<s>[INST] Triangle angles. 
For $\vartriangle$ABC it is given that $$\frac1{b+c}+\frac1{a+c}=\frac3{a+b+c}$$ Find the measure of angle $C$.

This is a "challenge problem" in my precalculus book that I was assigned. How do I find an angle from side lengths like this? I have tried everything I can. I think I may need to employ the law of cosines or sines. Thanks.
 [/INST] Multiply both sides by $a+b+c$ to get $$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=3$$
We get 
$$\frac{a}{b+c}+\frac{b}{a+c}=1$$
$$a(a+c)+b(b+c)=(a+c)(b+c)$$
$$a^2+ac+b^2+bc=ab+ac+bc+c^2$$
$$a^2+b^2=ab+c^2$$
$$a^2-ab+b^2=c^2$$
By the cosine law, we have $a^2-2\cos(C)ab+b^2=c^2$. Hence $\cos(C)=\frac12$, so $C=60^\circ$.
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