Finding the Maximum Viewing Angle

I want to find the maximum viewing angle of a person looking at a picture that is four feet tall and his eye level is two feet below.
  To be able to do this I need to find a formula expressing the value of that angle given the surrounding information.  This information
includes the height of the picture (4.00 cm or ft), the distance from eye level to the base of the picture, and the distance from the
picture to the viewer, which is our variable.

The easiest way (I think) to find values of angles is with right triangles and trigonometric functions.  Specifically, I am going
to use tan(ø) = opposite/adjacent.  I can not use this formula right away because the angle I am looking for is not a right triangle,
but I can find the difference between two right triangles.  The first right triangle is the triangle whose height goes
from the viewer's eye level to the top of the picture.  This triangle is highlighted below.  The formula for this triangle is
 tan(α) = 6/x where x is the distance from A to D.  We solve this for our angle and get α = arctan (6/x)

 
The second right triangle is formed from the lower triangle where the height is the distance from eye level to the bottom
of the picture (see below).  The formula for this is tan (β) = 2/x or β = arctan (2/x).


 
Now we are ready to find the angle we are really looking for.  θ = α - β = arctan (6/x) - arctan (2/x).
This is the angle of the triangle you see below:

 

Now that we have the equation for finding our angle, I can now show the graph, which is below.
The x values of this graph represent the distance that the viewer is from the picture, the y values
respresent the viewing angle.  You can see that there is one point where the angle is more than
anywhere else.  That is the point we want to find.
NOTE: our graph has negative values.  Suppose this picture was suspended from the ceiling
rather than hanging from a wall.  Then someone could approach it from either side.  However,
I will be only focusing on positive x values.

y = arctan (6/x) - arctan (2/x)
 
 
Deriving My Solution:  I used derivatives to find my solution.  Go to this link to find out how I did that.

It turns out that the largest angle occurs when the viewer is standing 2√(3) cm (or ft) away from the picture, which is approximately 3.46 ft.
The angle at this point is 30 degrees.
You can see below where this occurs.  Unfortunately, I could not get the exact measurements, but they are close enough to see:
 
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