Prove that in a triangle,other than an equilateral triangle,the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.

(A) Let the sides of the triangle be $a$,$b$,and $c$,and the angles opposite to them be $A$,$B$,and $C$ respectively.
Assume $a$ is the longest side,so $a > b$ and $a > c$.
By the property of triangles,the angle opposite to the longest side is the largest angle,so $A > B$ and $A > C$.
We know that the sum of angles in a triangle is $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we have $A + A + A > A + B + C$,which implies $3A > 180^{\circ}$.
Therefore,$A > 60^{\circ}$.
$A$ right angle is $90^{\circ}$,and $\frac{2}{3}$ of a right angle is $\frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Thus,$A > 60^{\circ}$ proves that the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.