 | Elias Loomis - 1859 - 372 páginas
...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.... | |
 | Charles Davies, Adrien Marie Legendre - 1869 - 470 páginas
...-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 páginas
...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 to the rectangle of the tangents of the adjacent extremes. part is equal to the rectangle of the cosines of the opposite extremes.... | |
 | Charles Davies - 1872 - 464 páginas
...C) = tan (90°-a) tan b • ' • • (10.) Comparing these formulas with the figure, we see that, The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier'a rales for Circular Parts, and they... | |
 | Charles Ramsay Drinkwater Bethune - 1872 - 102 páginas
...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.... | |
 | Adrien Marie Legendre - 1874 - 500 páginas
...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.... | |
 | William Guy Peck - 1876 - 376 páginas
...(5) Comparing these formulas with the diagram, we see that the following rule is always true : 1st. The sine of the middle part is equal to the rectangle of the cosine of the opposite parts. 2°. From formulas (7), (6), (8), (10), and (9), of Art. 46, we have,... | |
 | Charles Davies, Adrien Marie Legendre - 1885 - 538 páginas
...sin (90° - C) = cose cos (90° - B). .... (5.) Comparing these formulas with the figure, we see that The 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 parts adjacent.... | |
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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. 
