| John Playfair - 1819 - 333 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** rettangle under the tangents of the adjacent parts ; or to the rectangle under the cosines of the opposite,... | |
| Lant Carpenter - 1820 - 477 sider
...for the solution of right-angled spherical triangles. Here the short sentence, " the rectangle of the **radius and the sine of the middle part, is equal to the rectangle of the** tangents of the extremes conjunct, or, of the cosines of the extremes disjunct," enables the calculator... | |
| 1821
...„';,, '„_ These equations, when applied to-right-angled spheric triangles, signify as before thtit **the sine of the middle part is equal to the rectangle of the** tangents of the adjacent parts, or to the rectangle of the co-sines of the opposite parts ; but when... | |
| Rev. John Allen - 1822 - 494 sider
...under the cosines of AB and AC. PROP. XXIV. THEOR. The same things being supposed, the rectangle under **radius and the sine of the middle part, is equal to the rectangle** under the tangents of the parts, which, the right angle being excluded, are adjacent to the middle... | |
| James Mitchell - 1823 - 576 sider
...anplesof aright angled spherical triangle. Napier's general rule is this: the rectangle, under the **radius and the sine of the middle part, is equal to the rectangle** under the tangents of the adjacent parts, and to the rectangle under the cosines of the opposite parts.... | |
| Edward Riddle - 1824
...terms, Napier's Rules for the solution of the different cases of right angled spherical triangles 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... | |
| Nathaniel Bowditch - 1826 - 617 sider
...parts. , ''These equations, when applied to right-angled «phenc triangles, signify as before, H. that **the sine of the middle part is equal to the rectangle of the** tangents of the adjacent parts, 01 to the rectangle of the co-eiiies of the opposite parts : but when... | |
| Nathaniel Bowditch - 1826 - 617 sider
...Sine П1Ш< „art These equations, when applied to right-angled spheric triangles, signify as that **the sine of the middle part is equal to the rectangle of the** tangents of the ¡ parts, or to the rectangle of the co-sines of the opposite parts ; but when appui... | |
| Robert Simson - 1827 - 513 sider
...rectangle contained by the tangents of the adjacent parts. RULE II. The rectangle contained by the **radius and the sine of the middle part, is equal to the rectangle** contained, by the cosines of the opposite parts. These rules are demonstrated in the following manner... | |
| Dionysius Lardner - 1828 - 317 sider
...equal to the rectangle under the tangents of the adjacent extremes.1' 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.... | |
| |