<p>With the substitutions:
<span class="math-container">$$
\begin{align}
a &amp;= \cos{23.43928°} \approx 0.9174821
\\b &amp;= \sin{192.8583°}\sin{23.439289°} \approx -0.0885216
\\\theta &amp;= 60.2°
\end{align}
$$</span></p>
<p>I believe the solution works out this way:</p>
<p><span class="math-container">\begin{align}
\arccos(a \sin \delta - b \cos\delta) &amp;= \theta
\\ a \sin \delta - b \cos\delta &amp;= \cos \theta
\\\sqrt{a^2 + b^2} \sin(\delta - \arctan(b/a)) &amp;= \cos\theta
\\\sin(\delta - \arctan(b/a)) &amp;= \frac{\cos\theta}{\sqrt{a^2 + b^2}}
\\\delta&amp; =\arcsin\left(\frac{\cos\theta}{\sqrt{a^2 + b^2}}\right) + \arctan(b/a)
\end{align}</span></p>
<p>Plugging in the substituted values gives <span class="math-container">$\delta = 27.1°$</span> ,which is the declination of the North Galactic Pole to three significant figures, according to Wolfram Alpha.</p>

CC BY-SA 4.0
2023-04-06T05:23:22.390Z
Well, that's brilliant if you don't mind me saying so. I've been looking for this calculation for several years.
McAllister
CC BY-SA 4.0
2023-04-13T03:42:09.410Z
Does b = sin 192.8583° sin 23.439289° mean
(sin 192.8583°*sin 23.439289°)?
That's where I'm getting messed up trying to duplicate your calculations.
McAllister
CC BY-SA 4.0
2023-04-13T10:48:09.937Z
@user46620 It means $b=(\sin192.8583°) * (\sin23.439289°) \approx -0.08852$
notovny
CC BY-SA 4.0
2023-04-13T19:43:18.617Z
Thank you. I don't know how to use MathML, so I had to convert your equation to an Excel-friendly format. I finally got δ = asin(cosθ/(sqrt(a^2+b^2)))+((atan(b/a)))
and plugging in the values of a, b and θ, I got δ = 27.1°. Using a slightly more accurate value of θ (60.19°), I got δ = 27.13°. The declination of the North Galactic Pole in Stellarium is 27°07'41.7" which converts to 27.13°.
McAllister
CC BY-SA 4.0
2023-04-05T02:09:08.190Z
2023-04-13T13:07:20.633Z
2023-04-13T13:07:20.633Z
notovny
<p>I’m trying to back-calculate a value in astronomy using spherical trigonometry. (This is not a homework problem, I'm retired).</p>
<p>Using the following formula, where δ = the Declination of the Galactic North Pole in degrees (i.e., the angle between the North Celestial Pole and the Galactic Equator), and</p>
<p>Given that</p>
<p>acos(sin(δ)cos(23.43928°)−sin(192.8583°)cos(δ)sin(23.439289°)) = ~60.2°</p>
<p>where</p>
<p>23.43928°= obliquity of the Ecliptic (tilt of Earth's axis relative to Ecliptic North Pole)</p>
<p>192.8583°= right ascension of Galactic North Pole in degrees</p>
<p>60.18894°= angle between the Galactic and Ecliptic North Poles (or the angle between the Galactic and Ecliptic Planes)</p>
<p>δ= declination of Galactic North Pole using Equatorial Coordinates</p>
<p>Can you solve for δ?</p>

<coordinate><positional-astronomy><celestial-mechanics><ecliptic>
Calculate declination of Galactic N. Pole given obliquity of Ecliptic, rt. ascension of Galactic N. Pole, & angle between Galactic & Ecliptic Pole
McAllister
# Calculate declination of Galactic N. Pole given obliquity of Ecliptic, rt. ascension of Galactic N. Pole, & angle between Galactic & Ecliptic Pole

(Asked by: username_0 on 2023-04-04)

I’m trying to back-calculate a value in astronomy using spherical trigonometry. (This is not a homework problem, I'm retired).

Using the following formula, where δ = the Declination of the Galactic North Pole in degrees (i.e., the angle between the North Celestial Pole and the Galactic Equator), and

Given that

acos(sin(δ)cos(23.43928°)−sin(192.8583°)cos(δ)sin(23.439289°)) = ~60.2°

where

23.43928°= obliquity of the Ecliptic (tilt of Earth's axis relative to Ecliptic North Pole)

192.8583°= right ascension of Galactic North Pole in degrees

60.18894°= angle between the Galactic and Ecliptic North Poles (or the angle between the Galactic and Ecliptic Planes)

δ= declination of Galactic North Pole using Equatorial Coordinates

Can you solve for δ?

**Tags:** coordinate, positional-astronomy, celestial-mechanics, ecliptic

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## 1 answers

### Answer by username_1 on 2023-04-05. Score: 2 | Accepted answer

With the substitutions:
$$
\begin{align}
a &= \cos{23.43928°} \approx 0.9174821
\\b &= \sin{192.8583°}\sin{23.439289°} \approx -0.0885216
\\\theta &= 60.2°
\end{align}
$$

I believe the solution works out this way:

\begin{align}
\arccos(a \sin \delta - b \cos\delta) &= \theta
\\ a \sin \delta - b \cos\delta &= \cos \theta
\\\sqrt{a^2 + b^2} \sin(\delta - \arctan(b/a)) &= \cos\theta
\\\sin(\delta - \arctan(b/a)) &= \frac{\cos\theta}{\sqrt{a^2 + b^2}}
\\\delta& =\arcsin\left(\frac{\cos\theta}{\sqrt{a^2 + b^2}}\right) + \arctan(b/a)
\end{align}

Plugging in the substituted values gives $\delta = 27.1°$ ,which is the declination of the North Galactic Pole to three significant figures, according to Wolfram Alpha.

> **Comments:**
> **username_0**: Well, that's brilliant if you don't mind me saying so. I've been looking for this calculation for several years.  
> **username_0**: Does b = sin 192.8583° sin 23.439289° mean (sin 192.8583°*sin 23.439289°)? That's where I'm getting messed up trying to duplicate your calculations.  
> **username_1**: @user46620 It means $b=(\sin192.8583°) * (\sin23.439289°) \approx -0.08852$  
> **username_0**: Thank you. I don't know how to use MathML, so I had to convert your equation to an Excel-friendly format. I finally got δ = asin(cosθ/(sqrt(a^2+b^2)))+((atan(b/a))) and plugging in the values of a, b and θ, I got δ = 27.1°. Using a slightly more accurate value of θ (60.19°), I got δ = 27.13°. The declination of the North Galactic Pole in Stellarium is 27°07'41.7" which converts to 27.13°.  

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