log₁₀(A/B) + log₁₀(A/C)

log_cb = a    <=>    cª = b

Declinartion or R:.A:. ov a Planet

The Declination { or R:.A:. } ov a planet like Venus or Mercury cannot be obtianed by meridian observations, and the Dec/R:.A:. ov an outer planet like Mars or Jupiter can be determined by meridian observations only at periods when it transits during the hours ov darkness. So it is impossible to trace the positions ov any planet relative to the fixed stars in all parts ov its course unless W/we can determine their Declinations and Right Ascensions by some other method.

The local postition ov ever star at any instant can be placed at the corner ov a spherical triangle. ( Graphs and Charts to come soon ). One side is the arc between the celestial pole and the zenith along the prime meridian. The elevation ov the pole is the latitude (L) ov the observer. So b = 90° - L:.

cos ( 90° - declin:. ) = cos ( 90° - lat:. ) cos z:.d:. + sin ( 90° - lat:. ) sin z:.d:. cos azim:.

This may be rewritten...

sin ( declin:. ) = sin lat:. cos z:.d:. + cos lat:. sin z:.d:. cos azim:.

In applying this you have to remember the W/we have reckoned azimuth from the north point. If the star transits south, the azimuth may be greater than 90°, and W/we reckon it from the south point. Since...

c² = a² + b² + 2ab ( 180° - C )

when C > 90°, the formula for the spherical triangle becomes...

sin declin:. = sin lat:. cos z:.d:. - cos lat:. sin z:.d:. cos azim:.
   becomes   

This means that if Y/you know the azimuth and zenith distance ov a heavenly body at one the same time, Y/you can calculate its declination without waiting for it to reach the meridian.

How the Declination ov a planet changes is much less interesting then how its R:.A:. changes, because the way in which its R:.A:. changes is so much easier to explain if W/we assume that the earth and the planets revolved around the Sun, as Aristærchus belived and Copernicus taught.

For flat triangles the formula is

sin C = c [{sin A } / a ]
or...
sin C = sin c [{ sin B } / b ] iff : B

For spherical triangles the corresponding sine formulæ are...

sin C = [{ sin c · sin A } / sin a ]
or...
sin C = [{ sin c · sin B } / sin b]

Recalling the rule :

cos² x = 1 - sin² x

First, however, you should notice that W/we can get a when b, c and A are known from:

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

Also:

cos c = cos a · cos b + sin a · sin b · cos C

The demonstration ov the sine formula is as follows, Rearranging the first ov these, and then square both sides:

-cos A · sin b · sin c = cos b cos · c - cos a
cos² A · sin² b · sin² c = cos² b · cos² c -2 cos a · cos b · cos c + cos² a

Now make the substitution :

( 1 - sin² A ) sin² b · sin² c = ( 1 - sin² b)( 1 - sin² c ) - cos a · cos b · cos c + ( 1 - sin² a)
:.
sin² b sin² · c - sin² A · sin² c = 1 - sin² b - sin² c + sin² b · sin² c - 2 cos a · cos b · cos c + 1 - sin² a

After taking away sin² b · sin² c from both sides this becomes:

-sin² A · sin² b · sin² c = 2 - sin² a - sin² b - sin² c - 2 cos a · cos b · cos c

Hence:

-sin² C · sin² a · sin² b = - sin² A · sin² b · sin² c

Dividing by - sin² b...

sin C = ± [{sin A · sin c} / sin a]

With A as azimuth, c as zenith distance, and a the polar distance ( 90° - declin:. ) ov the star, i.e. sin a = cos declin:. . Hence :

sin hour angle = [{ sin azim:. · sin z:.d:. } / cos declin:. ]

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