Q. 34.0( 4 Votes )

# In Fig. 16.195, *ABCD* is a cyclic quadrilateral in which ∠*BAD*=75°, ∠*ABD*=58° and ∠*ADC*=77°, *AC* and *BD* intersect at *P.* Then, find ∠*DPC*.

Answer :

∠DBA = ∠DCA = 58° (Angles on the same segment)

In triangle DCA

∠DCA + ∠CDA + ∠DAC = 180°

58° + 77° + ∠DAC = 180°

∠DAC = 45°

∠DPC = 180° - 58° - 30°

= 92°

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