Question Number 133069 by Dwaipayan Shikari last updated on 18/Feb/21
Commented by Dwaipayan Shikari last updated on 18/Feb/21
$${One}\:{can}\:{stretch}\:{and}\:{reform}\:{sides}\:{AB}\:\:{and}\:{AC}\:,\:{by}\: \\ $$$${considering}\:{length}\:\:{BC}\:{as}\:{constant}.\:{the}\:{angle}\: \\ $$$$\phi\:{can}\:{change}.\:{Find}\:{the}\:{relationship}\:{of}\:{changes}\:{between}\:{Side}\: \\ $$$${AB}\:,\:{AC}\:{and}\:{angle}\:\phi \\ $$$$\left({Assuming}\:{that}\:{the}\:{triangle}\:{will}\:{not}\:{break}\:,{and}\:{lengths}\:{are}\right. \\ $$$$\left.{interelated}\right) \\ $$
Commented by mr W last updated on 18/Feb/21
$${BC}={a}={constant} \\ $$$${AC}={b}={variable} \\ $$$${AB}={c}={variable} \\ $$$$\angle{A}=\phi={variable} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\phi={a}^{\mathrm{2}} \\ $$$${or} \\ $$$$\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{2}{bc}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)={a}^{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 18/Feb/21
$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bccos}\phi={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{b}\overset{.} {{b}}+\mathrm{2}{c}\overset{.} {{c}}−\mathrm{2}\overset{.} {{b}ccos}\phi−\mathrm{2}{c}\overset{.} {{b}cos}\phi+\mathrm{2}{bcsin}\phi\overset{.} {\phi}=\mathrm{0} \\ $$$$\Rightarrow{b}\overset{.} {{b}}+{c}\overset{.} {{c}}+{bcsin}\phi\overset{.} {\phi}={cos}\phi\left({b}\overset{.} {{c}}+{c}\overset{.} {{b}}\right) \\ $$$${Can}\:{we}\:{do}\:{further}\:{like}\:{this}? \\ $$
Commented by mr W last updated on 18/Feb/21
$${yes} \\ $$