The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32°, ∠AOB = 70°, then ∠DBC is equal to:
<DBC = 38°
ABCD is a parallelogram .
AC,BD diagonals intersecting at "O".
<DAC = 32°, <AOB = 70°
i) AB//DC , AC is a transversal.
<DAC = <BCA = 32°
/* Alternate angles */
ii) In ∆BCO , CO extended to A.
Sum of interior opposite angles = Exterior angle at O
=> <BCA + <OBC = <AOB
=> <BCA + <DBC = <AOB
=> 32° + <DBC = 70°
=> <DBC = 70°-32°
=> <DBC = 38°
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[co interior angles]
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