| William Mitchell Gillespie - Surveying - 1856 - 478 pages
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
| Cambridge univ, exam. papers - 1856 - 200 pages
...Prove that all the internal angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides; and that all the external angles are together equal to four right angles. In what sense are these propositions... | |
| Henry James Castle - Surveying - 1856 - 220 pages
...angles are the exterior angles of an irregular polygon ; and as the sum of all the interior angles are equal to twice as many right angles, as the figure has sides, wanting four ; and as the sum of all the exterior, together with all the interior angles, are equal... | |
| Euclides - 1856 - 168 pages
...with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. XVI. If two triangles have two sides of the one equal to two sides of the other, each to each, and... | |
| William Mitchell Gillespie - Surveying - 1857 - 538 pages
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
| W. Davis Haskoll - Civil engineering - 1858 - 422 pages
...and in an irregular polygon they may be all unequal. The interior angles of a polygon are together equal to twice as many right angles as the figure has sides, less four. On this is based the theory of the traverse, of which further explanation will be given... | |
| Elias Loomis - Conic sections - 1858 - 256 pages
...these triangles, is equal to tw» right angles (Prop. XXVII.) ; therefore the sum of the angles of all the triangles, is equal to twice as many right angles as the polygon has sides. But the same angles are equal to the angles of the polygon, together with the angles... | |
| Horatio Nelson Robinson - Geometry - 1860 - 470 pages
...triangles is equal to two right angles, (Th. 11) ; and the sum of the angles of all the triangles must be equal to twice as many right angles as the figure has sides. But the sum of these angles contains the sum of four right angles about the point p ; taking these away, and the remainder... | |
| Moffatt and Paige - 1879 - 474 pages
...together with four right angles. But it has been proved that all the angles of all these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure, together with four right angles, are equal to twice as many... | |
| William Mitchell Gillespie - Surveying - 1880 - 540 pages
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
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