Question

1. (\triangle cde) is isosceles.

a. (angle ccong)
b. write and solve and equation to find the value of
c. (mangle c=)
d. (mangle e=)
e. (mangle d=)

Explanation:

Step1: Identify congruent angles in isosceles triangle

In an isosceles triangle \( \triangle CDE\) with \(CD = DE\), the base - angles are congruent. So, \( \angle C\cong\angle E\).

Step2: Set up the equation

Since \( \angle C=(2x + 17)^{\circ}\) and \( \angle E=(8x-13)^{\circ}\), and \( \angle C=\angle E\), we have the equation \(2x + 17=8x-13\).

Step3: Solve the equation for \(x\)

Subtract \(2x\) from both sides: \(17 = 6x-13\).
Add \(13\) to both sides: \(30=6x\).
Divide both sides by \(6\): \(x = 5\).

Step4: Find \(m\angle C\)

Substitute \(x = 5\) into \(m\angle C=(2x + 17)^{\circ}\).
\(m\angle C=(2\times5 + 17)^{\circ}=(10 + 17)^{\circ}=27^{\circ}\).

Step5: Find \(m\angle E\)

Since \( \angle C\cong\angle E\), \(m\angle E = 27^{\circ}\).

Step6: Find \(m\angle D\)

Use the triangle - angle sum theorem (\(m\angle C+m\angle E+m\angle D = 180^{\circ}\)).
Substitute \(m\angle C = 27^{\circ}\) and \(m\angle E = 27^{\circ}\): \(27^{\circ}+27^{\circ}+m\angle D=180^{\circ}\).
\(m\angle D=180^{\circ}-(27^{\circ}+27^{\circ})=126^{\circ}\).

Answer:

a. \( \angle E\)
b. Equation: \(2x + 17=8x-13\), Solution: \(x = 5\)
c. \(27^{\circ}\)
d. \(27^{\circ}\)
e. \(126^{\circ}\)