Question

The vertical angle of an isosceles triangle is twice the sum of its base angles. Find each angle of
the triangle.​

Answer 1

Answer:

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°.