2)

To solve the problem of finding the base angle of an isosceles triangle when the vertex angle is given:

1. Let's denote the vertex angle as \( \theta \).
2. The base angles of an isosceles triangle are equal. Let's denote each base angle as \( \alpha \).
3. The sum of all angles in a triangle is \( 180^\circ \).

Given: \( \theta = 84^\circ \).

Using the formula for the sum of angles: \[ \theta + 2\alpha = 180^\circ \]

Substitute the given angle: \[ 84^\circ + 2\alpha = 180^\circ \]

Solve for \( \alpha \): \[ 2\alpha = 180^\circ - 84^\circ \] \[ 2\alpha = 96^\circ \] \[ \alpha = 48^\circ \]

Therefore, the base angle is \( 48^\circ \).