Question 1127466
Find &#952;, 0° &#8804; &#952; < 360°, given the following information.
cos &#952; = &#8722; squr(3)/2 and &#952; in QII 
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Sketch a 30-60 degree right triangle
Let the side opposite the 30 deg angle be 1
Then the hypotenuse is 2
And the base is sqrt(3)
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So the cos(30) = sqrt(3)/2
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For your problem the reference angle is arccos(sqrt(3)/2) = 30 degrees
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In QII theta = 180+30 = 210 degrees
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Cheers,
Stan H.