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

\(\angle C = \angle 3B = 2(\angle A + \angle B)\)
Taking \( 3\angle B = 2 ( \angle A + \angle B ) \)
\(\angle B = 2\angle A\)

\(2\angle A – \angle B = 0\) …….(1)

We know that the sum of the measures of all angles of a triangle is 180°.

\(\angle A + \angle B + \angle C = 180°\)
\(\angle A + \angle B + 3\angle B = 180°\)
\(\angle A + 4\angle B = 180°\) …….(2)

Multiplying equation (1) by 4:
\(8\angle A – 4\angle B = 0\) …….(3)

Adding equations (2) and (3), we get
\(9\angle A = 180° \)
\(\angle A = 20°\)

From eq. (2), we get,
\(20° + 4\angle B = 180°\)
\( \angle B = 40°\)
And \( \angle C = 3 (40°) = 120°\)

Hence the measures of \(\angle A, \angle B \) and \(\angle C\) are respectively.