614 SOLUTION OF THE EQUATIONS. 



Also 



[ -0043626]^ -[ -0058168]#./ + [9-9103610]^ -[9-7656872]# 4 l + [9-5910501J0, 1 

 -[9'3962097]# li 1 + [9-1868105]#/- [8-9664160]^ + [87374212]# 9 '-[8-5015145]# 10 ' = a', 



[ -0043626JV-E -0058168>.; + [9-9103610] A 3 1 -[97656872]/* 4 1 + [9-591050lJ4 s 1 

 -[9-3962097]/C + [9'1868105]/> 7 1 -[8-9664160]// 8 1 + [87374212]^' -[8-5015l45]h w l =V, 



and 



-[0-3053926]^ +[0-3068468]^- [0-21 13910]# 3 + [0-0667172]# 4 -[9-892080l]0 6 



+ [9-6972397]^ -[9-4878405]^ + [9-2674460]^- [9-0384512]^; + [8-8025445]^ 10 = c / . 



Hence the equations for the polar element to be added to the equations 

 of condition previously found are 



-[ -0058168] & 1 - [97656872] #; -[9-3962097J& 1 -[8'9664160j^ -[S'5015145]^ 1 = \(a + a!) 



L 



-[ -0043626]^' -[9-9103610]^ -[9-5910501]^ -[9-1868105]^ -[87374212]^ = \(a -a') 



Li 



-[ -0058168] A 2 1 -[9-7656872]/i 4 1 - [9-3962097] A, 1 - [8'9664160] A 8 1 -[8'5015145]/t IQ 1 = 5(6 + 6') 



A 



-[ -0043626] /;/ -[9-9103610] A :i 1 -[9-591050l]A 5 I -[9-1868105]/< 7 1 -[8-7374212]A 9 1 = ^ (6 -6') 



m 



[0-3068468] (/ 2 + [0-OGG7172](7; + [9-6972397K + [9-2674460K + [8-8025445]^; = i(c + c') 



ft 



The weight of each of these equations taken on the same scale as 

 before is proportional to the area of the circular segment of the surface 

 over which the observations extend. When the observations extend over 

 a zone of 5 in breadth, the area of the zone will contain the factor 



2 sin-, where is the circular measure of 5. We must divide by this 

 factor to find the weight, as we have done in our former calculations. 



Now the area bounded by the small circle of radius 2 30' round the 

 pole is very approximately 



