Question 242755
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Two.  Since for a 30° angle, BC would have to be one-half of AC if the triangle were a right triangle, and 9 is not one-half of 11, we know this is not a right triangle.  Hence there is a possible obtuse angle solution for B and a possible acute angle solution for B.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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