| John Dougall - 1810
...be equal to twice as many right angles as there are triangles, that is, as the figure has sides, but **the sum of all the angles of all the triangles is equal to** the sum of all the angles of the pol ygor^ together with the angles at the central point G, which by... | |
| Charles Hutton - Mathematics - 1812
...three angles of each of these triangles, is equal to two right angles (th. 17) ; therefore the snm «f **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 about the point *p> which are so many many of the angles... | |
| Thomas Leybourn - Mathematics - 1814 - 420 pages
...Hence, by adopting the notation in the question, we have But the sum of the angles of any polygon being **equal to twice as many right angles as the polygon has sides,** less four; the sum of all the angles of the polygon will be equal to an even number of right angles,... | |
| Charles Hutton - Mathematics - 1816
...three angles of each of these triangles, is equal to two right angles (th 17*; therefore the sum of **the angles of all the triangles is equal to twice as many** iig:>t angles as the figure has sides. But the sum of all the angles about the point r, which are so... | |
| Charles Hutton - Mathematics - 1822 - 618 pages
...three angles of each of these triangles, is equal to two right angles (th. 17) ; 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 about the point p, which are so / many of the angles... | |
| John Radford Young - Euclid's Elements - 1827 - 208 pages
...to say, the sum of the angles of the polygon, together with those about the point within . it, are **equal to twice as many right angles as the polygon has sides ; but** those angles which are' about the point, amount to four right angles, (Prop. VI. Cor. 2.) deducting... | |
| Adrien Marie Legendre - Geometry - 1836 - 359 pages
...figure has sides, wanting four right angles. Hence, the interior angles plus four right .. -i angles, **is equal to twice as many right angles as the polygon has** aides, and consequently, equal to the sum of the interior angles plus the exterior angles. Taking from... | |
| Nathan Scholfield - 1845 - 896 pages
...of each as there arc sides of the polygon ; hence, the sum of all the interior and exterior angles **is equal to twice as many right angles as the polygon has sides.** Again, the sum of all the interior angles is equal to two right angles, taken as many times, less two,... | |
| William Scott - Measurement - 1845 - 290 pages
...end ought to coincide. Also, the sum of all the angles, together with four right angles, ought to be **equal to twice as many right angles as the polygon has sides** (Eue. i. 32. cor.). To find the angle contained by two straight lines conceived to be drawn from a... | |
| Anna Cabot Lowell - Geometry - 1846 - 161 pages
...into as many triangles as it has sides. The sum of all the angles of each triangle = 2 RA, therefore **the sum of all the angles of all the triangles is equal to** 2 RA multiplied by the number of sides in the figure. But the angles (equal to 4 RA) about the point... | |
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