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Last week, one of my exercise tasks was about determining the an observer's position on Earth, when in their zenith is a certain star with known right ascension and declination. We were supposed to use this relation for the hour angle $\tau$ to determine the longitude $\lambda$: $\tau = \text{GST} - \lambda - \alpha$, which would transform to $\lambda = \text{GST}-\alpha-\tau$. Everything to here is clear to me, including calculating $\lambda$.

My problem is that in the case of our exercise, the calculated longitude is bigger than 180°. For simplicity, let's say $\lambda=230°\text{E}$. My naive thought process last week was, that this would mean that the longitude would simply continue going an additional 50° into the Western hemisphere, such one could say the true $\lambda$ on the Western side is 360° minus the calculated value. In my example: $\lambda=130° \text{W}$.
However, this week, I was told this is wrong and that we should still calculated 360° minus the calculated value, but instead on the Western hemisphere, we still would be on the Eastern one. Thus, $\lambda=130°\text{E}$. Can someone explain to me, why this is the case? I've asked our tutor, but they couldn't explain this, without confusing me (and my fellow students) even more.

EDIT: Exercise task
After an emergency landing, you find yourself on an unknown beach. Directly above you, you can locate the star $\Psi$ Leo. You know that for this star $\alpha=9^\text{h} 43^\text{m} 54^\text{s}$ and $\delta=14^\circ1'15"$. The time is $17^\text{h} 17^\text{m} 12^\text{s}$ UT. In addition, you know that for the current day at 0 UT, we have $8^\text{h} 19^\text{m} 50^\text{s}$ GMST. Where are you located? Determine the longitude and latitude, as well as the name of the location.

EDIT 2: Detailed Calculations
Relation between hour angle and right ascension is given by $$\tau = \theta_\text{G}(t=0) + t\left(\frac{366.24}{365.24}\right) - \lambda - \alpha.$$ Since the star is directly above us, it's in the zenith and thus, $\tau=0$. It follows for the longitude that $$\lambda = \theta_\text{G}(t=0) + t\left(\frac{366.24}{365.24}\right) - \alpha.$$ It is given from the exercise that $$\theta_\text{G}(t=0) = 8^\text{h} 19^\text{m} 50^\text{s} = 8.33056\text{h},$$ $$t = 17^\text{h} 17^\text{m} 12^\text{s} = 17.28667\text{h},$$ $$\alpha = 9^\text{h} 43^\text{m} 54^\text{s} = 9.73167\text{h}.$$ Plugging everything into the equation for $\lambda$ leads to $$\lambda \approx 15.93289\text{h}=238.99335^\circ E.$$ Until this point everything was marked as correct (and matches with the sample solutions of the lecturer).

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    $\begingroup$ It is fairly common to use longitudes greater than 180 to represent Western longitudes. But, as you expected, they are in the Western hemisphere. $\endgroup$ Commented Nov 1, 2023 at 5:21
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    $\begingroup$ Your idea sounds correct to me, 230 E and 130 W are the same place. But maybe there's a detail of the exercise itself that makes it different? Maybe if you provide the question as written it might be easier to see where the issue is. $\endgroup$ Commented Nov 1, 2023 at 13:58
  • $\begingroup$ Hi, thanks for answering me! I've added the exercise as written into my question. $\endgroup$ Commented Nov 3, 2023 at 21:22
  • $\begingroup$ Did you take into account that the GMST is given for 0 UT and not for 17h17m12s UT? $\endgroup$ Commented Nov 4, 2023 at 0:03
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    $\begingroup$ @PierrePaquette Yeah, I think, I did take this into account c: To make more obvious where my value > 180 comes into play, I added my calculations for the value of the longitude to the question. $\endgroup$ Commented Nov 4, 2023 at 9:06

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