Question Number 14438 by ajfour last updated on 31/May/17
$$\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta},\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi},\:{and} \\ $$$$\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:;\:{where}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}} \\ $$$${are}\:{sides}\:{of}\:\bigtriangleup{ABC}\:{such}\:{that} \\ $$$$\boldsymbol{\phi}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{A}}, \\ $$$$\boldsymbol{\psi}−\boldsymbol{\theta}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{B}},\:{and} \\ $$$$\boldsymbol{\theta}−\boldsymbol{\psi}+\frac{\pi}{\mathrm{3}}=\angle\boldsymbol{{C}}\:. \\ $$$${Find}\:{at}\:{least}\:{one}\:{feasible} \\ $$$${solution}\:{set}\:{of}\:\boldsymbol{\theta},\boldsymbol{\phi},\:{and}\:\boldsymbol{\psi}\:{in} \\ $$$${terms}\:{of}\:\angle\boldsymbol{{A}},\:\angle\boldsymbol{{B}},\:{and}\:\angle\boldsymbol{{C}} \\ $$$${such}\:{that}\:{all}\:{angles}\:{and}\:{sides} \\ $$$${are}\:{positive}\:{with}\:\boldsymbol{{a}}\neq\boldsymbol{{b}}\neq\boldsymbol{{c}}\:, \\ $$$${and}\:\angle\boldsymbol{{A}}\neq\angle\boldsymbol{{B}}\neq\angle\boldsymbol{{C}}\:\:\neq\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\: \\ $$$${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:{even}\:{if}\:{you}\: \\ $$$${you}\:{please}.. \\ $$
Commented by ajfour last updated on 31/May/17
$${this}\:{is}\:{again}\:{related}\:{to}\:{your} \\ $$$${Q}.\mathrm{14157}\:{this}\:{is}\:{how}\:{we}\:{can} \\ $$$${find}\:{x},{y},\:{and}\:{z}\:,\:{given}\:\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17
$${mr}\:{Ajfour}!{it}\:{is}\:{a}\:{very}\:{difficult}\:{Q}. \\ $$
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 . .........................................](https://www.tinkutara.com/question/Q14461.png)
$$\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}, \\ $$$${s}={x}+{y}+{z},\:{then} \\ $$$${s}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\boldsymbol{{x}}=\frac{\mathrm{2}\boldsymbol{{a}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{as}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{a}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\theta} \\ $$$$\:\:\:\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{b}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{bs}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{b}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\phi} \\ $$$$\:\:\:\:\boldsymbol{{z}}=\frac{\mathrm{2}\boldsymbol{{c}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{cs}}}\right)\right]\:=\frac{\mathrm{2}\boldsymbol{{c}}}{\:\sqrt{\mathrm{3}}}\mathrm{sin}\:\boldsymbol{\psi}\:. \\ $$$$………………………………….. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
$${great}\:{job}\:{mr}\:{Ajfour}. \\ $$
Commented by ajfour last updated on 01/Jun/17
$${thanks}\:{sir},\:{but}\:{i}\:{wish}\:{to}\:{solve} \\ $$$${it}\:{by}\:{pythagoras}\:{theorem}\:{coz} \\ $$$${there}\:{is}\:{a}\:{vertical}\:{and}\:{horizontal} \\ $$$${shift}\:{from}\:{complete}\:{symmetry} \\ $$$${as}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}\:\:{become}\:{unequal}.. \\ $$
Commented by ajfour last updated on 01/Jun/17
$${see}\:{Q}.\mathrm{14502}\:{my}\:{solution}\:{i}\:{am} \\ $$$${delaying}\:{till}\:'\:{night}\:{times}\:{are} \\ $$$${all}\:{my}\:{own}\:'.. \\ $$