Let the angle be the angle that the vector makes with the +x-axis, measured counterclockwise from that axis. Find the angle for a vector that has the following components.
First one is 1.30 m, -3.40.
I've read the section, but it's not helping. It lists a couple of equations such as , but I don't see how that helps me.
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 . There is no Fourth Quadrant in there. It is the end user's responsibility to translate the result to the proper quadrant if 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.
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).
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.