The angle of a quadrilateral are in A.P. and the greatest angle is 120°.

Let the angles of the quadrilateral be $(a-3 d)^{\circ},(a-d)^{\circ},(a+d)^{\circ}$ and $(a+3 d)^{\circ}$.

We know:

$a-3 d+a-d+a+d+a-2 d=360$

$\Rightarrow 4 a=360$

$\Rightarrow a=90$

We have:

Greatest angle $=120^{\circ}$

Now,

$a+3 d=120$

$\Rightarrow 90+3 d=120$

$\Rightarrow 3 d=30$

$\Rightarrow d=10$

Hence, $(a-3 d)^{\circ},(a-d)^{\circ},(a+d)^{\circ}$ and $(a+3 d)^{\circ}$ are $60^{\circ}, 80^{\circ}, 100^{\circ}$ and $120^{\circ}$, respective

Angles of the quadrilateral in radians $=\left(60 \times \frac{\pi}{180}\right),\left(80 \times \frac{\pi}{180}\right),\left(100 \times \frac{\pi}{180}\right)$ and $\left(120 \times \frac{\pi}{180}\right)$$=\frac{\pi}{3}, \frac{4 \pi}{9}, \frac{5 \pi}{9}$ and $\frac{2 \pi}{3}$