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cos theta = 5/7 ,(270degree
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Think about a right triangle OAB where O=(0,0) is the origin, B=(0,5), A=(5, - 2\(\sqrt 6\)) and \(\theta = 2 \pi - AOB)\) then \[ \sin(\theta) =- \frac {2\sqrt 6}{7} \] Since you have sin and cos you can find the rest.
Notice that \[ OA^2 = (2 \sqrt 6)^2 + 5^2 = 49 \\ OA=7 \]
As always, practice trig on my site http://www.math.missouri.edu/escgi/mucgi-bin/munew.cgi?variable=trig
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