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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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