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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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