 | Euclid, John Keill - 1733 - 444 páginas
...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 Part, viz.... | |
 | David Stewart Erskine Earl of Buchan, Walter Minto - 1787 - 164 páginas
...fpherical triangle, The product of the tangents of half the fum and half the difference of the fegments of the middle part is equal to the product of the tangents of half the fum and half the difference of the oppofite parts. Dem. For fince cof BA: cof BC :: cof DA:... | |
 | 1801 - 658 páginas
...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. This... | |
 | Thomas Kerigan - 1828 - 776 páginas
...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 equal to the product... | |
 | Benjamin Peirce - 1836 - 92 páginas
...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 the cosines of... | |
 | Benjamin Peirce - 1836 - 84 páginas
...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 the cosines of... | |
 | Thomas Kerigan - 1838 - 804 páginas
...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, is equal to the... | |
 | Henry W. Jeans - 1842 - 138 páginas
...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 of the cosines... | |
 | James Hann - 1849 - 82 páginas
...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 BC, the angle B, and the side А В to be... | |
 | Benjamin Peirce - 1852 - 382 páginas
...and the other two parts are called the opposite parts. The two theorems are as follows. I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the two... | |
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I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
