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Question-117481




Question Number 117481 by I want to learn more last updated on 12/Oct/20
Answered by mr W last updated on 12/Oct/20
let AC=1  ((OC)/(sin 20))=((AC)/(sin (10+20)))  ⇒OC=((sin 20)/(sin 30))=2 sin 20  ((BC)/(sin 40))=((AC)/(sin (30+10+20+20)))  ⇒BC=(1/(2 cos 40))  ((BC)/(sin (x+30)))=((OC)/(sin x))  ((sin (x+30))/(sin x))=((BC)/(OC))=(1/(4 cos 40 sin 20))  cos 30+((sin 30)/(tan x))=(1/(4 cos 40 sin 20))  (1/(tan x))=((1/(2 cos 40 sin 20))−(√3))  ⇒x=80°
letAC=1OCsin20=ACsin(10+20)OC=sin20sin30=2sin20BCsin40=ACsin(30+10+20+20)BC=12cos40BCsin(x+30)=OCsinxsin(x+30)sinx=BCOC=14cos40sin20cos30+sin30tanx=14cos40sin201tanx=(12cos40sin203)x=80°
Commented by I want to learn more last updated on 12/Oct/20
Thanks sir. I appreciate
Thankssir.Iappreciate
Commented by I want to learn more last updated on 12/Oct/20
sir, please help with  Q117480
sir,pleasehelpwithQ117480
Answered by 1549442205PVT last updated on 12/Oct/20
Commented by 1549442205PVT last updated on 12/Oct/20
Construct the equilateral triangle  BCD such that D lie on halfplane  with the edge being the line  AC  don′t  containing B.Then  BCD^(�) =BDC^(�) =CBD^(�) =60°.From the  hypothesis BAC^(�) =BCA^(�) =40°we get  ACD^(�) =20°,ABD^(�) =100°−60°=40°  BA=BD=BC=CD⇒ΔABD is  isosceles at B⇒BAD^(�) =BDA^(�) =70°  ⇒CAD^(�) =70°−40°=30°(∗)  Draw the ray Cx such that BCx ^(�) =30°  M=(AB)∩Cx.ThenAMC^(�) =180°−(40+70)  =70°.Suppose O′ is the point symetry  to M through BC,E=BC∩MO′then  BME^(�) =BMC^(�) −CME^(�) =70−60=10°  ⇒CBM^(�) =80°= CBO′^(�) (the property of  two symetric figures).We will prove  that O≡O′.Indeed,we have ABO^(�) =  ABC^(�) −CBO′^(�) =100−80=20°=(1/2)ABD^(�)   O′∈CO since BCO^(�) =BCO′^(�) =30°  CO′D^(�) =CO′B^(�) =70°and CDO′^(�) =CBO′^(�) =80°  (since CO′Dand CO′B are two symetric  figures through CO)⇒O′DA^(�) =ADC^(�)   −CDO′^(�) =130−80=50°.(1)  ΔBO′A=ΔBO′D(s.a.s)⇒BO′A^(�) =BO′D^(�)   =140°⇒AO′D^(�) =360−2.140=80°(2)  From (1)(2)we get O′AD^(�) =180−(50+80)  =50°⇒O′AC^(�) =50−30=20°=OAC^(�) (3)  but O′CA^(�) =10°(4).From(3)(4)we infer  O′≡O.That shows that x=CBO^(�) =80°
ConstructtheequilateraltriangleBCDsuchthatDlieonhalfplanewiththeedgebeingthelineACdontcontainingB.ThenBCD^=BDC^=CBD^=60°.FromthehypothesisBAC^=BCA^=40°wegetACD^=20°,ABD^=100°60°=40°BA=BD=BC=CDΔABDisisoscelesatBBAD^=BDA^=70°CAD^=70°40°=30°()DrawtherayCxsuchthatBCx^=30°M=(AB)Cx.ThenAMC^=180°(40+70)=70°.SupposeOisthepointsymetrytoMthroughBC,E=BCMOthenBME^=BMC^CME^=7060=10°CBM^=80°=CBO^(thepropertyoftwosymetricfigures).WewillprovethatOO.Indeed,wehaveABO^=ABC^CBO^=10080=20°=12ABD^OCOsinceBCO^=BCO^=30°COD^=COB^=70°andCDO^=CBO^=80°(sinceCODandCOBaretwosymetricfiguresthroughCO)ODA^=ADC^CDO^=13080=50°.(1)ΔBOA=ΔBOD(s.a.s)BOA^=BOD^=140°AOD^=3602.140=80°(2)From(1)(2)wegetOAD^=180(50+80)=50°OAC^=5030=20°=OAC^(3)butOCA^=10°(4).From(3)(4)weinferOO.Thatshowsthatx=CBO^=80°
Commented by I want to learn more last updated on 12/Oct/20
Thanks sir, i appreciate.
Thankssir,iappreciate.
Commented by I want to learn more last updated on 12/Oct/20
sir, please help with  Q117480
sir,pleasehelpwithQ117480

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