Question

Suppose $ABCD$ is a rectangle whose diagonals $AC$ and $BD$ intersect at $O$. If $\angle OAB={62}^{\circ }$, find $\angle OBC$

Answer 1

The diagonals of a rectangle are equal and bisect each other. So,
$OA=OB$

$⟹\angle OBA=\angle OAB={62}^{\circ }$ (angles opposite to equal sides are equal)
Since the measure of each angle of rectangle is ${90}^{\circ }$
$⟹\angle ABC={90}^{\circ }$
$⟹\angle ABO+\angle OBC={90}^{\circ }$
$⟹{62}^{\circ }+\angle OBC={90}^{\circ }$
$⟹\angle OBC={90}^{\circ }-{62}^{\circ }$$={28}^{\circ }$