Finding an angle in a kite I feel like this shouldn't be this hard, but nothing on the internet seems to have helped at this point. I'm using a leg stretcher (which has 3 bars, 2 holding my legs, and a center one to pull to my crotch) with a belt to train my splits and wanted to know the angle between my legs (and see how I've improved). All given units are in centimeters.
Using my legs and the stretcher, I get a kite:
$\overline{AB} = \overline{AD} = a = 76$
(Stretcher Bars)
$\overline{BC} = \overline{CD} = b = 80$
(Legs)
$\overline{AC} = f = 57 + s$
(Center bar + belt)
$s = \{35, 30.5, 27, 25\}$
(Belt lengths)
I'm trying to find the angle $\gamma$ (located at C, so between the 2 b sides) and the length of $\overline{BD}$
I'd prefer an explanation or formula as I want to use the formula again later ($s$ - the belt - will get smaller and smaller after all).
Any help is appreciated. (in the picture I used $\mu$ and $d$ for the desired angle and side ($\mu$ only since I don't have the greek letters on my keyboard, but I do have this)

 A: The figure is left right symmetric in $AC$
Cosine Law for triangle $ABC$ gives...
$$ 
\cos( \frac \mu 2)= \frac{b^2+(s+f)^2-a^2}{ 2b(s+f)}
$$
Let $O$ be the point in the middle.
Then $COB$ is a right angled triangle with hypotenuse $b$ for which...
$$
\sin( \frac \mu 2)= \frac d{2b}
$$
A: Since $AB = AD$ and $CB = CD$ there is a reflection symmetry of the kite with respect to the vertical diagonal $AC$ and so the two triangles $\Delta \, ACD$ and $\Delta\, ACB$ are congruent and the diagonal $AC$ is perpendicular to the diagonal $BD$. $ $ Also $\angle\, DCA = \angle\,BCA = \frac{1}{2}\,\angle\, DCB = \frac{\mu}{2}$. If $Q = AC \, \cap \, BD$ then $DQ$ is the height of the triangle $\Delta\, ACD$.
In the right-angled triangle $\Delta \, ACQ$ you have $$ \sin\left(\frac{\mu}{2}\right) \, = \,\frac{DQ}{CD}$$
Since you have $AD = a, \, CD = b$ and $AC = f$, then you can use Heron's formula to calculate the area of $\Delta \, ACD$
$$\text{Area}(\Delta \, ACD) \, =\, \frac{1}{4} \sqrt{\,(a + b + f)(a + b - f)(a - b + f)(-a + b + f)\,}$$
and write the area equation $$\text{Area}(\Delta \, ACD) \, = \, \frac{1}{2} \, AC \cdot DQ \, = \, \frac{1}{2} \,f \,DQ $$ so you get to calculate the height segment $DQ$ as follows
$$DQ \, =\, \frac{2 \, \text{Area}(\Delta \, ACD)}{f} \, =\, \frac{\sqrt{\,(a + b + f)(a + b - f)(a - b + f)(-a + b + f)\,}}{2\,f}$$
From here, you can calculate the angle $\mu$ as follows
$$\sin\left(\frac{\mu}{2}\right) \, = \,\frac{DQ}{CD} \, =\, \frac{DQ}{b} \, = \, \frac{\sqrt{\,(a + b + f)(a + b - f)(a - b + f)(-a + b + f)\,}}{2\,b\,f}$$
Finally,
$$\mu \, =\, 2\,\arcsin\left(\, \frac{\sqrt{\,(a + b + f)(a + b - f)(a - b + f)(-a + b + f)\,}}{2\,b\,f}\,\right)$$
$$BD \, =\, \frac{\sqrt{\,(a + b + f)(a + b - f)(a - b + f)(-a + b + f)\,}}{f}$$
I wrote this python script that calculates what you want
import numpy as np

def Heron(a,b,c):
    return np.sqrt((a+b+c)*(a+b-c)*(a-b+c)*(-a+b+c))/4

def angle_and_length(a,b,f):
    Area = Heron(a,b,f)
    BD = 2 * Area / f
    sin_half = BD / b
    return 2*np.arcsin(sin_half)*180/np.pi, 2*BD

a = 76
b = 80
s = np.array([35, 30.5,27,25])
f = 57+s

print('angles (in degrees): ')
print(angle_and_length(a,b,f)[0])
print('')
print('diagonals: ')
print(angle_and_length(a,b,f)[1])


And the output is
angles (in degrees): 
[103.7482341  107.48056304 110.30019084 111.87997057]

diagonals: 
[125.86499203 129.01508721 131.30428906 132.55222221]

