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Question Number 14438 by ajfour last updated on 31/May/17

x=((2a)/(√3))sin 𝛉, y=((2b)/(√3))sin 𝛗, and  z=((2c)/(√3))sin 𝛙 ; where a,b, and c  are sides of △ABC such that  𝛗−𝛙+(π/3)=∠A,  𝛙−𝛉+(π/3)=∠B, and  𝛉−𝛙+(π/3)=∠C .  Find at least one feasible  solution set of 𝛉,𝛗, and 𝛙 in  terms of ∠A, ∠B, and ∠C  such that all angles and sides  are positive with a≠b≠c ,  and ∠A≠∠B≠∠C  ≠ (𝛑/2)   Find x,y, and z even if you   you please..

$$\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta},\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi},\:{and} \\ $$$$\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:;\:{where}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}} \\ $$$${are}\:{sides}\:{of}\:\bigtriangleup{ABC}\:{such}\:{that} \\ $$$$\boldsymbol{\phi}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{A}}, \\ $$$$\boldsymbol{\psi}−\boldsymbol{\theta}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{B}},\:{and} \\ $$$$\boldsymbol{\theta}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{C}}\:. \\ $$$${Find}\:{at}\:{least}\:{one}\:{feasible} \\ $$$${solution}\:{set}\:{of}\:\boldsymbol{\theta},\boldsymbol{\phi},\:{and}\:\boldsymbol{\psi}\:{in} \\ $$$${terms}\:{of}\:\angle\boldsymbol{{A}},\:\angle\boldsymbol{{B}},\:{and}\:\angle\boldsymbol{{C}} \\ $$$${such}\:{that}\:{all}\:{angles}\:{and}\:{sides} \\ $$$${are}\:{positive}\:{with}\:\boldsymbol{{a}}\neq\boldsymbol{{b}}\neq\boldsymbol{{c}}\:, \\ $$$${and}\:\angle\boldsymbol{{A}}\neq\angle\boldsymbol{{B}}\neq\angle\boldsymbol{{C}}\:\:\neq\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\: \\ $$$${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:{even}\:{if}\:{you}\: \\ $$$${you}\:{please}.. \\ $$

Commented by ajfour last updated on 31/May/17

this is again related to your  Q.14157 this is how we can  find x,y, and z , given a,b, and c.

$${this}\:{is}\:{again}\:{related}\:{to}\:{your} \\ $$$${Q}.\mathrm{14157}\:{this}\:{is}\:{how}\:{we}\:{can} \\ $$$${find}\:{x},{y},\:{and}\:{z}\:,\:{given}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

mr Ajfour!it is a very difficult Q.

$${mr}\:{Ajfour}!{it}\:{is}\:{a}\:{very}\:{difficult}\:{Q}. \\ $$

Commented by ajfour last updated on 01/Jun/17

Answered by ajfour last updated on 01/Jun/17

cos C=((a^2 +b^2 −c^2 )/(2ab)),  s=x+y+z, then  s^2 =a^2 +b^2 +2abcos (C−(π/3))    x=((2a)/(√3))sin [cos^(−1) (((a^2 +s^2 −c^2 )/(2as)))] =((2a)/(√3))sin 𝛉     y=((2b)/(√3))sin [cos^(−1) (((b^2 +s^2 −a^2 )/(2bs)))] =((2b)/(√3))sin 𝛗      z=((2c)/(√3))sin [cos^(−1) (((c^2 +s^2 −b^2 )/(2cs)))] =((2c)/(√3))sin 𝛙 .  .........................................

$$\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}, \\ $$$${s}={x}+{y}+{z},\:{then} \\ $$$${s}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{as}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{a}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta} \\ $$$$\:\:\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{bs}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{b}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi} \\ $$$$\:\:\:\:\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{cs}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{c}}}{\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:. \\ $$$$......................................... \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17

great job mr Ajfour.

$${great}\:{job}\:{mr}\:{Ajfour}. \\ $$

Commented by ajfour last updated on 01/Jun/17

thanks sir, but i wish to solve  it by pythagoras theorem coz  there is a vertical and horizontal  shift from complete symmetry  as a, b, and c  become unequal..

$${thanks}\:{sir},\:{but}\:{i}\:{wish}\:{to}\:{solve} \\ $$$${it}\:{by}\:{pythagoras}\:{theorem}\:{coz} \\ $$$${there}\:{is}\:{a}\:{vertical}\:{and}\:{horizontal} \\ $$$${shift}\:{from}\:{complete}\:{symmetry} \\ $$$${as}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}\:\:{become}\:{unequal}.. \\ $$

Commented by ajfour last updated on 01/Jun/17

see Q.14502 my solution i am  delaying till ′ night times are  all my own ′..

$${see}\:{Q}.\mathrm{14502}\:{my}\:{solution}\:{i}\:{am} \\ $$$${delaying}\:{till}\:'\:{night}\:{times}\:{are} \\ $$$${all}\:{my}\:{own}\:'.. \\ $$

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