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Question Number 187808 by LowLevelLump last updated on 22/Feb/23

Answered by a.lgnaoui last updated on 22/Feb/23

△ABC′   { ((AB=AC=BC=a)),((S_1 =((a^2 (√3))/2))) :}  △ACB′    { ((AC=AB′=CB′=b)),((S_2 =((b^2 (√3))/2))) :}  △BCC′    { ((BC=BC′=CC^′ =c)),((S_3 =((c^2 (√3))/2))) :}  sin B=((AH)/(AB))=(1/3)⇒AH=asin B  S_1 −S_2 +S_3 =(((a^2 −b^2 +c^2 )(√3))/2)  =((√3)/2)(1)  S(△ABC)=((c×((AH)/(AB)))/2)=((ac)/6)=     (2)   { ((a^2 −b^2 +c^2 =1)),((c=acos B+(√(b^2 −a^2 sin^2 B)))) :}  c=((2a(√2))/3)+(1/3)(√( 9b^2 −a^2 ))   { ((c^2 =b^2 −a^2 +1                     (3))),((c=((2a(√2) +(√(9b^2 −a^2 )))/3)        (4))) :}  b^2 = (((9−16a^2 )^2 )/(288a^2 ))+(a^2 /9)  (3)⇒c^2 =1+(((9−16a^2 )^2 )/(288a^2 ))−((8a^2 )/9)  c^2 =((81)/(288a^2 ))   ⇒          c=((9(√2))/(24a))     (5)        a)    AreaABC=((ac)/6)=   ((3(√2))/(48))          Area(ABC)=((√2)/(16))  b)If  sin Asin C=((√2)/3)          ((sin A)/c)=((sin B)/b)=((sin C)/a)⇒asin A=csin C  ((a(√2))/(3sin C))=csin C   ⇒3csin^2  C=a(√2)  sin^2 c=((a(√2))/(3c)); sin c=((AH)/b)=((asin B)/b)=(a/(3b))(sin B=(1/3))  =(a^2 /(9b^2 )) =((a(√2))/(3c))     b^2 =((ac(√2))/6)=  (5) ⇒b^2 =((9(√2))/(24)) ×((√2)/6)     ⇒  b=((√2)/4)

ABC{AB=AC=BC=aS1=a232ACB{AC=AB=CB=bS2=b232BCC{BC=BC=CC=cS3=c232sinB=AHAB=13AH=asinBS1S2+S3=(a2b2+c2)32=32(1)S(ABC)=c×AHAB2=ac6=(2){a2b2+c2=1c=acosB+b2a2sin2Bc=2a23+139b2a2{c2=b2a2+1(3)c=2a2+9b2a23(4)b2=(916a2)2288a2+a29(3)c2=1+(916a2)2288a28a29c2=81288a2c=9224a(5)a)AreaABC=ac6=3248Area(ABC)=216b)IfsinAsinC=23sinAc=sinBb=sinCaasinA=csinCa23sinC=csinC3csin2C=a2sin2c=a23c;sinc=AHb=asinBb=a3b(sinB=13)=a29b2=a23cb2=ac26=(5)b2=9224×26b=24

Commented by a.lgnaoui last updated on 22/Feb/23

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