Triangle $ABC$, with $M$ the midpoint of $BC$, $\angle BAM=\angle C$, and $\angle MAC=15^\circ$. Find $\angle C$

Question

In triangle $ABC$, $M$ is the midpoint of $BC$, $\angle BAM=\angle C$, $\angle MAC=15^{\circ}$, what is $\angle C$?

I've been stuck on this question for awhile now. What I've tried so far:

I let $BM=MC=a$ and $AM=b$, then applied Law of Sines on $\triangle BAM$ and $\triangle AMC$ to get:

$$\frac{a}{\sin x}=\frac{b}{\sin(165^{\circ}-2x)}$$

$$\frac{a}{\sin 15^{\circ}}=\frac{b}{\sin x}$$

and manipulated this equations to end up with:

$$\sin^2 x=\sin(165^{\circ}-2x)\cdot \sin 15^{\circ}$$

but I don't know what to do with this equation. Maybe I'm going in the wrong direction with trig...?

Answer 1

Note

$$\sin^2x=\sin(15+2x) \sin15$$

$$1-\cos2x = \cos2x - \cos(2x+30)$$

$$\cos(2x+30)=2\cos2x-1$$

$$\sqrt3\cos2x-\sin2x=4\cos2x-2$$

Let $t= \tan x$. Then, $\cos2x=\frac{1-t^2}{1+t^2}$, $\sin2x=\frac{2t}{1+t^2}$, and the quadratic equation in $t$

$$(-6+\sqrt3)t^2+2t+2-\sqrt3=0$$

which yields the valid solution $t=\frac1{\sqrt3}$. Thus, $x= 30^\circ$.