Question

if \\( \triangle cde \\) is isosceles, what is the measure of \\( \angle d \\)?

Explanation:

Step1: Use the property of isosceles triangle

In an isosceles triangle \( \triangle CDE\), if \(CD = DE\), then \(\angle C=\angle E\). So, \(2x + 17=8x-13\).

Step2: Solve the equation for \(x\)

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

Step3: Find the measure of \(\angle C\) and \(\angle E\)

Substitute \(x = 5\) into \(\angle C=2x + 17\), we get \(\angle C=2\times5+17=27^{\circ}\), and \(\angle E=\angle C = 27^{\circ}\).

Step4: Use the triangle - angle sum theorem

The sum of angles in a triangle is \(180^{\circ}\). Let \(\angle D=y\). Then \(y+\angle C+\angle E=180^{\circ}\). Substitute \(\angle C = 27^{\circ}\) and \(\angle E=27^{\circ}\) into the equation: \(y+27 + 27=180\).

Step5: Solve for \(\angle D\)

Simplify the left - hand side: \(y+54 = 180\). Subtract \(54\) from both sides: \(y=\angle D=126^{\circ}\).

Answer:

\(126^{\circ}\)