# Angle of Declination and Culminations

Is it true that angle subtended by the Zenith Distance at Upper and Lower culminations is equal to the declination of the star ?If It's true then why? Is the motion of Earth make the solid angle of the cone subtended at the centre by circumpolar stars equal to the declination of the star?.

• What is the source of this task? (It's considered bad form to do somebody's homework form them) Dec 26, 2021 at 12:48
• @James K Do you think such a drawing could be homework question. It's my personal observation I don't even know this is correct or not . I just framed it like that ,apologies for that. Dec 26, 2021 at 13:07
• Yes, I do think the question is homework. Have you been measuring the zenithal distance of various stars (at lower and higher culimnation) and noticed a pattern? I think it far more likely that you have an assignment to "Prove that the angle subtended by the Zenith Distance at Upper and Lower culminations is equal to the declination of the star and justify the answer." which you have copied directly into the question box. Dec 26, 2021 at 13:19
• Now that's not necessarly bad. But it needs to be acknowledged. Dec 26, 2021 at 13:20
• Why should I acknowledge if that's not the case. First thing, Astronomy is not taught in my school so from where will I get the assignmemt to copy? As I said I don't even know that statement is correct or not. It may be wrong. I was solving an example from the book by AE ROY and I saw this pattern in it. Dec 26, 2021 at 13:30

The declination ($$\delta$$) of the celestial pole is 90°. Therefore the angle $$\alpha=90-\delta$$ and 2 times that value would be $$180-2\delta$$. Two Alpha is not the declination.
The zenith distance at upper culmination would be $$(90-\delta_z)-(90-\delta) = \delta-\delta_z$$ where $$\delta_z$$ is the declination of the zenith. At lower culmination the zenith distance would be $$(90-\delta_z)+(90-\delta) = 180 -\delta-\delta_z$$