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)))
