The angles of a triangle are in A.P. such that the greatest is 5 times the least.

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:

Greatest angle $=5 \times$ Least angle

or, $\frac{\text { Greatest angle }}{\text { Least angle }}=5$

or, $\frac{a+d}{a-d}=5$

or, $\frac{60+d}{60-d}=5$

or, $60+d=300-5 d$

or, $6 d=240$

or, $d=40$

Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $20^{\circ}, 60^{\circ}$ and $100^{\circ}$, respectively.

$\therefore$ Angles of the triangle in radians $=\left(20 \times \frac{\pi}{180}\right),\left(60 \times \frac{\pi}{180}\right)$ and $\left(100 \times \frac{\pi}{180}\right)$

$=\frac{\pi}{9}, \frac{\pi}{3}$ and $\frac{5 \pi}{9}$