| Edward Atkins - 1874 - 428 pages
...inscribed in a 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... | |
| Edward Atkins - 1874 - 426 pages
...inscribed in a 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... | |
| 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... | |
| 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... | |
| William Guy Peck - Conic sections - 1876 - 412 pages
...angles of each triangle is two right angles ; hence, the sum of all the angles of all the triangles is equal to twice as many right angles as the polygon has sides. But, the sum of the angles of the polygon is equal to the sum of all the angles of all the triangles... | |
| Dublin city, univ - 1876 - 420 pages
...EUCLID AND TRIGONOMETRY. i . Prove that the sum of the internal angles of any polygon, together with four right angles, is equal to twice as many right angles as the figure has sides. Trisect an angle of 45o. 2. Construct a square equal to a given rectilinear figure.... | |
| Euclides, James Hamblin SMITH - 1876 - 382 pages
...Edition of Euclid. COB. 1. The sum of the interior angles of any rectilinear figwre together withfow right angles is equal to twice as many right angles as the figure has sides. Let ABCDE be any rectilinear figure. Take any pt. F within the figure, and from F... | |
| Edward Olney - Geometry - 1877 - 272 pages
...one re-entrant angle. PROPOSITION XV. 233. TJieorem, — The sum of the interior angles of a polygon is equal to twice as many right angles as the polygon has sides, less four right angles. FIo. 187. DEM. — Let n be the, number of sides of any polygon; then the sum... | |
| Elias Loomis - Conic sections - 1877 - 458 pages
...triangles is equal to two right angles (Pr. 27) ; therefore the sum of the angles of all the triangles is equal to twice as 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,... | |
| William Frothingham Bradbury - Geometry - 1880 - 260 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. \ DEFINITIONS. 126.. Every proposition has an hypothesis (19), and a conclusion. Thus in... | |
| |