The angles of a triangle are in A.P.

Solution:

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

We know:

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

$\Rightarrow 3 a=180$

$\Rightarrow a=60$

Given:

$\frac{\text { Number of degrees in the least angle }}{\text { Number of degrees in the mean angle }}=\frac{1}{120}$

or, $\frac{a-d}{a}=\frac{1}{120}$

or, $\frac{60-d}{60}=\frac{1}{120}$

or, $\frac{60-\mathrm{d}}{1}=\frac{1}{2}$

or, $120-2 d=1$

or, $2 \mathrm{~d}=119$

or, $d=59.5$

Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $0.5^{\circ}, 60^{\circ}$ and $119.5^{\circ}$.

$\therefore$ Angles of the triangle in radians $=\left(0.5 \times \frac{\pi}{180}\right),\left(60 \times \frac{\pi}{180}\right)$ and $\left(119.5 \times \frac{\pi}{180}\right)$

$=\frac{\pi}{360}, \frac{\pi}{3}$ and $\frac{239 \pi}{360}$