geometrytrigonometry
In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50ยบ$. 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?
Using trigonometry, the angle in question is approximately equal, in degrees, to $$ 10.558536057412143196227467316938626443256567512439 $$ which, given the lack of an apparent repeating block, is almost certainly not a rational number.
Hence you should not expect a solution vis synthetic geometry.
Reflect point $A$ about $BC$. Then by midpoint theorem, $BM \parallel A'E$
Now note that $\triangle A'CE \sim \triangle BCD $ and so we have $\angle CA'E = \angle CBD$
Also, $\angle ABM = \angle AA'E = 45^\circ - \angle CA'E = 45^\circ - \angle CBD$
That shows $\angle DBM = 45^\circ$
As a side note, if you also reflect point $E$ about $CD$, you can easily show $AE' = A'E$ and that means $BM = DM$. So, $\triangle DMB$ is indeed an isosceles triangle, right at $M$.
Best Answer
\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}