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 $OA,OB,OC$.

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@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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## Comments

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TopNewestThe 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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@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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@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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