by means of Two known Stars. 89 
cos V tang h _ sin A’ sin F- . 
tang (F—é) \ sinh sin 3 cos (F — 8) 
(13) cos W = 
bee 
h 
I make cos W = tang h'cos D fis anda 1) = tangh 
Ss 
sin h cos x 
Beery. tang (2 +t) (= Asin o'sin( +2) _ ): 
Our equations are thus again identical, notwithstanding. 
the apparent difference. I have sought tor the arch D se- 
parately, but it is clear I could have done without: my 
calculation is somewhat the shortest. 
If A =X’, as in the example chosen by M. Gauss, 
sin h’ 
‘sinh ! 
methods: but the trigonometrical method shows a more 
considerabie simplification. 
= 1, two logarithms are saved in each of the two 
My formula becomes cos W = tang h cot D(—5 = 1) 
__ tang kh cot D D tangh2sin? 5D __ tangh,2sin?2D 
cos D (1—~ cos D) = sin D ~ 9sinZDcosd D D 
= tang fh tani D: a formula which the inspection of the 
trianyle to the zenith, which in that case is isosceles, gives 
immediately. 
I draw the same simplification from M. Gauss’s for- 
mula; but it was not evident at first sight: his formula (s) 
tang / sin V 
then becomes cos W = —°,——.,, (1 — sin é sin &’ — cos ¢ 
sin § cos d 
‘tan A sin V 
cos & cos ———- — cos D 
cos 0) = sin § cos © (1 ). 
By making sin?@ sin & + cos 4 cos cos = cos D. Thus 
we tang h. sin V 2 sin? zD tang A 2 sin* 3D 
cos sin sin § cc cos FAP es ET ; “sind cos ies 
sinV 
tang 2 sin? iD 
~— 2 — = tang h tang 1 D, as heretofore. 
sin D 3 2 
If we continue the comparison : 
(14) cot G tang h 1 iy 
cat G= ———__ = ——=—" = cot 2 , 
) cos u cos u.cot h » Men my arc 
z= 90-—G. 
cos G tan u sin Z tang u 
(15) tanga = Sin(Ge8) eos (dteye our equations 
aré identical. 
(16) tang g = cos ~ cos (G — ¢) = cos A tang @ +2). 
Thus 
