Solution:

Consider that the angles of the triangle are: (a – d) ^o, a^o and (a + d) ^o.

We know that the sum of the angles of a triangle is 180°. Therefore, we can write:

$ a-d+a+a+d=180{}^\circ  $

$ 3a=180{}^\circ  $

$ a=180{}^\circ /3 $

$ a=\text{ }{{60}^{o}} $

It is given that the greatest angle is equal to five times of the least angle

Or, we can write that:

Greatest angle / least angle = 5

Upon putting values, we get:

$ \left( a+d \right)/\left( a-d \right)\text{ }=\text{ }5 $

$ \left( 60+d \right)/\left( 60-d \right)\text{ }=\text{ }5 $

Upon cross-multiplication we get,

$ 60\text{ }+\text{ }d\text{ }=\text{ }300\text{ }\text{ }5d $

$ 6d\text{ }=\text{ }240 $

$ d\text{ }=\text{ }240/6 $

$ d=\text{ }40 $

Hence, angles are:

(a – d) ° = 60° – 40° = 20°

a° = 60°

(a + d) ° = 60° + 40° = 100°

Therefore, the angles of triangle in radians will be:

$ \left( 20\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }\pi /9 $

$ \left( 60\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }\pi /3 $

$ \left( 100\text{ }\times \text{ }\pi /180 \right)\text{ }rad\text{ }=\text{ }5\pi /9 $