Question

is the figure, ABCD is a rectangle and $\angle BOC={56}^0$ .Find $\angle ADO$.

Answer 1

$ABCD$ is a rectangle

$\angle BOC={56}^o$

$\angle AOD={56}^o$ [vertically opposite $\angle S$]

$AO=DO$ [diagonals]

$\therefore \angle OAD=\angle ADO=x$

Now $△AOD$ is $\alpha △\therefore$

$\therefore \angle ADO+\angle OAD+{56}^o=180$

$x+x+{56}^o=180$

$\therefore x={62}^o\therefore \angle ADO={62}^o$