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If $\alpha,\beta,\gamma,a,b,c \in \mathbb{R}^+$, $\alpha+\beta+\gamma=180^\circ$, and $$ \frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b} = \frac{\sin(\gamma)}{c} \qquad \text{ (1)}$$ then there exists a triangle in 2-space with angle-side pairs $(\alpha,a),(\beta,b),(\gamma,c)$.

I know the Law of Sines says if we have a triangle, then $(1)$ is satisfied. I know the converse statement must be true, it seems like it would be. I am having trouble of how would I show it is true if it is?

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Are you looking for a proof of the Law of Sines or it's converse? Here's a proof of L.o.S. en.wikipedia.org/wiki/Law_of_sines#Proof –  Sammy Black Jun 5 '13 at 20:58
    
Yeah a proof of the converse –  Starlight Jun 5 '13 at 20:59
    
I clarified your question in the title. –  Sammy Black Jun 5 '13 at 21:07
    
Thank you, I came up with this while thinking about some problem and it looks like a converse but I wasn't confident enough to call it that for sure ;) –  Starlight Jun 5 '13 at 21:10

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Recall that for any $\alpha,\beta,\gamma>0$ (measured in degrees) with $\alpha+\beta+\gamma=180$ there is a triangle with angles $\alpha,\beta,$ and $\gamma$. By the law of sines the side lengths satisfy $\sin\alpha/s_1 = \sin\beta/s_2 = \sin\gamma/s_3$. Now multiply the side lengths by a factor $t$ so that $s_1t = a$. We claim that $s_2t = b$ and $s_3t = c$. Indeed, $$ s_2t = \frac{s_2}{s_1} s_1t = \frac{\sin \beta}{\sin \alpha} s_1 t = \frac{\sin \beta}{\sin \alpha} a $$ $$ = \frac{a}{\sin \alpha} \sin \beta = \frac{b}{\sin \beta} \sin \beta = b. $$ The result for $s_3t=c$ is similar. Thus the claim is proved.

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Cool thank you. I didn't think of that. So the existence of the triangle follows from Law of Cosines? –  Starlight Jun 5 '13 at 21:16
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No law of cosines. There are, in fact, infinitely many triangles. Draw a line segment and draw rays at angles $\alpha$ and $\beta$ from its endpoints. Since these rays aren't parallel (why?), they intersect. Presto! A triangle with angles $\alpha$, $\beta$, and $\gamma$. But notice that you obtain one such triangle (up to congruence) for every initial line segment, which can have any length you want. –  Ted Shifrin Jun 5 '13 at 21:56
    
Okay that sounds good. Is there any other kind of argument, analytically coming up with specific numbers or some kind of like "base" triangle which all the others are dilations of? I think this is convincing enough for myself though thank you –  Starlight Jun 6 '13 at 23:25

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