\qquad{ }$$
By this, we get $\angle CMN=180^\circ-\angle BMN=180^\circ-60^\circ$, 
$$\text{therefor }\angle CMN=120^\circ\  (2)$$
$\angle ABN=\angle CBN-\angle CBA=60^\circ-15^\circ=45^\circ$. So, the other angle of the right triangle $\triangle ABN$, i.e. $\angle BAN$, should be $180^\circ-90^\circ-45^\circ=45^\circ{}^\dagger$, therefor $\triangle ABN$ is isosceles.
So we would have $AN=BN$. By (1) we get $MN=AN$, so $\triangle AMN$ is isosceles. Therefor $$\angle AMN=\angle MAN=\frac{180^\circ-\angle ANM}2=\frac{180^\circ-30^\circ}2=75^\circ$$
Now, by using (2) we can find the size of $\angle AMC$, which is 
$$\angle AMC=\angle CMN-\angle AMN=120^\circ-75^\circ=45^\circ$$
${}^\dagger$let's note that one may also know that $\angle BAN$ is equal to $45^\circ$, since it is an external angle of the triangle  $\triangle ABC$
