Let #s# represent the side length of the square.
From the right side, we see that #s = sqrt(A)+bar(ED)#
From the area of the triangle, we have #A=(s*bar(ED))/2#. Solving for #bar(ED)# gives us #bar(ED) = (2A)/s#. Substituting this into the above leaves us with #s = sqrt(A)+(2A)/s#.
With our new equation in #s# and #A#, we can multiply both sides by #s# and gather the terms on one side to obtain the quadratic
#s^2-sqrt(A)s-2A = 0#
Applying the quadratic formula gives us
#s = (sqrt(A)+-3A)/2#
As we know #s > sqrt(A)# we can discard #(sqrt(A)-3A)/2#, leaving us with
#s = (sqrt(A)+3A)/2#
We can substitute this value for #s# into the equation obtained from the right side of the square to obtain
#(sqrt(A)+3A)/2 = sqrt(A)+bar(ED)#
#=> bar(ED) = (3A-sqrt(A))/2#
Now, as #triangleAED# is a right triangle, we have
#tan(theta)=bar(ED)/bar(AD)#
#=bar(ED)/s#
#=(3A-sqrt(A))/2-:(sqrt(A)+3A)/2#
#=(3A-sqrt(A))/(3A+sqrt(A))#
Thus, taking the inverse tangent function of both sides, we get the result
#theta = arctan((3A-sqrt(A))/(3A+sqrt(A)))#
