# Largest angle of a parallelogram in a plane

What is the measure of one of 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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Look at $C$, third point (3,5).

$$\Delta y = 4, \, \Delta x = 1$$

Required angle

$$\pi= \tan^{-1} (4/1)$$

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But why it came up so long after? – Narasimham Sep 14 '15 at 15:04