I have a box of height h and width w. I rotate it to r degrees. Now I resize it so that it can original box in it. What will be the size of newly box.
Original Box:
Box after rotating some degrees.
New box after rescaling.
So my question is what should be the formula to calculate new size (width, height), position relative to old one.
What I have Width "w", height "h", position (x,y) and angle (t).
Assuming the old rectangle inscribed in the new one, we have the following picture:
Let $\theta$ ($0 \leq \theta \leq \frac{\pi}{2}$) the rotation angle, $w'$ the new width and $h'$ the new height, then we have the following equations: $$w' = w \cos \theta + h \sin \theta$$ $$h' = w \sin \theta + h \cos \theta$$
The new rectangle is not similar to the old one, except for $h = w$ when both rectangles are in fact squares.
Edit:
Considering $O$ (the center of both rectangles) as the origin of the coordinate system, the points $E$, $F$, $G$, and $H$ can be calculated by the following equations: $$E=\left(\frac{w}{2}-w \cos^2 \theta,-\frac{h}{2}-w \sin \theta \cos \theta \right)$$ $$F=\left(\frac{w}{2}+h \sin \theta \cos \theta,-\frac{h}{2}+h \sin^2 \theta \right)$$ $$G=\left(-\frac{w}{2}+w \cos^2 \theta,\frac{h}{2}+w \sin \theta \cos \theta \right)$$ $$H=\left(-\frac{w}{2}-h \sin \theta \cos \theta,-\frac{h}{2}+h \cos^2 \theta \right)$$
