 | 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. "... | |
 | 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 ;... | |
 | 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... | |
 | 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... | |
 | Royal college of surgeons of England - 1860 - 332 pages
...two right angles ; and all the angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 6. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects... | |
 | Robert Potts - Geometry, Plane - 1860 - 380 pages
...there are as many triangles as the figure has sides, therefore all the angles of these triangles are equal to twice as many right angles as the figure has sides ; but the same angles of these triangles are equal to the interior angtef of the figure together with... | |
 | 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... | |
 | William Schofield Binns - 1861 - 238 pages
...Euc. I., 32, Cor. 1, "All the angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides." From this corollary, we can deduce a formula for finding the angle of any polygon. Let x equal the... | |
 | Isaac Sharpless - Geometry - 1879 - 282 pages
...But ACD+ACB = 2R; BAC+ABC+ACB = 2R. Corollary 1.—All the interior angles of a polygon are together equal to' twice as many right angles as the figure has sides, minus four right angles. Let ABODE be a polygon, and let n represent the number of its sides. Draw... | |
 | University of Madras - 1879 - 674 pages
...I. Prove 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. II. Prove the proposition to which the following is a corollary : The difference of the squares on... | |
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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. 
