# If triangle's angles are $\alpha \leq \beta \leq \gamma$, then respective opposite sides are $a \leq b \leq c$. How to handle obtuse case?

Knowing that $a$, $b$ and $c$ are the lengths of the sides opposite to (respective) angles $\alpha \le \beta \le \gamma$, prove that $a\le b \le c$.

I use the Law of Sines: $$a = 2R\sin(\alpha)$$ $$b = 2R\sin(\beta)$$ $$c= 2R\sin(\gamma)$$
Now, I subsitute these values into the initial inequality and divide by $2R$: $$\sin(\alpha) \le \sin(\beta) \le \sin(\gamma)$$

As for an acute triangle, it works great, but how do I prove that this inequality holds for — for example — an obtuse triangle?

• What does $R$ mean here? – Namaste Apr 16 '17 at 21:28
• The radius of the circumscribed circle. – ILoveChess Apr 16 '17 at 21:30
• @amWhy The mention of the law of sines seems to make clear that $\;R=$ the radius of the circumscribing circle. – DonAntonio Apr 16 '17 at 21:30
• @DonAntonio, is there any difference between a circumscribing and a circumscribed circle? I have never come across the gerund form. – ILoveChess Apr 16 '17 at 21:32
• What about "inscribed" and "circumscribed"? en.wikipedia.org/wiki/Circumscribed_circle – ILoveChess Apr 16 '17 at 21:36

We have the angles $\alpha<\beta<\gamma$. If all are acute, your analysis applies, because the sine is increasing in $[0,\pi/2]$.
If not all are acute, then only $\gamma$ is obtuse, and since $\gamma+\beta<\pi$, we also have $\beta<\pi-\gamma$, both acute. Then, by your argument, we have $\sin\alpha<\sin\beta<\sin(\pi-\gamma)$. Since $\sin\gamma=\sin(\pi-\gamma)$, the result $a<b<c$ follows in this case as well.
• If $\gamma$ is obtuse, then $\beta$ and $\alpha$ are acute. Is it always obvious that the sine of this obtuse angle will be greater than the sine of the acute angles? – ILoveChess Apr 16 '17 at 21:58