82 A Method for ascertaining Latitude and Time 
the algebraic signs. The two solutions are then applied 
to the same example, after which we are enabled to draw 
certain consequences. 
This is what we are going to do :—we shall reduce the 
trigonometrical solution to general tormulas: we shall then 
compare these formulas with M. Gauss’s, and shall caleu- 
late the example he has chosen. 
The difference of the two stars in right ascension, re- 
duced by an angle proportionate to the diurnal motion du- 
ring the interval of the observatious, gives first am-angle to 
the pole between two circles of declination passing through 
the points that were occupied by the stars at the moment 
of the observations: the two declinations and this angle 
being known, it it easy to resolve the triangle. 
Let d and % be the two declinations, and 4 the angle; the 
formula (¢ot 0 cos 9 = tang x)....(1) gives the first seg- 
ment of one side equal to (g0—8). 
The second segment will then be go -—3—x=90—(3+2). 
Suppose now that V is the angle to the figst star, you. 
: tang 6. sin. « 
5 T Lo Wade 
will have tang V oe “donb odor bhp: wih ae 
‘ = 
The are D, which joins the two points observed, would 
| sin & sin (0-42) 
—+_— + -——..... .(3): 
. cos © ae 
and the solution of this first triangle would require eleven 
different logarithms, but you may leave out one by making 
cot D = cos V tang (i-+ 2ys). .2. 38) ae ea 
be found by the formula cos D = 
Thus this first triangle requires ten or eleven logarithms, © 
aceording fo the choice you may make. If in these for- _ 
mulas ¢ 1s changed into @’, and & into 6, you will have the - 
angle V’ to the second star, D being always the same di- 
stance. : ; 
Then the three sides of the triangle between the zenith 
and the two points observed will be known. “Let W_ be the 
angle opposite to the second distance to the zenith, sin 2 W 
' , 5 
(eee an Oe 2) we) eeeee (4), h and K be- 
éosisnD  — 
ing the two observed altitudes. 
W: may be either positive or negative. The cireum- 
stances will almost always indicate the choice that is to be 
made, . 
By changing h into h’, and reciprocally, we would obtain - 
the angle W’ opposite to the first distance to the zenith. 
In all cases the solution may thus be obtained in two dif- 
ferent manners, which prove each other mutually, 
— 
—— 
This 
