 | 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... | |
 | 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... | |
 | Elias Loomis - Conic sections - 1857 - 242 pages
...angles of a polygon, is equal to twice as many right angles, wanting four, as the figure has sides. Let ABCDE be any polygon ; then the sum of all its...B, C, D, E is equal to twice as many right angles, wanting four, as the figure has sides (see next page). For, from any point, F, within it, draw lines... | |
 | Adrien Marie Legendre, Charles Davies - Geometry - 1857 - 442 pages
...equal to twice as many right angles as the polygon has sides. Again, the sum of all the interior angles is equal to twice as many right angles as the figure has sides, less four right angles (p. 26). Hence, the interior angles plus four right angles, is equal to twice as many right angles... | |
 | 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. " Calculate... | |
 | Elias Loomis - Conic sections - 1858 - 256 pages
...angles of a polygon, is equal to twice as many right angles, wanting four, as the figure has sides. Let ABCDE be any polygon ; then the sum of all its...B, C, D, E is equal to twice as many right angles, wanting four, as the figure has sides (see next page). For, from any point, F, within it, draw lines... | |
 | 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 in another... | |
 | Charles Hutton - Mathematics - 1860 - 1020 pages
...Hence it lotIons that the sum of all the inward angles of the polygon alone, A -f- В — -f. D -f. E, is equal to twice as many right angles as the figure has side*, «am¡ng the said tour right angles- Q. !•'- D. THEOREM xx. When every side of any figure... | |
 | John Henry Robson - 1880 - 116 pages
...proved that " All the Interior angles of any Rectilineal figure, "together with four right angles, are equal to "twice as many right angles as the figure has " sides." If, therefore, we suppose the polygon to have n sides, All its interior angles + 4.90 .= 272.90 . -.... | |
 | Sir Norman Lockyer - Electronic journals - 1880 - 668 pages
...XXVI. of the syllabus, that the interior angles of any polygon, together with four right angles, are equal to twice as many right angles as the figure has sides. In the new notation we would say that the sum of the interior angles of the polygon is equal to a number... | |
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All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 
