 | Thomas Kerigan - 1828 - 776 pàgines
...the product of the tangents of the extremes conjunct2d. — The product of radius and the sine of the middle part, is equal to the product of the co-sines of the extremes disjunct. Since these equations are adapted to the complements of the hypotheiiuse and angles,... | |
 | Benjamin Peirce - 1836 - 84 pàgines
...middle part is equal to the product of the tangents of the two adjacent parts. (47e) II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. A-jLsAoi ' Demonstration. To demonstrate the preceding rules, it is only necessary... | |
 | Benjamin Peirce - 1836 - 92 pàgines
...middle part is equal to the product of the tangents of the two adjacent parts. (475) II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. Demonstration. To demonstrate the preceding rules, it is only necessary to compare... | |
 | Roswell Park - 1841 - 722 pàgines
...sine of the middle part, will be equal to the product of the tangents of the adjacent parts, and also equal to the product of the cosines of the opposite parts. These rules, called Napier's Analogies, may be applied to oblique angled spherical triangles ; by dividing them... | |
 | Henry W. Jeans - 1842 - 138 pàgines
...part is equal to the product of the tangents of the two parts adjacent to it. RULE II. 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,... | |
 | Benjamin Peirce - 1845 - 498 pàgines
...equal to Ike product of the tangents of the two adjacent parts. Napier's Rules. II. The 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 necessary to... | |
 | Benjamin Peirce - 1845 - 498 pàgines
...are called the opposite parts. The two theorems are as follows. Napier's Rules. II. The 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 necessary to... | |
 | Roswell Park - 1847 - 632 pàgines
...sine of the middle part, will be equal to the product of the tangents of the adjacent parts, and also equal to the product of the cosines of the opposite parts. These rules, called Napier' '* Analogies, may be applied to oblique angled spherical triangles ; by dividing them... | |
 | James Hann - 1849 - 84 pàgines
...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 to take the... | |
 | William Chauvenet - 1852 - 268 pàgines
...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 parts as middle part, and comparing... | |
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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. 
