Vector Problem - Angle between two vectors

[jaredm2012]

Homework Statement

Vectors A-> and B-> have equal magnitudes of 55.0. If the sum of A-> and B-> is the vector 17.7 j, determine the angle between A-> and B->

Homework Equations

cos(theta) = {[A]^2 + ^2 - [A-B]^2}/2[A]

(sorry, not sure how to type out that formula!)

tan(theta) = (Ay + By)/(Ax + Bx)

R =[(Ax + Bx)^2 + (Ay + By)^2]^1/2

The Attempt at a Solution

This one I am not sure about how to start, at all. I plugged what I could into those formulas, but any answers I got were incorrect. I don't understand how I can find out any more information with what is given, namely the other components of the vector (the i for each, as well as discerning which values for j each vector has).

 

[rl.bhat]

Since resultant is 17.7j, it is along y-axis. So its magnitude is 17.7. Magnitude of A ans B 55.0 each.
Use the formula R = [ A^2 + B^2 + 2ABcos(theta)]^1/2 and find theta.

 

[thomate1]

First, let's define the vectors A-> and B-> in terms of their components. Since we know that they have equal magnitudes of 55.0, we can represent them as:

A-> = 55.0cos(theta) i + 55.0sin(theta) j
B-> = 55.0cos(phi) i + 55.0sin(phi) j

Note that theta and phi represent the angles that each vector makes with the x-axis.

Next, we can use the given information that the sum of A-> and B-> is the vector 17.7 j, which can be represented as:

A-> + B-> = 0i + 17.7 j

Since the i components of A-> and B-> will cancel out, we are left with:

55.0sin(theta) j + 55.0sin(phi) j = 0i + 17.7 j

This means that:

55.0sin(theta) + 55.0sin(phi) = 17.7

We also know that the magnitude of the vector resulting from the sum of A-> and B-> is 17.7 units, so we can set up the following equation using the Pythagorean theorem:

(55.0cos(theta) + 55.0cos(phi))^2 + (55.0sin(theta) + 55.0sin(phi))^2 = 17.7^2

Simplifying this equation, we get:

3025(cos^2(theta) + cos^2(phi) + sin^2(theta) + sin^2(phi) + 2cos(theta)cos(phi) + 2sin(theta)sin(phi)) = 312.09

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can simplify this equation further to:

3025(1 + 1 + 2cos(theta)cos(phi) + 2sin(theta)sin(phi)) = 312.09

Dividing both sides by 3025, we get:

1 + cos(theta)cos(phi) + sin(theta)sin(phi) = 0.103

Now, we can use the trigonometric identities cos(theta - phi) = cos(theta)cos(phi) + sin(theta)sin(phi) and sin(theta - phi) = sin(theta)cos(phi) - cos(theta)sin(phi