Prove Proposition III-46: Under the same assumptions as in III-45. angle ACF – angle DCG and angle CDF – angle BDG (Fig.4.34).

Solution
2) Calculate crd( ), crd( ), crd( ), and crd(


) using Hipparchus’s formula for crd(
).
Solution:
( ) (


)
Hipparchus used R = 3438 unit as the radius therefore
( ) ( ) (


)
( ) ( ) (


)
( ) ( ) (


)
(


) (

) (





)
14) Suppose that the maximum length of day at a particular location is known to be 15 hours. Calculate
the latitude of that location and the position of the sun at sunrise on the summer and winter solstices.
Solution:
Ideally the length of day should be 12 Hours. Therefore we have 3 hours of extra deviation from ideal
value. Suppose that deviation in sun rise and sun set is same then we are basically 1.5 hours deviated
from the actual value.
Using Napier’s rule we have
( ) ( ) ( )
( )
( )
( )

( )
( )
( )

We also know that 1 Hour = 15°
Therefore
Also
( )
( )
( )

( )
Now to calculate position of sun, we have following Napier’s formula
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( )
( )
One needs to solve a spherical trinagle with 'East' as a point (Az=90°). At this point the angle is the
latitude of the place . The side adjacent is the Sun's deviation or displacement on the horizon at sunrise
(away form East ); 'sunrise point'. The other adjaent side is Sun's 'declination' on the celestial meridian
(hour circle or RA; I call it 6AM meridian) passing through East. Max. declination is 23°.45.
The opposite side is the angle of Sun's traverse from the horizon (sunrise point) to the 6AM merdian. It
implies that the sunrise is before 6AM (assuming...