30 U. S. COAST AND GEODETIC SURVEY. 



Then the Z MOP will equal the Z0 of equation (76). 



Now, let X = ylP= latitude of place of observation, 

 5 = J/' if = declination of moon, 

 C= AAOM' = arc J. Jf' =hour angle of moon, 

 7= Z if/J. = inclination of moon's orbit to the Equator, 

 Z = /if == longitude of moon in its orbit reckoned from the 



intersection I, 

 ■^=.IA=- right ascension of meridian of place of observations 

 reckoned from the intersection 7. 



In this discussion the radian is considered as the angular unit. 

 In the spherical triangle GMF 



Cos Pi/= cos e = cos CP cos Cif+sin CP sin CM cos C 



= sinX sin 5 + cos X cos 8 cos J. if' (77) 



In the right spherical triangle MM' I, 



sin 5 = sin 7 sin Z (78) 



and in the right spherical triangle MM* A, 



cos h cos AM' = cos A M. (79) 



In the spherical triangle MAI 



cos A M= cos I cos X + sin Z sin X cos 7 (80) 



Substituting (78), (79), and (80) in (77) 



cos ^ = sin X sin 7 sin Z + cos X [cos Z cos x + cos 7 sin Z sin x] 

 = sin X sin 7 sini Z + i cos X [cos (Z — x) + cos (Z + x) 

 + cos7{cos (Z-x)-cos (Z + x)}] 

 = sin X sin 7 sin Z + cos X [cos^ ^ 7 cos (Z — x) 



+ sin2i7cos (Z + x)] (81) 



Then 



cos^ 6 = sin^ X sin^ 7 sin^ Z 



+ 2 sin X cos X sin 7 sin Z [cos^ ^ 7 cos (Z — x) 

 + si.n2 J :7cos (Z + x)] 

 + cos^ X [cos* i 7 cos^ (Z — x) 

 + 2 sin^ i 7 cos^ ^ 7 cos (Z-x) cos (Z + x) 

 + sin*^ 7cosM^ + x)] 

 = i cos2 X cos" i 7 cos (2Z- 2x) 

 ■. + i cos2 X sin* I 7 cos (2Z + 2x) .■: 



+ J cos^ X sin^ 7 cos 2x 

 + 1 siji 2X sin 7 cos^ | 7 sin (2Z — x) .. ' 

 + 1 sin 2X sin ,7 sin^ ^ 7 sin (2Z+ x) 

 + ^ sin 2X sin 7 cos 7 sin x 

 + (i cos2 X sin' 7- i sin2 X sin' 7) cos 2Z 

 + i (cos' X cos* i 7+ cos' X sin* ^ 7+ sin' X sin' 7) (82) 



The last two lines of (82) may be written 



(i- 3/2 sin' X) a sin' 7 cos 2Z) 

 + (^ - 3/2 sin' X) (1/3 - i sin' 7) + 1/3 (83) 



