Proving that larger side corresponds to larger angle

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Deviousfred
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I was asked by one of my professors to prove for the class why the largest side of the tringle corresponds to the largest angle of that triangle.

I was thinking of using law of sines to do so. Please let me know if I am wrong so that I do not look like a fool in front of class.

My triangle is ABC with vertices/angles a,b, and c respectively.

I am saying that angle a is the largest in that triangle, so;

Quick notes: L is my representation for angle.
mLa > mLb > mLc.

Law of Sines:

sin(a)/BC = sin(b)/AC = sin(c)/AB

therefore;

AC*sin(a)=BC*sin(b),
AC*sin(c)=AB*sin(b),
AB*sin(a)=BC*sin(c).

so;

AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).

once again;

mLa > mLb > mLc.

therefore;

sin(a) > sin(b) > sin(c)

so if;

AC/BC = sin(b)/sin(a),
AC/AB = sin(b)/sin(c),
AB/BC = sin(c)/sin(a).

then;

AC < BC
AC > AB
AB < BC

so;

BC > AC > AB

I don't know if I need to write anything more down from here. Take it easy on my, I threw this together last night while ignoring my mother-in-law's boring story.
 

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Hi Deviousfred! :smile:

I'm not sure, but I think you've assumed that if A > B, then sinA > sinB, which isn't necessarily true if one of them is > 90º …

how will you deal with that case?

However, this sine theorem assumes rather a lot of prior geometry …

how about a proof involving drawing a circle through the three vertices of the triangle? :wink:
 
tiny-tim said:
Hi Deviousfred! :smile:

I'm not sure, but I think you've assumed that if A > B, then sinA > sinB, which isn't necessarily true if one of them is > 90º …

how will you deal with that case?

However, this sine theorem assumes rather a lot of prior geometry …

how about a proof involving drawing a circle through the three vertices of the triangle? :wink:

I was thinking the exact same thing yesterday but I came to this conclusion:

In a triangle all three sides add up to 180 degrees. If we were to have a triangle and let's say the largest angle is 120 degrees then the second largest angle could only be 60-1=59 degrees.

sin (120) > sin (59)

I did several other conditions with 170, 160, 150, 140, 130, 120, 110, and 100 degrees and all resulted in the same manner.

I came to the conclusion that in the case of a triangle, the sin of the largest angle will always be greater than the sin of both other angles (individually of course.) I'm sure there is a better way to say/show this but I do not know it yet.

I am very interested in your idea though. Before I ask you to show me let me take a stab at it.
 
Hi Deviousfred! :smile:
Deviousfred said:
In a triangle all three sides add up to 180 degrees. If we were to have a triangle and let's say the largest angle is 120 degrees then the second largest angle could only be 60-1=59 degrees.

sin (120) > sin (59)

I did several other conditions with 170, 160, 150, 140, 130, 120, 110, and 100 degrees and all resulted in the same manner.

I came to the conclusion that in the case of a triangle, the sin of the largest angle will always be greater than the sin of both other angles (individually of course.) …

Yes … but why is sin (120) > sin (59)? … why is sin (170) > sin (9)? … etc …

you'll need a snappy reason to put before the class! :wink:​
 
tiny-tim said:
Hi Deviousfred! :smile:


Yes … but why is sin (120) > sin (59)? … why is sin (170) > sin (9)? … etc …

you'll need a snappy reason to put before the class! :wink:​

I know, I'm trying right now to find a way to prove that without values.
 
So I guess my new proof is:

Triangle ABC

If;
mLA > mLB > mLC

Prove;
sin A > sin B > sin C
 
ok … why, for example, is sin (120) > sin (59)?

because we chose 59 to be less than 180 - 120.

But sin120 = sin(180 - 120),

so all you have to prove is that if 59 < 180 - 120 then sin(59) < sin(180 - 120) :wink:
 
ok, I was ablt to prove it by other means using isoceles triangle and exterior angle inequality, but I'm hellbent on proving it this way.