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)?

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- Jun 1st 2009, 10:34 AMIronMan21angle of parallelogram
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)? - Jun 1st 2009, 10:44 AMmasters
Hi IronMan21,

Here's one approach to your problem. Extend the line through (3, 5) and (2, 1) until it crosses the x -axis. We can find the slope of this line to be 4.

To determine the angle made with the x-axis, we take the arctan of 4 which is approximately 76 degrees (75.9637).

The angle formed at (3, 5) is the larger of the two angles in the parallelogram and is supplementary to the angle we just found. - Jun 1st 2009, 11:54 AMHallsofIvy
Another way: The vector from (5,1) to (2,1) is <-3, 0>. The vector from (5,1) to (6,5) is <1, 4>. The length of each of those vectors is 3 and $\displaystyle \sqrt{17}$ respectively and their dot product is product is -3(1)+0(4)= -3. Now use the fact that $\displaystyle \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$ where $\displaystyle \theta$ is the angle between the vectors that you are looking for. (This is the angle at (5,1) which is the same as the angle at (3,5).)