 | James Hodgson - 1723 - 724 pàgines
...Geom. "Tbeor.) Axes, rtc~ct,¿f axct,l> a c. That is, the Radius multiplied into the Co-fine of the Middle Part, is equal to the Product of the Tangents of the Extreams Adjunft, for as the Co-fine of д с is the Sine of the Middle Part, fo the Co-tangents of... | |
 | Euclid, John Keill - 1733 - 444 pàgines
...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 Part, viz. the Complement... | |
 | 1801 - 658 pàgines
...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. This... | |
 | Thomas Kerigan - 1828 - 776 pàgines
...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 equal to the product... | |
 | Benjamin Peirce - 1836 - 84 pàgines
...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 the cosines of... | |
 | Benjamin Peirce - 1836 - 92 pàgines
...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 the cosines of... | |
 | Thomas Kerigan - 1838 - 804 pàgines
...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, is equal to the... | |
 | Henry W. Jeans - 1842 - 138 pàgines
...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 of the cosines... | |
 | James Hann - 1849 - 82 pàgines
...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 BC, the angle B, and the side А В to be... | |
 | William Chauvenet - 1852 - 268 pàgines
...considered, the two sides including it are regarded as adjacent. The rules are : I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. The correctness of these... | |
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I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
