# Calculate declination of Galactic N. Pole given obliquity of Ecliptic, rt. ascension of Galactic N. Pole, & angle between Galactic & Ecliptic Pole

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?

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}
\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.
• @user46620 It means $b=(\sin192.8583°) * (\sin23.439289°) \approx -0.08852$ Apr 13, 2023 at 10:48