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 \[\angle DEC={{27}^{o}}\]You need to login to perform this action.
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