 | James Hodgson - 1723 - 724 pągines
...Tangent of the Complement of 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... | |
 | John Keill - 1723 - 444 pągines
...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 - 1733 - 444 pągines
...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 - 1772 - 462 pągines
...will be, as Cof. C : Cof. BA : : S, B : R. And R xCof, C=Cof. BA XS, B; that is, Radius drawn into the Sine of the middle Part, is equal to the Rectangle under the Cofines of the oppofite Extremes. Cafe 2. Let the Complement of the Hypothenufe EC be the middle Part;... | |
 | John Keill - 1782 - 476 pągines
...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... | |
 | 1801 - 658 pągines
...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 - 1806 - 546 pągines
...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 manner:... | |
 | Euclid - 1810 - 554 pągines
...angled spherical 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 rectwpgle contained by the tangents of the adjaeent parts. '/ RULE II. The rectangle contained by the... | |
 | Thomas Simpson - 1810 - 168 pągines
...; 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 of the middle... | |
 | Francis Nichols - 1811 - 162 pągines
...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 parts. 'Note.... | |
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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 The... 
