| James Hodgson - Astronomy - 1723
...t,¿z¿xct,í7<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... | |
| Euclid, John Keill - Geometry - 1733 - 393 pages
...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.... | |
| David Stewart Erskine Earl of Buchan, Walter Minto - Logarithms - 1787 - 134 pages
...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:... | |
| Mathematics - 1801
...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... | |
| Thomas Kerigan - Nautical astronomy - 1828 - 664 pages
...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... | |
| Benjamin Peirce - Spherical trigonometry - 1836 - 71 pages
...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... | |
| Benjamin Peirce - Spherical trigonometry - 1836 - 71 pages
...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... | |
| Thomas Kerigan - Nautical astronomy - 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... | |
| Henry W. Jeans - Trigonometry - 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... | |
| Benjamin Peirce - Plane trigonometry - 1845 - 449 pages
...other two parts are called the opposite parts. The two theorems are as follows. 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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