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°


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?


1 Answer 1


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.

  • $\begingroup$ Well, that's brilliant if you don't mind me saying so. I've been looking for this calculation for several years. $\endgroup$
    – McAllister
    Apr 6, 2023 at 5:23
  • $\begingroup$ 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. $\endgroup$
    – McAllister
    Apr 13, 2023 at 3:42
  • $\begingroup$ @user46620 It means $b=(\sin192.8583°) * (\sin23.439289°) \approx -0.08852$ $\endgroup$
    – notovny
    Apr 13, 2023 at 10:48
  • $\begingroup$ 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°. $\endgroup$
    – McAllister
    Apr 13, 2023 at 19:43

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