Inside an equilateral \(\triangle ABC \) lies a point \(O\). It is known that \(\angle AOB=113^{\circ} \) and \(\angle BOC=123^{\circ} \). Find the angles of the triangle whose sides are equal to segments \(OA,OB,OC\).

@Vilakshan Gupta
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Yep, 'm an IMO aspirant, hoping for it but it is never going to be easy at all. If you too are an IMO aspirant then would you like to join our RMO / INMO prperation team at Slack.

@Vilakshan Gupta
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Hi guys @Vilakshan Gupta@Satwik Murarka I would also like to join you guy's team. Here is the email imjabitg@gmail.com I had asked @Rohit Camfaron a different forum but he told that he is no more in the team and asked me to ask on this notice board.

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TopNewest@Rohit Camfar satwikmurarka@yahoo.com

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Ok you are invited you can join us now by going into your Id

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@Rohit Camfar Can I join the RMO preparation team?I am also an RMO aspirant.

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ok you can join us by giving your email.

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The angle in their correct order are 64 , 53 , 63.

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please explain the solution

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Draw \(AO'\) = \(OA\) such that \(\angle OAO' = 60^{\circ}\). Then \(\Delta OAB \cong \Delta O'AC\) => \(OB\) \(=\) \(O'C\) . Now, draw (join) \(OO'.\) Then clearly \(\Delta O'OA\) is equilateral. [=> all \(\angle\) = \(60^{\circ}\)] Therefore, \(OO' = OA\) Then \(\Delta OO'C\) is the required triangle with the side lengths \(OA , OB\) and \(OC\) respectively. Now, calculating angles, \(\angle O'OC = \angle AOC - \angle AOO'\) = \(124^{\circ} - 60^{\circ} = 64^{\circ}........................\)

Calculate the rest angles yourself.

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@Vilakshan Gupta @Satwik Murarka I would also like to join you guy's team. Here is the email imjabitg@gmail.com I had asked @Rohit Camfaron a different forum but he told that he is no more in the team and asked me to ask on this notice board.

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