GEODESIC INTESTIGATIONS. 167 



Problem 5. 



Griyen tlie latitude V of one station 8' , the azimuth A' of the other station 

 S" , as taken at 8' , and given also the length 5 and circular measure 2 of the 

 geodesic arc between the stations : to find the latitude I" of the station ;S", 

 the difference of longitude co of the stations, the azimuth A" of the station 8" 

 as if taken at 8' , and all the other entities. 



From the triangle P aS"^ J Ave have — 



tan \ (cp, + 18,) = '°'!^,^l~!^ i . tan 1 ^' (a) 



tan i {<p, - /3,) = '^^i^,f~!^i . tan i.A' (a) 



cos t (t + iSj 



from which to find the angles (p,, j8^. Then to find the angle that the chord 

 makes with the plane of the equator, we have — 



cos e = ^j^^L^RliJ 



sin <p^ 

 The chord k is given by — 



^^ 2^sini2 ^j^ 



And its projection p^ p^^ on the plane of the equator can be found from — 

 k • sin A' cos i 2 



^.= 



sin (p, 



we shall have- 



Now, in the plane triangle Cp^ p,„ we know that the side Cp^ = E^ cos I' ; 

 that the side p^p^, =• k^ ; and that the angle C :=: oi. And it is evident that 

 by finding two subsidiary angles h^, h^^, such that — 



tan Ay = -ffiy cos I' ... ... ... ... (c) 



, , k . sin A' cos i 2 z ,> 



tan A„ = , =— {a) 



sm cp, 



tan i {<p,, -o>)= ^i^Lg-JzA] . cot icp, ... (e) 



i (<p,^ + w) = 90° - i ^, ... (/) 



from which to find (p^, and the difference of longitude ai. 



The azimuth A" is given by — 



sin A" = sin A'. '!" '^" {g) 



sm <p, 



And from formula 6 we have — 



. o ,,/ sin- ^" -•, 2 7/ . «"\ «^ /7A 



tan- r = . ^ ^ . (tan^ " + tt I — r", ••• W 



" sm^ A 0- / 0- 



from which to find I". 



The other entities can be found as already indicated in the solution of 

 Problem 1. 



