 | John Playfair - 1842 - 332 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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 σελίδες
...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. 
