A) \[{{50}^{o}}\]
B) \[{{100}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{20}^{o}}\]
Correct Answer: A
Solution :
BD is diameter of circle. So, \[\angle DCB={{90}^{o}}.\text{ }QCP\]is a straight line, \[\therefore \] \[\angle QCP={{180}^{o}}\Rightarrow {{40}^{o}}+{{90}^{o}}+\angle PCB={{180}^{o}}\] \[\Rightarrow \] \[\angle PCB={{50}^{o}}\]
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