350 MAGNETIC SURVEY OF A PART OF THE SOUTHERN HEMISPHERE. 
. ^' 32 be the observed azimuths, which of course are 11° 15', 22° 30', &c. 
Then we have 
sin S 0 =A +C +E 
sin \ = A+B sin ^ +C cos £'j +D sin 2£' 4 +E cos 2£' 4 
sin \ = A+B sin £' 2 +C cos £ 2 +D sin 2£' 2 +E cos 2£' 2 >. 
&c. &c. 
sin c$ 31 =A+B sin£' 31 +Ccos £31+ D sin2£' 31 +E cos 2^' 31 ^ 
• • (H.) 
Combining these equations by the method of least squares, we obtain by virtue of a 
well-known property of circular functions, 
A=^2 2sinS 
B = yg 2 sin & sin 
C= jg 2 sin & cos ?■ 
D= Yq ^ sin & sin 2£' 
E= ^ 2 sin & cos 2£' 
(15.) 
where 2 sin &= sin & 0 + sin + sin & 31 , 
2 sin & sin £'= sin & 0 sin £' 0 + sin & 4 sin £\+ &c. + sin & 31 sin g 31 
&c. &c. &c - 
<c ‘ If we represent sin S 0 , sin &c. by s 0 , s v &c., and remember that all the values of 
sin £', cos £', sin 2£', cos 2^' which occur in these formulae can be represented by the 
quantities s 1} s 2 , s 3 , .s 4 , s 3 , s 6 , s 7 , we shall find 
A=^2(5 0 +5 1 +^2 • • • • +%} ^ 16 '^ 
B = *0 1 22 (log = 2-086 1 1 ) { s 1 — j 31 + % - s l7 } , 
+•0239 (log =2-37872){5 2 -5 3 o+^ 14 -^ 18 }, 
+ •0347 (log =2*54062) {s 3 -s 29 +%-s 19 }, 
+*0442 (log =2'64536 ){s 4 — %+« 12 — %}, 
+•0520 (log =271572) {55—%+%—%}, 
+ •0577 (log =276149){s 6 — s 26 +s 10 — %}, 
+ •0613 (log =2"78745){s 7 — ^257^^9 ^23)5 
