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 $BC$.

Let $X$ be the point on $BC$ exactly below $A$ and $MX=z$ for comfort.
$BX=\frac{BC}{2}+z$ and $XC=\frac{BC}{2}-z$
Now:
$$\tan(\alpha )= \frac{AX}{BX}\implies \tan(\alpha )= \frac{AX}{\frac{BC}{2}+z}$$
$$\tan(AMC )= \frac{AX}{Z}$$
$$\tan(\beta )= \frac{AX}{XC} \implies \tan(\beta )= \frac{AX}{\frac{BC}{2}-z}$$
Now you have 3 questions with 3 variables $z,AX,\tan(AMC)$ and the rest is given. Let $AMC= \gamma$
$$AX=z \tan(\gamma)$$
$$(1)\tan(\alpha )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\alpha )}{2}=z(\tan(\gamma)-\tan(\alpha ))$$
$$(2)\tan(\beta )= \frac{z \tan(\gamma)}{\frac{BC}{2}+z}\implies \frac{BC\tan(\beta )}{2}=z(\tan(\gamma)+\tan(\beta ))$$
And by dividing (1)/(2) we get:
$$\frac{\tan(\alpha )}{\tan(\beta )}=\frac{\tan(\gamma)-\tan(\alpha )}{\tan(\gamma)+\tan(\beta )}$$
Hence:
$$\tan(\gamma)=\frac{2\tan(\alpha)\tan(\beta)}{\tan(\beta)-\tan(\alpha)}$$
$$\gamma=\arctan(\frac{2\tan(30^\circ)\tan(15^\circ)}{\tan(30^\circ)-\tan(15^\circ)})$$
