Q28 of 77 Page 16

The diagonals of a parallelogram ABCD intersect ay O. If ∠BOC =90° and ∠BDC=50°, then ∠OAB =

∠BOC is 90°

So,


∠COD and ∠AOB all should be 90° by linear pair


∠BDC is 50°,


So,


Now as in a parallelogram the opposite​ sides are equal


We say,


AB parallel to CD


∠DCA = 50°


So,


In triangle COA


∠C = 50° (Stated above)


∠COA = 90° (Proved above)


Therefore,


90° + 50° + x° = 180°


x = 40°

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