A) \[125{}^\circ \]
B) \[\angle 1+\angle 2+\angle 3+\angle 4<\angle 5+\angle 6+\angle 7+\angle 8\]
C) \[\angle 1+\angle 2+\angle 3+\angle 4>\angle 5+\angle 6+\angle 7+\angle 8\]
D) \[AB=2x+5,\]
Correct Answer: B
Solution :
Given: \[\therefore \] As we know, \[\Delta ABC\] \[\angle BAC\] \[\angle ECD={{30}^{o}}\] Now, APCB is a cyclic quadrilateral, and sum of opposite angles in quadrilateral is \[\angle BAC\] \[{{30}^{o}}\] \[{{40}^{o}}\].
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