| W.M. Gillespie, A.M., Civ. Eng - 1855
...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
...superposition. 3. 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 - 185 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
...straight lines from a point F within the figure to each of the angles. And by the last proposition all the **angles of these triangles are equal to twice as many right angles as** there are triangles or sides to the figure. And the same angles are equal to the internal angles of... | |
| William Mitchell Gillespie - Surveying - 1856 - 464 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. "... | |
| William Mitchell Gillespie - Surveying - 1857 - 524 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 - 234 pages
...F, 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 - 324 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... | |
| Surveying - 1878 - 508 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. "... | |
| Thomas Hunter - Geometry, Plane - 1878 - 132 pages
...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,... | |
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