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I first used the Pythagorean Theorem to get the hypotenuse needed for sine.

$$x^2 + y^2 = r^2$$ $$(-5)^2 + (1)^2 = r^2$$ $$25 + 1 = r^2$$ $$\sqrt{26} = \sqrt{r^2}$$ $$5.0990 \approx r$$

Once I had the hypotenuse I substituted into the sine formula, using alpha for my acute angle, letting theta represent my obtuse angle.

$$\sin\alpha = \frac yr$$ $$\sin\alpha = \frac{1}{5.0990}$$ $$\alpha = \sin^-1 \left(\frac{1}{5.0990}\right)$$ $$\alpha \approx 11.30998$$

However, I did not get the same result for cosine, and I'm not sure why. Instead, I got the obtuse angle. Normally I'd subtract 11.30998 from 180 since I'm in two quadrants, but I got slightly different results.

$$\cos\alpha = \frac xr$$ $$\cos\alpha = \frac {-5}{5.0990}$$ $$\alpha = \cos^-1 \left(\frac{-5}{5.0990}\right)$$ $$\alpha \approx 168.69116$$

But if I subtract 11.30998 from 180 since I'm in two quadrants, I get 168.69002 instead of 168.69116. What am I doing wrong?

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  • $\begingroup$ I think it's a rounding issue $\endgroup$ – imranfat Nov 26 '15 at 1:27
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    $\begingroup$ Yes it is. Notice: (-5/5.099)^2+(1/5.099)^2 = 25/25,999801+1/25,999801=26/25,999801 which is not 1. So the issue is rounding sqrt(26). Only round in your final answer $\endgroup$ – Pieter Rousseau Nov 26 '15 at 2:28
  • $\begingroup$ If you have used $\sqrt{26}=5.09902$, the results would have been much better. $\endgroup$ – Claude Leibovici Nov 26 '15 at 4:16

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