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I have solutions to two problems, but I just have a question about these solutions.

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So I am to solve the problem in the image above. For b iv), The ratios I got is sin=-5/13, cos=-12/13, and tan=5/12. The angle I got is 22.6, but because the point is in the 3rd quadrant, the angle will be more than 180 degrees, so I'd do 180 + 22.6 = 202.6 degrees. Converting that to radians I get (202.6 deg)*pi/180 = 3.536 radians, rounded to 3.54 radians.

However, if I were to solve a similar problem for a given point in the 2nd quadrant, I would have to do 180 - the principal angle. I was wondering why I don't add 90 degrees to the value I get because doesn't the angle in the second quadrant have to be above 90 degrees? I was wondering why I don't apply the same logic to the 2nd quadrant like I did with the 3rd quadrant.

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This is based on the facts that $tan{(\pi+\alpha)}=tan{\: \alpha}$ and $tan{(\pi-\alpha)}=-tan{\: \alpha}$. Also, if you draw 22.6° angle and 112.6° angle you'll see that $tan{\:112.6°} \ne -tan{\: 22.6°}$.

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