Question

The diagonals of a parallelogram ABCD intersect at O. If $BOC={90}^{\circ }and\angle BDC={50}^{\circ }$ then $\angle OAB=$

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${10}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A

${40}^{\circ }$

In || gm ABCD, diagonals AC and BD intersect each other at O.

$\angle BOC={90}^{\circ }$, $\angle BDC={50}^{\circ }$

$\because \angle BOC={90}^{\circ }$
$\therefore \angle COD={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

In $\Delta COD$, $\angle OCD={90}^{\circ }-{50}^{\circ }={40}^{\circ }$

But $\angle OAB=\angle OCD$ (Alternate angles)

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