84 4 Method for ascertaining Latitude and Time 
He only applies two triangles, while we have made use of 
three, which have an intimate connexion and necessary re- 
lations. M Gauss has thus proposed a much harder pro- 
blem ; but his analysis has set aside all difficulties in a very 
happy manner. ss 
From the equation (1) it is easy to obtain cos? h = 1 — 
sth = 1 — (sin dsin ¢ + cos cos ¢ cos A)? = 1 — sin? 
sin? ¢ — cos? cos? ¢ cos? A — 2 sin? d cos é sin ¢ cos ¢ cos 
A= 1 — 5704 57:9 cos? ¢ — cos? 8 cos? ¢ cos? A-— 2 sin 2 
Cos 6 sin ¢ cos ¢ cos A = cos? 8+ 5? ¢ cos* o — cos*d cus % 
++ cos é cos? ¢ sin? A — 2sin ? cos d sing cos¢ cosA = 
(cos o s A)? + (cos? sing — sind cos @ cos A)*; and 
cos P sin *) 3 
therefore dividing the whole by cos * hi=( oT 
glia san gin $5 
(ee d sin ¢ — s 8cos 6-cos x? 
Jeaving out sin? é, and substituting 1—cos? d; and even 
there is no very conclusive reason why the value of cos* & 
shouldbe sought for by means of that of sin A, given dir 
rectly by the problem; and in this generally consists the 
- inconveniency of analytical demonstrations. 
: 25).) cos @ sin Ay . : 
Tt is easy to perceive that ee eo h ) is the sine of the 
angle to. the star in the triangle ZPA to the pole, the zenith 
and the first star. Thus sine ZA: sin ZP :-: sin P: sin 
ge sin PZ sinP cos ¢ sin A 
asin ZA oS cosh * ™ 
The equation becomes, therefore, 1 = sin? A + 
cos dsin@—sindcosg@cosa\? t ‘ 
(a gt ) = sin? A + cos? A; from which 
at cos dsin 6—sind cos 6 cosa 
it is easy to conclude that cos A= (ae i aa 
‘ cos h 
(A) will then be in the first triangle the angle to the star. 
M.Gauss called it w,without any other explanation, and makes 
3 cos @ sina . i 
sin ws ane wre which is the formula (3); cos w= 
cos 6 sin @—sin cos 8 cos A mee 9 
ou), which is the formula (4.) 
cos h 
: sin u 
It might be concluded that tang vu = = 
COs u 
cos tang ¢—sin cos A 
ark » but as we have already found the 
value 
