| James Hodgson - Astronomy - 1723
...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 - Geometry - 1733 - 397 pages
...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... | |
| Mathematics - 1801
...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 - Nautical astronomy - 1828 - 776 pages
...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 - Spherical trigonometry - 1836 - 84 pages
...; 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... | |
| Thomas Kerigan - Nautical astronomy - 1838
...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,... | |
| Henry W. Jeans - Trigonometry - 1842 - 138 pages
...= 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 - Plane trigonometry - 1845 - 449 pages
...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 compare all the equations... | |
| Benjamin Peirce - Plane trigonometry - 1845 - 498 pages
...the other two parts are called the opposite parts. The two theorems are as follows. 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 compare all the equations... | |
| James Hann - Spherical trigonometry - 1849 - 84 pages
...two are called extremes disjunct*. These things being understood, the following is the general rule. **The sine of the middle part is equal to the product of the** tangents of the extremes conjunct. * Thus, if in figure page 12 we suppose В С, the angle B, and... | |
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