What is the Cosine Rule for Triangles with Operands Greater Than 1?

AI Thread Summary
The discussion centers on the application of the Cosine Rule in triangle problems, specifically when calculating angles in mechanics. A user initially struggled with finding the direction of the resultant force because the operand for Arccos exceeded 1, leading to confusion about the Cosine Rule's applicability. It was clarified that the Cosine Rule can be used for any angle, not just the one opposite the longest side, but the user mistakenly applied an incorrect formula. The correct formulas for the Cosine Rule were reiterated, emphasizing the relationships between the sides and angles of a triangle. Ultimately, the user resolved their issue by realizing the need to apply the Sine Rule instead.
W3bbo
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Homework Statement



As part of a Mechanics problem, I need to find the resultant of two forces. I was able to find F[Resultant]'s magnitude easily enough, but it's direction stumps me.

...because when I rearrange the Cosine rule to find angle A, the operand of Arccos is greater than 1.

Homework Equations



See attachment

The Attempt at a Solution



No idea :confused:
 

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Then you have erred in your calculations.
 
arildno said:
Then you have erred in your calculations.

Turns out my calculations were fine, I just forgot the Cosine rule was only of use when A is opposite the longest side in any given triangle.

I just needed to use regular Sine Rule to find the missing angle.

Problem solved.
 
W3bbo said:
Turns out my calculations were fine, I just forgot the Cosine rule was only of use when A is opposite the longest side in any given triangle.

Incorrect, wherever have you gotten that strange idea from?
 
yes largest angle is always opposite the longest side (common sense) BUT that doesn't mean you can only use cosine rule on that angle.
just reading the maths it is very hard for me to deduce what does your force diag looks like and what are you trying to solve...but just looking at the cosine rule it seems wrong (normal symbol usage assumed) should be
a^2=b^2+c^2-2bc \cos A where A is angle opposite side a.
 
There are actually three "cosine rules" for any triangle.
a^2= b^2+ c^2- 2bc cos(A)
b^2= a^2+ c^2- 2ac cos(B)
c^2= a^2+ b^2- 2ab cos(C)
Whate a is the length of the side opposite angle A, b is the length of the side opposite angle B, and c is the length of the side opposite angle C.

Your original formula was a^2= b^2+ c^2- 2ab cos(C), none of the above.
 
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