The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O.

If ∠DAC = 32° and ∠AOB = 70°, then find the measure of ∠DBC.

24°

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86°

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32°

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38°

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Solution

The correct option is D 38°
In given figure,
Quadrilateral ABCD is a parallelogram.
So, AD ∣∣ BC
∴ ∠DAC = ∠ACB --- ( Alternate angle)
∴ ∠ACB = 32°

∠AOB + ∠BOC = 180° --- (Linear pair)
⇒70° + ∠BOC = 180° ---- (∠AOB = 70°)
∴ ∠BOC = 180° - 70° = 110°

In △BOC,
$∠ O B C + ∠ B O C + ∠ O C B = 180 °$
$⟹ ∠ O B C + 110 ° + 32 ° = 180 ° (∠OCB = ∠ACB = 32° & ∠BOC = 110°)$
$⟹ ∠ O B C = 180 ° - 110 ° - 32 °$
$⟹ ∠ O B C = 38 °$
$∴ ∠ D B C = 38 °$

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The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32° and ∠AOB = 70°, then find the measure of ∠DBC.

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O.

If ∠DAC = 32° and ∠AOB = 70°, then find the measure of ∠DBC.