Question

$ABCD$ is a rectangle with $\angle BAC={48}^{\circ }$. Then $\angle DBC$ equals

• A. ${38}^{\circ }$
• B. ${48}^{\circ }$
• C. ${132}^{\circ }$
• D. ${42}^{\circ }$

Answer 1

The correct option is D ${42}^{\circ }$

In $\Delta AOB$,
$BO\approx OA$ (Diagonals of a rectangle are equal and bisects each other)
$\Rightarrow OAB=\angle OBA={48}^{\circ }$ (isos. $\Delta$ property)
$\Rightarrow DBA=\angle OBA={48}^{\circ }$
$\Rightarrow DBC=\angle CBA-\angle DBA$
$={90}^{\circ }-{48}^{\circ }={42}^{\circ }$