1. Componets of vectors

Let the angle $\theta$ be the angle that the vector $\vec{A}$ makes with the +x-axis, measured counterclockwise from that axis. Find the angle $\theta$ for a vector that has the following components.

First one is $A_{x}=$1.30 m, $A_{y}=$-3.40.

I've read the section, but it's not helping. It lists a couple of equations such as $\frac{A_{x}}{A}=\cos\theta$, but I don't see how that helps me.

2. Originally Posted by cinder
Let the angle $\theta$ be the angle that the vector $\vec{A}$ makes with the +x-axis, measured counterclockwise from that axis. Find the angle $\theta$ for a vector that has the following components.

First one is $A_{x}=$1.30 m, $A_{y}=$-3.40.

I've read the section, but it's not helping. It lists a couple of equations such as $\frac{A_{x}}{A}=\cos\theta$, but I don't see how that helps me.
I assume by $A_x$ you mean the horizontal displacement, and by $A_y$ you mean the vertical displacement. If so, draw a diagram of it. You will end up with a right-triangle (with legs 1.3 and 3.40), use the appropriate trig ratio to find the desired angle, which would end up being the one you were given i believe: $\cos \theta = \frac {A_x}{A}$, where A is the length of the hypotenuse. remember, you want to describe the angle as the angle obtained by going anti-clockwise from the positive x-axis

3. Ugh. Nevermind. I keep asking stupid questions and answering them myself.

4. Okay, I'm not sure where $\theta$ is.

$\frac{A_{x}}{A}=0.36=\cos\theta$, but what's next?

5. Originally Posted by cinder
Okay, I'm not sure where $\theta$ is.
the hypotenuse is in the fourth quadrant, correct? $\theta$ is the angle that goes from the positive x-axis counterclockwise (that is it goes from the 1st quad to the second to the third to the fourth and stops on the hypotenuse of the right-triangle.

6. I keep getting 1.20º, but I know that's not right.

7. The answer is 291º, but I have no idea how they got it.

8. You are having bioth Domain/Range and Calculator Setting problems.

1) Once you have determined the Quadrant, do NOT be dissuaded from that quadrant. The Range of the arccosine function is $[0,\pi]$. There is no Fourth Quadrant in there. It is the end user's responsibility to translate the result to the proper quadrant if $[0,\pi]$ is unsatisfactory. The inverse cosine function on your calculator WILL NOT give you a result in the Fourth Quadrant.

2) Your calculator is NOT telling you 1.20º. It is TRYING to tell you 1.20 Radians. Cconvert it to degrees. Better yet, if you want degrees, feel free to put your calculator in degrees mode. It should give you 69.075º, which still is not correct because that's in the First Quadrant. You have to fix it by point 1).

3) Why cosine when you've a perfectly good inverse tangent to solve the problem? Are you doing vectors and leaning on dot products defined as cosines? I guess that would motivate such behavior. It's still a little irritating to force a more difficult solution.

9. Well, I can see that 360º - 69º = 291º, but I really can't see why that gives me the correct result.

I had to do three more like these, and I'm given four tries each. I missed two completely and got the other two right after a few tries. The remaining three problems put a triangle into one of the remaining three quadrants (I, II, and III).

10. Inverse Functions give only hints. The end user muat get the answer in the correct quadrant.

I would do it this way (being forced to use the cosine).

y-component is negative
x-component is positive
The result is in Quadrant IV

acos(1.3/13.25) = 1.206 rad = 69.075º

This is a reference angle in Quadrant I.

The angle with identical reference in Quadrant IV is 360º - 69.075º = 290.925º

If you would like, you can define your own Inverse Cosine function. Give it whatever Range you need. You are not stuck with the standard definition.