Question

parallel lines ef and ab are shown with transversal cd and m∠adc = 130 and m∠fcd = 3x + 34°. what is the value of x? x =

Explanation:

Step1: Identify angle - relationship

Since \(EF\parallel AB\) and \(CD\) is a transversal, \(\angle ADC\) and \(\angle FCD\) are same - side interior angles. Same - side interior angles of parallel lines are supplementary, so \(m\angle ADC + m\angle FCD=180^{\circ}\).

Step2: Substitute the given angle measures

We know that \(m\angle ADC = 130^{\circ}\) and \(m\angle FCD=(3z + 34)^{\circ}\). Substituting into the supplementary - angle equation gives \(130+(3z + 34)=180\).

Step3: Simplify the left - hand side

First, combine like terms: \(130+3z + 34=164 + 3z\). So the equation becomes \(164+3z = 180\).

Step4: Solve for \(z\)

Subtract 164 from both sides: \(3z=180 - 164\), so \(3z = 16\). Then divide both sides by 3: \(z=\frac{16}{3}\approx5.33\).

Answer:

\(z=\frac{16}{3}\)