document.write( "Question 1060279: ABCDE is a regular pentagon. On AB, a square ABXY is drawn lying within the pentagon. Find the number of degrees in the angles CBX, DCX, and XBD. \n" ); document.write( "
Algebra.Com's Answer #675257 by KMST(5328)\"\" \"About 
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The sides of the pentagon and the sides of the square have all the same length,
\n" ); document.write( "so triangle BCX is isosceles, and DCB is isosceles.
\n" ); document.write( "\"ABX=90%5Eo\" because ABXY is a square.
\n" ); document.write( "\"ABC=108%5Eo\" because ABCDE is a pentagon,
\n" ); document.write( "with external angles measuring \"360%5Eo%2F5=72\" ,
\n" ); document.write( "and internal angles measuring \"180%5Eo-72%5Eo=108%5Eo\" .
\n" ); document.write( "So, \"ABC=BCD=108%5Eo\" .
\n" ); document.write( "\"CBX=ABC-ABX=108%5Eo-90%5Eo=highlight%2818%5Eo%29\"
\n" ); document.write( "Since BCX is isosceles, besides its \"CBX=18%5Eo\" vertex angle,
\n" ); document.write( "it has two base angles measuring
\n" ); document.write( "\"XCB=%28180%5Eo-CBX%29%2F2=%28180%5Eo-18%5Eo%29%2F2=162%5Eo%2F2=81%5Eo\" .
\n" ); document.write( "Since DCB is isosceles, besides its \"BCD=108%5Eo\" vertex angle,
\n" ); document.write( "it has two base angles measuring
\n" ); document.write( "\"CBD=%28180%5Eo-BCD%29%2F2=%28180%5Eo-108%5Eo%29%2F2=72%5Eo%2F2=36%5Eo\"
\n" ); document.write( "\"DCX=BCD-XCB=108%5Eo-81%5Eo=highlight%2827%5Eo%29\" .
\n" ); document.write( "\"XBD=CBD-CBX=36%5Eo-18%5Eo=highlight%2818%5Eo%29\"
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