The problem asks us to find the size of angle $BAC$ in the given triangle $ABC$. We are given that angle $ABC$ is $40^\circ$ and the exterior angle at $C$ is $120^\circ$.

1. Problem Description

The problem asks us to find the size of angle $BAC$ in the given triangle $ABC$. We are given that angle $ABC$ is $40^\circ$ and the exterior angle at $C$ is $120^\circ$.

2. Solution Steps

First, we need to find the measure of angle $ACB$. Since the exterior angle at $C$ is $120^\circ$, and the sum of an interior angle and its corresponding exterior angle is $180^\circ$, we have:

$ACB + 120^\circ = 180^\circ$

$ACB = 180^\circ - 120^\circ$

$ACB = 60^\circ$

Now we know two angles of the triangle $ABC$: angle $ABC = 40^\circ$ and angle $ACB = 60^\circ$. We also know that the sum of the angles in a triangle is $180^\circ$, so

$BAC + ABC + ACB = 180^\circ$

$BAC + 40^\circ + 60^\circ = 180^\circ$

$BAC + 100^\circ = 180^\circ$

$BAC = 180^\circ - 100^\circ$

$BAC = 80^\circ$

3. Final Answer

The size of angle $BAC$ is $80^\circ$.