A) \[60{}^\circ \]
B) \[110{}^\circ \]
C) \[70{}^\circ \]
D) \[80{}^\circ \]
Correct Answer: C
Solution :
\[\angle BAD=\frac{1}{2}\angle BOD=\frac{1}{2}\times {{140}^{o}}={{70}^{o}}\] \[\angle DCP=\angle BAD\][Since exterior angle of a cyclic quadrilateral is equal to the interior opp. angle.] \[\therefore \]\[\angle DCP=\angle BAD={{70}^{o}}\]
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