Let Δ ABC be isosceles where AB = AC and ∠ B = ∠ C
Given: Vertex angle A is twice the sum of the base angles B and C. i.e.,
∠ A = 2(∠ B + ∠ C)
∠ A = 2(∠ B + ∠ B)
∠ A = 2(2 ∠ B)
∠ A = 4(∠ B)
Now, We know that sum of angles in a triangle =180°
∠ A + ∠ B + ∠ C = 180°
4 ∠ B + ∠ B + ∠ B = 180°
6 ∠ B =180°
∠ B = 30°
Since, ∠ B = ∠ C
∠ B = ∠ C = 30°
And ∠ A = 4 ∠ B
∠ A = 4 x 30° = 120°
Therefore, angles of the given triangle are 30° and 30° and 120°.