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given that δabc = δdef m<e = x = y = z =

Question

given that δabc = δdef m<e = x = y = z =

Explanation:

Step1: Use angle - congruence property of congruent triangles

Since \(\triangle ABC\cong\triangle DEF\), corresponding angles are equal. \(\angle B=\angle E\). In \(\triangle ABC\), we know that the sum of angles in a triangle is \(180^{\circ}\). First, find \(\angle B\) in \(\triangle ABC\). The sum of angles in \(\triangle DEF\) is also \(180^{\circ}\), and \(\angle D = 60^{\circ}\), \(\angle F=70^{\circ}\), so \(\angle E=180-(60 + 70)=50^{\circ}\). Then \(2x + 10=50\).

Step2: Solve for \(x\)

Subtract 10 from both sides of the equation \(2x+10 = 50\): \(2x=50 - 10=40\). Divide both sides by 2: \(x=\frac{40}{2}=20\).

Step3: Use side - congruence property of congruent triangles

Corresponding sides of congruent triangles are equal. \(AB = DE\) and \(BC=EF\). \(AB = 22\) and \(DE=3z + 1\), so \(3z+1=22\). Subtract 1 from both sides: \(3z=22 - 1 = 21\), then \(z=\frac{21}{3}=7\). Also, \(BC = 20\) and \(EF=4y-4\), so \(4y-4=20\). Add 4 to both sides: \(4y=20 + 4=24\), then \(y=\frac{24}{4}=6\).

Answer:

\(m\angle E = 50^{\circ}\), \(x = 20\), \(y = 6\), \(z = 7\)