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What is the measure of one of the larger angles of a parallelogram in the xy-plane that has vertices with coordinates (2,1),(5,1),(3,5) and (6,5)? (A) 93.4 (B) 96.8 (C) 104.0 (D) 108.3 (E) 119.0 Now the answer strip says C, but it doesn't explain how to find the angle. So how would I go about solving this question? I've tried drawing in a plane but I couldn't relate my drawing to the angles. |
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If you have vectors at your disposal, the easiest thing to do is: (1) find the vectors from one vertex to the two adjacent vertices, call them $\vec{u}$ and $\vec{v}$; (2) the angle formed by these two vectors is $\arccos(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|})$. If this angle is acute, then its supplement is the measure of the obtuse angle in the parallelogram. Without vectors, you could use the Law of Cosines: (1) find the lengths of two adjacent sides of the parallelogram (call them $a$ and $b$) and the length of the longer diagonal (call it $c$; if you used the shorter diagonal, you'd get the acute angle); (2) the angle is $\arccos(\frac{a^2+b^2-c^2}{2ab})$. |
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One idea is to find the cartesian coordinates of the edges of the parallilogram and think of the slope of them(use the difference of their slopes). |
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