Question

In a given rectangle ABCD, diagonals AD and BC intersect at O. If $\angle COD={120}^{\circ }$, what is the value of $\angle OBA$?

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

Answer 1

The correct option is D ${30}^{\circ }$

Given that $\angle COD={120}^{\circ }$.
Since ABCD is a rectangle, diagonals bisect each other and are equal.
i.e., AO = OC = DO = OB.
In $\Delta ODC,\angle DOC+\angle OCD+\angle ODC={180}^{\circ }$ [Angle sum property of $\Delta$]
${120}^{\circ }+2\angle OCD={180}^{\circ }$ [equal sides subtend equal angles]
$2\angle OCD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
$\angle OCD={30}^{\circ }=\angle ODC$
$\angle OCD=\angle OBA$ (Alternate interior angle)
$\therefore \angle OBA={30}^{\circ }$.