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