Wentworth, G. A. (George Albert), 1835-1906 · Boston, Ginn & company · leaf 153
, ZPZM=a = azimuth; to find PM= 90° - d = polar distance, ZZPM= t = hour angle. Draw MQJL NS, and put ZQ = m, then, if a < 90°, PQ = 90° - (I + m), and if a>90°, PQ = 90° -\l- m) ; and, by Napiers Rules, cos a = ± tan m tan A,sin d = cos PQ cos MQ,sin A = cosm cos MQ; whence, tanm = ± cot A… … § 67. Problem. To find the hour angle of a heavenly body when its declina-tion, its altitude, and the lati-tude of the place are known. In the triangle ZPM (Fig. PZ= 90° -I,PM=90°-d = p,ZM=90°-h; required Z ZPM= t.If, in the first formula of [46], sin \ A = Vsin (s — b) sin (s — c) esc b esc c,we…
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