*spherical trigonometry*. This is merely an extension of plane trig, but to the sort of angles (many > 90 degrees) one finds in astronomical applications, since distances, angles are of spherical measure (derived from spherical triangles.) A simple illustration of spherical geometry is shown in Fig. 1. In the diagram, the angle Θ denotes the longitude measured from some defined meridian on the sphere, while the angle φ denotes a zenith distance, or the measured angle from an object to the zenith.

Fig.
2 shows a spherical

*right triangle*from which a host of different angle relationships can be obtained, which can then be used to find astronomical measurements, etc.
Fig.3
shows a diagram of the celestial sphere, such as used in many practical
astronomy applications,
and some of the key angles with reference to a particular object (star)
referenced within a given coordinate system:

In some applications, the coordinate system may not need to be changed, but in
others it must - for example, when going from the coordinate system applied to
sky objects (Right Ascension, Declination) to the observer's own coordinates
(altitude, azimuth). In this way, coordinate transformations will also enter
and are straightforward to perform, for example via use of matrices.

We consider first a simple angle relation in Fig.
1, say to find the

Let's now examine Fig. 2 and see what spherical trig relationships we can infer.

Two of the key ones embody the law of sines and law of cosines for spherical triangles, which are the analogs of the law of sines and cosines in plane trig.

We have for

Sin A/ sin a = sin B/ sin b = sin C/ sin c

where A, B, C denote ANGLES and a,b,c denote measured arcs. (Note: we could also have written these by flipping the numerators and denominators).

We have for the law of cosines:

cos a = cos b cos c + sin b sin c cos A

**, a. Then if we have the basic geometrical relationship: a + φ = 90 degrees, then***altitude***a**= (90 - φ).Let's now examine Fig. 2 and see what spherical trig relationships we can infer.

Two of the key ones embody the law of sines and law of cosines for spherical triangles, which are the analogs of the law of sines and cosines in plane trig.

We have for

**:***the law of sines*Sin A/ sin a = sin B/ sin b = sin C/ sin c

where A, B, C denote ANGLES and a,b,c denote measured arcs. (Note: we could also have written these by flipping the numerators and denominators).

We have for the law of cosines:

cos a = cos b cos c + sin b sin c cos A

Where
a, b, c have the same meanings, and of course, we could write the same
relationship out for any included angle.

Now, we use Fig. 3, for a celestial sphere application, in which we use the spherical trig relations to obtain an astronomical measurement.

Using the angles shown in Fig. 3 each of the angles for the law of cosines (given above) can be found. They are as follows:

cos a = cos (90

where d = declination

cos b = cos (90

where 'Lat' denotes the latitude. (Recall from Fig. 1 if φ is polar distance (which can also be zenith distance) then φ = (90 - Lat))

cos c = cos z

where z here is the zenith distance.

sin b = sin (90 deg - Lat)

sin c = sin z

and finally,

cos A = cos A

Where A is the

Let's say we want to find the declination of the star if the observer's latitude is 45

cos a = cos (90

cos (90

cos (90

+ sin (45

We know, or can use tables or calculator to find:

cos 45

cos 30

sin 45

sin 30

cos 60

Then:

Now, we use Fig. 3, for a celestial sphere application, in which we use the spherical trig relations to obtain an astronomical measurement.

Using the angles shown in Fig. 3 each of the angles for the law of cosines (given above) can be found. They are as follows:

cos a = cos (90

^{o}- d)where d = declination

cos b = cos (90

^{ o}- Lat)where 'Lat' denotes the latitude. (Recall from Fig. 1 if φ is polar distance (which can also be zenith distance) then φ = (90 - Lat))

cos c = cos z

where z here is the zenith distance.

sin b = sin (90 deg - Lat)

sin c = sin z

and finally,

cos A = cos A

Where A is the

*azimuth.*

*Example Problem*:

Let's say we want to find the declination of the star if the observer's latitude is 45

^{ o}N, the azimuth of the star is measured to be 60

^{ o}, and its zenith distance z = 30

^{ o}. Then one would solve for cos a:

cos a = cos (90

^{ o}- d)=

cos (90

^{ o}- Lat) cos z + sin (90

^{ o}- Lat) sin z cos (A)

cos (90

^{ o}- d) = cos (90

^{ o}- 45

^{ o}) cos 30

^{ o}

+ sin (90

And:

cos (90

^{ o}- 45^{ o}) sin 30^{ o}cos 60^{ o}And:

cos (90

^{ o}- d) = cos (45^{ o}) cos 30^{ o}^{ o}) sin 30^{ o}cos 60^{ o}We know, or can use tables or calculator to find:

cos 45

^{ o}= Ö2 / 2cos 30

^{ o}= Ö3/ 2sin 45

^{ o}= Ö2/ 2sin 30

^{ o}= ½cos 60

^{ o}= ½Then:

cos
(90 - d)= {(Ö2/ 2 )( Ö3/ 2)} + {Ö2/ 2} Ö (½) }

cos (90 - d)= Ö6/ 4 + Ö2/ 8 = {2Ö6 + Ö2}/ 8

cos (90 - d) = 0.789

arc cos (90 - d)= 37.

Then:

d = 90

Or, in more technical terms:

d (star) = + 52.1 degrees

A more detailed image of the celestial sphere appears below with key aspects not found in the simpler version (Fig. 3):

^{ }cos (90 - d)= Ö6/ 4 + Ö2/ 8 = {2Ö6 + Ö2}/ 8

cos (90 - d) = 0.789

arc cos (90 - d)= 37.

^{o}9Then:

d = 90

^{o}- 37.^{ o}9 = 52.^{ o}1Or, in more technical terms:

d (star) = + 52.1 degrees

A more detailed image of the celestial sphere appears below with key aspects not found in the simpler version (Fig. 3):

In this detailed version we see the Earth's north pole is projected to the

*North Celestial Pole*, the equator is projected to the*celestial equator*, and all latitude lines are projected to become declination lines, while longitude lines become*Right Ascension*lines. Thus, just as every geographical location on Earth has a latitude and longitude so also every sky location has a declination and Right Ascension. The "vernal equinox" position, for example, is at 0 degrees Declination and 0 hours RA. (The vernal equinox marks the*first day of spring*.)
The oblique red circle projected onto the celestial sphere defines the

*ecliptic*or the projected (apparent) path of the Sun onto the celestial sphere through the year. If we follow the red circle - the ecliptic - UP from the vernal equinox we come to the northernmost point at +23.5 degrees declination and 6h RA. This coincides with the summer solstice - or when the Sun appears over that latitude on Earth. This marks the longest day of the year in the northern hemisphere
We are led then to consider how to compute star positions on this sphere (which had the designated coordinates of R.A. and declination) and also how to transform between coordinate systems, say between the horizon system and the celestial sphere (equatorial) system.

For example, the coordinate system depicted in the color graphic above - the

*equatorial system*- is based on the projections of the Earth's own equator and N. and S. poles onto the sky sphere. The poles then become the*North and South Celestial poles*, and the equator becomes the*celestial equator*. If these poles are defined respectively at +90 degrees (NCP) and -90 degrees (SCP) and the celestial equator at 0 degrees, then a system of celestial latitude can be constructed.
Once the vernal equinox position is fixed at 0 hours R.A. then the celestial longitude emerges and spans 24 hours across the same celestial sphere. (Refer again to color graphic for direction of celestial longitude circles.) These can be used in conjunction with celestial latitude (declination) to locate any celestial object. It is then possible to make computations translating one system's coordinates to those of others (horizon, ecliptic etc.)

Example: φ = 51.5 degrees N, for London . Now, for the
December (winter) Solstice the Sun is directly over the Tropic of Capricorn
(φ = 23.5 S) therefore we do know its
declination is - 23.

cos (A) = sin (-23.

which gives approximately, 130

40 degrees SOUTH of due East. (90

Now, on

cos (A) = sin (23.

And A = 50

^{o}5. We have then for the Sun's azimuth at sunrise in London on Dec. 21:cos (A) = sin (-23.

^{o}5)/ cos (51.^{o}5)which gives approximately, 130

^{o}.
Where
is this on our directional reference circle for azimuth? We know that 180
degrees is

*due South*so that this must be:40 degrees SOUTH of due East. (90

^{ o}+ 40^{ o}= 130^{ o})Now, on

*the longest day of the year*(say June 21), the Sun is over the Tropic of Cancer at 23.5 N latitude, so the Sun's declination is + 23.^{ o}5 . Then the azimuth for that date is:cos (A) = sin (23.

^{ o}5)/ cos (51.^{ o}5)And A = 50

^{o}
This puts the Sun's rising position
North of due E. or specifically 40 degrees

*North of due East*.__Problem__:

*The altitude of a star as it transits your meridian is found to be 45*.

^{o}along a vertical circle at azimuth 180^{o}, the south point. Find the**declination**of the star
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