Question

The diagonals of a rectangle ABCD intersect at o. if $\angle$ BOC =${70}^{\circ }$. Find $\angle$ODA.

Answer 1

Since in a rec diagonal are equal and bisect each other

$\therefore OD=OA$

and $\angle S$ opp to equal sides are equal

$\therefore \angle 1=x$

Now, In $\Delta ODA$

$\angle DOA={70}^0$ (vertically opp. $\angle S$)

$x+\angle 1+70={180}^0$ ($\angle$ sum prop)

and $\angle 1=x$

$\therefore x+x+70=180$

$2x=110$

$x=\frac{110}{2}$

$x={55}^0$

$\Rightarrow \angle ODA={55}^0$