 | Elias Loomis - 1855 - 192 sider
...value of the 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.... | |
 | Elias Loomis - 1859 - 372 sider
...value of the 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.... | |
 | Adrien Marie Legendre - 1863 - 464 sider
...(7) = tan (90°- a) tan 6 • - • - (10.) Comparing these formulas with the figure, we see that, of the middle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier's rules for Circular Parts, and they are sufficient to solve... | |
 | Charles Davies, Adrien Marie Legendre - 1869 - 470 sider
...-C) = cose cos (90° -2?) • • • • (5.) Comparing these formulas with the figure, we see that, The sine of the middle part is equal to the rectangle of cosines of the opposite parts. Formulas (8), (7), (4), (6), and (3), of Art. 72. may bo written as... | |
 | Enoch Lewis - 1872 - 238 sider
...extremes; and the other two are termed the opposite extremes. Then Napier's rules are: 1. The^rectangle of radius and the sine of the middle part is equal...rectangle of the tangents of the adjacent extremes. part is equal to the rectangle of the cosines of the opposite extremes. These rules may be explained... | |
 | Charles Ramsay Drinkwater Bethune - 1872 - 102 sider
...if only two are adjacent, they are extremes, and the opposite part is the middle part. The product of Radius and the Sine of the middle part is equal to the products of the tangents of the adjacent extremes, or of the cosines of the opposite extremes : (tan.... | |
 | Charles Davies - 1872 - 464 sider
...c cos (90°—J?) • • • • (5.) Comparing tLese formulas with the figure, we see that, TTte sine of the middle part is equal to the rectangle of the cosines of the opposite parts. Let us now take the same middle parts, and the other parta adjacent.... | |
 | Adrien Marie Legendre - 1874 - 500 sider
...cos c cos (90°— 2?) • • • • (5.) Comparing tLese formulas with the figure, we see that. The sine of the middle part is equal to the rectangle of Ike cosines of the opposite parts. Let us now take the same middle parts, and the other parts adjacent.... | |
 | Charles Davies - 1874 - 464 sider
...• • • • (10.) Comparing these formulas with the figure, we seo that, The sine of the m1ddle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier's rules for Circular Parts, and they are sufficient to solve... | |
 | William Guy Peck - 1876 - 412 sider
...(10) Comparing these formulas with the diagram, we see that the following rule is always true : 2d. The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. Discussion. 48. In applying Napier's rules, the required part is always determined by means... | |
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In a right angled spherical triangle, the rectangle under the radius and the sine of the middle part, is equal to the rectangle under the tangents of the adjacent parts ; or', to the rectangle under the cosines of the opposite parts. 
