Given that RS and RT are tangent lines to the circle with center O, the radii OS and OT are perpendicular to the tangents at the points of tangency S and T, respectively.

Given that $RS$ and $RT$ are tangent lines to the circle with center $O$, the radii $OS$ and $OT$ are perpendicular to the tangents at the points of tangency $S$ and $T$, respectively.
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 the equation:

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

Substitute the known angle measures:

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

Subtract $220^\circ$ from both sides:

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

The measure of a central angle is equal to the measure of its intercepted arc. Since $\angle SOT$ is a central angle subtending arc $ST$:

$$m(\text{arc } ST) = m(\angle SOT) = 140^\circ$$

The measure of an inscribed angle is half the measure of its intercepted arc. $\angle TUS$ is an inscribed angle that subtends arc $ST$.

$$m(\angle TUS) = \frac{1}{2} m(\text{arc } ST)$$

Substitute the measure of arc $ST$:

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

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