Question
In Fig. $\angle\text{OAB} = 30^\circ$ and $\angle\text{OCB} = 57^\circ.$ Find $\angle\text{BOC}$ and $\angle\text{AOC}.$ 
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Answer
Given, $\angle\text{OAB}=30^\circ$ and $\angle\text{OCB}=57^\circ$ In
$\triangle\text{AOB},\ \ \ \text{AO}=\text{OB}$ [both are the radius of a circle]
$\Rightarrow\angle\text{OBA}=\angle\text{BAO}=30^\circ$ [angle opposite to equal sides are equal] In $\triangle\text{AOB},$
$\Rightarrow\angle\text{AOB}+\angle\text{OBA}+\angle\text{BAO}=180^\circ$ [by angle sum property of a triangle]
$\therefore\angle\text{AOB}+30^\circ+30^\circ=180^\circ$
$\therefore\angle\text{AOB}=180^\circ-2(30^\circ)$
$=180^\circ-60^\circ=120^\circ\ \ \ ...(\text{i})$
Now, in $\triangle\text{OCB}, OC = OB[ $ both are the radius of a circle$]$
$\Rightarrow\angle\text{OBC}=\angle\text{OCB}=57^\circ$ [angle opposite to equal sides are equal] In $\triangle\text{OCB}$
$\angle\text{COB}+\angle\text{OCB}+\angle\text{CBO}=180^\circ[$ by angle sum property of triangle$]$ $\therefore\angle\text{COB}=180^\circ-(\angle\text{OCB}+\angle\text{OBC})$
$=180^\circ-(57^\circ+57^\circ)$
$=180^\circ-114^\circ=66^\circ\ \ ...(\text{ii})$ From Eq. $(i),$
$\angle\text{AOB}=120^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=120^\circ$
$\Rightarrow\angle\text{AOC}+66^\circ=120^\circ[$ from Eq. $(ii)]$
$\therefore\angle\text{AOC}=120^\circ-66^\circ=54^\circ$
$\triangle\text{AOB},\ \ \ \text{AO}=\text{OB}$ [both are the radius of a circle]
$\Rightarrow\angle\text{OBA}=\angle\text{BAO}=30^\circ$ [angle opposite to equal sides are equal] In $\triangle\text{AOB},$
$\Rightarrow\angle\text{AOB}+\angle\text{OBA}+\angle\text{BAO}=180^\circ$ [by angle sum property of a triangle]
$\therefore\angle\text{AOB}+30^\circ+30^\circ=180^\circ$
$\therefore\angle\text{AOB}=180^\circ-2(30^\circ)$
$=180^\circ-60^\circ=120^\circ\ \ \ ...(\text{i})$
Now, in $\triangle\text{OCB}, OC = OB[ $ both are the radius of a circle$]$
$\Rightarrow\angle\text{OBC}=\angle\text{OCB}=57^\circ$ [angle opposite to equal sides are equal] In $\triangle\text{OCB}$
$\angle\text{COB}+\angle\text{OCB}+\angle\text{CBO}=180^\circ[$ by angle sum property of triangle$]$ $\therefore\angle\text{COB}=180^\circ-(\angle\text{OCB}+\angle\text{OBC})$
$=180^\circ-(57^\circ+57^\circ)$
$=180^\circ-114^\circ=66^\circ\ \ ...(\text{ii})$ From Eq. $(i),$
$\angle\text{AOB}=120^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=120^\circ$
$\Rightarrow\angle\text{AOC}+66^\circ=120^\circ[$ from Eq. $(ii)]$
$\therefore\angle\text{AOC}=120^\circ-66^\circ=54^\circ$
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