Diagonals of a parallelogram ABCD intersect at O. If ∠AOB = 90° and ∠BDC = 40°, then ∠OAB is __________.

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Solution

Let ∠OAB be x.

Since, AB || DC
Therefore, ∠DBA = ∠BDC = 40°

In ∆AOB,
∠OAB + ∠ABD + ∠AOB = 180°
⇒ x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° − 130°
⇒ x = 50°

Hence, ∠OAB is 50°.

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Diagonals of a parallelogram ABCD intersect at O. If ∠AOB = 90° and ∠BDC = 40°, then ∠OAB is __________.

Let ∠OAB be x. Since, AB || DC Therefore, ∠DBA = ∠BDC = 40° In ∆AOB, ∠OAB + ∠ABD + ∠AOB = 180° ⇒ x + 40° + 90° = 180° ⇒ x + 130° = 180° ⇒ x = 180° − 130° ⇒ x = 50° Hence, ∠OAB is 50°.

Let ∠OAB be x.

Since, AB || DC
Therefore, ∠DBA = ∠BDC = 40°

In ∆AOB,
∠OAB + ∠ABD + ∠AOB = 180°
⇒ x + 40° + 90° = 180°
⇒ x + 130° = 180°
⇒ x = 180° − 130°
⇒ x = 50°

Hence, ∠OAB is 50°.