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Question Number 206922 by mr W last updated on 01/May/24
solve for x, y, z ∈R^+   x^2 +y^2 −2xy cos γ=c^2   y^2 +z^2 −2yz cos α=a^2   z^2 +x^2 −2zx cos β=b^2   with α+β+γ=360°    example:   a=12, b=8, c=10  α=120°, β=90°, γ=150°
solveforx,y,zR+x2+y22xycosγ=c2y2+z22yzcosα=a2z2+x22zxcosβ=b2withα+β+γ=360°example:a=12,b=8,c=10α=120°,β=90°,γ=150°
Answered by mr W last updated on 01/May/24
Commented by mr W last updated on 01/May/24
Commented by mr W last updated on 01/May/24
∠A=cos^(−1) ((b^2 +c^2 −a^2 )/(2bc))  ∠B=cos^(−1) ((c^2 +a^2 −b^2 )/(2ca))  ∠C=cos^(−1) ((a^2 +b^2 −c^2 )/(2ab))  π−2φ_1 =2π−2α   ⇒φ_1 =α−(π/2)  similarly  φ_2 =β−(π/2)  φ_3 =γ−(π/2)  r_1 =(a/(2 cos φ_1 ))=(a/(2 sin α))  similarly  r_2 =(b/(2 sin β))  r_3 =(c/(2 sin γ))  QR^2 =r_2 ^2 +r_3 ^2 −2r_2 r_3  cos (∠A+φ_2 +φ_3 )  ∠A+φ_2 +φ_3 =∠A+β+γ−π=π−(α−∠A)  QR^2 =(b^2 /(4 sin^2  β))+(c^2 /(4 sin^2  γ))+((bc cos (α−∠A) )/(2 sin β sin γ))  ((x×QR)/2)=r_2 r_3  sin (∠A+φ_2 +φ_3 )=2[AQR]  ((x×QR)/2)=((bc sin (α−∠A))/(4 sin β sin γ))  ⇒x=((bc sin (α−∠A))/(sin β sin γ (√((b^2 /(sin^2  β))+(c^2 /(sin^2  γ))+((2bc cos (α−∠A) )/(sin β sin γ))))))  similarly  ⇒y=((ca sin (β−∠B))/(sin γ sin α (√((c^2 /(sin^2  γ))+(a^2 /(sin^2  α))+((2ca cos (β−∠B) )/(sin γ sin a))))))  ⇒z=((ab sin (γ−∠C))/(sin α sin β (√((a^2 /(sin^2  α))+(b^2 /(sin^2  β))+((2ab cos (γ−∠C) )/(sin α sin β))))))  example:  a=12, b=8, c=10  α=120°, β=90°, γ=150°  cos ∠C=((12^2 +8^2 −10^2 )/(2×12×8))=(9/(16))  sin (γ−∠C)=((9+5(√(21)))/(32))  cos (γ−∠C)=((−9(√3)+5(√7))/(32))  z=((12×8×((9+5(√(21)))/(32)))/(((√3)/2) (√(((12^2 ×4)/3)+8^2 +((2×12×8×2 )/( (√3)))×((−9(√3)+5(√7))/(32))))))    =((9(√3)+15(√7))/( (√(37+5(√(21))))))≈7.141
A=cos1b2+c2a22bcB=cos1c2+a2b22caC=cos1a2+b2c22abπ2ϕ1=2π2αϕ1=απ2similarlyϕ2=βπ2ϕ3=γπ2r1=a2cosϕ1=a2sinαsimilarlyr2=b2sinβr3=c2sinγQR2=r22+r322r2r3cos(A+ϕ2+ϕ3)A+ϕ2+ϕ3=A+β+γπ=π(αA)QR2=b24sin2β+c24sin2γ+bccos(αA)2sinβsinγx×QR2=r2r3sin(A+ϕ2+ϕ3)=2[AQR]x×QR2=bcsin(αA)4sinβsinγx=bcsin(αA)sinβsinγb2sin2β+c2sin2γ+2bccos(αA)sinβsinγsimilarlyy=casin(βB)sinγsinαc2sin2γ+a2sin2α+2cacos(βB)sinγsinaz=absin(γC)sinαsinβa2sin2α+b2sin2β+2abcos(γC)sinαsinβexample:a=12,b=8,c=10α=120°,β=90°,γ=150°cosC=122+821022×12×8=916sin(γC)=9+52132cos(γC)=93+5732z=12×8×9+5213232122×43+82+2×12×8×23×93+5732=93+15737+5217.141
Commented by mr W last updated on 01/May/24
geometrically it is to find the  distances from a point inside a  triangle to the vertexes of the triangle.  when α=β=γ=120°, this point is  the so−called fermat point.
geometricallyitistofindthedistancesfromapointinsideatriangletothevertexesofthetriangle.whenα=β=γ=120°,thispointisthesocalledfermatpoint.

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