Question

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

पर्याय

• 40°

• 50°

• 10°

• 90°

Answer 1

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).