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.
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.
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 $A$:
Carrying out these transformations on $A$ gives $F\approx(-90.7144, 42.8782).$
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
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.