Question

In a parallelogram ABCD, diagonals AC and BD intersect at O.

If $\angle DAC={41}^{\circ }$ and $\angle AOB={70}^{\circ }$. Then $\angle DBC=$

• A. ${39}^{\circ }$
• B. ${29}^{\circ }$
• C. ${41}^{\circ }$
• D. ${51}^{\circ }$

Answer 1

The correct option is B

${29}^{\circ }$

$\angle AOB=\angle DOC$ (vertically opposite angles)

$\angle AOD=\angle BOC$ (vertically opposite angles)

$\angle AOB+\angle DOC+\angle AOD+\angle BOC={360}^{\circ }$

$\angle AOD={110}^{\circ }$

$\angle OAD+\angle AOD+\angle ADO={180}^{\circ }$ (sum of angles in a triangle is ${180}^{\circ }$)

$\angle ADO={29}^{\circ }$

$\angle ADO=\angle DBC$ (alternate interior angles)

$\angle DBC={29}^{\circ }$