I assumed a central angle within the circle A to create an additional isosceles triangle and find the base N, as well as find the total height.
Because this is a unit circle, the radius is 1.
With the central angle A, we create the isosceles triangle with base N and with base angles X and with height K.
Drawing a perpendicular bisector down from the vertex of A creates two right triangles; each has height K, base angle X, upper angle (for lack of a better term) a/2, and hypotenuse of 1 (because the radius is 1). With this, we can create some functions/equations based on the variables:
X = 90 - (A/2)
sin(x) = K
cos(x) = n/2
2cos(x) = n
The total height of the isosceles triangle in question (the one with the largest area) is given by K+1.
The base of the isosceles triangle with base angle X is given by N (as is the largest possible isosceles triangle).
The area of a triangle is given by ((base*height)/2).
Therefore, we can perform substitutions, and get that
(2cos(x)(sin(x) + 1))/2 = Area.
More simply, cos(x)(sin(x) + 1) = Area.
However, we established these relations based on an assumed central angle; therefore, we need to substitute.
cos(90 - (a/2))(sin(90 - (a/2)) + 1) = Area.
Proceed from there.