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?

  • $\begingroup$ What does $R$ mean here? $\endgroup$
    – amWhy
    Apr 16 '17 at 21:28
  • $\begingroup$ The radius of the circumscribed circle. $\endgroup$
    – ILoveChess
    Apr 16 '17 at 21:30
  • $\begingroup$ @amWhy The mention of the law of sines seems to make clear that $\;R=$ the radius of the circumscribing circle. $\endgroup$
    – DonAntonio
    Apr 16 '17 at 21:30
  • 2
    $\begingroup$ @DonAntonio, is there any difference between a circumscribing and a circumscribed circle? I have never come across the gerund form. $\endgroup$
    – ILoveChess
    Apr 16 '17 at 21:32
  • 1
    $\begingroup$ What about "inscribed" and "circumscribed"? en.wikipedia.org/wiki/Circumscribed_circle $\endgroup$
    – ILoveChess
    Apr 16 '17 at 21:36

My original answer was inadequate. Try this:

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.

  • $\begingroup$ 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? $\endgroup$
    – ILoveChess
    Apr 16 '17 at 21:58
  • $\begingroup$ Sorry, this was a too-partial or inadequate answer. I'm away from my keyboard now, will correct and expand soon $\endgroup$
    – Lubin
    Apr 16 '17 at 22:23

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