Wentworth, G. A. (George Albert), 1835-1906 · Boston, Ginn & company · leaf 185
e pole of the ecliptic. Then, in the triangle PMQ, given PQ = e = 2,3° 27, . PM = 90° - d, ZMPQ = 90° + r; required QM= 90°- u, andwhere Z PQM = 90° - v ; r = the right ascension = TV?,u = the latitude of M = MT, and v = the longitude of M = VT. Q PH Illlllli % l - ■ llll! r ^i-^^7^^ aV w ■ :^-=^ /… … By Napiers Rules, sin r = tan n tan d, (1) sin u = cos (e -f- ri) cos Mff, (2)sin d = cos rc cos MH, (3) sin (e + n) = tan t? tan MH, (4) sin w = tan r tan ilfiiT. (5) From (1), FlG122 tan n = cot d sin r. (A) From (3), cos MH = sin ^ sec n.Substitute in (2) the value of cos MH, sin u = cos (e + n)…
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