Question

suppose that \\( \triangle q r s \\) is isosceles with base \\( \overline{q s} \\).
suppose also that \\( m \angle q=(3 x+43)^{circ} \\) and \\( m \angle s=(4 x+37)^{circ} \\).
find the degree measure of each angle in the triangle.

Explanation:

Step1: Use the property of isosceles triangle

In an isosceles triangle \(\triangle QRS\) with base \(\overline{QS}\), \(\angle Q=\angle S\). So, \(3x + 43=4x + 37\).
Solve for \(x\):
\(43-37=4x - 3x\)
\(x = 6\)

Step2: Find \(m\angle Q\) and \(m\angle S\)

Substitute \(x = 6\) into \(m\angle Q=(3x + 43)^{\circ}\) and \(m\angle S=(4x + 37)^{\circ}\)
\(m\angle Q=(3\times6 + 43)^{\circ}=(18 + 43)^{\circ}=61^{\circ}\)
\(m\angle S=(4\times6 + 37)^{\circ}=(24 + 37)^{\circ}=61^{\circ}\)

Step3: Find \(m\angle R\)

Use the triangle - angle - sum theorem (\(m\angle Q+m\angle R+m\angle S = 180^{\circ}\))
\(m\angle R=180-(m\angle Q + m\angle S)\)
\(m\angle R=180-(61 + 61)\)
\(m\angle R = 58^{\circ}\)

Answer:

\(m\angle Q = 61^{\circ}\)
\(m\angle R = 58^{\circ}\)
\(m\angle S = 61^{\circ}\)