Question

The diagonals of a parallelogram $ABCD$ intersect at $O.$ If $\angle BOC={90}^{\circ }$ and $\angle BDC={50}^{\circ },$ then $\angle OAB$
(a) ${40}^{\circ }$
(b) ${50}^{\circ }$
(c) ${10}^{\circ }$
(d) ${90}^{\circ }$

Answer 1

The correct option is (a) ${40}^{\circ }$

We have, $ABCD$ is a parallelogram with diagonals $AC$ and $BD$ intersecting at $O.$

It is given that $\angle BOC={90}^{\circ }$ and $\angle BDC={50}^{\circ }$.

We need to find $\angle OAB$

Now, $\angle BOC+\angle COD={180}^{\circ }$ (Linear pair)

$\Rightarrow {90}^{\circ }+\angle COD={180}^{\circ }$

$\Rightarrow \angle COD={90}^{\circ }$

Since, $O$ lies on $BD.$

Therefore, $\angle ODC=\angle BDC$

$\angle ODC={50}^{\circ }$

By angle sum property of a triangle, we get

$\angle ODC+\angle COD+\angle OCD={180}^{\circ }$

$\Rightarrow {50}^{\circ }+{90}^{\circ }+\angle OCD={180}^{\circ }$

$\Rightarrow {140}^{\circ }+\angle OCD={180}^{\circ }$

$\Rightarrow \angle OCD={40}^{\circ }$

Since, $O$ lies on $AC.$

Therefore,

$\angle ACD=\angle OCD$

$\angle ACD={40}^{\circ }$

Also, $DC\parallel AB$

Therefore,

$\angle OAB=\angle ACD$ [ $\because \angle OAB=\angle CAB$]

$\angle OAB={40}^{\circ }$