| James Hodgson - 1723 - 724 páginas
...the Angle с the other Extream ; wherefore, &c, as was to be proved. Rule i. The Rectangle under the Radius and the Sine of the Middle Part, is equal to the Produft of the Co-fines of the Extreams Disjunct, thus if the Complement of ac be taken for the Middle... | |
| Euclid, John Keill - 1733 - 444 páginas
...Part. Thefe Things premifed. RULE I. In any right-angled fpherical Triangle, the ReBangle under the Radius, and the Sine of the middle Part, is equal to the Reftangle under the Iangents of the adjacent Parts, RULE RULE II. Ibe ReEf angle under the Radius,... | |
| John Keill - 1782 - 476 páginas
...Part. Thcfc Things premifed, RULE I. In any Right-angled Spherical Triangle, the Re£langle under the Radius, and the Sine of the middle Part, is equal to the ReSlangle under the Tangents of the adjactnt Parts. RULE \ RULE II. Reffangle under the Radius, and... | |
| 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
...are sufficient for the 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. This... | |
| Robert Simson - 1806 - 546 páginas
...rectangle contained by the tangents o' the adjacent parts. RULE IT. The rectangle contained by the radius, and the sine of the middle part is equal to the rectangle contained by the co-sines of the opposite parts. These rules are demonstrated in the following... | |
| Samuel Webber - 1808 - 520 páginas
...are sufficient for the solutions of all the cases of right.angled spheric triangles. THEOREM VII. . The product of radius and the sine of the middle part is equal to tht- product of the tangents* of the adjacent extremes, or to that of the cosines of the apposite extremes.f... | |
| Thomas Simpson - 1810 - 152 páginas
...solved ; and which, being easily remembered, are frequently used in practice. 1 Theor. 1. The rectangle of radius and the sine of the middle part is equal to the rectangle of the tangents of the adjacent extremes. Theor. 2. The rectangle of the radius and sine... | |
| Francis Nichols - 1811 - 162 páginas
...in the following proposition. 100. In a right-angled spherical triangle, the rectan* gle under the radius and the sine of the middle part is equal to the rectangle under the tangents of the adjacent parts, or to the rectangle under the cosines of the opposite... | |
| Euclides - 1816 - 588 páginas
...angled spherics! triangles are resolved with the greatest ease. RULE I. The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the tangents of the adjacent parts. RULE II. The rectangle contained by the... | |
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