In the given figure, RS and RT are tangent to the circle with center O. If angle SRT is 40 degrees, find the measure of angle TUS.

Since $RS$ and $RT$ are tangent lines to the circle at points $S$ and $T$ respectively, and $O$ is the center of the circle, the radii $OS$ and $OT$ are perpendicular to the tangents $RS$ and $RT$.
Therefore, we have:

$$m(\angle OSR) = 90^\circ$$ $$m(\angle OTR) = 90^\circ$$

Consider the quadrilateral $OSRT$. The sum of the interior angles of a quadrilateral is $360^\circ$.
We are given $m(\angle SRT) = 40^\circ$.
So, we can write:

$$m(\angle SOT) + m(\angle OSR) + m(\angle OTR) + m(\angle SRT) = 360^\circ$$

Substitute the known values:

$$m(\angle SOT) + 90^\circ + 90^\circ + 40^\circ = 360^\circ$$ $$m(\angle SOT) + 220^\circ = 360^\circ$$

Subtract $220^\circ$ from both sides:

$$m(\angle SOT) = 360^\circ - 220^\circ$$ $$m(\angle SOT) = 140^\circ$$

The angle $\angle TUS$ is an inscribed angle that subtends arc $ST$. The angle $\angle SOT$ is the central angle that subtends the same arc $ST$.
According to the inscribed angle theorem, the measure of an inscribed angle is half the measure of the central angle that subtends the same arc.

$$m(\angle TUS) = \frac{1}{2} m(\angle SOT)$$

Substitute the value of $m(\angle SOT)$:

$$m(\angle TUS) = \frac{1}{2} \times 140^\circ$$ $$m(\angle TUS) = 70^\circ$$

The measure of $\angle TUS$ is \boxed{\text{70^\circ}}.