1.

In figure, 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° …(1) [Angles in same segment] 

ABCD is a cyclic quadrilateral : 

Sum of opposite angles = 180 degrees 

∠A +∠C = 180° 

75° + ∠C = 180° 

∠C = 105° 

Again, 

∠ACB + ∠ACD = 105° 

∠ACB + 58° = 105° 

or ∠ACB = 47° …(2) 

Now, 

∠ACB = ∠ADB = 47° [Angles in same segment] 

Also, ∠D = 77° (Given) 

Again From figure, ∠BDC + ∠ADB = 77° 

∠BDC + 47° = 77° 

∠BDC = 30° 

In triangle DPC 

∠PDC + ∠DCP + ∠DPC = 180° 

30° + 58° + ∠DPC = 180° 

or ∠DPC = 92° . Answer!!



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