<p>Assuming the observer is stationary, in 532 minutes (9h2m-10m) the star's GHA will have changed by 133 degrees. The star's GP will have changed by 11.8 degrees. A spherical triangle with vertices at the pole and the two GPs of the star is solvable because you know it's isosceles. Leaving its calculation as an exercise for you, the answer is 83 34.7 North or South.</p>
<p>At this point (2 days later) OP must have heard a complete answer from whomever asked. Analysis in last paragraph is from spherical law of cosines.
<span class="math-container">$\cos(a) = \cos(b)\cos(c) + \sin(b)\sin(c)\cos(A)$</span>. Here, upper case is for angles, and lower case is for sides opposite angles as arcangles. For this question the only known angle is the angle at the pole, (<span class="math-container">$A$</span>), and it's 133 degrees. The only known side, (<span class="math-container">$a$</span>) is the one opposite: 11.8 degrees. The star's declination and the observer's latitude are the same, since the star passed through (presumably exact) zenith. Neither OP's latitude or star's will change. So law of cosines for this case becomes <span class="math-container">$\cos(a) = \cos^2(b) + \sin^2(b)\cos(A)$</span>. Remember <span class="math-container">$\sin^2(x) + \cos^2(x) = 1$</span>. If there's any interest, I can show the steps to the answer.</p>
<p>Triangle is small enough that solving as if it were planar will give good approx results.</p>

CC BY-SA 4.0
2024-03-29T11:59:58.340Z
"a" is the side opposite the pole. It's the arc distance between the Sun's GP at 0H10M and its GP 532 minutes later.
stretch
CC BY-SA 4.0
2024-03-29T22:02:38.967Z
Oops! Changed it. Didn't make the same mistake in the calculations - still the same numerical answer.
stretch
CC BY-SA 4.0
2024-03-27T13:38:21.453Z
2024-03-29T22:00:02.507Z
2024-03-29T22:00:02.507Z
stretch
<p>A star passes through the zenith at 0h10m of sidereal time, whereas its altitude
on the horizon is 78◦12′ at 9h2m of sidereal time. Calculate the latitude of the observer.</p>

CC BY-SA 4.0
2024-03-27T00:32:10.260Z
What's your question? Where are you stuck? Do you know the PZX triangle? How is the declination of this star related to the observer's latitude?
PM 2Ring
CC BY-SA 4.0
2024-03-27T14:37:24.930Z
Hello.  You have posted your *teacher's* question.  What is *your* question?  We are not here to do your homework (or your test) for you.
James K
<spherical-astronomy>
Celestial Coordinates / Spherical Trigonometry problem
Huy Tran
# Celestial Coordinates / Spherical Trigonometry problem

(Asked by: username_0 on 2024-03-26)

A star passes through the zenith at 0h10m of sidereal time, whereas its altitude
on the horizon is 78◦12′ at 9h2m of sidereal time. Calculate the latitude of the observer.

**Tags:** spherical-astronomy
> **Comments:**
> **username_1**: What's your question? Where are you stuck? Do you know the PZX triangle? How is the declination of this star related to the observer's latitude?  
> **username_2**: Hello.  You have posted your *teacher's* question.  What is *your* question?  We are not here to do your homework (or your test) for you.  

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## 1 answers

### Answer by username_3 on 2024-03-27. Score: 1

Assuming the observer is stationary, in 532 minutes (9h2m-10m) the star's GHA will have changed by 133 degrees. The star's GP will have changed by 11.8 degrees. A spherical triangle with vertices at the pole and the two GPs of the star is solvable because you know it's isosceles. Leaving its calculation as an exercise for you, the answer is 83 34.7 North or South.

At this point (2 days later) OP must have heard a complete answer from whomever asked. Analysis in last paragraph is from spherical law of cosines.
$\cos(a) = \cos(b)\cos(c) + \sin(b)\sin(c)\cos(A)$. Here, upper case is for angles, and lower case is for sides opposite angles as arcangles. For this question the only known angle is the angle at the pole, ($A$), and it's 133 degrees. The only known side, ($a$) is the one opposite: 11.8 degrees. The star's declination and the observer's latitude are the same, since the star passed through (presumably exact) zenith. Neither OP's latitude or star's will change. So law of cosines for this case becomes $\cos(a) = \cos^2(b) + \sin^2(b)\cos(A)$. Remember $\sin^2(x) + \cos^2(x) = 1$. If there's any interest, I can show the steps to the answer.

Triangle is small enough that solving as if it were planar will give good approx results.

> **Comments:**
> **username_3**: "a" is the side opposite the pole. It's the arc distance between the Sun's GP at 0H10M and its GP 532 minutes later.  
> **username_3**: Oops! Changed it. Didn't make the same mistake in the calculations - still the same numerical answer.  

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