A) \[{{40}^{{}^\circ }}\]
B) \[{{65}^{{}^\circ }}\]
C) \[{{50}^{{}^\circ }}\]
D) \[{{60}^{{}^\circ }}\]
Correct Answer: B
Solution :
(b):- Given, \[\angle BAD={{50}^{{}^\circ }},\text{ }\angle ADC={{115}^{{}^\circ }}\] in cyclic quadrilateral ABCD. \[\angle BAD+\angle BCD={{180}^{{}^\circ }}\] \[\Rightarrow \] \[\angle BCD={{180}^{{}^\circ }}-{{50}^{{}^\circ }}={{130}^{{}^\circ }}\] Now \[\angle BCD+\angle DCP={{180}^{{}^\circ }}\] (Straight line) \[\angle DCP={{180}^{{}^\circ }}-{{130}^{{}^\circ }}={{50}^{{}^\circ }}\] And \[\angle ADC+\angle CDP={{180}^{{}^\circ }}\] (Straight line) \[{{115}^{{}^\circ }}+\angle CDP={{180}^{{}^\circ }}\Rightarrow \angle CDP={{65}^{{}^\circ }}\] Now in \[\Delta CPD\]. \[\angle DCP+\angle CDP+\angle DPC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[{{50}^{{}^\circ }}+{{65}^{{}^\circ }}+\angle DPC={{180}^{{}^\circ }}\] \[\angle DPC={{180}^{{}^\circ }}-{{115}^{{}^\circ }}={{65}^{{}^\circ }}\]
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