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## Homework Statement

Vector A and B have the same magnitude. Given that the magnitude of

**A+B**is 73 times greater than the magnitude of

**A-B**, find the angle between them.

## Homework Equations

[itex]sqrt[(A+Acosθ)^2+(ASinθ)^2)]=73sqrt[(A-Acosθ)^2+(-Asinθ)^2][/itex]

## The Attempt at a Solution

I managed to start the question using the equation above, which was derived from the question details. I numbered my steps as follows in an effort to keep track of the process:

1)Square both sides-this eliminates the square root over the majority of the terms, and also squares the "73" coefficient.

2)Expand all brackets. This is where the mess comes in, and results in an equation as follows:

[tex](A^2+2A^2cosθ+Acos^2θ)+(A^2sin^2θ)=5329[(A^2-2A^2cos+Acos^2θ)+(A^2sin^2θ)][/tex]

3)Here's where I began to have a little difficulty. Firstly, I worked on the left half of the equation without the coefficient. Specifically, I dropped the brackets and factored out "A", giving me [tex]A+2Acosθ+cos^2θ+A+2Asinθ+sin^2θ[/tex] on the left side of the equation. I noticed that there's a trigonometric identity for "[itex]cos^2θ-sin^2θ=2cosθ"[/itex], and since those two terms appear in a positive form on either side of the equation, I'm guessing I can switch the "[itex]sin^2θ[/itex]" terms on either side and get "[itex]cos2θ[/itex]" on both sides, then move one over and cancel them both.

But from there I'm unsure of how to make the equation simplify any further, barring distributing the 5329 amongst all the left-side terms, then dividing the entire equation by 5329 to remove them (which probably has to be done before switching the [itex]sin^2θ[/itex] on either side). I get the feeling cancelling further will be the key here, as well as distribution and factor addition, but I'm running into difficulty proceeding with calculations due to the mass number of terms and variables, so I would greatly appreciate some assistance in how to proceed without making an even bigger mess.

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