document.write( "Question 1162398: ABCDE is a regular pentagon. Find the angles of triangle ADC. \n" ); document.write( "
Algebra.Com's Answer #786194 by Edwin McCravy(20056)\"\" \"About 
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document.write( "The sum of the interior angles of an n-sided polygon is\r\n" );
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document.write( "SUM = (n-2)∙180°\r\n" );
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document.write( "So for a pentagon, the sum is\r\n" );
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document.write( "SUM = (5-2)∙180° = 3∙180° = 540°\r\n" );
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document.write( "Since all interior angles of a regular pentagon are equal, we\r\n" );
document.write( "divide that by 5, and get 540°÷5 = 108°\r\n" );
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document.write( "So each of the interior angles of the pentagon measures 108°.\r\n" );
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document.write( "Since ∠B = 108° and ΔABC is isosceles, its base angles are 36° each, and the\r\n" );
document.write( "same for ΔAED.  \r\n" );
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document.write( "Since all three angles at A add to 108°, ∠DAC also = 36°.\r\n" );
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document.write( "Since ΔADC is isosceles, its base angles are equal and 72° each.\r\n" );
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document.write( "Now you have all the angles of ΔADC.\r\n" );
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document.write( "Edwin
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