An Elementary Treatise on Plane & Spherical Trigonometry: With Their Applications to Navigation, Surveying, Heights, and Distances, and Spherical Astronomy, and Particularly Adapted to Explaining the Construction of Bowditch's Navigator, and the Nautical Almanac |
From inside the book
Results 1-5 of 21
Page 273
... 15 ' ; what was the latitude , if the moon's declination was 1 ° 12 ′ N. , and her increase of decli- nation per hour 18.5 ? Ans . 66 ' 1 ' N. Obliquity . Equinoxes . CHAPTER V. Signs . THE ECLIPTIC $ 52. ] 273 LATITUDE .
... 15 ' ; what was the latitude , if the moon's declination was 1 ° 12 ′ N. , and her increase of decli- nation per hour 18.5 ? Ans . 66 ' 1 ' N. Obliquity . Equinoxes . CHAPTER V. Signs . THE ECLIPTIC $ 52. ] 273 LATITUDE .
Page 274
... Obliquity . Equinoxes . CHAPTER V. Signs . THE ECLIPTIC . 53. The careful observation of the sun's motion shows this body to move nearly in the circumfer- ence of a great circle . This great circle is called the ecliptic . [ B. p . 48 ...
... Obliquity . Equinoxes . CHAPTER V. Signs . THE ECLIPTIC . 53. The careful observation of the sun's motion shows this body to move nearly in the circumfer- ence of a great circle . This great circle is called the ecliptic . [ B. p . 48 ...
Page 277
... obliquity of ecliptic E , ( 457 ) in which the declination and latitude are positive when north , and negative when south , and E has the same sign with R. A. 12h . The present problem does not , then , differ from that of 28 , and if ...
... obliquity of ecliptic E , ( 457 ) in which the declination and latitude are positive when north , and negative when south , and E has the same sign with R. A. 12h . The present problem does not , then , differ from that of 28 , and if ...
Page 281
... obliquity of the ecliptic 23 ° 27 ′ 45 ′′ . Solution . 27 ° 21 ′ 58 ′′ N. 4h 42m 56s tang . 0.45650 tang . 9.71400 cosec . 0.02503 A 28 ° 44 ′ 12 ′′ N. E - 23 ° 27 ' 45 " S. sec . 0.05708 tang . 9.73903 B = 5 ° 16 ′ 27 ′′ N. Cos ...
... obliquity of the ecliptic 23 ° 27 ′ 45 ′′ . Solution . 27 ° 21 ′ 58 ′′ N. 4h 42m 56s tang . 0.45650 tang . 9.71400 cosec . 0.02503 A 28 ° 44 ′ 12 ′′ N. E - 23 ° 27 ' 45 " S. sec . 0.05708 tang . 9.73903 B = 5 ° 16 ′ 27 ′′ N. Cos ...
Page 282
... obliquity 23 ° 27 ' 40 " . Ans . A 0.07965 B9.47565 C9.85327 6. Calculate the longitude and altitude of the nonagesi- mal , when the obliquity of the ecliptic is 23 ° 27 ′ 40 ′′ , the latitude 42 ° 12 ′ 2 ′′ N. , and the R. A. of the ...
... obliquity 23 ° 27 ' 40 " . Ans . A 0.07965 B9.47565 C9.85327 6. Calculate the longitude and altitude of the nonagesi- mal , when the obliquity of the ecliptic is 23 ° 27 ′ 40 ′′ , the latitude 42 ° 12 ′ 2 ′′ N. , and the R. A. of the ...
Contents
3 | |
6 | |
7 | |
27 | |
37 | |
47 | |
65 | |
79 | |
200 | |
202 | |
207 | |
212 | |
226 | |
239 | |
274 | |
288 | |
81 | |
93 | |
94 | |
97 | |
104 | |
119 | |
131 | |
137 | |
143 | |
145 | |
149 | |
Other editions - View all
Common terms and phrases
A₁ aberration altitude and azimuth angle given ascension and declination azimuth celestial equator celestial sphere centre circle computed Corollary corr correct central altitude corresponding cosec cosine cotan diff difference of latitude difference of longitude dist earth eclipse of April equal to 90 formula gives Greenwich Hence horizon horizontal parallax hour angle hypothenuse included angle interval latitude and longitude lunar distance mean meridian altitude method middle latitude moon's motion N₁ Napier's Rules Nautical Almanac Navigator Nutation obliquity obtuse perpendicular plane polar triangle prime vertical Problem R₁ radius reduced right ascension sailing Scholium second member semidiameter sideral sideral day solar eclipse Solution solve the triangle spherical right triangle spherical triangle star's sun's Table XXIII tang tangent Theorem transit triangle ABC Trig true latitude tude vernal equinox whence
Popular passages
Page 156 - I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.
Page 145 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
Page 48 - As the sine of the angle opposite the given side is to the sine of the angle opposite the required side, so is the given side to the required side. Thus, if a (fig.
Page 50 - The third side is found by the proportion. As the sine of the given angle is to the sine of the angle opposite the required side, so is the side opposite the given angle to the required side.
Page 41 - Since, when an angle is acute its supplement is obtuse, it follows from the preceding proposition, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant, are negative.
Page 53 - The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference.
Page 182 - But a' = 180° - A, b' = 180° - ß, c' = 180° - C. and A' = 180° - a. Therefore, — cos A = (— cos B)(— cos C) + sin B sin C(— cos a...