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Cosine rule and operands > 1

  • Thread starter W3bbo
  • Start date
31
0
1. 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.

2. Homework Equations

See attachment

3. The Attempt at a Solution

No idea :confused:
 

Attachments

Answers and Replies

arildno
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Then you have erred in your calculations.
 
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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.
 
arildno
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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?
 
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mjsd
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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
[tex]a^2=b^2+c^2-2bc \cos A[/tex] where A is angle opposite side a.
 
HallsofIvy
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There are actually three "cosine rules" for any triangle.
[tex]a^2= b^2+ c^2- 2bc cos(A)[/tex]
[tex]b^2= a^2+ c^2- 2ac cos(B)[/tex]
[tex]c^2= a^2+ b^2- 2ab cos(C)[/tex]
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 [itex]a^2= b^2+ c^2- 2ab cos(C)[/itex], none of the above.
 

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