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Trigonometry Finding the greatest angle

Pranav

Well-known member
Nov 4, 2013
428
Problem:

In $\Delta$ABC, $\displaystyle \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \cos \left( \frac{C}{2} \right)\leq \frac{1}{4}$, then greatest angle of triangle

A)lies in $\left(0,\frac{\pi}{2}\right)$

B)lies in $\left(\frac{2\pi}{3},\frac{5\pi}{6}\right)$

C)lies in $\left(\frac{5\pi}{6},\pi\right)$

D)lies in $\left(\frac{\pi}{2},\frac{2\pi}{3}\right)$

Attempt:

I haven't been able to proceed anywhere with this problem. I could only simplify the given inequality to
$$\sin A+\sin B+\sin C\leq 1$$

(The above can be proved by using $C=\pi-(A+B)$)

But I am not sure if the above helps.

Any help is appreciated. Thanks!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,902
Problem:

In $\Delta$ABC, $\displaystyle \cos \left( \frac{A}{2} \right) \cos \left( \frac{B}{2} \right) \cos \left( \frac{C}{2} \right)\leq \frac{1}{4}$, then greatest angle of triangle

A)lies in $\left(0,\frac{\pi}{2}\right)$

B)lies in $\left(\frac{2\pi}{3},\frac{5\pi}{6}\right)$

C)lies in $\left(\frac{5\pi}{6},\pi\right)$

D)lies in $\left(\frac{\pi}{2},\frac{2\pi}{3}\right)$

Attempt:

I haven't been able to proceed anywhere with this problem. I could only simplify the given inequality to
$$\sin A+\sin B+\sin C\leq 1$$

(The above can be proved by using $C=\pi-(A+B)$)

But I am not sure if the above helps.

Any help is appreciated. Thanks!
Hi!

Suppose A is the greatest angle.
That means A is at least 60 degrees and at most 180.
$$\frac \pi 3 \le A < \pi \qquad \qquad (1)$$

You've got:
$$\sin A+\sin B+\sin C\leq 1$$
That means that:
$$-1 \le \sin A \le \frac 1 3 \qquad \qquad (2)$$

Combine (1) and (2) to find:
$$0 < \sin A \le \frac 1 3 \wedge A \ge \frac \pi 3$$

With $\sin(\frac {5\pi}{6}) = \frac 1 2$, we can conclude that...
 

Pranav

Well-known member
Nov 4, 2013
428
Hi ILS! :)

With $\sin(\frac {5\pi}{6}) = \frac 1 2$, we can conclude that...
I understand what you have done before this but isn't 1/2 out of the range of $\sin A$?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
... isn't 1/2 out of the range of $\sin A$?
If $\Delta$ is the area of the triangle then $\Delta = \frac12bc\sin A$ and so $\sin A = \frac{2\Delta}{bc}$, and similarly $\sin B = \frac{2\Delta}{ca}$ and $\sin C = \frac{2\Delta}{ab}$. The inequality $\sin A + \sin B + \sin C \leqslant 1$ then becomes $2\Delta(a+b+c) \leqslant abc.$ But (see here for example) $abc = 4R\Delta$, where $R$ is the radius of the circumcircle. Therefore $2\Delta(a+b+c) \leqslant 4R\Delta$, or $s\leqslant R$, where $s = \frac12(a+b+c)$ is the semi-perimeter of the triangle. In particular, we must have $a\leqslant R$, where $a = |BC|$ is the longest side. It follows that the angle $BOC$ in the diagram below ($O$ being the centre of the circumcircle) must be at most $\pi/3$, and hence the angle $A$ must be at least $5\pi/6$. Therefore $\sin A \leqslant \frac12.$

circumcircle.png
 

Pranav

Well-known member
Nov 4, 2013
428
Hi Opalg!

...... $a\leqslant R$, where $a = |BC|$ is the longest side.
Sorry if this is obvious but I can't quite follow this. How do you get $a\leq R$? Why not $a\leq 2R$? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,902
Hi ILS! :)

I understand what you have done before this but isn't 1/2 out of the range of $\sin A$?
Yep. It's out of range. I'll get to that in a sec.

What is the solution of:
$$0 < \sin A \le \frac 1 3$$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi Opalg!



Sorry if this is obvious but I can't quite follow this. How do you get $a\leq R$? Why not $a\leq 2R$? :confused:
$a$ must be less than $s$ because $a\leqslant b+c$ (any side of a triangle must be less than half the perimeter, because the sum of the other two sides is larger then it). So $a = \frac12(a+a) \leqslant \frac12(a+b+c) = s \leqslant R.$
 

Pranav

Well-known member
Nov 4, 2013
428
$a$ must be less than $s$ because $a\leqslant b+c$ (any side of a triangle must be less than half the perimeter, because the sum of the other two sides is larger then it). So $a = \frac12(a+a) \leqslant \frac12(a+b+c) = s \leqslant R.$
Thanks Opalg but I am still stuck at your previous post.

Why do you say the least angle is $5\pi/6$ when it is clearly greater than $\pi/3$. :confused:

Yep. It's out of range. I'll get to that in a sec.

What is the solution of:
$$0 < \sin A \le \frac 1 3$$
$$A \in \left(0,\arcsin\left(\frac{1}{3}\right)\right] \cup \left[\pi-\arcsin\left(\frac{1}{3}\right),\pi \right)$$

Correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,902
$$A \in \left(0,\arcsin\left(\frac{1}{3}\right)\right] \cup \left[\pi-\arcsin\left(\frac{1}{3}\right),\pi \right)$$

Correct?
Yes. Correct. :)

Since, \(\displaystyle \sin(\pi/6) = 1/2 > 1/3\), we can conclude that the angle corresponding to $\arcsin(1/3) < \pi/6$.

Therefore a weaker, but still true, condition is:
$$A \in \left(0,\pi/6 \right) \cup \left(\pi-\pi/6,\pi \right)$$

Combine with the fact that $A > \pi /3$, and we find that \(\displaystyle A \in \left(\frac {5\pi}{6},\pi \right)\).
 

Pranav

Well-known member
Nov 4, 2013
428
Yes. Correct. :)

Since, \(\displaystyle \sin(\pi/6) = 1/2 > 1/3\), we can conclude that the angle corresponding to $\arcsin(1/3) < \pi/6$.

Therefore a weaker, but still true, condition is:
$$A \in \left(0,\pi/6 \right) \cup \left(\pi-\pi/6,\pi \right)$$

Combine with the fact that $A > \pi /3$, and we find that \(\displaystyle A \in \left(\frac {5\pi}{6},\pi \right)\).
Thanks. :)

But I am again confused. I was looking at your previous and found the following confusing.

You've got:
$$\sin A+\sin B+\sin C\leq 1$$
That means that:
$$-1 \le \sin A \le \frac 1 3 \qquad \qquad (2)$$

$1/3=0.33$. I can have $\sin A=0.9$ and $\sin B=\sin C=0.05$ which satisfies the inequality but not the range you have mentioned for $\sin A$. Sorry if I am being dumb. :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,902
$1/3=0.33$. I can have $\sin A=0.9$ and $\sin B=\sin C=0.05$ which satisfies the inequality but not the range you have mentioned for $\sin A$. Sorry if I am being dumb.
Good point. I made a mistake there. :eek:
 

Pranav

Well-known member
Nov 4, 2013
428
Good point. I made a mistake there. :eek:
I am sorry, the values I mentioned won't satisfy A+B+C=$\pi$.

But I still don't see how you get $-1\leq \sin A \leq \frac{1}{3}$, can you please elaborate that?

And I am still stuck at Opalg's post. I understand how $a\leq R$. When a=R, I see that A is $5\pi /6$. I don't see how A=$\pi/6$ is achieved. :(

EDIT: I think I have arrived at the answer but I am still interested to know about Opalg's method. Can you please explain a bit more, it would really help. Many thanks!

From triangle inequality, b+c>a $\Rightarrow \sin B+\sin C>\sin A \Rightarrow \sin A+\sin B+\sin C>2\sin A$. Hence $\sin A<1/2$. This gives the answer. Thanks a lot both of you. :D
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
And I am still stuck at Opalg's post. I understand how $a\leq R$. When a=R, I see that A is $5\pi /6$. I don't see how A=$\pi/6$ is achieved. :(
Given that $a\leqslant R$, it follows that either $A\geqslant 5\pi/6$ or $A\leqslant \pi/6$. But $A$ is supposed to be the greatest angle in the triangle, so it cannot be $< \pi/6$. If the greatest angle in the triangle is $\pi/6$ then the triangle would have to be equilateral, in which case $\sin A + \sin B + \sin C = 3\sqrt3/2 >1$, so we can certainly rule out that case.
 

Pranav

Well-known member
Nov 4, 2013
428
Given that $a\leqslant R$, it follows that either $A\geqslant 5\pi/6$ or $A\leqslant \pi/6$. But $A$ is supposed to be the greatest angle in the triangle, so it cannot be $< \pi/6$. If the greatest angle in the triangle is $\pi/6$ then the triangle would have to be equilateral, in which case $\sin A + \sin B + \sin C = 3\sqrt3/2 >1$, so we can certainly rule out that case.
I realise my mistake, thanks a lot Opalg! :eek: