We are given a triangle ABC with angle ABC = $40^{\circ}$ and angle ACB = $120^{\circ}$. We need to find the size of angle BAC.

1. Problem Description

We are given a triangle ABC with angle ABC = $40^{\circ}$ and angle ACB = $120^{\circ}$. We need to find the size of angle BAC.

2. Solution Steps

We know that the sum of the angles in a triangle is $180^{\circ}$. Therefore,

$$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$$

We are given that $\angle ABC = 40^{\circ}$ and $\angle ACB = 120^{\circ}$. Substituting these values into the equation, we have

$$\angle BAC + 40^{\circ} + 120^{\circ} = 180^{\circ}$$

$$\angle BAC + 160^{\circ} = 180^{\circ}$$

Subtracting $160^{\circ}$ from both sides of the equation gives

$$\angle BAC = 180^{\circ} - 160^{\circ}$$

$$\angle BAC = 20^{\circ}$$

3. Final Answer

The size of angle BAC is $20^{\circ}$.