 | John Playfair - 1842 - 332 páginas
...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 - 1844 - 240 páginas
...the adjacent extremes ; and the other two are termed the opposite extremes. Then 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 adjacent extremes. 2. The rectangle of radius and the sine of the middle part is equal... | |
 | Euclid, John Playfair - 1846 - 334 páginas
...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 - 1849 - 372 páginas
...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 - 1851 - 536 páginas
...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 - 1852 - 436 páginas
...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 - 1854 - 630 páginas
...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... | |
 | Charles Davies - 1854 - 436 páginas
...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... | |
 | Thomas Jefferson - 1854 - 636 páginas
...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... | |
 | Elias Loomis - 1855 - 192 páginas
...value of the part required may then be found by the following RULE OF NAPIER. (211.) The product of the radius and the sine of the middle part, is equal to the product of the tangents of the adjacent parts, or to the product of the cosines of the opposite parts.... | |
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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. 
