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Proving that larger side corresponds to larger angle

  1. Mar 12, 2009 #1
    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.
     
  2. jcsd
  3. Mar 12, 2009 #2

    tiny-tim

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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:
     
  4. Mar 13, 2009 #3
    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 lets 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.
     
  5. Mar 13, 2009 #4

    tiny-tim

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    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:
     
  6. Mar 13, 2009 #5
    I know, I'm trying right now to find a way to prove that without values.
     
  7. Mar 13, 2009 #6

    tiny-tim

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    I know you know :biggrin:

    I'm just encouraging you! :wink:
     
  8. Mar 13, 2009 #7
    So I guess my new proof is:

    Triangle ABC

    If;
    mLA > mLB > mLC

    Prove;
    sin A > sin B > sin C
     
  9. Mar 22, 2009 #8
    I'm stumped.
     
  10. Mar 23, 2009 #9

    tiny-tim

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    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:
     
  11. Apr 2, 2009 #10
    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.
     
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