I think I got Steve's problem worked out. See attatched Deltacad file. The text is reproduced below. z is azimuth. x is an intermediate value needed for the calculation:
Law of Sines: sin(x)=sin(90-lat)*sin(180-z)/sin(90-dec) sin(x)=cos(lat)*sin(z)/cos(dec) This gives 2 possible angles for x in the ranges of -90 to 90, and (90 to 180, -90 to -180) To get the 2 possible solutions for ha, enter each value for x into the half-cotangent formula: cot(ha/2)=tan((x+(180-z))/2)*cos((90-dec+90-lat)/2)/cos((90-dec-90+lat)/2) tan(ha/2)=cos((lat-dec)/2)/[tan((x+180-z)/2)*cos(90-(dec+lat)/2)] And, finally, your two answers: ha=2*atan(cos((lat-dec)/2)/[tan((x+180-z)/2)*sin((dec+lat)/2)]) I ran these through a spreadsheet, and at first glance, they look pretty good. Bill Gottesman Burlington, VT 44.4674 N, 73.2027 W Content-Type: application/octet-stream; name="M STEVE'S ANSWER.DC" Content-Disposition: attachment; filename="M STEVE'S ANSWER.DC" Attachment converted: Macintosh HD:M STEVE'S ANSWER.DC (????/----) (0001FD1C)
