 | Frederick Walter Simms - Leveling - 1856 - 258 pages
...together all the internal angles, marked by dotted segments of circles; and subtract their sum from twice as many right angles as the figure has sides, less four, for the angle db e. Example. — Let the angles denoted by the dotted segments at the different letters... | |
 | 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... | |
 | Adrien Marie Legendre, Charles Davies - Geometry - 1857 - 442 pages
...hence, in the equiangular hexagon, each angle is the sixth part of eight right angles, or £ of one. Cor. 4. In any equiangular polygon, any interior angle is equal to twice as many right angles, less four, as the figure has sides, divided by the number of angles. Scholium. When this proposition... | |
 | 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... | |
 | 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... | |
 | Elias Loomis - Conic sections - 1858 - 256 pages
...that is, together with four right angles (Prop. V., Cor. 2). Therefore the angles of the polygon are equal to twice as many right angles as the figure has sides, wanting four right angles. Cor. 1. The sum of the angles of a quadrilateral is four right angles ;... | |
 | 1860 - 462 pages
...must be aliquot parts of the circle or of four right angles. All the angles of any such figure are equal to twice as many right angles as the figure has sides minus four right angles, or if « be the number of sides, the sum of all the angles is (2n — 4) right... | |
 | 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... | |
 | 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... | |
 | Isaac Todhunter - Euclid's Elements - 1880 - 426 pages
...the foregoing Corollary all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Therefore all the interior angles of the figure, together with all its exterior angles, are equal to... | |
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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. 
