 | Enoch Lewis - 1844 - 240 σελίδες
...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... | |
 | Charles Davies - 1849 - 372 σελίδες
...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 - 524 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... | |
 | Adrien Marie Legendre - 1852 - 436 σελίδες
...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 - 1854 - 630 σελίδες
...EXTREMES disjunct from the middle or 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... | |
 | Thomas Jefferson - 1854 - 636 σελίδες
...EXTREMES disjunct from the middle or 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... | |
 | Charles Davies - 1854 - 436 σελίδες
...which, it must be remembered, apply to the circular parts, as already defined. 1st. Radius inlo 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 tidangle... | |
 | Charles Davies - 1872 - 464 σελίδες
...= tan (90°-a) tan b • ' • • (10.) Comparing these formulas with the figure, we see that, The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier'a rales for Circular Parts, and they... | |
 | Adrien Marie Legendre - 1874 - 500 σελίδες
...tan (90°- a) tan 1> • • • • (10.) Comparing these formulas with the figure, wo seo that, The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. These two rules are called Napier's rules for Circular Parts, and they... | |
 | William Guy Peck - 1876 - 376 σελίδες
...(10) Comparing these formulas with the diagram, we see that the following rule is always true: 2d. The sine of the middle part is equal to the rectangle of the tangents of the adjacent parts. Discussion. 48. In applying Napier's rules, the required part is always... | |
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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. 
