We are asked to find the size of angle $BAC$ in the given triangle. The triangle has angle $ABC = 40^\circ$ and the exterior angle at vertex $C$ is $110^\circ$.

1. Problem Description

We are asked to find the size of angle $BAC$ in the given triangle. The triangle has angle $ABC = 40^\circ$ and the exterior angle at vertex $C$ is $110^\circ$.

2. Solution Steps

Let the angle $BCA$ be $x$. Since the exterior angle at $C$ is $110^\circ$, we have:

$x + 110^\circ = 180^\circ$

Subtract $110^\circ$ from both sides:

$x = 180^\circ - 110^\circ$

$x = 70^\circ$

So, angle $BCA$ is $70^\circ$.

Now, we know that the sum of angles in a triangle is $180^\circ$. Let the angle $BAC$ be $y$. We have:

$y + 40^\circ + 70^\circ = 180^\circ$

$y + 110^\circ = 180^\circ$

$y = 180^\circ - 110^\circ$

$y = 70^\circ$

Thus, the angle $BAC$ is $70^\circ$.

3. Final Answer

The size of angle $BAC$ is $70^\circ$.