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Question
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
Options
40°
50°
10°
90°
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Solution
ABCD is a parallelogram with diagonals AC and BD intersect at O.
It is given that ∠BOC = 90° and∠BDC = 50°.
We need to find ∠OAB
Now,
∠BOC +∠COD = 180° (Linear pair)
90° + ∠COD = 180
∠COD = 90°
Since, O lies on BD.
Therefore,
∠ODC = ∠BDC
∠ODC = 50°
By angle sum property of a triangle, we get:
∠ODC + ∠COD + ∠OCD = 180°
50° + 90° + ∠OCD = 180°
140° + ∠OCD = 180°
∠OCD = 40°
Since, O lies on AC.
Therefore,
∠ACD = ∠OCD
∠ACD = 40°
Also, DC || AB
Therefore,
∠PAB = ∠ACD
∠OAB = 40°
Hence the correct choice is (a).
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