 | Popular educator - 1854 - 922 pages
...into three equal parts. *"'t 3Fig. .42. No. 3. interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with all the exterior angles are equal (Ax. 1) to all the... | |
 | William Mitchell Gillespie - Surveying - 1855 - 436 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... | |
 | Euclides - 1855 - 270 pages
...and there are as many triangles in the figure as it has sides, all the angles of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz. ABС, BСD,... | |
 | William Mitchell Gillespie - Surveying - 1856 - 478 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... | |
 | Euclides - 1856 - 168 pages
...with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. XVI. If two triangles have two sides of the one equal to two sides of the other, each to each, and... | |
 | 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... | |
 | 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... | |
 | Henry James Castle - Surveying - 1856 - 220 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... | |
 | Adrien Marie Legendre, Charles Davies - Geometry - 1857 - 442 pages
...triangles in the figure ; that is, as many times as there are sides, less two. But this product is equal to twice as many right angles as the figure has sides, less four right angles. Cor. 1. The sum of the interior angles in a quadrilateral is equal to two right angles multiplied by... | |
 | 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... | |
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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. 
