190 Mathematical Problems, by Mr. Adams. [Manrcu, 
Or by Problem 1, 
2sin. A. sin. B. cos. S H.cos, MR 
cos. mR. eos. s H 
Or, 
2sin. A. sin. B. cos. S H.cos.M R 
cos. S M = (l ar cos. m R. cos.s H. cos. C ) cos. C 
Then by Problem 2 (when the apparent distance is less than 
90°), we have 
Rene Be LIOR 5 aes Su «Palatal oss olwln erg ans piste 
SIN. YA... sod h fe ore alateieeolils Pee 
MAE ES. iad vases: « miarerais part SFuiee 
Gas. Stars true. alts eo.ng\tdialaraty 
cos. moon’s true alt. .......4. 
cos. star’s app. alt. (ar. com.) . 
cos. moon’s app. alt. (ar. coin.) 
pos, GC Par CO) 6s ie Beinpinides « 
cos. S M = cos. C — 
Sum of logarithms att rele 
Halfsum, corresponding to sin.D 
DOR GOAT: v's sive inl Roe see 
FE AOR A cian gad'ty Gees eerie te 
cos. true distance. .....¢.s00. 2 
And by Problem 3 (when the apparent distance is greater than 
90°), we have 
Heute 2 LORS Pe aS aan s ae a 
SALhy Daisy « SHR UNe ne kb aee hha 
BANAL sO REL in » aoa: we Re 
cos? star's true salty 2; /2aetes 
cos. moon’s true alt. ........ 
cos. star’s app. alt. (ar. com.).. 
cos moon’s app. alt. (ar. com.) . 
cose Ci(asiicome) was hci Be s1siald 
Hit ded dew 
I 
Sum of logarithms .......... 
Half ar. com. of sum corres- 
ponding to sin. D........ 
PLOT, GOR AN Ss sat dba ers a ayahe = 
BG Rr Pee ec ots cde eee 
sum of logs. as above..... aS 
cos. sup. of true distance. .... 
-_——-— 
vote 1.—The difference between the apparent and true dist- 
ance can never exceed one degree. (Emerson’s Miscellanies, 
age 206.) 
Note 2.—The error of one second in calculating the moon’s 
distance will produce an error of half a mile in longitude. 
