Question

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 ∠DBC is equal to:

• A. 24°
• B. 86°
• C. 38°
• D. 32°

Answer 1

The correct option is C

38°

$Given,\angle AOB={70}^{\circ }and\angle DAC={32}^{\circ }$

$\therefore \angle ACB={32}^{\circ }$ [AD ∥ BC and AC is transversal]

Now, $\angle AOB+\angle BOC={180}^{\circ }\left[Linearpairaxiom\right]$

$\Rightarrow \angle BOC={180}^{\circ }-\angle AOB={180}^{\circ }-{70}^{\circ }={110}^{\circ }$

$Now,in\Delta BOC,wehave$

$\angle BOC+\angle BCO+\angle OBC={180}^{\circ }$ [by angle sum property of a triangle]

$\Rightarrow {110}^{\circ }+{32}^{\circ }+\angle OBC={180}^{\circ }[\therefore \angle BCO=\angle ACB={32}^{\circ }]$

$\Rightarrow \angle OBC={180}^{\circ }-({110}^{\circ }+{32}^{\circ })={38}^{\circ }$

$\therefore \angle DBC=\angle OBC={38}^{\circ }$