Now it’s time to determine the small angles that get applied as corrections. As with before, we’ll start with the first opposition, again using the same diagram.
As with before, I’ll be using modern math to speed things up, and it’s all being done in a Google Sheet.
We’ll start off by noting that $arc \; XE$ we determined to be $42;40º$ in the last post (where it was called $arc \; AL$). The central angle $\angle E \Theta X$ has the same measure, as does its vertical angle, $\angle D \Theta F$.
We’ll also quickly note that
$$\overline{D \Theta} = \overline{DN} = \frac{1}{2} \cdot \overline{\Theta N} = \frac{1}{2} \cdot 11;50^p = 5;55^p.$$
We can then directly determine $\overline{DF}$ using some trig:
$$\overline{DF} = 5;55^p \cdot sin(42;40º) = 4;01^p.$$
And we can do the same for $\overline{F \Theta}$:
$$\overline{DF} = 5;55^p \cdot cos(42;40º) = 4;21^p.$$
We now have two sides of $\triangle FAD$. We just determined $\overline{DF}$ and $\overline{DA} = 60^p$ as it’s a radius. Thus, we can use the Pythagorean theorem to solve for $\overline{FA}$:
$$\overline{FA} = \sqrt{60^2 – 4;21^2} = 59;52^p.$$
Next, $\overline{QF}$ is the same length as $\overline{F \Theta}$ which is $4;35^p$.
That can be added to $\overline{FA}$ to determine $\overline{QA} = 64;13^p$.
Similarly, $\overline{QN} = 2 \cdot \overline{FD} = 8;01^p$.
That gives us two sides in right triangle $\triangle NAQ$. Thus, we can again use a bit of trig to solve for $\angle NAQ$:
$$\angle NAQ = tan^{-1} (\frac{8;01^p}{64;13^p}) = 7;07º.$$
We’ll quickly note that $\overline{\Theta E} = 60^p$ since it’s a radius.
That allows us determine $\overline{QE}$ which is:
$$\overline{QE} = \overline{\Theta E} + \overline{QF} + \overline{F \Theta} = 60^p + 4;35^p + 4;35^p = 68;42^p.$$
This now gives us two sides in $\triangle NEQ$: $\overline{QN}$ and $\overline{QE}$. So a bit more trig allows us to determine $\angle NEQ$:
$$\angle NEQ = tan^{-1} (\frac{8;01^p}{68;42^p}) = 6;40º.$$
Lastly,
$$\angle ANE = \angle NAQ – \angle NEQ = 7;07º – 6;40º = 0;28º.$$
This is the same as $arc \; KS$ and is what we set about to find as it’s the correction for the first opposition.
Again, the value derived here agrees with the values found by Ptolemy and Toomer.