| Lant Carpenter - 1820 - 477 sider
...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... | |
| Edward Riddle - 1824
...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
...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 - 617 sider
..., ''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
...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 - 359 sider
...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 - 307 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 part = the rectangle of... | |
| Charles William Hackley - 1838 - 307 sider
...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 - 228 sider
...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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