Question

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

Answer 1

EXPLANATION -

Let each base angle be x in an isosceles △ABC

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4x

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180° ⇒x=30°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180° ⇒x=30°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180° ⇒x=30° ∴ Angles are 30° , 30° and 120°

Let each base angle be x in an isosceles △ABCThen vertex angle be 2(x+x)=4xSince sum of angles of a triangle is 180 Hence 4x+x+x=180° ⇒6x=180° ⇒x=30° ∴ Angles are 30° , 30° and 120°

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