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 $$ 42^{\circ} $$.

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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 $$ 110^{\circ} $$.**
The vertex angle is $$ 110^{\circ} $$.

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 \( 42^{\circ} \).

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