Inside an equilateral △ABC\triangle ABC △ABC lies a point OOO. It is known that ∠AOB=113∘\angle AOB=113^{\circ} ∠AOB=113∘ and ∠BOC=123∘\angle BOC=123^{\circ} ∠BOC=123∘. Find the angles of the triangle whose sides are equal to segments OA,OB,OCOA,OB,OCOA,OB,OC.
Note by Vilakshan Gupta 4 years, 2 months ago
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The angle in their correct order are 64 , 53 , 63.
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please explain the solution
Draw AO′AO'AO′ = OAOAOA such that ∠OAO′=60∘\angle OAO' = 60^{\circ}∠OAO′=60∘. Then ΔOAB≅ΔO′AC\Delta OAB \cong \Delta O'ACΔOAB≅ΔO′AC => OBOBOB === O′CO'CO′C . Now, draw (join) OO′.OO'.OO′. Then clearly ΔO′OA\Delta O'OAΔO′OA is equilateral. [=> all ∠\angle∠ = 60∘60^{\circ}60∘] Therefore, OO′=OAOO' = OAOO′=OA Then ΔOO′C\Delta OO'CΔOO′C is the required triangle with the side lengths OA,OBOA , OBOA,OB and OCOCOC respectively. Now, calculating angles, ∠O′OC=∠AOC−∠AOO′\angle O'OC = \angle AOC - \angle AOO'∠O′OC=∠AOC−∠AOO′ = 124∘−60∘=64∘........................124^{\circ} - 60^{\circ} = 64^{\circ}........................124∘−60∘=64∘........................
Calculate the rest angles yourself.
@Vishwash Kumar Γξω – Thank You Rohit. U are really good in geometry.Are You Really 14? And it seems that u are an IMO aspirant
@Vilakshan Gupta – 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.
@Vishwash Kumar Γξω – Of Course, I would surely like to join your team .Please Invite me , my email id is- [email protected]
@Vilakshan Gupta – You have been invited on the team. Check your email and join us.
@Vilakshan Gupta – Hi guys @Vilakshan Gupta @Satwik Murarka I would also like to join you guy's team. Here is the email [email protected] 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.
@Rohit Camfar Can I join the RMO preparation team?I am also an RMO aspirant.
ok you can join us by giving your email.
@Rohit Camfar [email protected]
Ok you are invited you can join us now by going into your Id
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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 ∠OAO′=60∘. Then ΔOAB≅ΔO′AC => OB = O′C . Now, draw (join) OO′. Then clearly ΔO′OA is equilateral. [=> all ∠ = 60∘] Therefore, OO′=OA Then ΔOO′C is the required triangle with the side lengths OA,OB and OC respectively. Now, calculating angles, ∠O′OC=∠AOC−∠AOO′ = 124∘−60∘=64∘........................
Calculate the rest angles yourself.
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[email protected]
Of Course, I would surely like to join your team .Please Invite me , my email id is-Log in to reply
@Vilakshan Gupta @Satwik Murarka I would also like to join you guy's team. Here is the email [email protected] 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.
Hi guys@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.
@Rohit Camfar [email protected]
Log in to reply
Ok you are invited you can join us now by going into your Id