SECT. 2.] TO POSITION IN SPACE. 63 



these formulas, both for the solution of our present problem, and for other pur 

 poses, hold good in all cases generally. 



55. 



Designating as above, the longitude of the ascending node of the orbit upon 

 the ecliptic by 8, the inclination by i ; also, the longitude of the ascending node 

 of the new plane upon the ecliptic by n, the inclination by t ; the distance of the 

 ascending node of the orbit upon the new plane from the ascending node of the 

 new plane upon the ecliptic (the arc nQ, in fig. 2) by 8 , the inclination of the 

 orbit to the new plane by i ; finally, the arc from 8 to 8 in the direction of the 

 motion by A: the sides of our spherical triangle will be & n, 8 , A, and the 

 opposite angles,/, 180 i, e. Hence, according to the formulas of the preceding 

 article, we shall have 



sin i sin ( 8 -\- A] = sin ^ ( 8 ) sin (i -\- e) 

 sin i i cos ( 8 -j- A) = cos i ( & n) sin J (i e) 

 cos /sin i (8 //) = sin k (8 ?z)cos (i-\- e) 

 cos H cos i (8 A} =cos i (8 w)cosi (/ ). 



The two first equations will furnish i (8 -)-//) and sin i /; the remaining two, 

 i(S --J) and cos it&quot;; from ^Q -j-//) and J(8 z/) will follow 8 and J ; 

 from sin i / and cos & y (the agreement of which will serve to prove the calcula 

 tion) will result i . The uncertainty, whether ( 8 + ^) and ( Q, -- A) should 

 be taken between and 180 or between 180 and 360, will be removed in this 

 manner, that both sin \ i , cos J i , are positive, since, from the nature of the case, i 

 must fall below 180. 



56. 



It will not prove unprofitable to illustrate the preceding precepts by an 

 example. Let 8 = 172 28 13&quot;. 7, i = 3438 l&quot;.l ; let also the new plane be 

 parallel to the equator, so that n = 180 ; we put the angle e, which will be the 

 obliquity of the ecliptic = 2327 55&quot;.8. We have, therefore, 



