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 $BC$.

Another geometric solution:

Let $E$ be the foot of the altitude through $A$. Let $D$ be the midpoint of $\overline{AC}$, so $\triangle EDC$ is isosceles with $\angle DEC=\angle ECD=30$ and $ED = DC$.
Both $M$ and $D$ are midpoints, so $\triangle ABC\sim\triangle DMC$. Conclude $\angle DMC=\angle ABC=15$. Knowing $\angle DEC=30$, compute $\angle EDM=15$ so $\triangle MED$ is isosceles. Hence $ME=ED$.
Since $D$ is the midpoint of $AC$, we know $DC=\frac12 AC$. But $\triangle AEC$ is a 30-60-90 triangle so also $AE=\frac12 AC$.
Putting it all together, we've shown
$ME=ED=DC=\frac12 AC=AE$. So $\triangle MEA$ is isosceles. Since $\angle MEA$ is right, it follows that $\angle AME=45$.
