Find the rotation of the object? 
Image 1: The object's original height and width are 9 and 4. Here the object is not tilted, so the bounding box would cover the same dimension.
Image 2: If we tilt the object, the bounding box, as you can observe from the image, has the height and width of 10 and 8.
From this information, can we find out how much degree is the object rotated in the second image?


As you can see, $\triangle ABC \sim \triangle BED$
So we have
$\frac{BC}{AB} = \frac{DE}{BE}$
$BC = \sqrt{16 - x^2}$
$\displaystyle \frac{\sqrt{16 - x^2}}{4} = \frac{8-x}{9}$
Squaring and solving the quadratic, we get $x \approx 3.4516$
Angle by which the object has been rotated is
$\angle BAC = \arccos \frac{3.4516}{4} \approx 30.6^0$
EDIT:
On the comment that the height does not add up to $10$, it will not. When the box width is $8$ due to rotation, the required height will be $\approx 9.78$ and so if the box height is $10$, it will have some room on the top. Please see below diagram that explains when the required height will be maximum.

The maximum required height will be the length of the diagonal and will be the case when the diagonal $AE$ is vertical.
So the maximum required height = $\sqrt{9^2 + 4^2} \approx 9.85$
This happens when we rotate by angle $23.95^0 (= \arctan \frac{4}{9})$. The required width at this point is $\approx 7.31$.
Please also consider that as we keep rotating, there will be a point $(90^0 - 23.95^0)$ when the required width is $9.85$ and the required height is $7.31$. Given the dimension of the object, you will never need box to go beyond $9.85$ on either side (width or height).
Here is a plot of how width and height of the box will change with the angle of rotation between $(0 \leq \theta \leq \frac{\pi}{2}) \,$ which is taking it from the current vertical position to the horizontal position. As you can see, at angle of rotation of $45^0$, the box requires the same width and height of $\approx 9.2$.

