Question

The difference between the two values of $c$ is:

Answer 1

Step- 1: Solve for the values of $c$:

We know that, for a triangle with angles $A,B,C$ and sides $a,b,c$; $\cos \left(A\right)=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow c^2-2bc\cos \left(A\right)+\left(b^2-a^2\right)=0$ is a quadratic equation in $c$

For a Quadratic equation $p{x}^2+qx+r=0$,

Sum of roots $=\frac{-q}{p}$ and product of roots $=\frac{r}{p}$

Let $c_1,c_2$ be the roots of the quadratic equation in $c$ and here $p=1,q=-2b\cos \left(A\right),r=\left(b^2-a^2\right)$

$\begin{array}{crcl}\therefore & c_1+c_2& =& 2b\cos \left(A\right)\\ & c_1·c_2& =& \left(b^2-a^2\right)\end{array}$

Step- 2: Solve for the required value:

$\begin{array}{cccll}& \left|c_1-c_2\right|& =& \sqrt{{\left(c_1+c_2\right)}^2-4c_1·c_2}& \mathbf{}\left[\because {\left(p-q\right)}^2={\left(p+q\right)}^2-4pq\right]\\ & & =& \sqrt{{\left(2b\cos \left(A\right)\right)}^2-4\left(b^2-a^2\right)}& \\ & & =& \sqrt{4b^2{\cos }^2\left(A\right)-4b^2+4a^2}& \\ & & =& \sqrt{-4b^2\left(1-{\cos }^2\left(A\right)\right)+4a^2}& \\ & & =& \sqrt{4a^2-4b^2{\sin }^2\left(A\right)}& \left[\because si{n}^2\left(\theta \right)+co{s}^2\left(\theta \right)=1\right]\\ \Rightarrow & \left|c_1-c_2\right|& =& 2\sqrt{a^2-b^2{\sin }^2\left(A\right)}& \end{array}$

Hence the difference between two values of $c$ is $2\sqrt{a^2-b^2{\sin }^2\left(A\right)}$.