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

Solution:

Let ΔABC be isosceles such that AB = AC.

∠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 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 × 30° = 120°

Therefore, angles of the given triangle are 120°, 30° and 30°.

= 428 and LB = LC]