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II. The sine of the middle part is equal to the product of the cosines of the opposite parts.
First Part of an Elementary Treatise on Spherical Trigonometry - Page 8
by Benjamin Peirce - 1836 - 71 pages
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## Treatise on Geometry and Trigonometry: For Colleges, Schools and Private ...

Eli Todd Tappan - Geometry - 1868 - 436 pages
...next to it are the adjacent parts, and the remaining two are the opposite parts. Napier's rule is : The sine of the middle part is equal to the product of the tangents of the adjacent parts, also to the product of the cosines of the opposite parts. The words sine and middle...
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## Nature, Volume 5

Sir Norman Lockyer - Electronic journals - 1872 - 540 pages
...included in the two rules known as Napier's rules, that the sine of the middle (that is, of any chosen) part is equal to the product of the tangents of the two adjacents, as well as to the product of the cosines of the two opposite parts. As a rule to assist...
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## A Treatise on Plane and Spherical Trigonometry

William Chauvenet - Trigonometry - 1871 - 268 pages
...considered, the two sides including it are regarded as adjacent parts. The rules are : I. The tine 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...
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## Plane and Spherical Trigonometry, Part 1

Henry W. Jeans - 1873 - 272 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, in the , P right-angled...
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## Surveying and Navigation, with a Preliminary Treatise on Trigonometry and ...

Aaron Schuyler - Measurement - 1873 - 508 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. BDE is a right...
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## Plane and spherical trigonometry. [With] Solutions of problems. [Followed by ...

Henry William Jeans - 1873 - 288 pages
...substituted for cos. со. A ; cos. A for sin. со. A ; cot. A for tan. со. A, to. (See Part II.) 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...
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## Surveying and Navigation, with a Preliminary Treatise on Trigonometry and ...

Aaron Schuyler - Measurement - 1864 - 506 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...
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## Surveying and Navigation: With a Preliminary Treatise on Trigonometry and ...

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...
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## Plane and Spherical Trigonometry and Mensuration

Aaron Schuyler - Measurement - 1875 - 284 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. BDE is a right...
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## Key to Robinson's New Geometry and Trigonometry, and Conic Sections and ...

Horatio Nelson Robinson - 1875 - 288 pages
...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 of the cosines of the opposite parts. The parts are the two sides, the complements of the hypotenuse, and...
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