1819.] Mathematical Problems, by Mr. Adams. 189 
Problem 2.—To find (A + B) — B, by logarithms. 
(A+ B)-B=(- x - 5) (A + B). Pind an arc corres=- 
B 
ponding to the log. sin. / Ap Which denote by C; 
Then 2 log. cos. C + log.(A + B) = log. { (A + B)— BY. 
This solution depends on the property sin. A + cos.?A = rad.® 
Problem 3.—Yo find 1 — (1 + m) by logarithms. 
1-( +m)=—{0+m)—1}=-U-, )d +m). 
lim 
: se 1 ; 
Find an arc corresponding to the log. sin. 4 / ——, which re- 
1 +m 
present by D; vm 
Then — § 2 log. cos. D + log.(1 + m)} = log.§ 1—(1+m)}. 
This solution also depends on the property sin.* A + cos. A= 
rad.?. 
Problem 4.—To reduce the observed distance of the moon and 
sun, or moon and star, to the true, by the addition of log. sines, 
and cosines, only. 
Let M and S represent the true places of the moon and sun, 
or star, m and s the apparent places, m s the apparent distance, 
M S the true distance, Z the zenith, and H R the horizon of the 
place of observation. 
Put half the sum of the apparent distance and differ-) _ Z 
ence of apparent altitudes.........ccceesescenecs Ps 
Flalt themditterence sere Oe Os os a's vale es =B 
And the difference of the true altitudes .......... =C 
Then per Simpson’s Trigonometry, p. 74. 
ens.(Zm — Zs) —cos,sm __ cos. (ZS —ZM) —cos. SM 
———$—<—$—$——— $$ rrr 
= . 
sin, Zm.sin. Zs sin. ZS.sin. ZM 
From whence 
cos. (Zm— Zs) — cos, sm 
cos. § M = cos.(Z8 — ZM) — §[-" ot 
sin.Z S$. sin. Z M 
Or, 
cos. SM =cos.(MR — S$ H) — pee 
cos.S H.cos. MR 
