Cosine rule and operands > 1

1. Feb 3, 2007

W3bbo

1. The problem statement, all variables and given/known data

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. Relevant equations

See attachment

3. The attempt at a solution

No idea

File size:
4.2 KB
Views:
54
2. Feb 3, 2007

arildno

Then you have erred in your calculations.

3. Feb 3, 2007

W3bbo

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.

4. Feb 3, 2007

arildno

Incorrect, wherever have you gotten that strange idea from?

5. Feb 3, 2007

W3bbo

This posting: https://www.physicsforums.com/showpost.php?p=1226780&postcount=3

And if my calculations are wrong, where? Just saying I'm wrong without further detail isn't helping. (Or isn't my attachment showing?)

6. Feb 3, 2007

mjsd

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.

7. Feb 5, 2007

HallsofIvy

Staff Emeritus
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.