What is the Correct Theta Angle for Vector Degrees Problem?

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The correct theta angle for the vector degrees problem, given Ax = -4.00 and Ay = -8.00, is found by recognizing that the vector lies in the third quadrant. The inverse tangent calculation initially yields 63.4 degrees, which is incorrect because it does not account for the quadrant. To find the correct angle, 180 degrees is added to the arctangent result, resulting in a final answer of 243.4 degrees for theta. This adjustment is necessary because the arctangent function only provides angles in the first and second quadrants. Understanding the quadrant placement is crucial for determining the correct angle.
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Homework Statement



Ax= -4.00 , Ay= -8.00
Please give your answer in the interval (0,360) degrees for theta.




Homework Equations



Theta = ______ degrees

The Attempt at a Solution



I tried inverse of tangent (-8/-4) = 63.4 degrees for theta, but its telling me I am wrong. Tried the degree in negative and the computer showed up wrong also. Any Ideas?
 
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Bottomsouth said:
Ax= -4.00 , Ay= -8.00


I tried inverse of tangent (-8/-4) = 63.4 degrees for theta, but its telling me I am wrong. Tried the degree in negative and the computer showed up wrong also. Any Ideas?

Hmmm... well just looking at Ax and Ay, what quadrant does A lie in? Does theta=63.4 degrees lie in that quadrant? :wink:

The problem is that the inverse tangent of (Ay/Ax) always returns a value between 0 and 180 degrees. But the tangent function has a period of 180 degrees, so tan(theta)=tan(theta+n*180)=Ay/Ax implies that theta=ArcTan(Ay/Ax)-n*180 for some integer value of n and not just theta=ArcTan(Ay/Ax).
 
Remember the unit circle. The angle is in the third quadrant.
 
It is in the 3rd quadrant making it positive for tangent. Just what would the variable n be?

Thanks for the help, appreciate it
 
Got it now, since its int he 3rd quadrant we add 180 degrees. I knew it was something so simple. so its 243.4 degrees for theta.

Thanks,
 
Choose n such that your angle is in the correct quadrant. In this case, what are the limits of n?
 
Bottomsouth said:
It is in the 3rd quadrant making it positive for tangent.

Exactly; A is in the 3rd quadrant.

The tangent is positive in both the third quadrant and the first quadrant, and the arctangent always returns a value in the first to quadrants. That is why you are getting an angle that is in the first quadrant.

Just what would the variable n be?

Since A is in the 3rd quadrant, 'n' will be any integer that gives a theta in between 180 degrees and 270 degrees. In this case, n=-1 should do nicely.
 
Bottomsouth said:
Got it now, since its int he 3rd quadrant we add 180 degrees. I knew it was something so simple. so its 243.4 degrees for theta.

Thanks,

Welcome :smile:
 

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