| 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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