TITLE: Prove the converse of the Law of Sines
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: 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.