It really isn't that hard to do a lot better with these measurements.
The most direct way to determine the sun elevation during an Apollo lunar EVA is to find a picture that includes both the sun and the level horizon taken with a camera whose image size and lens focal length are known, and just measure the angle between the two. Yes, the sun will be vastly overexposed and its image will bloom enormously. But it'll remain reasonably symmetrical, and with lens flares and other artifacts as additional clues it is usually easy to determine the sun's true position in the frame.
The Apollo surface Hasselblad cameras used 70mm film with 60 mm focal length lenses. The fiducials inscribed on the focal plane are at 1 cm spacings, with a larger cross marking the center of the photo. Therefore, the subtended angle between two adjacent fiducial marks is about 10 mm/60 mm = 1/6 radian or 9.55 degrees. (I do not know about the behavior of the lens on off-axis points, but taking or not taking the arcsin has a minimal effect of only about 1/2% so I ignored it).
I can't find any Apollo 11 EVA pictures with the sun actually in frame, so I looked for the next best - a picture showing both the level horizon and the camera's shadow. The angle between these two will also be the sun's elevation.
AS11-40-5882HR, taken by Buzz Aldrin as part of a panorama, is one such picture. Although the camera's shadow is hidden in Aldrin's own shadow, I can accurately estimate its position from the brightness of the Heiligenschein surrounding his head and from the orientations of small nearby rock shadows. (see below)
Using The Gimp, I measure 448 pixels vertically between the two fiducials to Aldrin's left. That works out to 9.55 deg/448 = .02132 degrees/pixel.
Also using The Gimp, I measure 795 pixels vertically between the camera's shadow and the horizon (which isn't quite straight because of a crater rim). That's an angle of 16.9 degrees, not terribly far from the known sun elevation angle of 14-15.4 degrees during the Apollo 11 EVA and certainly a lot closer than the 30 degrees claimed by profmunkin with his erroneous methods.
Besides possible errors in locating the camera's shadow, it's also possible (probable, even) that the horizon was not level. If the horizon to the west rose above level, as I believe it did, my result will be too large. Another possible error is the apparent drop in the horizon with the height of the camera, but even on the moon this seems too small to worry about given Aldrin's < 2m height.
(Regarding determining camera position from rock shadows: a line drawn through a shadow and the object casting it will necessarily pass through the camera's shadow, whether or not the latter is in frame. Do this for a whole bunch of object/shadow pairs and your lines will all cross at the camera's shadow. It is very important to ensure that each line joins a particular point on a shadow with the corresponding point on the object that cast it. You can't just draw lines through the centers of rocks and their shadows, especially when the shadows are short. Pick distinctive points that are visible both on the rock and in its shadow.)