92 A Method for ascertaining Latitude and Time. 
dh sin (A—A’—A) +d’ sin A  —dhsin A’ + dh' sin A 4 
sin (A—A’) sil sin (A—A’) Uh 
This is M. Gauss’s formula, who computing the azi- 
muths outwardly on the triangle, writes (A’—A) which 
comes to the same. 
This formula requires that A and A’ be known. 
cos 0 sin u 
3 Or, 
cos 
Now, cos ¢:sinw::cosd:sinA = 
sin 0— sin h sin 6 
cos h cos @ 
to compute, but it leaves no doubt on the sort of the angle 
A which will be obtuse. If sin } Z sin Asin 9; we shall 
sin &’ — sin A’ sin @ 
cos A = 
. The second value is longer 
likewise have cos A = ———__— 
cos h! cos 
After having calculated A and A’, we shall have (A—A’), 
and we may determine dg without going through du. 
We have, lastly; sin «cot A = tang h cos ¢ — sin 6 cos 
da sin u dh cos ¢ 
u—Cu cosucotA——..— = ———— + dusinu sind 
SA cos? h iT: ; ? 
ms dh cos 3 uy } dd sin u 
du cos u cotang A — ——.—- — du sin usin 0 = ———— 
cos® h sin® A_ 
eee ducotu cosasinA dhcosdsin® aA dusin dsinr2a 
ee sin u cos*h sinu 1 a7 
dh cos 6 sind sin (5) 
coshcoshcosu ~" 
F cos ucos A—sin dé sin Asin uw dh sin A sin > 
= dusina(————___- | ate Spo ae i 
sin u cos A sin u 
"du sin aA ,. 3 *dasin A 
——— (sin A sin z sin  — cos A cos vu) — ——— 
sin u cos 2 
du cos h dh sin A cos h cos A 
— ———— (cos A) = ———— 1. (5*) = 
cos cos @ f cos 
at y—dh ) dhsin A oy dh eos A cos(A—A’)—dhcosA 
cos A sin(A—A’) A RA 
du sin A (cos u cos A — sin é sin A) — 
~ “cos (a) cos @ sin (A—A’) 
dh sin A dhcos A cos (A—A’) . dh cos A 
—"cos@ ~—~—«s Gos. @ sin (A— A’) cos @ sin(A—A’) ~~ 
dhsin A sin’ A— A’) cos Acos(A—A) +8, A sin (A—A)\—dl'c. A 
cos @ sin (A—A’) — ara ( cos @ sin (A—A’) 
Pe dh cos (A—A +A’) — dh’ cos A — dh cos A’+ dh! cos A (6) 
cos @ sin (A—A’) “cos @sin (A — A’) ays 5 
If we take the A’s outwardly on the triangle, the for- 
mula will be the same as M. Gauss’s, and will become 
+ dhcos A’— dh’ cos A 
cos ¢sin (A—A‘) , 
Thus my formulas 1, 3,5, can be brought back to 
M. Gauss’s 4 and 6, who has found them by other means« 
The 
= 
d= 
