Astronomical and Nautical Collections. 



Example I. 



March 1, 1824, in N. latitude, at P 20'" P.M., the true altitude 

 of the sun's centre was 47° 50' ; at 4'' 40°" the same afternoon his 

 altitude was 13° 40', the mean declination being 7° 26'. Required 

 the latitude. 



To find arc I. 

 1. cosin. D' - 9.996335 

 1. sin. i E =; 9.625948 



24° 46' 32" 9.622283 1. s. 

 I = 49 33 4 



To find arc II. 

 I. sin. E = 9.884254 

 1. cos. d ~ 9.996362 

 1. cosec. I r= 0.118625 

 93 23 9.999241 1. s. 

 11=46 41 30 



To find arc HI. 

 1. cosed z= 0.118625 

 1. secant A=: 0.173090 



l.cosin.i sTa = 9.752846 

 l.sin. i.S^ = 9.824320 

 19.868881 

 III. 59 18 13 9,934440 1.3. 

 II. 46 41 30 



IV. = 12 36 43 



To find arc V. 



I. cosin. A :=: 9.826910 



I. cosin. D = 9.996308 



2. 1. sin. IV " 8.678294 



constant log. 0.301030 



N t=: 63466 8.802542 



nat. cos.A + D= 569375 

 - N = 63466 

 Lat = 30° 23' 54" 505909 u.s. 



I £= 49 33 4 

 A = 47 50 

 S = 97~2 3~4 

 a = 13 40 



111 3 4 



83 43 4 



i-H + a 5^3 1 32 



i.S-a=41 51 32 



In this example the latitude and declination are of differen 

 names ; therefore no distinction of case is necessary by Rule I. 



