Question

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Answer 1

$\huge\bold{B-\:A-\:C-\:K}$

$<u><b>HEY BUDDY!!!!!$

$\huge\bold{SOLUTION}$

from the figure

given= <OAB= 30° & <BOC = 57°

REFER ATTACHMENT..

$<b><u>RGRDSM@STYLGG$

Answer 2

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
$\angle{OAB} = 30^{\circ}$ ————[1]

$\angle{OCB} = 57^{\circ}$ ————[2]

$\angle{BOC} = ? ^{\circ}$

$\angle{AOC} = ? ^{\circ}$

$\underline{\large{\textsf{Step-1:}}}$
Note the Radius of the given Circle

As $\textbf{O}$ is the $\textsf{Centre of the circle}$

OA , OB, OC represents the radius of the circle.

$\therefore$
$\textbf{OA = OB = OC}$ ————[3]

$\underline{\large{\textsf{Step-2:}}}$
Apply the theorem,
$\textbf{Opposite\: Angles\: to\: equal\: sides\: of\: a\: triangle\: are\: equal}$

$\angle{OBA} = \angle{BAO} = 30^{\circ}$ ————[4]

($\angle{BAO}= \angle{OAB}$ and from[1] & [3])

$\angle{OBC} = \angle{BCO} = 57^{\circ}$ ————[5]

($\angle{BCO}= \angle{OCB}$ and from[2] & [3])

$\underline{\large{\textsf{Step-3:}}}$
Apply the Angle Sum property in $\triangle{AOB}$ and $\triangle{OCB}$

$\textsf{Angle sum property of triangle states that sum of angles in a triangle is 180^{\circ} }$

In $\triangle{AOB}$,
$\implies \angle{AOB}+\angle{OBA}+\angle{BAO}= 180^{\circ}$

Substituting [4]

$\implies \angle{AOB} + 30^{\circ} + 30^{\circ} = 180^{\circ}$

$\implies \angle{AOB} + 60^{\circ} = 180^{\circ}$

$\implies \angle{AOB} = 180^{\circ} - 60^{\circ}$

$\implies \angle{AOB} = 120^{\circ}$ ————[6]

In $\triangle{OCB}$,
$\implies \angle{COB}+\angle{OBC}+\angle{BCO}= 180^{\circ}$

Substituting [5]

$\implies \angle{COB}+ 57^{\circ} + 57^{\circ} = 180^{\circ}$

$\implies \angle{COB}+ 114^{\circ}= 180^{\circ}$

$\implies \angle{COB} = 180^{\circ} - 114^{\circ}$

$\implies \angle{COB} = 66^{\circ}$ ————[7]

$\underline{\large{\textsf{Step-4:}}}$
Express $\angle{AOB}$ in terms of $\angle{AOC}$ and $\angle{COB}$

From the figure,
$\angle{AOB} = \angle{AOC} + \angle{COB}$

Substituting [6] & [7]

$\implies 120^{\circ} = \angle{AOC} + 66^{\circ}$

$\implies \angle{AOC} = 120^{\circ} - 66^{\circ}$

$\implies \angle{AOC} = 54^{\circ}$

$\therefore$

$\bigstar \underline{\mathbf{\angle{BOC} = \angle{COB} = 66^{\circ}}}$

$\: \: \: \underline{\mathbf{\angle{AOC} = 54^{\circ}}}$