In $∆$ABC , a, c and A are given and b₁,b₂ are two values of the third side b such that b₂ = 2b₁ then sin A=

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

 ${\mathrm{b}}^2-2\mathrm{bccos}\mathrm{A}+{\mathrm{c}}^2-{\mathrm{a}}^2=0$
Now ${\mathrm{b}}_1,{\mathrm{b}}_2$ are the roots of this equation
$\begin{array}{l}\therefore {\mathrm{b}}_1+{\mathrm{b}}_2=2\mathrm{ccos}A and b_1{\mathrm{b}}_2={\mathrm{c}}^2-{\mathrm{a}}^2\\ \Rightarrow 3{\mathrm{b}}_1=2\mathrm{ccos}\mathrm{A} \mathrm{and} 2{\mathrm{b}}_1^2={\mathrm{c}}^2-{\mathrm{a}}^2(\because {\mathrm{b}}_2=2{\mathrm{b}}_1)\\ 2{\left(\frac{2\mathrm{c}}{3}\cos \mathrm{A}\right)}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ \Rightarrow 8{\mathrm{c}}^2\left(1-{\sin }^2\mathrm{A}\right)=9{\mathrm{c}}^2-9{\mathrm{a}}^2\\ \Rightarrow 1-{\sin }^2\mathrm{A}=\frac{9{\mathrm{c}}^2-9{\mathrm{a}}^2}{8{\mathrm{c}}^2}\\ \Rightarrow \sin \mathrm{A}=\sqrt{\frac{9{\mathrm{a}}^2-{\mathrm{c}}^2}{8{\mathrm{c}}^2}}\end{array}$