The diagonals AC and BD of a parallelogram AB | vedclass

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(A) In parallelogram $ABCD,$ $AD \parallel BC$ and $AC$ is a transversal.Therefore,$\angle DAC = \angle ACB$ (Alternate interior angles).Given $\angle

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) In parallelogram $ABCD,$ $AD \parallel BC$ and $AC$ is a transversal.Therefore,$\angle DAC = \angle ACB$ (Alternate interior angles).Given $\angle DAC = 32^{\circ},$ so $\angle ACB = 32^{\circ}.$Now,consider $\Delta BOC.$ The angle $\angle AOB$ is an exterior angle to $\Delta BOC$ at vertex $O.$By the exterior angle theorem,the exterior angle of a triangle is equal to the sum of the two interior opposite angles.Therefore,$\angle AOB = \angle OCB + \angle OBC.$Substituting the known values,$70^{\circ} = 32^{\circ} + \angle OBC.$Thus,$\angle OBC = 70^{\circ} - 32^{\circ} = 38^{\circ}.$Since $\angle DBC$ is the same as $\angle OBC,$ we have $\angle DBC = 38^{\circ}.$

The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point $O.$ If $\angle DAC = 32^{\circ}$ and $\angle AOB = 70^{\circ},$ then $\angle DBC$ is equal to: (in $^{\circ}$)

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