 | W. Davis Haskoll - 1858 - 422 Seiten
...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... | |
 | Robert Potts - 1860 - 380 Seiten
...the interior angles. But 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, that is, if ir be assumed to designate two right angles, .'. »9 + 2ir = mr, and nfl = mr — 2ir =... | |
 | Royal college of surgeons of England - 1860 - 332 Seiten
...equal to 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... | |
 | 1860 - 462 Seiten
...figure 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... | |
 | John Daniel Runkle - 1860 - 460 Seiten
...figure 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 (2 n — 4)... | |
 | Horatio Nelson Robinson - 1860 - 470 Seiten
...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... | |
 | Euclides - 1860 - 142 Seiten
...polygon be produced to meet, the sum of the salient angles thus formed, with eight right angles, will be equal to twice as many right angles as the figure has sides. Let ABCDE be a polygon, and let its sides produced meet in F, G, H, T, K ; then the sum of the salient... | |
 | William Schofield Binns - 1861 - 238 Seiten
...pentagon. From 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... | |
 | 1878 - 534 Seiten
...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. "... | |
 | Thomas Hunter - 1878 - 142 Seiten
...other, the remaining angles must be equal. Cor. 2. The sum of all the interior angles of a polygon is equal to twice as many right angles as the figure has sides, minus four right angles. In the case of the triangle, this corollary has just been demonstrated; for,... | |
| |
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. 
