 | Thomas Kerigan - Nautical astronomy - 1828 - 776 pages
...equal to 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 - Spherical trigonometry - 1836 - 84 pages
...the 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 - Spherical trigonometry - 1836 - 92 pages
...the 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... | |
 | Henry W. Jeans - Trigonometry - 1842 - 138 pages
...middle 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 - Plane trigonometry - 1845 - 498 pages
...parts 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... | |
 | Benjamin Peirce - Plane trigonometry - 1845 - 498 pages
...part is 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... | |
 | James Hann - Spherical trigonometry - 1849 - 80 pages
...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 to take the... | |
 | Charles Davies, William Guy Peck - Mathematics - 1855 - 628 pages
...middle pari is equal to the products of the tangents of the adjacent parís. 2. The tine of the muidle part is equal to the -product of the cosines of the opposite part* : thus, sin (90° - «) = tan (90° - B) tan (90° - C), and sin (90° — it) = cos e cos 4.... | |
 | George Roberts Perkins - Geometry - 1856 - 460 pages
...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 apply these Eules, we shall obtain... | |
 | Henry William Jeans - 1858 - 106 pages
...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 to the case,... | |
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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. 
