In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Let each base angle be x in an isosceles `/_\ABC`
Then vertex angle be `2(x+x)=4x`
Since sum of angles of a triangle is `180^0`
Hence,`4x+x+x=180^0`
`6x=180^0`
`=>x=30^0`
Angles are `30^0,30^0,120^0`

Explanation

To solve the problem step by step, we can follow this approach:

### Step 1: Define the angles
Let each base angle of the isosceles triangle be represented as \( x \). Since it is an isosceles triangle, the two base angles are equal.

### Step 2: Express the vertex angle
According to the problem, the vertex angle (let's call it angle \( A \)) is twice the sum of the base angles. Since there are two base angles, we can express this as:
\[
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