Question

In a rectangle ABCD, diagonals intersect at O. If $\angle$OAB$={30}^o$ find: $\angle$ACB, $\angle$ABO, $\angle$COD, $\angle$BOC.

Answer 1

In $△ABC,$

$\Rightarrow$ $\angle CAB+\angle ABC+\angle ACB={180}^o.$

$\Rightarrow$ ${30}^o+{90}^o+\angle ACB={180}^o.$

$\Rightarrow$ ${120}^o+\angle ACB={180}^o.$

$\therefore$ $\angle ACB={60}^o$

We know that, diagonals of rectangle are equal and bisect each other equally.

$\therefore$ $AO=OC=BO=OD$

In $△ABO$,

$\Rightarrow$ $AO=BO$

$\Rightarrow$ $\angle OAB=\angle ABO$ [ Angle opposite to equal side are also equal ]

$\Rightarrow$ $\angle OAB=\angle ABO={30}^o$

$\Rightarrow$ $\angle OAB+\angle ABO+\angle BOA={180}^o$

$\Rightarrow$ ${30}^o+{30}^o+\angle BOA={180}^o.$

$\Rightarrow$ $\angle BOA={120}^o.$

$\Rightarrow$ $\angle BOA=\angle COD$ [ Vertically opposite angle ]

$\therefore$ $\angle COD={120}^o$

$\Rightarrow$ $\angle COD+\angle BOC={180}^o$ [ Linear pair ]

$\Rightarrow$ ${120}^o+\angle BOC={180}^o$

$\therefore$ $\angle BOC={60}^o.$

$\Rightarrow$ $\angle ACB={60}^o,\angle ABO={30}^o,\angle COD={120}^o$ and $\angle BOC={60}^o.$