Question

The diagonals AC and BD of a rectangle ABCD intersect each other at the point O. If $\angle DAC={32}^{\circ },\angle AOB={70}^{\circ },$ then $\angle DBC$ is equal to :

• A. ${24}^{\circ }$
• B. ${88}^{\circ }$
• C. ${38}^{\circ }$
• D. ${32}^{\circ }$

Answer 1

Answer: (C)

${38}^{\circ }$

Given in the rectangle ABCD $\angle DAC=32°$

$\Rightarrow \angle AOB=70°$

$\Rightarrow \angle A=90°$

$\Rightarrow \angle DAC+\angle BAC=90°$

$\Rightarrow \angle BAC=90°-32°=58°$

From $△AOB$

$\Rightarrow \angle AOB+\angle OAB+\angle OBA=180°$ [ Sum of angle property ]

$\Rightarrow {58}^o+{70}^o+\angle OBA=180°$

$\therefore \angle OBA=180°-58°-70°=52°$

We know that $\angle B=90°$

$\Rightarrow \angle ABO+\angle DBC=90°$

$\Rightarrow \angle DBC=90°-52°=38°$

Hence, the answer is $38°.$