Finding Latitude and Longitude of a person or of where a picture was taken
EDIT: It turns out I answered this question earlier (in a slightly different format):

Cancelling out earth rotation speed, Altazimuth mount

Imagine you are at 35N,106W at 5h local sidereal time.

The celestial north pole is at (0,90) in the celestial frame and
(0,35) in your frame.

The point (5h,0) is (75,0) in the celestial sphere (1h = 15 degrees),
and (180,55) in your frame

The point (11h,0) is (165,0) in the celestial sphere and (90,0)
(setting in the east) in your frame.

Converting to rectangular coordinates, we need the matrix that does this:

{0,0,1} -&gt; {Cos[35 Degree], 0, Sin[35 Degree]}

{Cos[75*Degree],Sin[75*Degree],0} -&gt; {-Sin[35 Degree], 0, Cos[35 Degree]}

{Sin[-75*Degree],Cos[75*Degree],0} -&gt; {0,1,0}


or, generalizing a bit (lst = local sidereal time as an angle, lat =
latitude, notice that lon is irrelevant since we're using local
sidereal time)

{0,0,1} -&gt; {Cos[lat], 0, Sin[lat]}

{Cos[lst],Sin[lst],0} -&gt; {-Sin[lat], 0, Cos[lat]}

{-Sin[lst],Cos[lst],0} -&gt; {0,1,0}


Solving (Mathematica):

m0 = Table[a[i,j],{i,1,3},{j,1,3}]

(* Some simplifying assumptions, intentionally avoiding corner cases *)

conds = {-Pi/2 &lt; dec &lt; Pi/2, -Pi/2 &lt; lat &lt; Pi/2, 0 &lt; ra &lt; 2*Pi, -Pi &lt; lst &lt; Pi}

m = Simplify[m0 /. Solve[{
 m0.{0,0,1} == {Cos[lat], 0, Sin[lat]},
 m0.{Cos[lst],Sin[lst],0} == {-Sin[lat], 0, Cos[lat]},
 m0.{-Sin[lst],Cos[lst],0} == {0,1,0}
},Flatten[m0]],conds]

m = m[[1]]


This gives us

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) &amp; -\sin (\text{lat}) \sin (\text{lst})
  &amp; \cos (\text{lat}) \\
-\sin (\text{lst}) &amp; \cos (\text{lst}) &amp; 0 \\
\cos (\text{lat}) \cos (\text{lst}) &amp; \cos (\text{lat}) \sin (\text{lst}) &amp;
  \sin (\text{lat}) \\
   \end{array}
   \right)
$

The inverse of this matrix:

$
   \left(
   \begin{array}{ccc}
-\cos (\text{lst}) \sin (\text{lat}) &amp; -\sin (\text{lst}) &amp; \cos
  (\text{lat}) \cos (\text{lst}) \\
-\sin (\text{lat}) \sin (\text{lst}) &amp; \cos (\text{lst}) &amp; \cos (\text{lat})
  \sin (\text{lst}) \\
\cos (\text{lat}) &amp; 0 &amp; \sin (\text{lat}) \\
   \end{array}
   \right)
$

will transform rectangular azimuth and elevation back to right
ascension and declination and right ascension. This would be an
interesting alternative approach to this problem, but I won't be using
it.

OK, if we convert right ascension and declination to rectangular
coordinates, multiply by the first matrix above and convert the result
back to spherical coordinates, we will have azimuth and
elevation. Let's do that...

The spherical coordinates for (ra,dec) are:

{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]}


Multiplying by the matrix, we get

$
   \{\sin (\text{dec}) \cos (\text{lat})-\cos (\text{dec}) \sin (\text{lat})
\cos (\text{lst}-\text{ra}),-\cos (\text{dec}) \sin
(\text{lst}-\text{ra}),\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat})\}
$

(note that lst-ra is sometimes called the "hour angle", which would
simplify the above)

Converting back to spherical coordinates and simplifying:

{x,y,z} = Simplify[m.{Cos[dec] Cos[ra], Cos[dec] Sin[ra], Sin[dec]},conds]

(* NOTE: the use of single argument arctangent here introduces
ambiguity, and the final answer must compensate for this *)

az = Simplify[ArcTan[y/x],conds]

el = Simplify[ArcSin[z],conds]


yielding:

$
   \text{azimuth}=-\tan ^{-1}\left(\frac{\cos (\text{dec}) \sin
(\text{lst}-\text{ra})}{\sin (\text{dec}) \cos (\text{lat})-\cos
(\text{dec}) \sin (\text{lat}) \cos (\text{lst}-\text{ra})}\right)
$

$
   \text{elevation}=\sin ^{-1}(\cos (\text{dec}) \cos (\text{lat}) \cos
(\text{lst}-\text{ra})+\sin (\text{dec}) \sin (\text{lat}))
$

Assuming I haven't made any mistakes, the next steps would be:


Converting local time to local sidereal time
Inverting these equations to get latitude and longitude from
azimuth, elevation, and time.


I may or may not edit this answer to add those steps, but this
hopefully provides a start.