In this problem, we are given that $m\angle ACD = 40^\circ$, and we need to determine $m\angle DCB$.

From the image, it's clear that points $A$, $C$, and $B$ are collinear, forming a straight line. Therefore, the angles $\angle ACD$ and $\angle DCB$ form a linear pair. A linear pair of angles always sums up to $180^\circ$.

Since $m\angle ACD = 40^\circ$, the measure of $m\angle DCB$ can be found using the following equation:

$m\angle ACD + m\angle DCB = 180^\circ$

Substituting the known value:

$40^\circ + m\angle DCB = 180^\circ$

Solving for $m\angle DCB$:

$m\angle DCB = 180^\circ - 40^\circ = 140^\circ$

Thus, $m\angle DCB = 140^\circ$.

Do you want more details or have any further questions?

Here are some related questions for further practice:

1. What is the measure of the third angle in a triangle if two angles are $40^\circ$ and $80^\circ$?
2. If two angles form a linear pair and one of them is $65^\circ$, what is the measure of the other angle?
3. What is the sum of the interior angles of a quadrilateral?
4. How can you determine if two angles are supplementary?
5. In a straight angle, if one of the angles is $75^\circ$, what is the measure of the other?

Tip: When two angles form a straight line (linear pair), their measures always add up to $180^\circ$.