Question

Question 12
The diagonals AC and BD of a parallelogram ABCD intersect each other. The point O. If $\angle DAC={32}^{\circ }and\angle AOB={70}^{\circ },then\angle DBC$ is equal to
(A) ${24}^{\circ }$
(B) ${86}^{\circ }$
(C) ${38}^{\circ }$
(D) ${32}^{\circ }$

Answer 1

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

$\therefore \angle ACB={32}^{\circ }$ [AD || BC and AC is transversal]
Now, $\angle AOB+\angle BOC={180}^{\circ }$ [Linear pair axiom]
$\Rightarrow \angle BOC={180}^{\circ }-\angle AOB={180}^{\circ }-{70}^{\circ }={110}^{\circ }$
Now, in $\Delta BOC$, we have
$\angle BOC+\angle BCO+\angle OBC={180}^{\circ }$ [ by angle sum property of a triangle]
$\Rightarrow {110}^{\circ }+{32}^{\circ }+\angle OBC={180}^{\circ }[\because \angle BCO=\angle ACB={32}^{\circ }]\Rightarrow \angle OBC={180}^{\circ }–(110+32{)}^{\circ }={38}^{\circ }$
$\therefore \angle DBC=\angle OBC={38}^{\circ }$