| Aaron Schuyler - Measurement - 1864 - 512 pages
...OZi = OE= cos A = sin (90°— A). . • . (5) sin (900— A) = tan (90°— Б) tan (90°— P). 2. The sine of the middle part is equal to the product of the cosines of the opposite parts. OE=cos EOD X OD, or cos A = cos b cos p. ... (6) sin (90°— A) = cos b cos p. DB... | |
| Aaron Schuyler - Navigation - 1873 - 536 pages
...cosEODxOD=OE=coa h = sin (90°— A). . • . (5) sin (90°— A) = tan (90°— B) tan (90°— P). 2. Sine of the middle part is equal to the product of the co-sines of the opposite parts. = cos EOD X OD, or cos h = cos 6 cos p. ... (6) sin (90°— A) = cos 6 cos p. DB=EB... | |
| Henry William Jeans - 1873 - 292 pages
...selected is called the middle part. Eule A will apply to the former case, Eule B to the latter. EULE A. The sine of the middle part is equal to the product of the tangents of the two parts adjacent to it. * The complement of an angle is what it wants of 90° ; thus,... | |
| Aaron Schuyler - Measurement - 1873 - 506 pages
...90°— h, b. 9. 90°— h, 90°— P, b. 10. 90°— P, 90°— B, p. 126. Napier's Principles. 1. The sine of the middle part is equal to the product of the tangents of the adjacent parts. Draw BD and DE, respectively perpendicular to OH and OP, and draw BE.... | |
| Aaron Schuyler - Measurement - 1864 - 504 pages
.../< = sin (90°- A). . . . (5) sin (9<P— A) = tan (90°— B) tan (90°- P). 2. TVtf si.nf of </ie middle part is equal to the product of the co-sines of the opposite parts. OE=cos EOD X #£>, or cos h — cos b cos p. .-. (6) sin (90°— h) = cos 6 cos p.... | |
| Horatio Nelson Robinson - 1875 - 288 pages
...will be two adjacent parts, and two opposite parts. Then we can easily remember, that 1st. Radius into the sine of the middle part, is equal to the product of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part, is equal to the product... | |
| Aaron Schuyler - Measurement - 1875 - 276 pages
...90°— Л, b. 9. 90°— A, 90'— P, b. 10. 90°— P, 90°— B, p. 126. Napier's Principles. 1. The sine of the middle part is equal to the product of the tangents of the adjacent parts. Draw BD and DE, respectively perpendicular to OH and OP, and draw BE.... | |
| Benjamin Greenleaf - Trigonometry - 1876 - 204 pages
...are called the opposite parts. Then, whatever be the middle part, we have as THE RULES OP NAPIER. 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... | |
| Horatio Nelson Robinson - Navigation - 1878 - 564 pages
...that it corresponds to one of the following invariable and comprehensive rides. 1. The radius into the sine of the middle part is equal to the product of the tangents of the adjacent parts. 2. The radius into the sine of the middle part is equal io> the product... | |
| Eugene Lamb Richards - Trigonometry - 1879 - 232 pages
...adjacent parts are a, 90°—b; opposite parts are c, 90°— A. 115. Napier's rule of the Circular Parts. The sine of the middle part is equal to the product of the tangents of the adjacent parts; and the sine of the middle part is equal to the product of the cosines... | |
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