 | John Playfair - 1819 - 350 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 rettangle under the tangents of the adjacent parts ; or to the rectangle under the cosines of the opposite,... | |
 | Lant Carpenter - 1820 - 514 σελίδες
...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 - 708 σελίδες
...„';,, '„_ 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 - 516 σελίδες
...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 - 666 σελίδες
...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 - 572 σελίδες
...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 - 732 σελίδες
...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 - 764 σελίδες
...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 - 546 σελίδες
...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 - 434 σελίδες
...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.... | |
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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. 
