What is the measure of an acute base angle of an isosceles trapezoid that is part of an isosceles triangle with a 34-degree vertex angle?

To find the measure of the acute base angles of the isosceles trapezoid, we first need to analyze the relationship between the trapezoid and the isosceles triangle.

Let’s denote the isosceles triangle as ΔABC, where vertex A is at the top. The vertex angle at A measures 34 degrees. Consequently, the two base angles at B and C must be equal since the triangle is isosceles.

To find the measure of these base angles, we can use the fact that the sum of all angles in a triangle equals 180 degrees:

Angle A + Angle B + Angle C = 180°

Substituting the known values into this equation, we get:

34° + 2 * Angle B = 180°

Now, let’s solve for Angle B:

2 * Angle B = 180° - 34° 
2 * Angle B = 146°
Angle B = 73°

So, each base angle (B and C) of the isosceles triangle measures 73 degrees.

The isosceles trapezoid is formed by cutting the isosceles triangle at the base, where the bases of the trapezoid are parallel to the base of the triangle. In an isosceles trapezoid, the angles adjacent to each base are equal. Therefore, the acute angles of the trapezoid correspond to the base angles of the triangle.

Thus, the measure of an acute base angle of the trapezoid is:

73°

In conclusion, the acute base angles of the isosceles trapezoid are each 73 degrees.

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