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?


\begin{align} 
\triangle ADC:\quad
|AD|=4\cos50^\circ
,\quad
|CD|&=4\sin50^\circ
,\\
\triangle BCD:\quad
|BC|&=\sqrt{(4\sin 50^\circ)^2+(7-4\cos 50^\circ)^2}
\\
&=\sqrt{16\sin^2 50^\circ
+49-56\cos 50^\circ+16\cos^2 50^\circ}
\\
&=\sqrt{65-56\cos50^\circ}
,\\
\triangle BCD,\triangle BME:\quad
|ME|&=\tfrac12|CD|
=2\sin 50^\circ
,\\
|DE|&=\tfrac12|BD|
=\tfrac72-2\cos 50^\circ
,\\
|AE|&=|AD|+|DE|
=\tfrac72+2\cos 50^\circ
,\\
\triangle AME:\quad
|AM|&=\sqrt{|AE|^2+|ME|^2}
\\
&=\sqrt{(\tfrac72+2\cos 50^\circ)^2
+(2\sin 50^\circ)^2
}
\\
&=\sqrt{\tfrac{49}4+14\cos 50^\circ
+4\cos^2 50^\circ+4\sin^2 50^\circ}
\\
&=\tfrac12\sqrt{65+56\cos 50^\circ}
.
\end{align} 
