Question

The diagonals of a rectangle ABCD meet at O. If $\angle BOC={44}^o$, find $\angle OAD$

• A. ${68}^o$
• B. ${58}^o$
• C. ${62}^o$
• D. ${67}^o$

Answer 1

The correct option is A ${68}^o$

We have
$\angle BOC+\angle BOA={180}^o$ [Linear pairs]

$\Rightarrow {44}^o+\angle BOA={180}^o$

$\Rightarrow \angle BOA={136}^o$
Since diagonals of a rectangle are equal and they bisect each other.

Therefore, in $\Delta OAB$, We have
$OA=OB$
$\Rightarrow \angle 1=\angle 2[\therefore$ Angle opp. to equal sides are equal]

In $\Delta OAB$, we have
$\angle 1+\angle 2+\angle BOA={180}^o$
$\Rightarrow 2\angle 1+{136}^o={180}^o$
$\Rightarrow 2\angle 1={44}^o$

$\Rightarrow \angle 1={22}^o$

Since each angles of a rectangle is a right angle.
$\angle BAD={90}^o$

$\Rightarrow \angle 1+\angle 3={90}^o$
$\Rightarrow {22}^o+\angle 3={90}^o$

$\Rightarrow \angle 3={68}^o$

Hence, $\angle OAD={68}^o$