Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Are you looking for a proof of the Law of Sines or it's converse? Here's a proof of L.o.S. – 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
up vote 2 down vote accepted

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.

share|cite|improve this answer
Cool thank you. I didn't think of that. So the existence of the triangle follows from Law of Cosines? – Frudrururu Jun 5 '13 at 21:16
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 – Frudrururu Jun 6 '13 at 23:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.