In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Let the isosceles triangle be ΔABC as shown with base as BC and equal sides are AB and AC

In an isosceles triangle the base angles are equal let them be ‘x°’

Thus ∠B = x° and ∠C = x°

Given that ∠A = 2 × (∠B + ∠C)

⇒ ∠A = 2 × (x° + x°)

⇒ ∠A = 2 × 2x°

⇒ ∠A = 4x°

Now the sum of all angles of a triangle is 180°

⇒ ∠A + ∠B + ∠C = 180°

⇒ 4x° + x° + x° = 180°

⇒ 6x° = 180°

⇒ x = 30°

Therefore, angles of triangle are ∠B = 30°, ∠C = 30° and ∠A = 4x = 4 × 30° = 120°