The yellow parallelogram is the parallelogram that we want to have.
To make it from the square, we have to decrease horizontal and vertical sizes
of initial square to p and q correspondingly (we shall
m
find them), rotate on φ, squeeze
------
times (φ, m and n we have to find, too) and rotate
n
back on σ (it is marked on pic.5 by dotted arc, we shall determine it, too).
Let's calculate p and q. From similarity of the triangles ACD and CDK ([DK] is perpendicular dropped from the point named by D on the diagonal d):
d
q
------
=
------,
q
g
where g is the length of the segment [KC]. Hence:
q =
√
d g
and g we can find from the triangle ACD1:
g = cos τ.
Here is:
d2 = l2 + w2 - 2 l w cos γ,
where d is the biggest diagonal of the parallelogram, l and w are lengths of its sides,
and cos γ is the angle between w and l. Let's calculate smaller diagonal f of the parallelogram:
f2 = l2 + w2 - 2 l w cos (180° - γ) = l2 + w2 + 2 l w cos γ,
and here is from the triangle-quater of the parallelogram:
a2
d2
------
=
------
+ w2 - d w cos τ,
4
4
and here is from this:
d2
w2
------
+ w2 -
------
4
4
cos τ =
---------------------------.
d w
So, τ we already have found. Let's find the angle named by φ:
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