Question

suppose that \\( \triangle l m n \\) is isosceles with base \\( \overline{n m} \\).
suppose also that \\( m \angle l=(4 x+30)^{\circ} \\) and \\( m \angle n=(3 x+40)^{\circ} \\).
find the degree measure of each angle in the triangle.

Explanation:

Step1: Use the property of isosceles triangle

Since \(\triangle LMN\) is isosceles with base \(\overline{NM}\), then \(\angle N=\angle M\). So \(m\angle M=(3x + 40)^{\circ}\).

Step2: Apply the triangle - angle - sum theorem

The sum of the interior angles of a triangle is \(180^{\circ}\). So \(m\angle L+m\angle M+m\angle N = 180^{\circ}\). Substitute \(m\angle L=(4x + 30)^{\circ}\), \(m\angle M=(3x + 40)^{\circ}\), and \(m\angle N=(3x + 40)^{\circ}\) into the equation:

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Step3: Solve for \(x\)

Subtract \(110\) from both sides of \(10x + 110=180\): \(10x=180 - 110=70\). Then divide both sides by \(10\): \(x = 7\).

Step4: Calculate the measure of each angle

• For \(\angle L\): \(m\angle L=(4x + 30)^{\circ}\). Substitute \(x = 7\), \(m\angle L=(4\times7+30)^{\circ}=(28 + 30)^{\circ}=58^{\circ}\).
• For \(\angle M\) and \(\angle N\): \(m\angle N=(3x + 40)^{\circ}\). Substitute \(x = 7\), \(m\angle N=(3\times7+40)^{\circ}=(21+40)^{\circ}=61^{\circ}\). Since \(m\angle M=m\angle N\), \(m\angle M = 61^{\circ}\).

Answer:

\(m\angle L = 58^{\circ}\), \(m\angle M=61^{\circ}\), \(m\angle N = 61^{\circ}\)