Question

ABCD is a quadrilateral whose diagonals intersect each other at the point O such that OA = OB = OD. If $\angle OAB={30}^{\circ }$, then the measures of $\angle ODA$ is:

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Given, $OA=OB$
$\therefore \angle OAB=\angle OBA$ (opp. angle of equal side is equal)
Thus, $\angle OAB=\angle OBA={30}^{\circ }$
In $\Delta OAB$
$\angle OAB+\angle AOB+\angle OBA={180}^{\circ }$ (Angle Sum Property)
${30}^{\circ }+\angle AOB+{30}^{\circ }={180}^{\circ }$
$\angle AOB={120}^{\circ }$
As DOB is a straight line
$\therefore \angle DOA={180}^{\circ }$ (Linear Pair)
$\angle DOA+\angle AOB={180}^{\circ }$
$\angle DOA={60}^{\circ }$
Now, in $\Delta AOD$
$\angle ODA+\angle DOA+\angle DAO{180}^{\circ }$
$2\angle ODA+{60}^{\circ }={180}^{\circ }[\because \angle ODA=\angle DOAasOA=OD]$
$2\angle ODA={120}^{\circ }$
$\angle ODA={60}^{\circ }$
Hence, option 'C' is correct.