What is the Correct Theta Angle for Vector Degrees Problem?

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Homework Help Overview

The discussion revolves around determining the correct angle theta for a vector given its components Ax and Ay, specifically in the context of vector degrees and trigonometry.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the calculation of theta using the inverse tangent function and question the quadrant in which the vector lies. There is discussion about the periodic nature of the tangent function and how it affects the angle calculation.

Discussion Status

Participants are actively engaging with the problem, providing insights into the quadrant analysis and the implications for the angle calculation. Some guidance has been offered regarding the use of the unit circle and the adjustment of the angle based on the quadrant.

Contextual Notes

There is an emphasis on ensuring the angle is expressed within the interval (0,360) degrees, and participants are considering the implications of the vector's position in the third quadrant.

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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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