question_answer

ED is chord parallel to the diameter AC of a circle. A point B is on the perimeter of the circle such that angle CBE =\[63{}^\circ \]. The angle DEC is equal to

A) \[63{}^\circ \]

B) \[42{}^\circ \]

C) \[31.5{}^\circ \]

D)        \[27{}^\circ \]  

Correct Answer: D

Solution :

 Since D, C, B, E are concyclic, therefore \[\angle A={{180}^{o}}-\angle B\] \[={{180}^{o}}-{{63}^{o}}={{107}^{o}}\] Also DE is parallel to AC, therefore                 \[\angle DEC=\angle ECA=y\,(say)\] Also                       \[\angle XCD=\angle DEC=y\] Let                          \[\angle DCE=x\] Since                      \[AC\bot XC\] therefore            \[+x+y={{90}^{o}}\] or                            \[x+2y={{90}^{o}}\] or                            \[x={{90}^{o}}-2y\] Also from \[\Delta \,CDE,\]                 \[{{107}^{o}}+x+y={{180}^{o}}\] or            \[{{90}^{o}}-2y+y={{180}^{o}}-{{107}^{o}}={{63}^{o}}\] or            \[y={{90}^{o}}-{{63}^{o}}={{27}^{o}}\] Hence   \[\angle DEC={{27}^{o}}\]