The angles of a triangle are in AP. The greatest angle is twice the least. All the angles of the triangle are ________.

Given, the angles of a triangle are in AP and the greatest angle is twice the least. 

We know that, the first term = $a$  and d = common difference. 

Let the interior angles of a triangle as: $\left(a-d\right)$, $a$, $a+d$.

As the sum of three interior angles of a triangle is ${180}^{°}$,

$\begin{array}{l}\left(a-d\right)+a+\left(a+d\right)={180}^{°}\\ a-d+a+a+d={180}^{°}\\ 3a={180}^{°}\end{array}$ 

$\begin{array}{l}a=\frac{{180}^{°}}{3}\\ ={60}^{°}\end{array}$

As the greatest angle is 2 times of the smallest angle i.e., $\left(a+d\right)=2\left(a-d\right)$. 

$\begin{array}{l}\left(a+d\right)=2\left(a-d\right)\\ \left(60+d\right)=2\left(60-d\right)\\ 60+d=120-2d \\ 2d+d=120-60 \end{array}$ 

$\begin{array}{l}3d=60 \\ d=\frac{60}{3}\\ d={20}^{°}\end{array}$ 

So, the three angles be,

$\begin{array}{l}\left(a-d\right), a, \left(a+d\right)=\left(60-20\right), 60, \left(60+20\right)\\ ={40}^{°}, {60}^{°}, {80}^{°}\end{array}$ 

Hence, the angles of the triangle are ${40}^{\circ }, {60}^{\circ }, {80}^{\circ }$.