- #1

joelio36

- 22

- 1

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter joelio36
- Start date

- #1

joelio36

- 22

- 1

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

steelphantom

- 159

- 0

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.

Last edited by a moderator:

- #3

stingray78

- 23

- 0

I think you'll have to calculate, no general way to do it. :)

- #4

PuzzledMe

- 34

- 0

Stick by the cosine rules and you'll be fine.

- #5

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

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

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?

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.

I think you'll have to calculate, no general way to do it. :)

- #6

kamerling

- 454

- 0

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

stingray78

- 23

- 0

Hehehe sorry, you're right. Didn't give it enough thinking. I was the one absolutely wrong. Sorry!!!

Share:

- Last Post

- Replies
- 7

- Views
- 5K

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 26

- Views
- 2K

- Last Post

- Replies
- 15

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 7

- Views
- 2K

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 17

- Views
- 3K