| Abel Flint - 1818 - 168 Seiten
...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
...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
...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
...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
...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 - 551 Seiten
...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 - 241 Seiten
...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 - 284 Seiten
...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 - 159 Seiten
...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 - 208 Seiten
...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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