Question 499849
I need to fine the sine and cosine of angle theta in standard position who's terminal side passes through the given point (-15,-12). I also need to find the measure of angle theta to the nearest tenth of a degree. I must assume that theta is greater than or equal to 0 degrees and is less than 360 degrees. 
I got as far as finding out the hypotenuse is the square root of 369, but I don't know how to use those trig. functions to find the reference angle in order to find theta.
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Quadrant III is the only quadrant where sin and cos functions are both negative. 
You are working a right triangle in quadrant III with a horizontal leg of -15 (cos), a vertical leg of -12 (sin) and a hypotenuse you correctly calculated to be=√(369). The reference angle is
the inverse of the sin theta or cos theta.  
sin theta=-12/√369
cos theta=-15/√369
..
Using a calculator to find reference angle:
arccos(15/√369)=38.7º
arcsin(12/√369)=38.7º
Theta=38.7+180=218.7º (quadrant III)