2. The vertex angle of an isosceles triangle is 30 less than 4 times the base angles. Find the vertex angle. The vertex angle is \( 138^{\circ} \). The vertex angle is \( 110^{\circ} \). The vertex angle is \( 35^{\circ} \). The vertex angle is \( 42^{\circ} \).

Let the measure of each base angle of the isosceles triangle be \( x \). Since the triangle is isosceles, the vertex angle can be expressed in terms of the base angles. According to the problem, the vertex angle is 30 degrees less than 4 times the base angle: \[ \text{Vertex angle} = 4x - 30 \] In any triangle, the sum of the angles is always 180 degrees. Therefore, we can write the equation: \[ \text{Vertex angle} + 2 \times \text{Base angle} = 180 \] Substituting the expression for the vertex angle into the equation gives: \[ (4x - 30) + 2x = 180 \] Now, combine like terms: \[ 6x - 30 = 180 \] Next, add 30 to both sides: \[ 6x = 210 \] Now, divide by 6: \[ x = 35 \] Now that we have the base angle, we can find the vertex angle: \[ \text{Vertex angle} = 4x - 30 = 4(35) - 30 = 140 - 30 = 110 \] Thus, the vertex angle is \( 110^{\circ} \). The answer is: **The vertex angle is \( 110^{\circ} \).**