Question

ABCD is a parallelogram in which $\angle DAO={40}^{\circ }$,$\angle BAO={35}^{\circ }and\angle COD={65}^{\circ }$ . Find $\angle ODC$.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${80}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${80}^{\circ }$

$\angle AOB=\angle COD={65}^{\circ }$ [Vertically Opposite Angles]
In $\Delta AOB,\angle OAB+\angle AOB+\angle ABO={180}^{\circ }$
$\Rightarrow {35}^{\circ }+{65}^{\circ }+\angle OBA={180}^{\circ }$
$\Rightarrow \angle OBA={80}^{\circ }$
$\angle ODC=\angle OBA={80}^{\circ }$ [Alternate interior angles as DC $\parallel$ AB.]