The isosceles trapezoid is part of an isosceles triangle with a 38° vertex angle. What is the measure of an acute base angle of the trapezoid? The diagram is not drawn to scale.
What is the measure of an obtuse base angle of the trapezoid?
Let's call the isosceles triangle ABC, with AB being the base and AC and BC being the congruent sides. Let's call the isosceles trapezoid ADEF, with AD being the top base, EF being the bottom base, and AE, DF being the legs.
Since triangle ABC is isosceles, angle BAC = angle BCA. Let's call this angle x.
Now, let's consider triangle AED. Since AE and DE are legs of the trapezoid, they are congruent. Thus, triangle AED is also isosceles. Therefore, angle AED = angle ADE.
Since triangle ABC and triangle ADE are similar, we know that angle BAC = angle AED and angle BCA = angle ADE.
Since angle BAC = angle AED = x, and angle BCA = angle ADE = x, we can conclude that angle ADE = angle AED = angle DEA = angle DAE = x.
Since we were given that angle BAC = 38 degrees, we have:
x + x + 38 = 180
2x + 38 = 180
2x = 142
x = 71
Therefore, the measure of an acute base angle of the trapezoid is 71 degrees.
Since triangle AED is isosceles with angle ADE = angle AED = x = 71 degrees, we can see that angle FED = angle AEF = 180 - (2x) = 180 - (2*71) = 38 degrees.
Therefore, the measure of an obtuse base angle of the trapezoid is 38 degrees.