Interior angles of a Parallelogram

**supersaiyan** · April 16th 2009, 04:57 PM

Hello, Ho do I find the interior angles of this

I know how to find for a triangle using
A.B = |A||B| cos theta

not sure how to find for a parallelogram

**mollymcf2009** · April 16th 2009, 05:14 PM

Originally Posted by supersaiyan

Hello, Ho do I find the interior angles of this

I know how to find for a triangle using
A.B = |A||B| cos theta

not sure how to find for a parallelogram

You do just that! Draw a line up from point P to make your triangle (or you can use point R) and using the coordinates for the points find the lengths of the sides of the triangle. Then you can find the measurement for angle PQR (which of course will be that same for angle PSR. Then using what you know about supplementary angles, you will be able to find the other two angles!

**Twig** · April 16th 2009, 10:17 PM

$\vec{PQ} = (-4,3) \, \Rightarrow |\vec{PQ}| = 5$
$\vec{PS} = (10,-1) \, \Rightarrow |\vec{PS}| = \sqrt{101}$

$\theta = arccos(\frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| \cdot |\vec{PS}|} \approx 148.84$

Giving the other angle: $\frac{360 -2 \cdot \theta}{2} \approx 31.15$

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