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Question -

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



Answer -

Given : In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)
 
To calculate: Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ Sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x== 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120

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