Mathematics - Properties of Triangle Question with Solution | TestHub

$b^2-2bc\cos A+c^2-a^2=0\Rightarrow b_1+b_2=2c\cos A,b_1{b}_2=c^2-a^2$

$$\begin{array}{rlrlr}& b_2=2b_1;b_1=\frac{2c}{3}\cos A,b_2=\frac{\sqrt{c^2-a^2}}{2}\Rightarrow \frac{2c}{3}\cos A=\frac{c^2-a^2}{2}& & & \end{array}$$

$$\Rightarrow \sin A=\frac{\sqrt{9a^2-c^2}}{8c^2}a=20-6\pi ;b=10\pi -30c=10-3\pi$$

$$\therefore a+b+c=\pi$$