Sum of all exterior angles of a polygon = 360 deg.

Legend : D = the; S = sum; A = angles; P = polygon; RP = regular polygon; IN = interior;

EX = exterior;  O = others;
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Using the key: The sum of the interior angles of a polygon is 1/4 [that of] a regular polygon with n exterior angles of 12 degrees. If one of the interior angles is 140 degrees and the others are each equal to 2.5X, find X.

The RP's exterior angle=12=360/n, making n=30, and interior angle=180-360/n. Sum of interior angles of regular polygon=n(180-360/n)=180n-360=180(n-2). 1/4 of this is 45(n-2)=45*28=1260. Let M=number of angles and sides of the irregular polygon. The irregular polygon has sum of interior angles=140+2.5X(M-1). Sum of interior angles is 140+2.5X(M-1)=45(n-2). Therefore 140+2.5X(M-1)=1260, and M-1=1120/2.5X=448/X the sum of the exterior angles of P=360=40+(M-1)(180-2.5X), or (M-1)(180-2.5X)=320 and M-1=320/(180-2.5X). Therefore, 448/X=320/(180-2.5X) or X/448=(180-2.5X)/320. 5X/7=180-2.5X, 45X/14=180, X=180*14/45=56 degrees. Also M-1=448/56=8, so M=9.

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