 | James Hodgson - 1723 - 724 σελίδες
...t,¿z¿xct,í7<r— Axes, bac; that is the Radius multiplied into the Sine of the Complement of the Angle a or Sine of the Middle Part, is equal to the Product of the Tangent of ab one of the Extreams, into the Tangent of the Complement of л с the other Extream. By... | |
 | Euclid, John Keill - 1733 - 444 σελίδες
...S, CF=Cof. BC and T, DF = Cot. B. Wherefore R x Cof. BC=Cot. Cx Cot. B ; that is, Radius drawn into the Sine of the. middle Part, is equal to the Product of the Tangents of the adjacent extreme Parts. : X And And BA, AC, are the oppofite Extremes to the faid middle... | |
 | 1801 - 658 σελίδες
...solutions of all the cases of right-angled spherical triangles. THEOREM VII. The product of radius and the sine of the middle part is equal to the product of the tangents of the conjunct extremes, or to that of the cosines of the disjunct extremes.* NOTE. * DEMONSTRATION.... | |
 | Thomas Kerigan - 1828 - 776 σελίδες
...parts are to be computed by the two following equations ; viz., 1st. — The product of radius and the sine of the middle part, is equal to the product of the tangents of the extremes conjunct2d. — The product of radius and the sine of the middle part, is... | |
 | Benjamin Peirce - 1836 - 84 σελίδες
...; and the other two parts are called the opposite parts. The two theorems are as follows. (474) I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. (47e) II. The sine of the middle part is equal to the product of... | |
 | Benjamin Peirce - 1836 - 92 σελίδες
...; and the other two parts are called the opposite parts. The two theorems are as follows. (474) I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. (475) II. The sine of the middle part is equal to the product of... | |
 | Thomas Kerigan - 1838 - 804 σελίδες
...parts are to be computed by the two following equations ; viz., 1st. — The product of radius and the sine of the middle part, is equal to the product of the tangents of the extremes conjunct. 2d. — Tlie product of radius and the sine of the middle part,... | |
 | Roswell Park - 1841 - 722 σελίδες
...sine of the middle part, will be equal to the product of the tangents of the adjacent parts, and also equal to the product of the cosines of the opposite parts. These rules, called Napier's Analogies, may be applied to oblique angled spherical triangles ; by dividing them... | |
 | Henry W. Jeans - 1842 - 138 σελίδες
...= cos- P/=cos. co. A. CP C/P/ PN P/N, tan. A = — = = cot. P,= cot. co. A Are. CN C,N, Gl RULE I. The sine of the middle part is equal to the product of the tangents of the two parts adjacent to it. RULE II. The sine of the middle part is equal to the product... | |
 | Benjamin Peirce - 1845 - 498 σελίδες
...the middle part is equal to Ike product of the tangents of the two adjacent parts. Napier's Rules. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only necessary to... | |
| |
I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
