 | Webster Wells - Trigonometry - 1896 - 236 pages
...I. The sine of the middle part is equal to the product of the tangents of the adjacent parís. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 142. Napier's rules may be proved by taking each circular part in succession... | |
 | William Chauvenet - Geometry - 1896 - 274 pages
...: I. The sine of the middle part is equal to the product of the tangents of the adjacent farts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. The correctness of these rules will be shown by taking each of the five... | |
 | George Albert Wentworth - Logarithms - 1897 - 384 pages
...parts immediately adjacent are called adjacent parts, and the other two are called opposite parts. Rule I. The sine of the middle part is equal to the product of the t&ngents of the Adjacent parts. Rule II. The sine of the middle, part is equal to the product of the... | |
 | English language - 1897 - 726 pages
...a C 1 The sine of the middle part is equal to the product of the tangents of the adjacent parts. 2 The sine of the middle part is equal to the product of the cosines of the opposite parts. Demonstration of the following : An angle and its opposite side are... | |
 | Andrew Wheeler Phillips, Wendell Melville Strong - Trigonometry - 1898 - 362 pages
...comp C. sin comp R = cos comp 6• cos b. sin comp C=cos compjS cose. Napier's rules may be stated : I. The sine of the middle part is equal to the product...parts. II. The sine of the middle part is equal to t he product of the cosines of the opposite parts. 84:. In the right spherical triangles considered... | |
 | James William Nicholson - Trigonometry - 1898 - 204 pages
...90°— A and 90°— Б are adjacent parts, and a and b opposite parts. 129. Rules. l. The sine of any part is equal to the product of the tangents of the adjacent parts. ll. The sine of any part is equal to the product of the cosines of the opposite parts. NOTE. — These... | |
 | Canada - 1899 - 1074 pages
...SPHERICAL TRIGONOMETRY. TIME, 3 HOUR?. 1. Prove with the aid of a figure the following Napier's principle : "The sine of the middle part is equal to the product of the tangents of the adjacent parts. 2. In a right angled spherical triangle is known p a. side adjacent to the right angle and P the angle... | |
 | Charles Hamilton Ashton, Walter Randall Marsh - Trigonometry - 1900 - 184 pages
...rules : The sine of the middle part is equal to the product of the tangents of the adjacent parts. The sine of the middle part is equal to the product of the cosines of the opposite parts. If we apply these rules, using each part successively as the middle... | |
 | George Albert Wentworth - Trigonometry - 1902 - 256 pages
...adjacent parts, and the other two are called opposite parts. Napier's Rules are RULE I. The sine of any middle part is equal to the product of the tangents of the Adjacent parts. RULE II. The sine of any middle part is equal to the product of the cosines of the opposite parts.... | |
 | Charles Hamilton Ashton, Walter Randall Marsh - Trigonometry - 1902 - 186 pages
...rules : The sine of the middle part is equal to the product of the tangents of the adjacent parts. The sine of the middle part is equal to the product of the cosines of the opposite parts. If we apply these rules, using each part successively as the middle... | |
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I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
