Converting radians to degrees, help

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Homework Help Overview

The discussion revolves around converting polar coordinates to Cartesian coordinates, specifically focusing on the conversion of angles from radians to degrees. The original poster seeks clarification on how to determine the angle associated with given Cartesian coordinates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between Cartesian coordinates and polar coordinates, questioning the conversion of a y-coordinate to degrees. There is discussion about using trigonometric functions to find angles, particularly the inverse tangent function.

Discussion Status

The conversation is actively exploring the correct interpretation of angles in relation to quadrants. Participants have provided guidance on checking the quadrant of the angle and ensuring that all angles correspond to the correct location of the point in the Cartesian plane.

Contextual Notes

There is a focus on ensuring that the angles derived from the calculations are valid within the context of the second quadrant, as the original Cartesian coordinates are located there. Participants are also considering the implications of negative angles and their corresponding positions in different quadrants.

StarkyDee
I know this is an easy problem, but I need to know 3 sets of polar coordinates for the Cartesian coordinates (-4,4[itex]\sqrt{3}}[/itex])

So I graphed the points and got the hypotenuse, r = 8.

How do I convert 4[itex]\sqrt{3}}[/itex]) to degrees?

(4[itex]\sqrt{3}}[/itex]) = 6.92820 Is this in radians?

and do I then multiply it by (180/pi) to get degrees?

thanks.
 
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[itex]4 \sqrt{3}[/itex] isn't an angle...
 
It sounds like you've already drawn a triangle -- so use trig to solve for the angle. The angle is the inverse tangent of y/x.

Why are you trying to convert the cartesian y-coordinate to degrees? That doesn't even make sense.

- Warren
 
Ah. Inverse Tan of y/x.

4[itex]\sqrt{3}}[/itex]/-4 = -1.7320 = -60 degrees.

Thanks ~Dave~
 
Don't forget to check that your answer is in the correct quadrant!
 
well -60 degrees would be in the 4th quadrant.
but when i graph (-4,4[itex]\sqrt{3}}[/itex]) it is only valid in the 2nd quadrant and you can't have a -60 degrees in the 2nd quad. so if i start in the 4 quadrant at -60

the 3 points would be: (8,-60);
add 360 (8,-240);
opposite angle (8,120). is this correct?

thanks again!
 
Well, tell me, how many of those are in the second quadrant?

All of your answers must be in the second quadrant because the desired point is in the second quadrant...
 
Ah. So -60 would not work because it's in the 4th quadrant. therefore (8,-60) is not a point. that makes sense.

but (8,-240) and (8,120) and (8,-600) would be in the 2nd quadrant.
 
right.
 
  • #10
Thanks for helping me out Hurkyl, appreciated much!
Studying for finals have turned my brain into mush.
 

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