In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Answers
Given: In an isosceles triangle, if the vertex angle is twice the sum of the base angles.
Let Δ ABC be an isosceles such that ∠ A be a vertical angle and AB = AC and base angles ∠ B = ∠ C.
Vertex angle A is twice the sum of the base angles B and C. i.e., ∠ A = 2(∠ B + ∠ C)
∠ A = 2(∠ B + ∠ B)
[∠ B = ∠ C]
∠ A = 2(2 ∠ B)
∠ A = 4∠ B) ………..(1)
We know that sum of angles in a triangle is 180°.
∠ A + ∠ B + ∠ C =180°
4 ∠ B + ∠ B + ∠ B = 180°
[From eq 1]
6 ∠ B = 180°
∠ B = 180°/6
∠ B = 30°
Since, ∠ B = ∠ C
∠ B = ∠ C = 30°
Therefore, ∠ A = 4 ∠ B
∠ A = 4 x 30° = 120°
Hence, angles of the given triangle are 30° and 30° and 120°.
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