 | William Frothingham Bradbury - Geometry - 1872 - 262 pages
...the sum of the angles of all the triangles, that is, the sum of the interior angles of the polygon, is equal to twice as many right angles as the polygon has sides minus two. PRACTICAL QUESTIONS. 1. Do two lines that do not meet form an angle with each other ? Two... | |
 | Edward Olney - Geometry - 1872 - 562 pages
...one re-entrant angle. PROPOSITION XT. 253. TJieorem. — Tlie sum of the inferior angles of a polygon is equal to twice as many right angles as the polygon has sides, less four right angles. DEM. — Let n he the number of sides of any polygon ; then the sum of its... | |
 | Eli Todd Tappan - Geometry - 1873 - 288 pages
...among the equal parts. SUM OF THE ANGLES. 423. Theorem. — The sum of all the angles of a polygon is equal to twice as many right angles as the polygon has sides, less two. For the polygon may be divided into as many triangles as it has sides, less two (417); and... | |
 | Edward Olney - Geometry - 1872 - 472 pages
...re.entrant angle. FIG. 186. PROPOSITION XT. 253. Theorem. — The sum of the inferior angles of a polygon s* equal to twice as many right angles as the polygon has sides, less four right angles. DEM. — Let n be the number of sides of any polygon; then the sum of its angles... | |
 | Edward Atkins - 1874 - 428 pages
...circle, the sum of the angles in the segments exterior to the polygon, together with two right angles, is equal to twice as many right angles as the polygon has sides. 20. Draw the common tangents to two given circles. 21. From a given point draw a straight line cutting... | |
 | Adrien Marie Legendre - Geometry - 1874 - 500 pages
...similar. For, the corresponding angles in each are equal, because any angle in F( >C either polygon is equal to twice as many right angles as the polygon has sides, less four right angles, divided by the number of angles (B. I, P. XXVI , C. 4); and further, the corresponding... | |
 | Edward Atkins - 1874 - 426 pages
...circle, the stun of the angles in the segments exterior to the polygon, together with two right angles, is equal to twice as many right angles as the polygon has aides. SO. Draw the common tangents to two given circles. 21. From a given point draw a straight line... | |
 | William Alexander Willock - Circle - 1875 - 196 pages
...corresponding external is equal to two right angles, the sum of all the internal and all the external is equal to twice as many right angles as the polygon has angles, or sides. If we take from this latter sum the four right angles to which the external angles... | |
 | Aaron Schuyler - Geometry - 1876 - 384 pages
...adjacent, it is evident that the polygon will be divided into n -- 2 triangles. It is also evident that the sum of all the angles of all the triangles is equal to the sum of all the angles of the p>Iygon. But the sum of all the angles of one triangle is '2Jt. Hence,... | |
 | Elias Loomis - Conic sections - 1877 - 458 pages
...three angles of each of these triangles is equal to two right angles (Pr. 27) ; therefore the sum of the angles of all the triangles is equal to twice...many right angles as the polygon has sides. But the same angles are equal to the angles of the polygon, together with the angles at the point F, that is,... | |
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... therefore the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles... 
