| Thomas Kerigan - 1828 - 664 sider
...premised, the required parts are to be computed by the two following equations ; viz., 1st. — The product **of radius and the sine of the middle part, is equal to the** product of the tangents of the extremes conjunct2d. — The product of radius and the sine of the middle... | |
| Thomas Curtis - 1829
...complement С are adjoining, and AB and В С opposite extremes. With these explanations Napier's rules are **1. The rectangle of radius and the sine of the middle part is equal to the rectangle of the** tangents of the adjoining extremes. 2. The rectangle of radius and the sine of the middle part is equal... | |
| 1832
...is equal to the rectangle under the tangents of the adjacent extremes." 2. "The rectangle under the **radius and the sine of the middle part is equal to the rectangle** under the cosines of the opposite extremes." The radius being unity does not appear in the formulae.... | |
| John Playfair - 1832 - 333 sider
...contained in the following PROPOSITION. In a right angled spherical triangle, the rectangle under the **radius and the sine of the middle part, is equal to the rectangle** under the tangents of the adjacent parts; or to the rectangle under the cosines of the opposite parts.... | |
| William Galbraith - 1834 - 428 sider
...obtained by the following THEOREM. In any right-angled spherical triangle, the rectangle under the **radius, and the sine of the middle part is equal to the rectangle** under the tangents of the adjacent parts ; or to the rectangle under the COSINES of the OPPOSITE parts.... | |
| Euclid, Robert Simson - 1835 - 513 sider
...angled spherical triangles are resolved with the greatest ease. RULE I. THE rectangle contained by the **radius and the sine of the middle part, is equal to the rectangle** contained by the tangents of the adjacent parts. RULE II. THE rectangle contained by the radius, and... | |
| Adrien Marie Legendre - 1836 - 359 sider
...Making A =90°, we have sin B sin C cos a=R cos B cos C, or R cos a=cot B cot C ; that is, radius into **the sine of the middle part is equal to the rectangle of the** tangent of the complement of B into the tangent ot the complement of C, that is, to the rectangle of... | |
| John Playfair - 1837 - 318 sider
...contained in the following PROPOSITION. In a right angled spherical triangle, the rectangle under the **radius and the sine of the middle part, is equal to the rectangle** under the tangents of the adjacent parts ; or', to the rectangle under the cosines of the opposite... | |
| Thomas Kerigan - 1838
...middle part, 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... | |
| Charles William Hackley - 1838 - 307 sider
...middle part is equal to the rectangle of the tangents of the adjacent parts. 2. Radius multiplied by **the sine of the middle part is equal to the rectangle of the** cosines of the opposite parts. Or both rules may be given thus : radius into the sine of the middle... | |
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