Arithmetic Angles

$A\,,\,B\,$ and $\,C$ are in arithmetic progression.

$\Rightarrow A + C = 2B\newline$ Also, $A + B + C = 180\degree\hspace{20pt} [$ Angle sum property of triangle $]$

Combining both the equation we get, $B = 60\degree$

Let $A = 60\degree - \alpha$ . Then $C = 60\degree + \alpha$ .

Now, $\text{sin }A\,,\,\text{sin}^2B\,$ and $\,\text{sin }C$ are also in arithmetic progression. So,

$\text{sin }A + \text{sin }C = 2\text{sin}^2B\newline \Rightarrow 2\text{sin}\Large\Big(\frac{A + C}{2}\Big)$ $\text{cos}\Large\Big(\frac{A - C}{2}\Big)$ $= 2\Large\Big(\frac{\sqrt{3}}{2}\Big)^2$ $\newline \Rightarrow \text{sin }60\degree \text{cos }\alpha =$ $\Large\frac{3}{4}$ $\newline\Rightarrow \text{cos }\alpha =$ $\Large\frac{\sqrt{3}}{2}$ $\newline \Rightarrow \alpha = 30\degree$ or $\alpha = -30\degree$

So possible sets of $A,B,C$ are $(30\degree,60\degree,90\degree)$ and $(90\degree,60\degree,30\degree)$ .

Hence $|A - B|$ in both the case is same which is $|30\degree - 60\degree| = |90\degree - 60\degree| = 30\degree$