Finding parallelogram dimensions inside a 15×20.5in rectangle

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Storminnorman
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A parallelogram exists within a rectangle which measures 15 inches tall by 20.5 inches wide.

A=15 inches
B=20.5 inches
C=1.5 inches
(C makes a 90° angle with X)

Solve for lengths X and Y and angle Z of the parallelogram and please tell me how you did it!

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Hello and welcome to MHB, Storminnorman! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

I think the way I would begin is to observe that the area of the parallelogram is:

$$A_P=1.5X$$

And then we can deconstruct the area of the rectangle into 2 right triangles and the parallelogram:

$$AB=(B-Y)A+1.5X$$

Now, by Pythagoras, we find:

$$B-Y=\sqrt{X^2-A^2}$$

Hence:

$$AB=A\sqrt{X^2-A^2}+1.5X$$

Now you have an equation with only one unknown...can you proceed?