<p>I’ll write how I got the 4th degree polynomial, which is mentioned in the comments. I used the same formulas from the <a href="https://astronomy.stackexchange.com/a/55661/48182">answer</a> to the previous formulation of the problem.
According to the conditions of the new problem, <span class="math-container">$OC = DC = R$</span>.</p>
<p>That is:</p>
<p><span class="math-container">$OC=\sqrt{\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2}=R$</span></p>
<p>Hence:</p>
<p><span class="math-container">$\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2=R^2$</span></p>
<p>We also know that:</p>
<p><span class="math-container">$\text{CE}=\sqrt{R^2-\text{DE}^2 \sin ^2(\text{CED})}+\text{DE} \cos
   (\text{CED})$</span></p>
<p>Let's substitute <span class="math-container">$CE$</span> into the formula for <span class="math-container">$OC$</span>, open all the brackets and get:</p>
<p><span class="math-container">$-2 \text{OE} \cos (\angle \text{OEC}) \sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}+2 \text{DE} \cos (\angle \text{CED})\sqrt{R^2-\text{DE}^2 \sin ^2(\angle \text{CED})}-\text{DE}^2\sin ^2(\angle \text{CED})+\text{DE}^2 \cos ^2(\angle \text{CED})-2\text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})+\text{OE}^2+R^2=R^2$</span></p>
<p>Let's make a replacement <span class="math-container">$\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}=S$</span> in the above formula:</p>
<p><span class="math-container">$-\text{DE}^2 \sin ^2(\angle \text{CED})+\text{DE}^2 \cos^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED})\cos (\angle \text{OEC})+2 \text{DE} \cdot S \cos(\angle \text{CED})+\text{OE}^2-2 \text{OE} \cdot S \cos(\angle \text{OEC})+R^2=R^2$</span></p>
<p>Let express <span class="math-container">$S$</span> from it, obtaining:</p>
<p><span class="math-container">$S = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$</span></p>
<p>Hence:</p>
<p><span class="math-container">$\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})} = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$</span></p>
<p>Let's square both sides, rewriting them in the following form:</p>
<p><span class="math-container">$R^2-\text{DE}^2 \sin^2(\angle \text{CED})=\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}$</span></p>
<p>Well, therefore:</p>
<p><span class="math-container">$R^2-\text{DE}^2 \sin ^2(\angle \text{CED})-\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}=0$</span></p>
<p>From the previous problem (and taking into account the new conditions) we know that:</p>
<p><span class="math-container">$\angle CED=\frac{3}{2}\tilde{\kappa}_0; \angle OEC=\pi-\tilde{\kappa}_0; DE=2e\cos(\frac{\tilde{\kappa}_0}{2}); OE=FE=FD=e$</span></p>
<p>Let’s substitute everything into the previous formula, open the brackets and take out the common factor:</p>
<p><span class="math-container">$\frac{\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}_0)+14 e^2 \cos (3 \tilde{\kappa}_0)+4 e^2 \cos (4\tilde{\kappa}_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}_0)-2 R^2 \cos (4 \tilde{\kappa}_0)-10 R^2}{2\cos (\tilde{\kappa}_0)+\cos (2 \tilde{\kappa}_0)}=0$</span></p>
<p>From the expression below we will find the polynomial we need:
<span class="math-container">$pol=\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}_0)+14 e^2 \cos (3 \tilde{\kappa}_0)+4 e^2 \cos (4\tilde{\kappa}_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}_0)-2 R^2 \cos (4 \tilde{\kappa}_0)-10 R^2$</span></p>
<p>Using <a href="https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Multiple-angle_and_half-angle_formulae" rel="nofollow noreferrer">formulas for multiple angles of trigonometric functions</a> and substitution <span class="math-container">$\sin(\tilde{\kappa}_0)=\sqrt{1-\cos(\tilde{\kappa}_0)^2}$</span>, we obtain the following expression:</p>
<p><span class="math-container">$16 \left(2 e^2-R^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4R^2\right) \cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos(\tilde{\kappa}_0)+(5 e^2-4 R^2)=0$</span></p>
<p>This is the required equation for the angle <span class="math-container">$\tilde{\kappa}_0$</span> at which <span class="math-container">$OC=DC=R$</span>.</p>
<p>For <span class="math-container">$e=3$</span> and <span class="math-container">$R=60$</span>, polynomial coefficients will be respectively: <span class="math-container">$16 \left(2 e^2-R^2\right)=-57312; 8 \left(7 e^2-4 R^2\right)=-114696;4 e^2=36; 16 \left(e^2-R^2\right)=-57456;(5e^2-4 R^2)=-14355$</span>.</p>
<p>Then, polynomial equation will look like this:<span class="math-container">$-57312X^4-114696X^3+36X^2+57456X-14355=0$</span>, where <span class="math-container">$X=cos(\tilde{\kappa}_0)$</span>. One of the solutions to this polynomial will be <span class="math-container">$X=cos(\tilde{\kappa}_0)=0.37887$</span>, and <span class="math-container">$acos(0.37887)=67.736^{\circ}$</span></p>
<p>Which coincides with the result of measuring the angle in a sketch constructed according to given geometric conditions in SolidWorks.</p>
<p><a href="https://i.sstatic.net/9qkUM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9qkUM.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-01-06T09:08:16.537Z
2024-01-06T09:13:41.530Z
2024-01-06T09:13:41.530Z
ayr
<p>Following <a href="https://astronomy.stackexchange.com/questions/55579/%CE%BA%E2%82%80-for-mercury-formula">my other question</a> about a specific “hidden” formula in Ptolemy’s model for Mercury, I am now looking for yet another “hidden” formula, this time the one used to find <span class="math-container">$\bar\kappa_0$</span> so that <span class="math-container">$OC=60$</span> in the figure below. We are given that <span class="math-container">$DC=60$</span> (in arbitrary units), that angles <span class="math-container">$AFD$</span> and <span class="math-container">$AEC$</span> are equal but of opposite direction (so <span class="math-container">$\angle AFD = -\angle AEC$</span>), and that <span class="math-container">$OE=EF=FD$</span> (which is equal to 3, but I’d like to keep 3 out of the equation and just symbolize it by <em>e</em> (for “eccentricity”).</p>
<p>(<span class="math-container">$A$</span> is the apogee of Mercury’s <em>deferent</em> centered on <span class="math-container">$D$</span>. <span class="math-container">$D$</span> rides on the small circle centered on <span class="math-container">$F$</span>. The Earth is at point <span class="math-container">$O$</span> [for “observer”]. Points <span class="math-container">$O$</span>, <span class="math-container">$E$</span>, and <span class="math-container">$F$</span> are always in a straight line with <span class="math-container">$A$</span>. Mercury itself rides on the <em>epicycle</em> centered on <span class="math-container">$C$</span>.)</p>
<p>While Ptolemy knew of absolutely no trigonometry and other modern mathematics, I don’t mind the answer to be expressed in modern terms, although simple trigonometry would be nice since math is not my forte.</p>
<p><a href="https://i.sstatic.net/4RkoB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4RkoB.png" alt="enter image description here" /></a></p>
<p>P.S. I’ll also ask this question on MathSE.</p>

CC BY-SA 4.0
2024-01-04T15:16:47.230Z
Please check whether the conditions $OC=60$ and $DC=60$ are formulated correctly. The fact is that if they are true, then the $\Delta OEC$ cannot be right-angled. By the way, in my calculation I simply substituted $OE = e$, and instead of $R = R_D$ (that is, I replaced numbers with symbols). And I got the following equation: $-6e^2\cos(\tilde{\kappa}_0)^3+(R_D^2-4e^2)\cos(\tilde{\kappa}_0)^2+2e^2\cos(\tilde{\kappa}_0)-e^2=0$
ayr
CC BY-SA 4.0
2024-01-04T18:04:26.903Z
@dtn: Yes, this is a different problem. Both $OC$ and $DC$ are here $60$ units and indeed, $\triangle OEC$ is not a right triangle. And awesome for the formula (which I guess applies to my first problem).
Pierre Paquette
CC BY-SA 4.0
2024-01-05T13:30:07.580Z
According to the conditions of this problem, I got this equation: $16 \left(2 R^2-e^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4 R^2\right)\cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos (\tilde{\kappa}_0)+(5e^2-4 R^2)=0$
ayr
CC BY-SA 4.0
2024-01-05T13:30:32.260Z
$\Delta OEC$ is not rectangular, but now $OC=DC$ :)) However, please note that the equation for such conditions will no longer have the 3rd, but the 4th degree.
ayr
CC BY-SA 4.0
2024-01-06T03:08:00.683Z
Thanks a bunch @dtn! Now I understand why Ptolemy had to solve it by trial and error!
Pierre Paquette
CC BY-SA 4.0
2024-01-06T04:37:23.830Z
Hmm with $=60$ and $=3$, I get $\cos \bar\kappa_0 ≈  −0.6483588$, which means $\bar\kappa_0 ≈ 130.42°$. The right value is $\bar\kappa_0 ≈ 67.736°$. The other possible values for $\bar\kappa_0$ aren’t any better. Am I doing something wrong?
Pierre Paquette
CC BY-SA 4.0
2024-01-06T05:50:41.603Z
For $e=3$ and $R=60$, polynomial coefficients will be respectively: $16 \left(2 R^2-e^2\right)=-57312; 8 \left(7 e^2-4 R^2\right)=-114696;4 e^2=36; 16 \left(e^2-R^2\right)=-57456;(5e^2-4 R^2)=-14355$.
ayr
CC BY-SA 4.0
2024-01-06T05:51:25.330Z
Then, polynomial equation will look like this:$-57312X^4-114696X^3+36X^2+57456X-14355=0$, where $X=cos(\tilde{\kappa}_0)$. One of the solutions to this polynomial will be $X=cos(\tilde{\kappa}_0)=0.37887$, and $acos(0.37887)=67.736^{\circ}$ https://www.wolframalpha.com/input?i=-57312x%5E4-114696x%5E3%2B36x%5E2%2B57456x-14355%3D0 Perhaps signs of coefficients were mixed up, check carefully
ayr
CC BY-SA 4.0
2024-01-06T05:56:15.620Z
How do you get $16(2R^2-e^2)=-57312$ ? $R^2=3600$ and $e^2=9$, so you get $16(7200-9)$ and that’s 113904…
Pierre Paquette
CC BY-SA 4.0
2024-01-06T05:57:34.163Z
Ha! It seems to be it should be $-16(R^2-2e^2)$, no?
Pierre Paquette
CC BY-SA 4.0
2024-01-06T05:58:09.260Z
Oh my...It seems I made a mistake. The first coefficient is written as $16(2e^2-R^2)$ I mixed up $e$ and $R$ :)) Yes, your version shows exactly that, you are right.
ayr
CC BY-SA 4.0
2024-01-06T06:00:07.890Z
No worries! *Errare humanum est*, so I know you’re not a bot or an alien! ;) THANK YOU again so much! :)
Pierre Paquette
CC BY-SA 4.0
2024-01-07T20:03:07.433Z
Last coefficient is likewise $-16(e^2-R^2)$ (sign difference).
Pierre Paquette
CC BY-SA 4.0
2024-01-09T05:39:46.450Z
I would like to know the following: are these problems that were published here and their formulation have practical application in astronomy (observations, calculations) or is it just a theoretical study of the Ptolemaic model in different specific “modes”? Tell me, please.
ayr
CC BY-SA 4.0
2024-01-09T23:27:09.923Z
@dtn: These problems are purely for my own curiosity. I haven’t found satisfaction in the explanations given my modern commentators of Ptolemy (e.g. Pedersen, Neugebauer, Toomer, or Van Brummelen) about this problem and the problem explained in my other question. I’m the kind of person who wants to *know* things, so that’s why I asked. Hopefully I didn’t make you feel like you were wasting your time on some futility.
Pierre Paquette
CC BY-SA 4.0
2024-01-11T08:47:39.833Z
Not at all, it was an interesting task, and I'm glad that it turned out to be useful for you. Also, someone else took a slightly different approach, but we got basically the same results. This makes me happy. And my question is an attempt to establish the relationship between theory and practice. Since you are clearly a much more experienced person in astronomy than me, when formulating your problems, you may have proceeded from some physical considerations. That's what I'm trying to find out.
ayr
<history><mercury><ptolemy>
κ₀ for Mercury—Formula (another definition)
Pierre Paquette
# κ₀ for Mercury—Formula (another definition)

(Asked by: username_0 on 2024-01-04)

Following [my other question](https://astronomy.stackexchange.com/questions/55579/%CE%BA%E2%82%80-for-mercury-formula) about a specific “hidden” formula in Ptolemy’s model for Mercury, I am now looking for yet another “hidden” formula, this time the one used to find $\bar\kappa\_0$ so that $OC=60$ in the figure below. We are given that $DC=60$ (in arbitrary units), that angles $AFD$ and $AEC$ are equal but of opposite direction (so $\angle AFD = -\angle AEC$), and that $OE=EF=FD$ (which is equal to 3, but I’d like to keep 3 out of the equation and just symbolize it by *e* (for “eccentricity”).

($A$ is the apogee of Mercury’s *deferent* centered on $D$. $D$ rides on the small circle centered on $F$. The Earth is at point $O$ [for “observer”]. Points $O$, $E$, and $F$ are always in a straight line with $A$. Mercury itself rides on the *epicycle* centered on $C$.)

While Ptolemy knew of absolutely no trigonometry and other modern mathematics, I don’t mind the answer to be expressed in modern terms, although simple trigonometry would be nice since math is not my forte.

[![enter image description here](https://i.sstatic.net/4RkoB.png)](https://i.sstatic.net/4RkoB.png)

P.S. I’ll also ask this question on MathSE.

**Tags:** history, mercury, ptolemy
> **Comments:**
> **username_1**: Please check whether the conditions $OC=60$ and $DC=60$ are formulated correctly. The fact is that if they are true, then the $\Delta OEC$ cannot be right-angled. By the way, in my calculation I simply substituted $OE = e$, and instead of $R = R_D$ (that is, I replaced numbers with symbols). And I got the following equation: $-6e^2\cos(\tilde{\kappa}_0)^3+(R_D^2-4e^2)\cos(\tilde{\kappa}_0)^2+2e^2\cos(\tilde{\kappa}_0)-e^2=0$  
> **username_0**: @dtn: Yes, this is a different problem. Both $OC$ and $DC$ are here $60$ units and indeed, $\triangle OEC$ is not a right triangle. And awesome for the formula (which I guess applies to my first problem).  
> **username_1**: According to the conditions of this problem, I got this equation: $16 \left(2 R^2-e^2\right) \cos ^4(\tilde{\kappa}_0)+8 \left(7 e^2-4 R^2\right)\cos ^3(\tilde{\kappa}_0)+4 e^2 \cos ^2(\tilde{\kappa}_0)-16 \left(e^2-R^2\right) \cos (\tilde{\kappa}_0)+(5e^2-4 R^2)=0$  
> **username_1**: $\Delta OEC$ is not rectangular, but now $OC=DC$ :)) However, please note that the equation for such conditions will no longer have the 3rd, but the 4th degree.  
> **username_0**: Thanks a bunch @dtn! Now I understand why Ptolemy had to solve it by trial and error!  
> **username_0**: Hmm with $=60$ and $=3$, I get $\cos \bar\kappa_0 ≈  −0.6483588$, which means $\bar\kappa_0 ≈ 130.42°$. The right value is $\bar\kappa_0 ≈ 67.736°$. The other possible values for $\bar\kappa_0$ aren’t any better. Am I doing something wrong?  
> **username_1**: For $e=3$ and $R=60$, polynomial coefficients will be respectively: $16 \left(2 R^2-e^2\right)=-57312; 8 \left(7 e^2-4 R^2\right)=-114696;4 e^2=36; 16 \left(e^2-R^2\right)=-57456;(5e^2-4 R^2)=-14355$.  
> **username_1**: Then, polynomial equation will look like this:$-57312X^4-114696X^3+36X^2+57456X-14355=0$, where $X=cos(\tilde{\kappa}_0)$. One of the solutions to this polynomial will be $X=cos(\tilde{\kappa}_0)=0.37887$, and $acos(0.37887)=67.736^{\circ}$ https://www.wolframalpha.com/input?i=-57312x%5E4-114696x%5E3%2B36x%5E2%2B57456x-14355%3D0 Perhaps signs of coefficients were mixed up, check carefully  
> **username_0**: How do you get $16(2R^2-e^2)=-57312$ ? $R^2=3600$ and $e^2=9$, so you get $16(7200-9)$ and that’s 113904…  
> **username_0**: Ha! It seems to be it should be $-16(R^2-2e^2)$, no?  
> **username_1**: Oh my...It seems I made a mistake. The first coefficient is written as $16(2e^2-R^2)$ I mixed up $e$ and $R$ :)) Yes, your version shows exactly that, you are right.  
> **username_0**: No worries! *Errare humanum est*, so I know you’re not a bot or an alien! ;) THANK YOU again so much! :)  
> **username_0**: Last coefficient is likewise $-16(e^2-R^2)$ (sign difference).  
> **username_1**: I would like to know the following: are these problems that were published here and their formulation have practical application in astronomy (observations, calculations) or is it just a theoretical study of the Ptolemaic model in different specific “modes”? Tell me, please.  
> **username_0**: @dtn: These problems are purely for my own curiosity. I haven’t found satisfaction in the explanations given my modern commentators of Ptolemy (e.g. Pedersen, Neugebauer, Toomer, or Van Brummelen) about this problem and the problem explained in my other question. I’m the kind of person who wants to *know* things, so that’s why I asked. Hopefully I didn’t make you feel like you were wasting your time on some futility.  
> **username_1**: Not at all, it was an interesting task, and I'm glad that it turned out to be useful for you. Also, someone else took a slightly different approach, but we got basically the same results. This makes me happy. And my question is an attempt to establish the relationship between theory and practice. Since you are clearly a much more experienced person in astronomy than me, when formulating your problems, you may have proceeded from some physical considerations. That's what I'm trying to find out.  

---

## 1 answers

### Answer by username_1 on 2024-01-06. Score: 3 | Accepted answer

I’ll write how I got the 4th degree polynomial, which is mentioned in the comments. I used the same formulas from the [answer](https://astronomy.stackexchange.com/a/55661/48182) to the previous formulation of the problem.
According to the conditions of the new problem, $OC = DC = R$.

That is:

$OC=\sqrt{\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2}=R$

Hence:

$\text{CE}^2-2 \text{CE} \cdot\text{OE} \cos(\angle\text{OEC})+\text{OE}^2=R^2$

We also know that:

$\text{CE}=\sqrt{R^2-\text{DE}^2 \sin ^2(\text{CED})}+\text{DE} \cos
(\text{CED})$

Let's substitute $CE$ into the formula for $OC$, open all the brackets and get:

$-2 \text{OE} \cos (\angle \text{OEC}) \sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}+2 \text{DE} \cos (\angle \text{CED})\sqrt{R^2-\text{DE}^2 \sin ^2(\angle \text{CED})}-\text{DE}^2\sin ^2(\angle \text{CED})+\text{DE}^2 \cos ^2(\angle \text{CED})-2\text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})+\text{OE}^2+R^2=R^2$

Let's make a replacement $\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})}=S$ in the above formula:

$-\text{DE}^2 \sin ^2(\angle \text{CED})+\text{DE}^2 \cos^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED})\cos (\angle \text{OEC})+2 \text{DE} \cdot S \cos(\angle \text{CED})+\text{OE}^2-2 \text{OE} \cdot S \cos(\angle \text{OEC})+R^2=R^2$

Let express $S$ from it, obtaining:

$S = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$

Hence:

$\sqrt{R^2-\text{DE}^2 \sin^2(\angle \text{CED})} = \frac{(\text{DE}-\text{OE})(\text{DE}+\text{OE})}{2 \text{DE} \cos (\angle \text{CED})-2\text{OE} \cos (\angle \text{OEC})}-\text{DE} \cos(\angle \text{CED})$

Let's square both sides, rewriting them in the following form:

$R^2-\text{DE}^2 \sin^2(\angle \text{CED})=\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}$

Well, therefore:

$R^2-\text{DE}^2 \sin ^2(\angle \text{CED})-\frac{\left(2 \text{DE}^2 \cos ^2(\angle \text{CED})-2 \text{DE} \cdot \text{OE} \cos (\angle \text{CED}) \cos(\angle \text{OEC})-\text{DE}^2+\text{OE}^2\right)^2}{4(\text{DE} \cos (\angle \text{CED})-\text{OE} \cos(\angle \text{OEC}))^2}=0$

From the previous problem (and taking into account the new conditions) we know that:

$\angle CED=\frac{3}{2}\tilde{\kappa}\_0; \angle OEC=\pi-\tilde{\kappa}\_0; DE=2e\cos(\frac{\tilde{\kappa}\_0}{2}); OE=FE=FD=e$

Let’s substitute everything into the previous formula, open the brackets and take out the common factor:

$\frac{\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}\_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}\_0)+14 e^2 \cos (3 \tilde{\kappa}\_0)+4 e^2 \cos (4\tilde{\kappa}\_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}\_0)-2 R^2 \cos (4 \tilde{\kappa}\_0)-10 R^2}{2\cos (\tilde{\kappa}\_0)+\cos (2 \tilde{\kappa}\_0)}=0$

From the expression below we will find the polynomial we need:
$pol=\left(26 e^2-8 R^2\right) \cos (\tilde{\kappa}\_0)+2 \left(9 e^2-4R^2\right) \cos (2 \tilde{\kappa}\_0)+14 e^2 \cos (3 \tilde{\kappa}\_0)+4 e^2 \cos (4\tilde{\kappa}\_0)+19 e^2-8 R^2 \cos (3 \tilde{\kappa}\_0)-2 R^2 \cos (4 \tilde{\kappa}\_0)-10 R^2$

Using [formulas for multiple angles of trigonometric functions](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Multiple-angle_and_half-angle_formulae) and substitution $\sin(\tilde{\kappa}\_0)=\sqrt{1-\cos(\tilde{\kappa}\_0)^2}$, we obtain the following expression:

$16 \left(2 e^2-R^2\right) \cos ^4(\tilde{\kappa}\_0)+8 \left(7 e^2-4R^2\right) \cos ^3(\tilde{\kappa}\_0)+4 e^2 \cos ^2(\tilde{\kappa}\_0)-16 \left(e^2-R^2\right) \cos(\tilde{\kappa}\_0)+(5 e^2-4 R^2)=0$

This is the required equation for the angle $\tilde{\kappa}\_0$ at which $OC=DC=R$.

For $e=3$ and $R=60$, polynomial coefficients will be respectively: $16 \left(2 e^2-R^2\right)=-57312; 8 \left(7 e^2-4 R^2\right)=-114696;4 e^2=36; 16 \left(e^2-R^2\right)=-57456;(5e^2-4 R^2)=-14355$.

Then, polynomial equation will look like this:$-57312X^4-114696X^3+36X^2+57456X-14355=0$, where $X=cos(\tilde{\kappa}\_0)$. One of the solutions to this polynomial will be $X=cos(\tilde{\kappa}\_0)=0.37887$, and $acos(0.37887)=67.736^{\circ}$

Which coincides with the result of measuring the angle in a sketch constructed according to given geometric conditions in SolidWorks.

[![enter image description here](https://i.sstatic.net/9qkUM.png)](https://i.sstatic.net/9qkUM.png)


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