Ambiguous Case
Author: Dave Matthews
Date: March 20, 2007

Adjust the angle using the slider, then drag the side lengths to find out how many triangles are possible in the various cases.  Note that when sin(alpha)*BD/DE > 1, no triangle is possible.  When it equals exactly one, a single (right) triangle is possible.  When the quantity is less than one, AND DE < BD, two triangles are possible (the ambiguous case), while when DE > BD, only one triangle is possible (the other one is on the wrong side of angle alpha.

δ = 0.8011061266653973 rad (slider from 0 to 1.5707963267948966, step 0.0017453292519943296)
A = Point(xAxis)
B = Intersect(xAxis, yAxis)
C = Rotate(A, δ, B)
α = Angle(A, B, C)
a = Segment(B, C)
D = Point(a)
b = Distance(B, D)
T1: "" + b
E = (8.133333333333333, 1.8666666666666667)
c = Circle(D, E)
d = Segment(D, E)
e = Line(B, A)
F = Intersect(c, e, 1.0)
G = Intersect(c, e, 2.0)
f = Segment(D, F)
g = Segment(D, G)
P, f_1, b_1, d_1 = Polygon(B, D, F)
Q, g_1, b_2, d_2 = Polygon(B, D, G)
T2: "sin(" + α + ")*" + b + "/" + d + " = " + (b / d * sin(α))
β = Angle(D, F, B)
γ = Angle(D, G, B)
T3: "ArcSin(" + (sin(α) * b / d) + ") = " + (asin(sin(α) * b / d) * 180.0 / 3.141592653589793) + "°"
T4: "" + d
T5: "Angle α = " + α