Cartesian Vectors Question

[JTJ]

I apologise if this is in the wrong section. This is a 12th grade Vectors problem, and I had no idea where to put it.

Here's the question I am having difficulty with:

The points P(-2, 1), Q (-6, 4), and R(4, 3) are the three vertices of parallelogram PQRS.

a) Find the coordinates of S

With the help of my teacher in class today, I've done that, and got [0,6]. (The back of the book says answers may vary)

It's part b I'm stuck on:
b) Find the measures of the interior angles of the parellelogram, to the nearest degree.

I tried using cos theta = (a1 b1 + a2 b2) / lal lbl but I can not get the answer in the back of the book. Am I just plugging the numbers in wrong? If that is the correct formula, which numbers should I be using?

The back of the book says the answer is 31 degrees and 149 degrees. I am getting 126 degrees.

Thanks in advance for any help.

 

[soroban]

Hello, JTJ!

The points P(-2, 1), Q (-6, 4), and R(4, 3) are the three vertices of parallelogram PQRS.

a) Find the coordinates of S

With the help of my teacher in class today, I've done that, and got [0,6].
(The back of the book says answers may vary.) . This is true.

Given those three points,
. . there are three possible parallelograms.

              S
              ♥
            *  *
          *     *
        *        *
    Q o           o R
       *        *
        *     *
         *  *
          o
          P

     Q             R 
      o * * * * * o
       *           *
        *           *
         *           *
          o * * * * * ♥
         P             S

           Q             R
            o * * * * * o
          *           *
        *           * 
      *           * 
    ♥ * * * * * o
   S             P

However, the problem said "parallelogram PQRS".
The convention is that the vertices are given
. . in order around the polygon.

If that is the case, the second diagram is the correct parallelogram.
Therefore, $\displaystyle S$ is $\displaystyle (8,0).$