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

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...?

• 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}
– dfnu
Feb 2 '20 at 18:15
• but how is that any better? :( Feb 2 '20 at 18:17
• Use half-angle formula on LHS and then the equation becomes linear in $\sin 2x$ and $\cos 2x$...
– dfnu
Feb 2 '20 at 18:19

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, $$x= 30^\circ$$.