Step 1: Identify the properties of tangents from an external point. 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: $$m(\angle OSR) = 90^\circ$$ $$m(\angle OTR) = 90^\circ$$ Step 2: Find the measure of the central angle $\angle SOT$. 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: $$m(\angle SOT) = 360^\circ - 220^\circ$$ $$m(\angle SOT) = 140^\circ$$ Step 3: Determine the measure of arc $ST$. The measure of a central angle is equal to the measure of its intercepted arc. Since $\angle SOT$ is a central angle subtending arc $ST$: $$m(\text{arc } ST) = m(\angle SOT) = 140^\circ$$ Step 4: Find the measure of the inscribed angle $\angle TUS$. 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 $\boxed{\text{70}^\circ}$. Send me the next one 📸