Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.
Neglecting atmospheric refraction, the altitude at culmination is
$$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta| $$

Here are some special cases:

• 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.