Question

In the given figure, the diagonals of rectangle $ABCD$ intersect at $O$. If $\angle AOB={64}^o$, find $\angle OCB$.

Answer 1

$\begin{array}{c}Given,\\ \angle AOB={64}^0\\ \angle AOB+\angle COB={180}^0\left[linearpair\right]\\ ={64}^0+\angle COB={180}^0\\ \angle COB={180}^0-{64}^0\\ \angle COB={116}^0\end{array}$

Hence,

We get $\angle COB={116}^0$.