 | John Playfair - Euclid's Elements - 1842 - 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 parts... | |
 | Enoch Lewis - Conic sections - 1844 - 240 pages
...the sine of the middle part is equal to the rectangle of the tangents of the adjacent extremes. 2. The rectangle of radius and the sine of the middle part is equal to the rectangle of the cosines of the opposite extremes. These rules may be explained and demonstrated in the following manner... | |
 | Euclid, John Playfair - Euclid's Elements - 1846 - 334 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... | |
 | Charles Davies - Trigonometry - 1849 - 372 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 of the complement of C, that is, to the rectangle of... | |
 | Charles William Hackley - Trigonometry - 1851 - 524 pages
...right angled triangle may be expressed in the two following rules of Napier : 1. Radius multiplied by the sine of the 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... | |
 | Charles William Hackley - Trigonometry - 1851 - 536 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... | |
 | Adrien Marie Legendre - Geometry - 1852 - 436 pages
...have, sin B sin 0 cos a — cos B cos G, or, cos a — cot B cot (7; that is, radius, which is 1, into the sine of the middle part is equal to the rectangle of the tangent of the complement of B, into the tangent of the complement of (7, that is, to the rectangle... | |
 | Thomas Jefferson - United States - 1854 - 630 pages
...EXTREMES DISJUNCT. He then laid down his catholic rule, to wit : " The rectangle of the radius, and sine of the middle part, is equal to the rectangle of the tangents of the two EXTREMES CONJUNCT, and to that of the cosines of the two EXTREMES DISJUNCT." And to aid our recollection... | |
 | Thomas Jefferson - United States - 1854 - 636 pages
...EXTREMES DISJUNCT. He then kid down his catholic rule, to wit : " The rectangle of the radius, and sine of the middle part, is equal to the rectangle of the tangents of the two EXTREMES CONJUNCT, and to that of the cosines of the two EXTREMES DISJUNCT." And to aid our recollection... | |
 | Charles Davies - Geometry - 1854 - 436 pages
...have, sin B sin C cos a = cos B cos C, or, cos a = cot B cot C ; that is, radius, which is 1, into the sine of the middle part is equal to the rectangle of the tangent of the complement of B, into the tangent of the complement of (7, that is, to the rectangle... | |
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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. 
