 | Thomas Kerigan - Nautical astronomy - 1828 - 776 pages
...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 - Aeronautics - 1829 - 878 pages
...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... | |
 | Encyclopedias and dictionaries - 1832 - 636 pages
...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 - Euclid's Elements - 1832 - 358 pages
...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 - Astronomy - 1834 - 454 pages
...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 - 1835 - 540 pages
...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 - Geometry - 1836 - 394 pages
...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 - Geometry - 1837 - 332 pages
...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 - Nautical astronomy - 1838 - 804 pages
...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 - Trigonometry - 1838 - 338 pages
...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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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. 
