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Would it be correct to assume that Mintaka's declination was zero?

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> On Oct 27, 2014, at 1:47 PM, Frank Reed  wrote:
> 
> When you're at the equator, all stars rise vertically from the eastern 
horizon. At other latitudes, stars due east (necessarily with declination 
equal to zero) rise at an angle from the vertical equal to the latitude. 
While driving east through dark, rural Rhode Island last night, I was 
enjoying the view of Orion's belt rising in front of me. I could trace 
Mintaka back at an angle from the vertical of 41° (near enough to my 
latitude) and know that this was almost exactly due east as I drove along. 
That got me wondering about the rising angle for other declinations or 
azimuths... At home I decided to sit down and work it out. It turns out that 
the relationship is simple. If theta, θ, is the angle with respect to the 
vertical for a rising (or setting) star, then 
> tan θ = tan L / sin Z
> where L is latitude and Z is azimuth. For example, if the azimuth is exactly 
either NE or SE as a star rises at latitude 45°, then the rising angle is 
54.7° away from the vertical. This is symmetrical north and south of east, 
which surprised me a bit, though it makes sense in retrospect. Naturally when 
Z=90°, the rising angle is equal to the latitude. At all latitudes, for 
azimuths from 75° to 105° (within 15° of due east), the rising angle is less 
than one degree greater than the latitude.
> -FER
>  
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