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# How To Apply Rotation/Construction In It

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 $$OA,OB,OC$$.

Note by Vilakshan Gupta
4 months, 2 weeks ago

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@Rohit Camfar satwikmurarka@yahoo.com · 4 months, 1 week ago

Ok you are invited you can join us now by going into your Id · 4 months ago

@Rohit Camfar Can I join the RMO preparation team?I am also an RMO aspirant. · 4 months, 1 week ago

The angle in their correct order are 64 , 53 , 63. · 4 months, 2 weeks ago

please explain the solution · 4 months, 2 weeks ago

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. · 4 months, 2 weeks ago

Thank You Rohit. U are really good in geometry.Are You Really 14? And it seems that u are an IMO aspirant · 4 months, 2 weeks ago

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. · 4 months, 2 weeks ago

Of Course, I would surely like to join your team .Please Invite me , my email id is- vilakshangupta2002@gmail.com · 4 months, 2 weeks ago

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. · 1 month, 4 weeks ago