| John Playfair - 1842 - 317 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... | |
| Enoch Lewis - 1844 - 228 sider
...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... | |
| John Playfair - 1846 - 317 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... | |
| Charles Davies - 1849 - 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 of the complement of C, that is, to the rectangle of... | |
| Charles William Hackley - 1851 - 372 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... | |
| A. M. LEGENDRE - 1852
...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, Henry Augustine Washington - 1854
...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, Adrien Marie Legendre - 1854 - 432 sider
...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, Henry Augustine Washington - 1854
...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 - 178 sider
...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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