SOLUTION: 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.

Algebra ->  Geometry-proofs -> SOLUTION: 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.      Log On


   

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.
Answer by KMST(5328) About Me  (Show Source):
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The sides of the pentagon and the sides of the square have all the same length,
so triangle BCX is isosceles, and DCB is isosceles.
ABX=90%5Eo because ABXY is a square.
ABC=108%5Eo because ABCDE is a pentagon,
with external angles measuring 360%5Eo%2F5=72 ,
and internal angles measuring 180%5Eo-72%5Eo=108%5Eo .
So, ABC=BCD=108%5Eo .
CBX=ABC-ABX=108%5Eo-90%5Eo=highlight%2818%5Eo%29
Since BCX is isosceles, besides its CBX=18%5Eo vertex angle,
it has two base angles measuring
XCB=%28180%5Eo-CBX%29%2F2=%28180%5Eo-18%5Eo%29%2F2=162%5Eo%2F2=81%5Eo .
Since DCB is isosceles, besides its BCD=108%5Eo vertex angle,
it has two base angles measuring
CBD=%28180%5Eo-BCD%29%2F2=%28180%5Eo-108%5Eo%29%2F2=72%5Eo%2F2=36%5Eo
DCX=BCD-XCB=108%5Eo-81%5Eo=highlight%2827%5Eo%29 .
XBD=CBD-CBX=36%5Eo-18%5Eo=highlight%2818%5Eo%29