Find the median in a triangle with trigonometry In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50º$. Let $M$ be the midpoint of $BC$. Determinate $AM$.
My try
I applied law of cosines $3$ times, first to find $BC$, then I let $\angle BCA=\alpha$ and $\angle ABC=130-\alpha$, and applied law of cosines in triangle $ACM$ and $ABM$, but it didn't work.
Any hints?

The Law of Cosines tells us that 
$$CB^2=7^2+4^2-2\cdot7\cdot 4\cdot\cos50=65-56\cos50$$
Finally, in virtue of Apollonius’s theorem, we obtain that 
\begin{align*}
AM^2&=-\frac{BC^2}{4}+\frac{AC^2}{2}+\frac{AB^2}{2}\\
&=\frac{56\cos50-65+2\cdot 4^2+2\cdot 7^2}{4}\\
&=\frac{56\cos50+65}{4}
\end{align*}
