 | John Keill - Logarithms - 1723 - 444 pages
...Part Thefe Things premifed. RULE. I, . In any Right-angled fpherical Triangle, the Reft angle tinder the Radius, and the Sine of the middle Part, is equal to theReftangle under the "tangents oftheadjatent Parts• . { -fc RULE 7" fre Reftangle under the Radius... | |
 | Euclid, John Keill - Geometry - 1733 - 444 pages
...Part, is equal to the Reftangle under the Iangents of the adjacent Parts, RULE RULE II. Ibe ReEf angle under the Radius, and the Sine of the middle Part, is equal to the Reff angle under the Cofines of the oppofite Parts. . . . Each of the Rules have three Cafes. For the... | |
 | John Keill - Geometry - 1782 - 476 pages
...Part, is equal to the ReSlangle under the Tangents of the adjactnt Parts. RULE \ RULE II. Reffangle under the Radius, and the Sine of the middle Part, is equal to the Reftangle un~ der tbe Cojine of the oppo/ite Parts. "Each of the Rules have three Cafes : For the middle... | |
 | Mathematics - 1801 - 658 pages
...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.*... | |
 | Robert Simson - Trigonometry - 1806 - 546 pages
...the 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... | |
 | Thomas Simpson - Trigonometry - 1810 - 168 pages
...; and which, being easily remembered, are frequently used in practice. Theor. 1 . The rectangle of radius and the sine of the middle part is equal to the rectangle of the tan gents of the adjacent extremes. Theor. 2. The rectangle of the radius and sine... | |
 | Francis Nichols - Plane trigonometry - 1811 - 162 pages
...contained 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 pages
...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... | |
 | John Playfair - Circle-squaring - 1819 - 350 pages
...circular part is contained in the following " * PROPOSITION. In a right angled spherical triangle, the rectangle under the radius and the sine of the middle part, is equal to the rettangle under the tangents of the adjacent parts ; or to the rectangle under the cosines of the opposite,... | |
 | Lant Carpenter - Education - 1820 - 514 pages
...canons for the solution of right-angled spherical triangles. Here the short sentence, " the rectangle of the radius and the sine of the middle part, is equal to the rectangle of the tangents of the extremes conjunct, or, of the cosines of the extremes disjunct," enables... | |
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In a right angled spherical triangle, the rectangle 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 parts. 
