calculating a point on circumference See the diagram

Known values are A: (-87.91, 41.98)
B: (-104.67, 39.85)
C: (-96.29, 40.92)
L: 14.63  // L is OC

Known angles
ADB: 60 deg
BAD: 60 deg
ADF: 10 deg

How to calculate Point F? that is 10 deg from point A.
A: Given that two of the angles in $\triangle ABO$ have measure $60°$, $\triangle ABO$ is equilateral.  It appears that $C$ is the midpoint of $\overline{AB}$, so $L$ is the length of an altitude of $\triangle ABO$ and the lengths of the sides of the triangle are $\frac{2}{\sqrt{3}}L\approx16.89$.  Now, $O$ is $16.89$ from both $A$ and $B$, which gives a system of equations that can be solved for the coordinates of $O$: $(-94.4456,26.4022)$ (as shown in your picture, so I'll use this one) or $(-98.1344,55.4278)$ (which would be above $\overline{AB}$).
Assuming that the arc shown is intended to be circular and centered at $O$, $F$ is the image of $A$ under a $10°$ rotation about $O$, which we can carry out by applying to $A$:


*

*a translation that takes $O$ to $(0,0)$, $(x,y)\to(x+98.1344,y-55.4278)$,

*a rotation of $10°$ about the origin, $(x,y)\to(x\cos10°-y\sin10°,y\cos10°+x\sin10°)$, and

*a translation that takes $(0,0)$ to $O$, $(x,y)\to(x-98.1344,y+55.4278)$.


Carrying out these transformations on $A$ gives $$F\approx(-90.7144, 42.8782).$$
A: if L is OE then u have the radius,
OA= L
AF = $\frac{\theta }{2\pi }2\pi r$ where r=L
u know A,O
so solve the 2 equations to get coordinates of F
A: C is exactly the center of AB, so COA = 30 deg. 
AOF is 10 deg (given), so EOF is 20 deg. 
Calculate R using tan(30)•L. Calculate d(F,EO) using R•sin(20). 
The entire shape is turned, so turn everything so that OE and the y-axis are parallel. 
