Determining the angle of a photograph containing known parallel objects/lines. I have a photograph of a house and a window taken at an angle. I'm trying to determine the angle at which the photograph was taken. The house has wooden siding that can be safely assumed to be parallel.
Normally, I'd try to derive something on paper, but I don't know how to approach this type of problem. I can't supply the photograph, but I can supply a stunning representation in MS Paint.
I've shift the origin such that the top left point is 0,0. The following are the coordinates and lengths between points:
Points:
A (0,0)
B (741, 49)
C (0, 844)
D (741,692)

Lengths (in pixels):
AB = 743
AD = 1014
AC = 844
BD = 643
BC = 1087
CD = 756


 A: The problem as stated is underdetermined in two ways:
1) Consider a pinhole camera aimed normal to a wall (with windows) so that the film is parallel to the wall. Then by simple geometry, the image is a proportional map of the wall (just shrunken and rotated 180 degrees). All the windows will be rectangular in the image. If you rotate the camera to point at a window to your right (so the film is no longer parallel to the wall), the window's image will appear trapezoidal as in your drawing. We would at least need to know where is the centre of the image (i.e. the spot on the wall the camera is aimed at) relative to the trapezoid.
2)If we know the distance $F$ from the pinhole to the film, we can measure the width $W$ of an image on the film and estimate the angular size of the object from the camera position as $W/F$. The diagram provided does not provide this information.
The ratio $R$ of the left and right sides of the trapezoid is inversely proportional to the ratio of the distances of the window sides from the camera.  Assume the camera is pointed in the horizontal plane (I think this follows from the fact that both sides of the window in the image are vertical, but I won't try to prove this here) so we have a two dimensional problem. Further assume that $W/F$  is small and the camera is pointed at the centre of the window. From a diagram and a short calculation I get $$ \tan \theta = \frac{2F}{W} \frac{R-1}{R+1}$$ where $\theta$ is the angle from the normal to the wall.
Note: In a previous version, I suggested that we need to know the distance from the camera to the wall. This is obviously incorrect by dimensional considerations. This distance occurs in my calculation, but cancels out.
