Question

In a $\Delta ABC,a,c,A$ are give and $b_1,b_2$ are two values of third side b such that $b_2=2b_1$. Then the value of $\sin A$

Answer 1

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

rearranging in form of quadratic equation
$b^2-2bc\cos A+(c^2-a^2)=0$
It is given that $b_1,b_2$ are the roots of this equation.

using properties of roots of quadratic equation

Therefore,$b_1+b_2=2c\cos A,$ and $b_1{b}_2=c^2-a^2$

solving
$3b_1=2c\cos A$ and $2b_1^2=c^2-a^2(\because b_2=2b_{1}given)$
$2{\left(\frac{2c}{3}\cos A\right)}^2=c^2-a^2$
$8c^2(1-{\sin }^{2}A)=9c^2-9a^2$
$\sin A=\sqrt{\frac{9a^2-c^2}{8c^2}}$