Solution:
Let the angles of quadrilateral be given by:
(a – 3d) °, (a – d) °, (a + d) ° and (a + 3d) °
This is known that the sum of angles of a quadrilateral is 360°. Therefore, we can write:
$ a-3d+a-d+a+d+a+3d=360{}^\circ $
$ 4a=360{}^\circ $
$ a=360/4 $
$ a=90{}^\circ $
It is given that the greatest angle is equal to 120°
$ a\text{ }+\text{ }3d\text{ }=\text{ }120{}^\circ $
$ 90{}^\circ \text{ }+\text{ }3d\text{ }=\text{ }120{}^\circ $
$ 3d\text{ }=\text{ }120{}^\circ \text{ }\text{ }90{}^\circ $
$ 3d\text{ }=\text{ }30{}^\circ $
$ d\text{ }=\text{ }30{}^\circ /3 $
$ d\text{ }=\text{ }{{10}^{o}} $
Therefore, the angles are:
$ \left( a-3d \right){}^\circ =\text{ }90{}^\circ -30{}^\circ \text{ }=\text{ }60{}^\circ $
$ \left( a-d \right){}^\circ =\text{ }90{}^\circ -10{}^\circ =80{}^\circ $
$ \left( a\text{ }+\text{ }d \right)\text{ }{}^\circ =90{}^\circ \text{ }+\text{ }10{}^\circ \text{ }=\text{ }100{}^\circ $
$ \left( a\text{ }+\text{ }3d \right)\text{ }{}^\circ \text{ }=\text{ }120{}^\circ $
The angles of quadrilateral in radians are as follows:
$ \left( 60\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }\pi /3 $
$ \left( 80\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }4\pi /9 $
$ \left( 100\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }5\pi /9 $
$ \left( 120\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }2\pi /3 $