# Prove the converse of the Law of Sines

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 – Frudrururu 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 ;) – Frudrururu Jun 5 '13 at 21:10

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.
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