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Problem 3rd

Page 6 of 13
[p. 10:] From the apparent altitude of the sun, together with the latitude [of the] place [of] observation and estimated Greenwich time; to find the true hour of the day. Directions - At any time when the sun is not less than 3 points from the meridian (but the nearer to the East or West points the better) take the altitude.
- From the apparent altitude find the true altitude.
- Find the declination for the estimated Greenwich time, which subtract from 90° when the latitude and declination are of the same name; but add to 90° when they are of different names, and the result will be the polar distance.
From the above data, the hour angle and thence the hour of day may be computed as in the following example. [p. 11:] Example Suppose the apparent altitude of sun's lower limb west of the meridian 22°35'. Latitude place observation 34°40°N. Estimated Greenwich time May 24, 1799 about 18 hours p.m. Required the time? 
Form III 
[p. 12:] Note, When the declination = 0; then to the secant of the latitude add the sine of the altitude, and the sum (abating 10 from the index) will be the cosine of the hour angle. When the latitude = 0; then to the secant of the declination add the sine of the altitude and the sum (abating 10 from the index) will be the cosine of the hour angle When the altitude = 0; then to the tangent of the latitude, add the tangent of the declination and the sum (abating 10 from the index) will be the cosine of the hour angle, taking the supplement to 180° when latitude and declination are of the same name. When both latitude and declination = 0; then the complement of the altitude or zenith distance will be the hour angle. --edited by Robert N. Bergantino, 05/95, 09/04, 06/05. Funded in part by a grant from the NPS Challenge-Cost Share Program.
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