 | Charles Hutton - Mathematics - 1831 - 632 pages
...right angles. THEOHEM xtx. IN any figure whatever, the sum of nil the inward angles, taken together, is equal to twice as many right angles, wanting four, as the figure has sidi s. Let ABCDE be any figure ; then the sum of all its inward angles, л 4- в + с -f- D + E, is... | |
 | William Grier - Mechanical engineering - 1832 - 366 pages
...(Fig. 17.) 19. Let ABCDE be any figure; then the sum of all its inward angles, AfBf- C -|- D -|- E, is equal to twice as many right angles, wanting four, as the figure has sides (Fig. 18.) 20. Let A, B, C, &c., be the outward angles of any polygon, made by producing all the sides;... | |
 | Robert Mudie - Mathematics - 1836 - 524 pages
...triangles are equal to all the angles of the figure ; therefore, again, all the angles of the figure are equal to twice as many right angles wanting four as the figure has sides. We are now in possession of all the more important elementary cases in which angles can be shown to... | |
 | Robert Mudie - Mathematics - 1836 - 542 pages
...exterior angles are always equal to four right angles, all the interior angles of every rectilineal figure are equal to twice as many right angles wanting four, as the figure has sides. It is only the sum of the angles that is known, except when they are all equal, and then each angle... | |
 | William Grier - Mechanical engineering - 1836 - 384 pages
...\p_cj 19. Let ABCDE be any figure ; then the sum of all its inward angles, A -f- B -f- C + D -f- E, is equal to twice as many right angles, wanting four, as the figure has sides. 20. Let A, B, C, &c., be the outward angles of any polygon, made by producing y all the sides ; then... | |
 | 1838 - 1014 pages
...be divided into as many triangles as it has sides. 2. The angles of any polygons taken together make twice as many right angles, wanting four, as the figure has sides. Thus, if the polygon has five sides, the double of that is ten, from which subtracting four leaves... | |
 | Charles Davies - Geometrical drawing - 1840 - 262 pages
...second, the similar triangles FHG, FBI, and F1K. 6. The sum of all the inward angles of any polygon is equal to twice as many right angles, wanting four, as the figure has sides. Thus, if the polygon has five sides, we have A + B + C+ D+E—W right angles — 4 right angles =6... | |
 | William Grier - Mechanical engineering - 1842 - 320 pages
...right angles. 19. Let ABCDE be any figure; then the sum of all its inward angles, A+B + C + D + E, is equal to twice as many right angles, wanting four, as the figure has sides. 20. Let A, B, C, &c., be the outward angles of any polygon, made by producing all the sides; then will... | |
 | Charles Davies - Geometrical drawing - 1846 - 254 pages
...of all the inward angles of a polygon equal to ? The sum of all the inward angles of any polygon is equal to twice as many right angles, wanting four, as the figure has sides. Thus, if the polygon has five sides, we have A + B+C+D + E=10 right angles — 4 right angles = 6 right... | |
 | Charles William Hackley - Geometry - 1847 - 248 pages
...in another, the triangles are equal. THEOREM XVI. The sum of all the inward angles of a polygon is equal to twice as many right angles, wanting four, as the figure has sides. Let ABODE be any figure ; then the sum of all its inward angles, A + B + C + D + E, is equal to twice... | |
| |
Angles, taken together, is equal to Twice as many Right Angles, wanting four, as the Figure has Sides. 
