The Equilateral Triangle Wizardry Part-2

We can solve it easily by rotations. But, there is also a formula that gives us the side $L$ of the equilateral triangle knowing $a=MA$ , $b=MB$ and $c=MC$ . That formula is:

$L=\sqrt{\dfrac{a^2+b^2+c^2+\sqrt{3((2ab)^2-(a^2+b^2-c^2)^2)}}{2}}$

Let $a=\sqrt{3}k$ , $b=2k$ and $c=\sqrt{5}k$ .

Substituting we get:

$L=\sqrt{6+\sqrt{33}}k$

Using Law of Cosines on $\triangle AMB$ , we get:

$\cos \angle AMB=\dfrac{a^2+b^2-L^2}{2ab}$

$\cos \angle AMB=\dfrac{7k^2-(6+\sqrt{33})k^2}{4\sqrt{3}k^2}$

$cos \angle AMB=\dfrac{\sqrt{3}-3\sqrt{11}}{12}$

$\angle AMB \approx \boxed{133.221°}$

If we make an equilateral $\triangle KMA$ then $\triangle MBK$ has side lengths of $MA, MB, MC$ . Then our $\angle AMB = \angle BMK + 60 = 133.221$