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The largest angle in a triangle:

  1. Apr 17, 2008 #1
    I have a triangle, with sides 8.5, 6.8, and 9.4, what is the largest angle?

    I think that it will always be the angle opposite the longest side, and used the cosine rule from there, along with trig identies to find the largest possible angle (<180).

    Is this a correct method?
  2. jcsd
  3. Apr 17, 2008 #2
    Just use the law of cosines to find each angle. Then you will be able to say which one is the biggest.
  4. Apr 17, 2008 #3
    You' re absolutely wrong! You can have a big side with a small angle as its opposite.
    I think you'll have to calculate, no general way to do it. :)
  5. Apr 17, 2008 #4
    Stick by the cosine rules and you'll be fine.
  6. Apr 17, 2008 #5


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    Yes, in a given triangle the longest side is always opposite the largest angle. Just use the cosine rule for that side.

    Sorry, but you are the one who is "absolutely wrong". You might be thinking that if the triangle is very large then a side opposite a small angle can be "big"- but in that case the other sides will be even larger.
  7. Apr 17, 2008 #6
    The largest side is always opposite the largest angle. consider a triangle with angles
    A,B and C and the side a opposite to the angle A etc.

    For acute triangles you can get this from the law of sines: [tex]\frac{a}{sinA} = \frac{b}{sinB} [/tex], so if a>b then a/b>1 and so [tex]\frac{sinA}{sinB} > 1[/tex] therefore sin(A) > sin(B). since sin is increasing from 0 to 90 degrees this implies A>B

    For obtuse triangles there can be only one obtuse angle, wich must be the largest. call this A. cos(A) < 0 sothe cosine rule tells us that a^2 > b^2 + c^2 and therefore a>b and a>c so the largest angle is also to the opposite of the largest side.

    P.S. Is there a way to get tex fractions lined up with the line of text they are in?
  8. Apr 19, 2008 #7
    Hehehe sorry, you're right. Didn't give it enough thinking. I was the one absolutely wrong. Sorry!!!
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