Given equilateral $\triangle ABC$ and $M$ at distances $3$, $5$, $4$ from $A$, $B$, $C$, find $\angle AMC$. 
Given that $\triangle ABC$ is equilateral triangle, and $M$ is a point inside $\triangle ABC$ such that 
  $$AM=3\;\text{units} \qquad BM=5\;\text{units} \qquad CM=4\;\text{units}$$ 
  what is the measure of $\angle AMC$ ?


I tried to use cosine rule since the sides are congruent, but it didn't help !
Any help is appreciated.
 A: Reflect $M$ in $BC$ to obtain $P$. Reflect $M$ in $CA$ to obtain $Q$. Reflect $M$ in $AB$ to obtain $R$.

Then $\triangle AQR$, $\triangle BRP$ and $\triangle CPQ$ are three $120^\circ$-$30^\circ$-$30^\circ$ triangles. $QR:PQ:RP=3:4:5$. Hence $\angle PQR=90^\circ$. $\angle AQR=\angle CQP=30^\circ$. So, $\angle AQC=150^\circ$.
$\angle AMC=\angle AQC=150^\circ$.
A: The hint.
Rotate $\Delta MBC$ on $-60^{\circ}$ around $C$.
I got $\measuredangle AMC=90^{\circ}+60^{\circ}=150^{\circ}.$
A: Hint:
This is a really long solution, but I think you'll get to the answer at the end.
Take the projections $M_1$, $M_2$, and $M_3$ of $M$ on $AB$, $BC$, and $AC$ respectively.
Now you have that $AM^2= AM_1^2+ M_1M^2$ and $BM^2= M_1M^2+BM_1^2$ and $AM_1+BM_1=a$ 
The same applies for all the 2 other triangles $AMC$ and $BMC$ (note that, $AM_1+BM_1=AM_3+CM_3=CM_2+BM_2= a$)
Now you get $a$ (the length of the side of the equileteral triangle) and then apply cosine law to get the angle.
(From the figure you can notice that $5<a<7$)
