Calculate rotated rectangle size from known bounding box coordinates

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Solution

Given bounding box dimensions bx by by and t being the anticlockwise rotation of rectangle sized x by y:

x = (1/(cos(t)^2-sin(t)^2)) * (  bx * cos(t) - by * sin(t))
y = (1/(cos(t)^2-sin(t)^2)) * (- bx * sin(t) + by * cos(t))

Derivation

Why is this?

First, consider that the length bx is cut in two pieces, a and b, by the corner of the rectangle. Use trigonometry to express bx in terms of x, y, and theta:

bx = b          + a
bx = x * cos(t) + y * sin(t)            [1]

and similarly for by:

by = c          + d
by = x * sin(t) + y * cos(t)            [2]

1 and 2 can be expressed in matrix form as:

[ bx ] = [ cos(t)  sin(t) ] * [ x ]     [3]
[ by ]   [ sin(t)  cos(t) ]   [ y ]

Note that the matrix is nearly a rotation matrix (but not quite – it’s off by a minus sign.)

Left-divide the matrix on both sides, giving:

[ x ] = inverse ( [ cos(t)  sin(t) ]    * [ bx ]                        [4]
[ y ]             [ sin(t)  cos(t) ] )    [ by ]

The matrix inverse is easy to evaluate for a 2×2 matrix and expands to:

[ x ] = (1/(cos(t)^2-sin(t)^2)) * [ cos(t) -sin(t) ] * [ bx ]           [5]
[ y ]                             [-sin(t)  cos(t) ]   [ by ]

[5] gives the two formulas:

x = (1/(cos(t)^2-sin(t)^2)) * (  bx * cos(t) - by * sin(t))             [6]
y = (1/(cos(t)^2-sin(t)^2)) * (- bx * sin(t) + by * cos(t))

Easy as pie!

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