 | Henry William Jeans - 1858 - 106 pages
...part is equal to the product of the tangents of the two parts adjacent to it. EULE B. The sine of the middle part is equal to the product of the cosines of the two parts opposite to, or separated from it. Having written down the equation according to the case,... | |
 | George Roberts Perkins - Geometry - 1860 - 472 pages
...the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. If now we take in turn each of the five parts as the middle part, and apply these Rules, we shall obtain... | |
 | Horatio Nelson Robinson - Geometry - 1860 - 470 pages
...part is equal to the product of the tangents of the adjacent parts. 2. The radius into the sine of the middle part is equal to the product of the cosines of the opposite parts. These rules are known as .Napier's Rules, because they were first given by that distinguished mathematician,... | |
 | Benjamin Greenleaf - Geometry - 1862 - 532 pages
...the middle part is equal to Hie product of tlte tangents of the adjacent parts. IL The sine of the middle part is equal to the product of the cosines of the opposite parts. 168. Napier's rules may be proved by showing that they agree with the results already established,... | |
 | Benjamin Greenleaf - Geometry - 1862 - 518 pages
...the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 168. Napier's rules may be proved by showing that they agree with the results already established,... | |
 | Benjamin Greenleaf - Geometry - 1863 - 504 pages
...the middle part is equal to the product of the tangents of the adjacent parts. . II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 168. Napier's rules may be proved by showing that they agree with the results already established,... | |
 | Gerardus Beekman Docharty - Geometry - 1867 - 474 pages
...to the product of the tangents of the adjacent parts. . 2. The product of sin. M and tabular radius is equal to the product of the cosines of the opposite parts. In order to make these rules clearly understood, we will show the manner in which they are applied... | |
 | Edwin Pliny Seaver - Mathematics - 1871 - 78 pages
...sine of any part is equal to the product of the tangents of the adjacent parts. II. The sine of any part is equal to the product of the cosines of the opposite parts. (с) SOLUTIONS. Spherical Bight Triangles. 185. Given an angle and the hypotenuse, A and A. sin a =... | |
 | Charles Davies, William Guy Peck - Mathematics - 1872 - 602 pages
...middle jtart is ci/ital to the products of the tangents of the adjacent parti. 2. The sine of the muidle part is equal to the product of the cosines of the opposite parts : thus, sin (90° - a) = tan (90° - B) tan (90° - C), and sin (90° — a) = cos с cos 6. The only... | |
 | Aaron Schuyler - Measurement - 1864 - 512 pages
...sin (90°— A). . • . (5) sin (900— A) = tan (90°— Б) tan (90°— P). 2. The sine of the middle part is equal to the product of the cosines of the opposite parts. OE=cos EOD X OD, or cos A = cos b cos p. ... (6) sin (90°— A) = cos b cos p. DB = EB sin DEB, or... | |
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I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 
