document.write( "Question 503773: in a regualr pentagon ABCDE draw diagonal BE and then find the measure of angle BAE,ABE and BED \n" ); document.write( "
Algebra.Com's Answer #339444 by Edwin McCravy(20056)\"\" \"About 
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in a regualr pentagon ABCDE draw diagonal BE and then find the measure of angle BAE,ABE and BED
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document.write( "The sum of the interior angles of a polygon is given\r\n" );
document.write( "by the this formula, with N=5\r\n" );
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document.write( "(N-2)×180° = (5-2)×180° = 3×180° = 540°\r\n" );
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document.write( "And since all 5 interior angles of a regular polygon\r\n" );
document.write( "have equal measure, each interior angle is \"%22540%B0%22%2F5\" = 108°\r\n" );
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document.write( "Angle A is an interior angle so the measure of angle A is 108².\r\n" );
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document.write( "Triangle ABE is isosceles, its base angles have sum 180°-108° = 72°\r\n" );
document.write( "Each base angle is half that, so angle ABE has measure 36°.\r\n" );
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document.write( "Angle AED is an interior angle of the regular pentagon, so\r\n" );
document.write( "it has measure 108°.  Angle AEB is the other base angle of the\r\n" );
document.write( "isosceles triangle ABE, so it is 36°.\r\n" );
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document.write( "Angle BED = Angle AED - Angle AEB = 108°-36° = 72°  \r\n" );
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document.write( "Edwin
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