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I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.
Surveying and Navigation, with a Preliminary Treatise on Trigonometry and ... - Pągina 110
per Aaron Schuyler - 1864 - 490 pągines
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A System of the Mathematics: Containing the Euclidean Geometry ..., Volum 2

James Hodgson - 1723
...t,æzæxct,ķ7<r— Axes, bac; that is the Radius multiplied into the Sine of the Complement of the Angle a or Sine of the Middle Part, is equal to the Product of the Tangent of ab one of the Extreams, into the Tangent of the Complement of л с the other Extream. By...
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Euclid's Elements of Geometry,: From the Latin Translation of Commandine. To ...

Euclid, John Keill - 1733 - 393 pągines
...S, CF=Cof. BC and T, DF = Cot. B. Wherefore R x Cof. BC=Cot. Cx Cot. B ; that is, Radius drawn into the Sine of the. middle Part, is equal to the Product of the Tangents of the adjacent extreme Parts. : X And And BA, AC, are the oppofite Extremes to the faid middle Part, viz. the Complement...
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An Account of the Life, Writings, and Inventions of John Napier, of Merchiston

David Stewart Erskine Earl of Buchan, Walter Minto - 1787 - 134 pągines
...fpherical triangle, The product of the tangents of half the fum and half the difference of the fegments of the middle part is equal to the product of the tangents of half the fum and half the difference of the oppofite parts. Dem. For fince cof BA: cof BC :: cof DA:...
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Mathematics: Compiled from the Best Authors and Intended to be the ..., Volum 2

1801 - 426 pągines
...solutions of all the cases of right-angled spherical triangles. THEOREM VII. The product of radius and the sine of the middle part is equal to the product of the tangents of the conjunct extremes, or to that of the cosines of the disjunct extremes.* NOTE. * DEMONSTRATION. This...
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The Complete Mathematical and General Navigation Tables: Including ..., Volum 1

Thomas Kerigan - 1828 - 664 pągines
...part, is equal to the product of the tangents of the extremes conjunct2d. — The product of radius and the sine of the middle part, is equal to the product of the co-sines of the extremes disjunct. Since these equations are adapted to the complements of the hypotheiiuse...
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First Part of an Elementary Treatise on Spherical Trigonometry

Benjamin Peirce - 1836 - 71 pągines
...sine of the middle part is equal to the product of the tangents of the two adjacent parts. (47e) II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. A-jLsAoi ' Demonstration. To demonstrate the preceding rules, it...
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First Part of an Elementary Treatise on Spherical Trigonometry

Benjamin Peirce - 1836 - 71 pągines
...sine of the middle part is equal to the product of the tangents of the two adjacent parts. (475) II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. Demonstration. To demonstrate the preceding rules, it is only necessary...
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The Complete Mathematical and General Navigation Tables: Including Every ...

Thomas Kerigan - 1838
...is equal to the product of the tangents of the extremes conjunct. 2d. — Tlie product of radius and the sine of the middle part, is equal to the product of the co- sines of the extremes disjunct. Since these equations are adapted to the complements of the hypothenuse...
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Plane and Spherical Trigonometry ...

Henry W. Jeans - 1842
...of the middle part is equal to the product of the tangents of the two parts adjacent to it. RULE II. 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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An Elementary Treatise on Plane & Spherical Trigonometry: With Their ...

Benjamin Peirce - 1845 - 449 pągines
...middle part is equal to Ike product of the tangents of the two adjacent parts. Napier's Rules. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only...
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