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

Question

Philoid · Accepted Answer

Let 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^o

∠A + ∠B + ∠C = 180^o

4∠B + ∠B + ∠B = 180^o (Therefore, ∠A = 4∠B, ∠C = ∠B)

6∠B = 180^o

∠B =

= 30^o

Since, ∠B = ∠C = 30^o

And, ∠A = 4∠B

= 4 * 30^o = 120^o

Therefore, the angles of the triangle are 120^o, 30^o, 30^o.