If the angles of a triangle are in A.P.,

(b) $\frac{\pi}{3}$

Let the angles of the triangle be $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$.

Thus, we have:

$a-d+a+a+d=180$

$\Rightarrow 3 a=180$

$\Rightarrow a=60$

Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $(60-d)^{\circ}, 60^{\circ}$ and $(60+d)^{\circ}$. $60^{\circ}$ is the only angle which is independent of $d$.

$\therefore$ One of the angles of the triangle (in radians) $=\left(60 \times \frac{\pi}{180}\right)=\frac{\pi}{3}$