The angles of a triangle are in A.P and the greatest angle is double the least angle, then sine of the third angle is

Let the angles of the triangle be \(A-D\), \(A\), and \(A+D\).

The sum of the angles of a triangle is \(180^{\circ}\).

\((A-D) + A + (A+D) = 180^{\circ}\)

\(3A = 180^{\circ} \Rightarrow A = 60^{\circ}\)

The greatest angle is double the least angle.

\(A+D = 2(A-D)\)

Substituting \(A = 60^{\circ}\):

\(60^{\circ}+D = 2(60^{\circ}-D)\)

\(60^{\circ}+D = 120^{\circ}-2D\)

\(3D = 60^{\circ} \Rightarrow D = 20^{\circ}\)

The angles of the triangle are \(40^{\circ}\), \(60^{\circ}\), and \(80^{\circ}\).

The third angle (middle angle) is \(60^{\circ}\).

The sine of the third angle is \(\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}\).

Answer: \(\dfrac{\sqrt{3}}{2}\)