 | James Hann - Spherical trigonometry - 1849 - 84 pages
...disjunct. This practical method will be useful to seamen, and requires very little effort of memory. The sine of the middle part, is equal to the product of the cosines of the extremes disjunct. From these two equations, proportions may be formed, observing always... | |
 | Benjamin Peirce - Trigonometry - 1852 - 398 pages
...adjacent parts ; and the other two parts are called the opposite parts. The two theorems are as follows. I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B.... | |
 | William Chauvenet - 1852 - 268 pages
...: 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. The correctness of these rules will be shown by taking each of the five... | |
 | Benjamin Peirce - Trigonometry - 1852 - 408 pages
...sine of the middle part is equal to the product of the tangents of the two adjacent parts. IL TJie sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only... | |
 | Elias Loomis - Trigonometry - 1855 - 192 pages
...part required may then be found by the following RULE OF NAPIER. (211.) The product of the radius and the sine of the middle part, is equal to the product of the tangents of the adjacent parts, or to the product of the cosines of the opposite parts. It will assist the learner... | |
 | George Roberts Perkins - Geometry - 1856 - 460 pages
...RULES, 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. If now we take in turn each of the five parts as the middle part, and... | |
 | Henry William Jeans - 1858 - 106 pages
...of the middle 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... | |
 | Elias Loomis - Logarithms - 1859 - 372 pages
...part required may then be found by the following RULE OF NAPIER. (211.) The product of the radius and the sine of the middle part, is equal to the product of the tangents of the adjacent parts, or to the product of the cosines of the opposite parts. It will assist the learner... | |
 | John Daniel Runkle - Mathematics - 1859 - 478 pages
...NAPIER'S RULES. BY TUI MAN HENRY 8AFFORD. IN the form in which they are usually given, the rules are — I. The sine of the middle part is equal to the product of tlie tangents of tJie adjacent parts. II. T/te sine of the middle part is equal to tJic product of... | |
 | 1860 - 462 pages
...RULE I. The sine of the middle pari equals the product of the cosines of the opposite parts. RULE II. The sine of the middle part is equal to the product of the tangents of the adjacent parts. That the second of these rules may be deduced from the first has been shown by Mr.... | |
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II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 
