 | Lant Carpenter - 1820 - 514 σελίδες
...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... | |
 | Edward Riddle - 1824 - 572 σελίδες
...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 - 764 σελίδες
...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... | |
 | Nathaniel Bowditch - 1826 - 732 σελίδες
..., ''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... | |
 | Thomas Curtis - 1829 - 878 σελίδες
...С opposite extremes. With these explanations 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 adjoining extremes. 2. The rectangle of radius and the sine of the middle part is equal... | |
 | Adrien Marie Legendre - 1836 - 394 σελίδες
...triangles, which it must be remembered apply to the circular parts, as already defined. 1st. Radius into the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part is equal to the rectangle... | |
 | Charles William Hackley - 1838 - 338 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 part = the rectangle of... | |
 | Charles William Hackley - 1838 - 328 σελίδες
...angled triangle may be expressed in the two following rules of Napier : 1. Radius multiplied by the sine of the 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... | |
 | Enoch Lewis - 1844 - 234 σελίδες
...the sine of the middle part is equal to the rectangle of the tangents of the adjacent extremes. 99 part is equal to the rectangle of the cosines of the opposite extremes. These rules may be explained and demonstrated in the following manner : c Let ABC be the triangle,... | |
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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. 
