1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What does $R$ mean here? $\endgroup$ – Namaste 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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