tan s = (-sqrt7)/2, sec s is greater than 0 Thanks

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Nicole -Tan is negative only in Quadrants II and IV.  The problem also states that sec is positive which can only happen in Quadrants III and IV.  So, looking at these two facts, angle s must me in Quadrant IV.So draw your triangle in Quad IV and label the angle opposite s is -sqrt(7) and the side adjacent s is 2.  From this info you should be able to solve for s.Good luck