Question

In Fig. 10.16 $\angle OAB={30}^{\circ }and\angle OCB={57}^{\circ }$ Find $\angle BOCand\angle AOC$

Answer 1

$\Rightarrow$ $\angle OAB={30}^o$ and $\angle OCB={57}^o$.

In $△AOB,$

$\Rightarrow$ $AO=OB$ [ Radius of a circle ]

$\Rightarrow$ $\angle OBA=\angle BAO={30}^o$ [ Angles opposite to equal sides are equal ]
In $△AOB,$

$\Rightarrow$ $\angle AOB+\angle OBA+\angle BAO={180}^o.$

$\therefore$ $\angle AOB+{30}^o+{30}^o={180}^o$

$\therefore$ $\angle AOB={180}^o-{30}^o-{30}^o$

$\therefore$ $\angle AOB={120}^o$ ----- ( 1 )

Now, in $△OCB,$

$\Rightarrow$ $OC=OB$ [ Radius of a circle ]

$\Rightarrow$ $\angle OBC=\angle OCB={57}^o.$ [ Angles opposite to equal sides are equal ]

In $△OCB,$

$\Rightarrow$ $\angle BOC+\angle OCB+\angle OBC={180}^o.$

$\Rightarrow$ $\angle BOC+{57}^o+{57}^o={180}^o$

$\Rightarrow$ $\angle BOC={180}^o-{114}^o$

$\therefore$ $\angle BOC={66}^o$ ------ ( 2 )

$\Rightarrow$ $\angle AOB={120}^o$ [ From ( 1 ) ]

$\Rightarrow$ $\angle AOC+\angle COB={120}^o.$

$\Rightarrow$ $\angle AOC+{66}^o={120}^o$

$\Rightarrow$ $\angle AOC={120}^o-{66}^o$

$\therefore$ $\angle AOC={54}^o$