 | Edward Albert Bowser - 1890 - 420 Seiten
...Zs (n — 2) = 2n rt. Zs — 4 rt. Z s. Therefore, the sum of the angles of a polygon is also equal to twice as many right angles as the figure has sides, less four right angles. 149. COR. 2. The sum of the angles of a quadrilateral is equal to two right angles taken... | |
 | Edward Albert Bowser - 1890 - 418 Seiten
...rt. Zs (n - 2) = 2» rt. Zs - 4 rt. Z s. Therefore, the sum of the angles of a polygon is also equal to twice as many right angles as the figure has sides, less four right angles. 149. COR. 2. TJie sum of the angles of a quadrilateral is equal to two right angles taken... | |
 | John Fry Heather - 1890 - 252 Seiten
...Сor. 1). Hence the angles of a regular polygon are each equal to the quotient obtained by dividing twice as many right angles as the figure has sides, less four right angles, by the number of sides : —Thus, for a regular pentagon, or five.sided figure, all the... | |
 | Edward Albert Bowser - 1891 - 424 Seiten
...rt. Zs (n - 2) = 2n rt. Zs - 4 rt. Zs. Therefore, the sum of the angles of a polygon is also equal to twice as many right angles as the figure has sides, less four right angles. 149. COB. 2. The sum of the angles of a quadrilateral is equal to two right angles taken... | |
 | John Cresson Trautwine - 1893 - 204 Seiten
...together all the internal angles, marked by dotted portions 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 arcs at the different letters be... | |
 | George D. Pettee - 1896 - 272 Seiten
...adj. 4] .-. BO = DO AO = CO Proposition XXX 111. Theorem. The sum of the angles of a polygon is equal to twice as many right angles as the figure has sides, less four right angles. Appl. Cons. Dem. ED Prove A + B + C, etc. = (2 n - 4) rt. A Draw diagonals from one vertex.... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1896 - 276 Seiten
...sides in more than two points. PROPOSITION XVI. THEOREM 66. The sum of all the angles of any polygon is twice as many right angles as the figure has sides, less four right angles. Given ABCDE, any polygon, having n sides. To PROVE—the sum of its angles is in —... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1896 - 574 Seiten
...sides in more than two points. PROPOSITION XVI. THEOREM 66. The sum of all the angles of any polygon is twice as many right angles as the figure has sides, less four right angles. GIVEN ABCDE, any polygon, having n sides. To PROVE—the sum of its angles is 2« —... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1896 - 276 Seiten
...sides in more than two_points. PROPOSITION XVI. THEOREM 66. The sum of all the angles of any polygon is twice as many right angles as the figure has sides, less four right angles. GIVEN ABCDE, any polygon, having n sides. To PROVE—the sum of its angles is in —... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1897 - 374 Seiten
...in more than two points. PROPOSITION XVI. THEOREM 6'tT. The sum of all the angles of any polygon is twice as many right angles as the figure has sides, less four right angles. GIVEN ABCDE, any polygon, having n sides. To PROVE—the sum of its angles is 2n—i,... | |
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add up to twice as many Right angles as the figure has sides, less four. 
