Question

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

If $\angle DAC={32}^o$ and $\angle AOB={70}^o$, then $\angle DBC$ is equal to?

• A. ${24}^o$
• B. ${86}^o$
• C. ${38}^o$
• D. ${32}^o$

Answer 1

The correct option is C ${38}^o$

In given figure,

Quadrilateral ABCD is a parallelogram.

So, AD $\left|\right|$ BC

$\therefore$ $\angle$DAC = $\angle$ACB --- ( Alternate angle)

$\therefore$ $\angle$ACB = 32${}^{\circ }$

$\angle$AOB + $\angle$BOC = 180${}^{\circ }$ --- (straight angle)

$\Rightarrow$70${}^{\circ }$ + $\angle$BOC = 180${}^{\circ }$

$\therefore$ $\angle$BOC = 110${}^{\circ }$

In $△$BOC,

$\angle$OBC + $\angle$BOC + $\angle$OCB = 180${}^{\circ }$

$\Rightarrow$$\angle$OBC + 110${}^{\circ }$ + 32${}^{\circ }$ = 180${}^{\circ }$

$\Rightarrow$ $\angle$OBC = 38${}^{\circ }$

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