| 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 - 400 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
...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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