If the angles $A, B$ and $C$ of $\triangle A B C$ are in arithmetic progression, then
If the angles $A, B$ and $C$ of $\triangle A B C$ are in arithmetic progression, then
$b^2=a^2+c^2-a c$
$c^2=b^2+a^2-a b$
$a^2=b^2+c^2-b c$
$c^2=a^2+b^2$
Solution
The angles of $\triangle A B C$ are in AP.
$\therefore \quad 2 B=A+C \quad\left[\because A+B+C=180^{\circ}\right]$
$\Rightarrow \quad 2 B=180^{\circ}-B \Rightarrow B=60^{\circ}$
Now, $\cos B=\frac{a^2+c^2-b^2}{2 a c}$
$\cos 60^{\circ}=\frac{a^2+c^2-b^2}{2 a c} \Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2 a c}$
$\therefore \quad b^2=a^2+c^2-a c$