 | James Hodgson - Astronomy - 1723 - 724 pages
...15?/; 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... | |
 | Euclid, John Keill - Geometry - 1733 - 444 pages
...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,... | |
 | Ralph Griffiths, G. E. Griffiths - Books - 1766 - 722 pages
...diftances of the Moon and ftar, and their diflance from each other being given nearly. . * Add together the tangents of. half the fum, and half the difference of the zenith diftances of the Moon and ftar, and the cotangent of half the diftance of the Moon from the... | |
 | Several Hands - 1766 - 596 pages
...diftances of the Moon and ftar, and their diilance from each other being given nearly. * Add together the tangents of half the fum, and half the difference of the zenith diftances of the Moon and ftar, and the cotangent of half the diftnnce of the Moon from the... | |
 | Mathematics - 1801 - 658 pages
...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
...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... | |
 | Benjamin Peirce - Spherical trigonometry - 1836 - 92 pages
...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... | |
 | Benjamin Peirce - Spherical trigonometry - 1836 - 84 pages
...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... | |
 | Thomas Kerigan - Nautical astronomy - 1838 - 804 pages
...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... | |
 | Henry W. Jeans - Trigonometry - 1842 - 138 pages
...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... | |
| |
I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
