The largest angle in a triangle:

  • Thread starter joelio36
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  • #1
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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?
 

Answers and Replies

  • #2
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Just use the http://en.wikipedia.org/wiki/Law_of_cosines" [Broken] to find each angle. Then you will be able to say which one is the biggest.
 
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  • #3
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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. :)
 
  • #4
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Stick by the cosine rules and you'll be fine.
 
  • #5
HallsofIvy
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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?
Yes, in a given triangle the longest side is always opposite the largest angle. Just use the cosine rule for that side.

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. :)
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.
 
  • #6
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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. :)
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?
 
  • #7
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Hehehe sorry, you're right. Didn't give it enough thinking. I was the one absolutely wrong. Sorry!!!
 

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