Question

The diagonals of a rectangle $ABCD$ intersect each other in $O$. If $\angle AOD$ is ${30}^{\circ }$, then find $\angle OCD.$

Answer 1

In $△OBC,$

$\angle AOD=\angle BOC={30}^{\circ }($vertically opposite angles$)$

As we know that diagonals of a rectangle are equal and bisect each other so,

$OB=OC$

In isoceles $△BOC$,

$\angle OBC=\angle OCB=x$

Now, by angle sum property of a triangle,

$\begin{array}{cc}{30}^{\circ }+x+x& ={180}^{\circ }\\ 2x& ={180}^{\circ }-{30}^{\circ }\\ & ={150}^{\circ }\\ x& ={75}^{\circ }\end{array}$

So, $\angle OCB={75}^{\circ }$

We know that each vertex angle of a rectangle is ${90}^{\circ }.$ So,

$\begin{array}{cc}\angle OCD+\angle OCB& ={90}^{\circ }\\ \angle OCD+{75}^{\circ }& ={90}^{\circ }\\ \angle OCD& ={90}^{\circ }-{75}^{\circ }\\ \angle OCD& ={15}^{\circ }\end{array}$

So we have angle $\angle OCD={15}^{\circ }$