# How to calculate observer's latitude from declination and culmination of a star?

From gcseastronomy.co.uk: A star culminates at 50°. It has a declination of +20°. What is the latitude from where it is observed?

What is the relationship between a culmination, declination, and latitude? How would one work this out?

This is the answer: 50 - 30 = 20° Latitude = 20°

• There is something wrong, either in the website's answer or the way you typed it. A star at +20 declination would culminate at the zenith (90 degrees) for an observer at +20 latitude. No wonder you are confused! – JohnHoltz May 8 '18 at 16:09

Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude. Neglecting atmospheric refraction, the altitude at culmination is $$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta|$$
• A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).
• A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.
For the example in the question, if we solve for $\phi$, $$\phi = \delta \pm (90^\circ - \mathsf{alt_{max}})$$
Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N. GCSE Astronomy's answer is incorrect.