SOLUTION: I'm trying to generate Penrose tiling in Photoshop. To get my rhomb shapes accurate I need to know the following. I skew a square of 400 units by 400 units to form a parallelog

Algebra ->  Pythagorean-theorem -> SOLUTION: I'm trying to generate Penrose tiling in Photoshop. To get my rhomb shapes accurate I need to know the following. I skew a square of 400 units by 400 units to form a parallelog      Log On


   



Question 29865: I'm trying to generate Penrose tiling in Photoshop. To get my rhomb shapes accurate I need to know the following.
I skew a square of 400 units by 400 units to form a parallelogram with angles 72, 108, 72, 108. I know the angles I'm aiming for but the only input data in Photoshop is the distance I push the square along the X axis.
As far as I can tell I could use Pythagoras theorem to calculate the length a by knowing that a right angle triangle is formed with an angle of 18 degrees as I push the square to form the angle of 72 degrees.
I think I calculated that the length of the hypotenuse would have to be sqrt of 400squared by 400 squared ie 566.
I pretty much failed maths at school and it's taken me hours to get this far. I'm stumped when I start reading about finding the length of a leg based on the angle when I didn't even know what cosecant was.
I'd like to understand this, not just have the answer, because I'm sure that I will be needing to redo the calculation to make my 'thin rhomb' and if I continue an interest in tesselation art this is going to be a regular need.
Many thanks if you can help.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
If I get the picture correctly, you are trying to create a rhomb(sic) (in the U.S. this is called a rhombus) by skewing a 400 X 400-unit square so that the interior angles of the rhombus are: 72, 108, 72, and 108 degrees.
Let's assume that your initial square is based on the x-axis and the left
side is coincident with the y-axis, so the bottom-left corner is located at the origin of the coordiante system.
Now you'll displace the top of the square some distance to the right in the x-direction, leaving the bottom fixed of course. Your question is: What should the displacement distance be to obtain the required rhombus?
After the displacement, if you drop a perpendicular line from the top-left corner of the newly-formed rhombus to the base, you will have formed a right-triangle whose hypotenuse (h) is the length of the original square (400 units) and whose base angle is 72 degrees.
You can find the displacement distance (d) using the cosine of 72 degrees. Recall that the cosine of the angle of interest is:
Cosine = (side adjacent to the angle)/(the hypotenuse). The side adjacent is the displacent (d). So:
cos%2872%29+=+d%2Fh But h = 400
cos%2872%29+=+d%2F400 Solve for d.
d+=+400cos%2872%29
d+=+123.606
The required displacement is 123.6 units (rounded to the nearest tenth)