In a $\Delta ABC$,$\angle C = 3 \angle B = 2(\angle A + \angle B)$. Find the three angles.

(A) Given that,$\angle C = 3 \angle B = 2(\angle A + \angle B)$.
From $3 \angle B = 2(\angle A + \angle B)$,we get $3 \angle B = 2 \angle A + 2 \angle B$,which simplifies to $\angle B = 2 \angle A$,or $2 \angle A - \angle B = 0 \dots (i)$.
We know that the sum of the angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Since $\angle C = 3 \angle B$,we substitute this into the sum: $\angle A + \angle B + 3 \angle B = 180^{\circ}$,which gives $\angle A + 4 \angle B = 180^{\circ} \dots (ii)$.
Multiplying equation $(i)$ by $4$,we get $8 \angle A - 4 \angle B = 0 \dots (iii)$.
Adding equations $(ii)$ and $(iii)$,we get $9 \angle A = 180^{\circ}$,which implies $\angle A = 20^{\circ}$.
Substituting $\angle A = 20^{\circ}$ into equation $(ii)$,we get $20^{\circ} + 4 \angle B = 180^{\circ}$,so $4 \angle B = 160^{\circ}$,which gives $\angle B = 40^{\circ}$.
Finally,$\angle C = 3 \angle B = 3 \times 40^{\circ} = 120^{\circ}$.
Thus,the angles are $\angle A = 20^{\circ}$,$\angle B = 40^{\circ}$,and $\angle C = 120^{\circ}$.