If  $a,b\text{\hspace{0.17em}and\hspace{0.17em}}A$ are given in a triangle and  $c_1,c_2$ are the possible values of the third side, then  $c_1^2+c_2^2-2c_1{c}_2\cos 2A=k{a}^2{\cos }^{2}A$, where  $k=$

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$c^2-2bc\cos A+b^2-a^2=0$ which is quadratic equation in  $\text{'}c\text{'}$

$\therefore c_1+c_2=2b\cos A$

$c_1{c}_2=b^2-a^2$

Now,  $c_1^2+c_2^2-2c_1{c}_2\cos 2A={(c_1+c_2)}^2-2c_1{c}_2-2c_1{c}_2\cos 2A$

$={\left(2b\cos A\right)}^2-2c_1{c}_2(1+\cos 2A)$

$=4b^2{\cos }^{2}A-2c_1{c}_2\cdot 2{\cos }^{2}A$

$=4b^2{\cos }^{2}A-4(b^2-a^2){\cos }^{2}A$

$=4a^2{\cos }^{2}A$

$\therefore k=4$