Let ss represent the side length of the square.
From the right side, we see that s = sqrt(A)+bar(ED)s=√A+¯¯¯¯¯¯ED
From the area of the triangle, we have A=(s*bar(ED))/2A=s⋅¯¯¯¯¯¯ED2. Solving for bar(ED)¯¯¯¯¯¯ED gives us bar(ED) = (2A)/s¯¯¯¯¯¯ED=2As. Substituting this into the above leaves us with s = sqrt(A)+(2A)/ss=√A+2As.
With our new equation in ss and AA, we can multiply both sides by ss and gather the terms on one side to obtain the quadratic
s^2-sqrt(A)s-2A = 0s2−√As−2A=0
Applying the quadratic formula gives us
s = (sqrt(A)+-3A)/2s=√A±3A2
As we know s > sqrt(A)s>√A we can discard (sqrt(A)-3A)/2√A−3A2, leaving us with
s = (sqrt(A)+3A)/2s=√A+3A2
We can substitute this value for ss into the equation obtained from the right side of the square to obtain
(sqrt(A)+3A)/2 = sqrt(A)+bar(ED)√A+3A2=√A+¯¯¯¯¯¯ED
=> bar(ED) = (3A-sqrt(A))/2⇒¯¯¯¯¯¯ED=3A−√A2
Now, as triangleAED△AED is a right triangle, we have
tan(theta)=bar(ED)/bar(AD)tan(θ)=¯¯¯¯¯¯ED¯¯¯¯¯¯AD
=bar(ED)/s=¯¯¯¯¯¯EDs
=(3A-sqrt(A))/2-:(sqrt(A)+3A)/2=3A−√A2÷√A+3A2
=(3A-sqrt(A))/(3A+sqrt(A))=3A−√A3A+√A
Thus, taking the inverse tangent function of both sides, we get the result
theta = arctan((3A-sqrt(A))/(3A+sqrt(A)))θ=arctan(3A−√A3A+√A)
