 | 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... | |
 | 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... | |
 | 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... | |
 | 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... | |
 | Elizabethan club - 1880 - 156 pages
...the obtuse angles. 3. All the angles of a rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. A floor has to be laid with tiles in the form of regular figures all equal and similar ; show what... | |
 | 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 . -.... | |
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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. 
