[GeeTwo]

When I saw θ^b, my mind read "theta flat" .

In any case, I found the answer a different way:
Assuming that r=θ^b, we know that dr/dθ = bθ^b-1, so that r/(dr/dθ) = θ/b. I therefore calculated the slope of r using the numbers given. The intercept of the line with r=0 is therefore (a good approximation to) θ₀, and the slope is (approximately) 1/b. My spreadsheet gave a θ₀ of 2.007223652 radians, or 115.0054438 degrees, and a b=0.499973651. These did not work out exactly, but a bit of tweaking found that 115 degrees and 0.5 worked "exactly".

Using the mathematical polar coordinate system centered on 1.16, -0.3 and rotated 115 degrees counterclockwise, the equation is simply:

r = √θ

Going back to calculate the original series:

D = [139..199]
θ₁ = D * π / 180
θ = (D-115) * π / 180
r = √θ
x = cosθ₁ + 1.16
y = sinθ₁ - 0.3

An updated spreadsheet with a "double check" tab is attached. The maximum error in the reconstructed x and y is less than 10^-14.