9th Class
Mathematics
Geometry
Question Bank
Geometry
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 \[\angle DEC={{27}^{o}}\]