 | Abel Flint - 1818 - 184 páginas
...40' 23 20 1 95 00 48 20 CASE IV. y Tht three Sides given to find the Angles. Fig. 50. The solution of this CASE depends on the following PROPOSITION....half the Sum of the Segments, that is to half the krigth of the longest Side will give the greatest Segment ; and this half Difference subtracted from... | |
 | Charles Hutton - 1818 - 646 páginas
...right-angled triangles : then the proportion will be, As , As the base, or sums of the segments, 1s to the sum of the other two sides ; ' So is the difference of those sides, To the diff. of the segments of the base. Then take half this difference of the segments,... | |
 | William Nicholson - 1821 - 356 páginas
...986786 To the side DC required 144.8 2.16U76 .Axiom If. In any plane triangle, as the base, or greater side, is to the sum of the other two sides ; so is the difference of the sides to the difference of the segments of the base, made by a perpendicular let fall from the... | |
 | William Nicholson - 1821 - 356 páginas
...2.03743 9-99116 To the side DC required 144.8 Aaeiom IV. In any plane triangle, as the base, or greater side, is to the sum of the other two sides ; so is the difference of the sides to the difference of the segments of the base, made by a perpendicular let fall from the... | |
 | Charles Hutton - 1822 - 616 páginas
...into two right-angled triangles : then the proportion will be. As the base, or sums of the segments, Is to the sum of the other two sides ; So is the difference of those sides, To the diff. of the segments of the base. Then take half this difference of ihe segments,... | |
 | Edward Riddle - 1824 - 572 páginas
...appropriate analogy in the first case. 3. When the three sides are given to find the angles. As the base, or longest side, is to the sum of the other two sides, so is the difference of those sides to the difference of the segments of the base, made by a perpendicular from the opposite... | |
 | Abel Flint - 1825 - 252 páginas
...TRIGONOMETRY. CASE IV. The three Sides given to find the Angles. Fig. 50. Fig. 50. A. 105 B 2J The solution of this CASE depends on the following PROPOSITION....half the length of the longest Side, will give the greatest Segment ; and this half difference subtracted from the half sum will be the lesser Segment.... | |
 | Abel Flint - 1830 - 322 páginas
...given tojlnd the angles. Fig. 49. The solution of this CASE depends on the following PROPOSITION. Itf EVERY PLANE TRIANGLE, AS THE LONGEST SIDE IS TO THE...half the length of the longest side, will give the greatest segment ; and this half difference subtracted from the half sum will be the lesser segment.... | |
 | Tobias Ostrander - 1833 - 172 páginas
...angles, or the diameter of a circle circumscribing the triangle. Rule—As the whole length of the base is to the sum of the other two sides, so is the difference of those sides to the difference of the segments of the base; half the said difference added to half... | |
 | James Hann, Isaac Dodds - 1833 - 234 páginas
...planes 50 inches ; required their common height, and the ratio of the weights. Rule. — As the base or longest side is to the sum of the other two sides, so is their difference to the difference of the segments of the base. And half the difference of the segments,... | |
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As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base. 
