Question

The diagonals of a rectangle ABCD intersect at O. If $\angle BOC={68}^{\circ }$, then $\angle ODA$ is

• A. ${56}^{\circ }$
• B. ${72}^{\circ }$
• C. ${112}^{\circ }$
• D. ${68}^{\circ }$

Answer 1

The correct option is A ${56}^{\circ }$


We have $\angle BOC={68}^{\circ }$
$\Rightarrow \angle AOD={68}^{\circ }$
[Vertically opposite angles]

Since the diagonals of a rectangle are equal and they bisect each other, $OA=OD.$

Thus, in $\Delta AOD$, we have OA = OD.

$\Rightarrow \angle ODA=\angle OAD$
[angles opp. to equal sides]

$\angle ODA+\angle OAD+\angle AOD={180}^{\circ }$
[using angle sum property ]

$\Rightarrow 2\angle ODA+{68}^{\circ }={180}^{\circ }$

$\Rightarrow 2\angle ODA={180}^{\circ }-{68}^{\circ }$

$\Rightarrow \angle ODA=\frac{{112}^{\circ }}{2}={56}^{\circ }$