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in Triangles by (56.3k points)

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

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Let Δ ABC be isosceles where AB = AC and ∠ B = ∠ C 

Given: 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 x 30° = 120° 

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

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