Calculate rotated rectangle size from known boundi

Question

I read the Calculate Bounding box coordinates from a rotated rectangle to know how to calculate bounding box coordinates from a rotated rectangle. But in a special case as follow image:

How to get the rotated rectangle size if had get the bounding box size, coordinates and rotate degree?

I try write code in javascript

//assume w=123,h=98,deg=35 and get calculate box size
var deg = 35;
var bw = 156.9661922099485;
var bh = 150.82680201149986;

//calculate w and h
var xMax = bw / 2;
var yMax = bh / 2;
var radian = (deg / 180) * Math.PI;
var cosine = Math.cos(radian);
var sine = Math.sin(radian);
var cx = (xMax * cosine) + (yMax * sine)   / (cosine * cosine + sine * sine);
var cy =  -(-(xMax * sine)  - (yMax * cosine) / (cosine * cosine + sine * sine));
var w = (cx * 2 - bw)*2;
var h = (cy * 2 - bh)*2;

But...the answer is not match w and h

Answer 1

You'll probably need something like affine transformation to discover point coordinates. And then using standard geometry formulas calculate the size.

Answer 2

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 2x2 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!