Question

suppose that \\( \triangle c d e \\) is isosceles with base \\( \overline{d e} \\).
suppose also that \\( m \angle d=(4 x+25)^{circ} \\) and \\( m \angle e=(2 x+41)^{circ} \\).
find the degree measure of each angle in the triangle.

Explanation:

Step1: Use the property of isosceles triangle

In an isosceles triangle \( \triangle CDE \) with base \( \overline{DE} \), \( \angle D=\angle E \). So, \( 4x + 25=2x+41 \).
Solve for \( x \):
\( 4x-2x=41 - 25 \)
\( 2x=16 \)
\( x = 8 \)

Step2: Calculate \( m\angle D \) and \( m\angle E \)

Substitute \( x = 8 \) into \( m\angle D=(4x + 25)^{\circ} \):
\( m\angle D=(4\times8 + 25)^{\circ}=(32 + 25)^{\circ}=57^{\circ} \)
Since \( \angle D=\angle E \), \( m\angle E = 57^{\circ} \)

Step3: Calculate \( m\angle C \)

Use the triangle - angle sum theorem (\( m\angle C+m\angle D+m\angle E = 180^{\circ} \))
\( m\angle C=180-(m\angle D + m\angle E) \)
\( m\angle C=180-(57 + 57) \)
\( m\angle C=180 - 114=66^{\circ} \)

Answer:

\( m\angle C = 66^{\circ} \), \( m\angle D = 57^{\circ} \), \( m\angle E = 57^{\circ} \)