Let ΔABC be isosceles
Such that,
AB = BC
∠B = ∠C
Given, that 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 the sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
4∠B + ∠B + ∠B = 180°(Therefore, ∠A = 4∠B, ∠C = ∠B)
6∠B = 180°
∠B = \(\frac{180}{6}\)
= 30°
Since,
∠B = ∠C = 30°
And,
∠A = 4∠B
= 4 x 30° = 120°
Therefore, the angles of the triangle are 120°, 30°, 30°.