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

Given:

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

To do:

We have to calculate the angles of the triangle.

Solution:

Let in $\triangle ABC, AB = AC$

$\angle A = 2(\angle B + \angle C)$

Let $\angle B = \angle C = x$

This implies,

$\angle A = 2(x + x)$

$ = 2 \times 2x$

$ = 4x$

Sum of the angles in a triangle is $180^o$.

This implies,

$4x + x + x = 180^o$

$6x = 180^o$

$x= \frac{180^o}{6}$

$x = 30^o$

Therefore,

$\angle B = \angle C = 30^o$

$\angle A = 4 \times 30^o = 120^o$

**The angles of the triangle are $30^o, 30^o$ and $120^o$.**

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