Question

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

Answer 1

∠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°