Angle Between Vectors given Ratio of Resultants, why is this wrong?

[Ordain]

Homework Statement

Vectors A and B have the same magnitude. Given that the magnitude of A + B is 75 times greater than the magnitude of A - B, find the angle between them?

Homework Equations

We know that A=B, so:

2AB+2ABCos$\theta$=75(2AB-2ABCos$\theta$)

Given that A=B

2A²+2A²Cos$\theta$=75(2A²-2A²Cos$\theta$)


3. Attempt at a solution

2A²+2A²Cos$\theta$=75(2A²-2A²Cos$\theta$)

Couldn't get latex to work for the fraction so:

2A²+2A²Cos$\theta$ divided by (2A²-2A²Cos$\theta$)=75

Factor out 2A²:

[itex]\frac{1+Cos}{1-cos}=75<br /> 1+Cos$\theta$=75-75$\theta$<br /> -74=-76Cos$\theta$<br /> Cos$\theta$=74/76<br /> $\theta$=Cos^-1(74/76)<br /> $\theta$=13.2 deg<br /> <br /> Apparently this is wrong, how so?[/itex]

 

[danielakkerma]

Hey,
Actually, the way I see it, verifying your logic and procedure, it seems fine!
Are you sure they're not asking for the solution in radians?
It might also help, to get abit of order in your calculations, to mark it as this:
Assuming the angle between the two is theta(like you said), but the ratio, for the sake of generality, we'll call alpha, that gives us:
$\large<br /> |A+B|=\alpha |A-B|, |A|=|B| \Longrightarrow A^2+2A^2\cos{\theta}+A^2 = \alpha(A^2-2A^2\cos{\theta}+A^2)$
Dividing by 2*A^2:
$\large<br /> 1+\cos{\theta} = \alpha(1-\cos{\theta}) \Longrightarrow \cos{\theta} = \frac{\alpha-1}{\alpha+1}$
Check to see that you have the proper answer key.
Are there any other considerations you may have overlooked?
Daniel

 

[Ordain]

Thanks for that, noted for next time. This is the question in exact form, I can't find anything I'm missing.

 

[danielakkerma]

Try entering it in radians, namely: 0.229 ~ 0.23.
What's the typical accuracy, digit significance asked? are two decimal places usually sufficient?
If so, try either 0.229, 0.23.
Daniel