Question

In the given figure, $ABCD$ is a rectangle and diagonals intersect at $O$.
If $\angle AOB={118}^{\circ }$, find
(i) $\angle ABO$
(ii) $\angle ADO$
(iii) $\angle OCB$

Answer 1

From the figure it is given that, $ABCD$ is a rectangle and diagonals intersect at $O$

$\angle AOB={118}^{\circ }$

(i) Consider the $\Delta AOB$

$\angle OAB=\angle OBA$

Let us assume $\angle OAB=\angle OBA=y^{\circ }$

We know that, sum of measures of interior angles of triangle is equal to ${180}^{\circ }$.

$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$

$y+y+{118}^{\circ }={180}^{\circ }$

$2x+{118}^{\circ }={180}^{\circ }$

By transposing we get, $2y={180}^{\circ }-{118}^{\circ }$

$2y={62}^{\circ }$

$y=62/2$

$y={31}^{\circ }$

$So,\angle OAB=\angle OBA={31}^{\circ }$

Therefore, $\angle ABO={31}^{\circ }$

(ii) We know that sum of liner pair angles is equal to ${180}^{\circ }.$

$\angle AOB+\angle AOD={180}^{\circ }$

${118}^{\circ }+\angle AOD={180}^{\circ }$

$\angle AOD={180}^{\circ }-{118}^{\circ }$

$\angle AOD={62}^{\circ }$

Now consider the $△AOD$

Let us assume the $\angle ADO=\angle DAO=x$

$\angle AOD+\angle ADO+\angle DAO={180}^{\circ }$

${62}^{\circ }+x+x={180}^{\circ }$

${62}^{\circ }+2x={180}^{\circ }$

By transposing we get, $2x={180}^{\circ }-62$

$2x={118}^{\circ }$

$x={118}^{\circ }/2$

$x={59}^{\circ }$

Therefore, $\angle ADO={59}^{\circ }$

(iii) $\angle OCB=\angle OAD={59}^{\circ }$... [because alternate angles are equal]