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given cosθ=9/13 and sin θ negative, find θ in degrees between 0 and 360 degrees.

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Kenneth S. | Algebra/Trig/Precalculus/Calculus Instructor Can Be Your TutorAlgebra/Trig/Precalculus/Calculus Instru...
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sine negative, cosine positive, quadrant IV
θ = arccos(9/13) = 46.19 degrees, IN FIRSTQUADRANT, so θ =360-46.19 = 313.81 deg.

Comments

given tan q =-1.5 and q in quadrant 4, find θ in degrees between 0 and 360 degrees.
can you help me with this one too?
all of this kind of problems begin by doing the INVERSE TANGENT
;;;sorry, comment chopped off...once you have the calculator answer, you must CORRECT for quadrant, if needed.
But in the case of arctan, your answer will always be a negative angle (quadrant IV) when the argument is negative.
So in this case, just add 360 degrees to the answer, and now that angle is positive &in quadrant IV.