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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...?

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  • $\begingroup$ You could use $$ \sin\alpha\sin\beta = \frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}2$$ to get \begin{eqnarray} 2\sin^2x &=& \cos(150^\circ-2x)-\cos(180^\circ-2x)\\ 2\sin^2x &=&-\frac{\sqrt 3}2\cos 2x+\frac12 \sin2x+\cos2x \end{eqnarray} $\endgroup$
    – dfnu
    Feb 2 '20 at 18:15
  • $\begingroup$ but how is that any better? :( $\endgroup$ Feb 2 '20 at 18:17
  • $\begingroup$ Use half-angle formula on LHS and then the equation becomes linear in $\sin 2x$ and $\cos 2x$... $\endgroup$
    – dfnu
    Feb 2 '20 at 18:19
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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, $x= 30^\circ$.

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