Dr. Robertson’s demonstrations > &c. 
1 39 
hope, will view them only as accidental oversights, and the 
most sincere regard for his memory will allow the propriety 
of correcting them. 
Problem I. 
“ The right ascension and declination of a celestial object, 
together with the obliquity of the ecliptic, being given, to find 
its longitude and latitude.” 
Let QAR, Fig. i, 2, 3, 4, 5, 6 , (Pi. VI.) be the equator, P its 
north, and its south pole. Let CAL be the ecliptic, E its 
north and e its south pole. In the first three figures, let P/>R 
be the first and P pQ the fourth quadrant of right ascension; 
Eeh the first and EeC the fourth quadrant of longitude, A in 
these figures being the first point of aries. In the last three 
figures, let PpR be the second and PpQ the third quadrant of 
right ascension ; E«?L the second and Ee C the third quadrant of 
longitude, A in these figures being the first point of libra. 
Let S be a celestial object, and let PSH or y>SH be a circle 
of declination, and ESF or eSF a circle of latitude passing 
through it, the angle LAR or QAC being the obliquity of 
the ecliptic. Then, reckoning from the first point of aries 
and according to the order of the signs, AH is the right 
ascension, SH the declination, AF the longitude, and SF the 
latitude of S. 
In each of the figures, let it be supposed that the arc of a 
great circle passes from A to S, and then SAH, SAF will be 
two right angled triangles. 
By trigonometry, sin. AH : R :: tan. HS : tan; HAS= 
R tan. HS _ 
sin. AH 
R tan. declination 
sin. " 
, north or south as the 
declination is. 
Let this first auxiliary angle be called A, and let O denote the 
