the angles of a triangle are …

Given the angles of the triangle are in A.P

So let them be a-d,a,a+d.

 We have sum of the three angles of a triangle is {tex}{ 180 }^{ \circ }{/tex}

{tex}\Rightarrow \left( a-d \right) +a+\left( a+d \right) ={ 180 }^{ \circ }\\ \Rightarrow 3a={ 180 }^{ \circ }\\ \Rightarrow a=\frac { { 180 }^{ \circ } }{ 3 } ={ 60 }^{ \circ }{/tex} 

According to the question the greatest angle is twice the smallest angle

{tex}\Rightarrow a+d=2\left( a-d \right) \\ \Rightarrow a+d=2a-2d\\ \Rightarrow 3d=a\\ \Rightarrow d=\frac { a }{ 3 } =\frac { 60 }{ 3 } ={ 20 }^{ \circ }{/tex}

{tex}\therefore a-d={ 60 }^{ \circ }-{ 20 }^{ \circ }={ 40 }^{ \circ },a={ 60 }^{ \circ },a+d={ 60 }^{ \circ }+{ 20 }^{ \circ }={ 80 }^{ \circ }{/tex}

So the angles of the triangle which are in A.P are {tex}{ 40 }^{ \circ },{ 60 }^{ \circ }{/tex}and {tex}{ 80 }^{ \circ }{/tex}