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Suppose that ACDE is isosceles with base EC.

Suppose also that mZD= (2x+42)° and mZE= (4x+14)°.
Find the degree measure of each angle in the triangle.
Check
-(4x + 14).
(2x + 42)
m2c=
mZD=
mZE =
X
D
0

Suppose that ACDE is isosceles with base EC. Suppose also that mZD= (2x+42)° and mZE-example-1

1 Answer

5 votes

Okay, here are the steps to solve this problem:

1) Since ACDE is isosceles with base EC, the angles at the base (mECD and mCEA) are equal. Let's call this common angle measure θ.

2) We know: mZD = (2x + 42)°

So, (2x + 42) + θ = 180° (angles sum to 180° in a triangle)

2x + 42 + θ = 180

=> 2x = 138

=> x = 69

3) Substitute x = 69 into mZE = (4x + 14)°

=> mZE = (4(69) + 14) = 278°

4) Now we have all 3 angles:

mECD = mCEA = θ (these are equal, common base angle)

mZD = (2)(69) + 42 = 174°

mZE = 278°

5) As a check:

174 + 278 + θ = 180

θ = 128

So the degree measures of the angles are:

mECD = mCEA = 128° (common base angle)

mZD = 174°

mZE = 278°

Let me know if you have any other questions! I'm happy to explain further.

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