1819.] Mathematical Problems, by Mr. Adams. 191 
The following examples will show how very simple the opera- 
tions by these rules are. 
Example 1.—Given the apparent altitudes of the moon and 
sun 27° 2’ 30% and 59° 11’ 52”; the true altitudes 27° 53’ 25”, 
59° 11’ 22”, and the apparent distance 59° 25’ 34” ; to find the 
true distance. 
From the data 
A = 45° 47’ 28, B = 13° 38’ 6”, and C = 31° 17’ 57”. 
Then per Rule 1. 
uly BISA eo es Oe Cn + = 0°3010300 
ime. A ieraenyh ir seen = ADIT, 28% salen corse = 9-8553995 
sine Bye. suv WaRRG gS ig fy o'F s/he © = 93724256 
cos. sun’s true alt... =59 11 22 ........ = 9-7094405 
cos. moon’s true alt. =27 53 25 ........ = 9-946376L 
cas. star’s app. alt. .=59 11 52 (ar. com.) = 0-2904536 
cos. moon’s app. alt.=27 2 30 (ar. com.) = 0-0502802 
Sale ai e.a scree s)0 =31 17 57 (ar. com.) = 0-0683050 
Sum of logarithms. ........eeeeeeeeee ee = 19°5937105 
Half sum corres-2 _ » wis 
Sreunse ones me a eee Ltt = 9-7968552 
Plage COS. Do. ois euswisle ape geieige se ens «. =, 97836276 
ON OF EE ae Dabiig a, ak alias n'a, ox >, obs = 9-9316950 
cos. true distance .. =54 43 20 ........ = 9°7153226 
The same as determined by Mr. Sanderson’s rule in the 
Ladies’ Diary for 1787, but 9” less than is given at page 47, 
Requisite Tables, whence the example is taken. 
Example 2.—Given the apparent altitudes of the moon and 
star 28° 29’ 44” and 45° 9’ 12”; the true altitudes 29° 17’ 45”, 
45° 8’ 15”; and the apparent distance 63° 35’ 13” ; to find the 
true distance. 
From the data : 
A = 40° 7’ 2017, B = 23° 277.521”, and. C= 15° 50° 30”. 
Then per Rule 1. 
iat nina 'aiicla's atu eialereheyt s Mielacis's aiftajemiets ¥.<« = 0°3010300 
RE CE ee ine A 7! 20 ec es ee, 0 OORT IRS 
Ue, Ent we INST SAIC, == OG0OCRIS 
cos. star’s true alt. =45°"68 16° P2 0. = 9°8484403 
cos. moon’s true alt.=29 17 45  ........ = 99405687 
cos. star’s app. alt..=45 9 12 (ar.com.) = 01516804 
cos. moon’s app.alt.=28 29 44 (ar.com.) = 0°0560832 
cos,C. .......,..=15 50 30 (ar.com.) = 0°0168160 
Sum of logarithms. ..........00.se0 eee = 19°7238708 
* There is no necessity for taking out the arc D ; for having found the half sum 
amongst the log. sines, the log. cosines even therewith may be easily seen, and its 
double taken out at once. 
