In the given figure, \(\overline{PQ}\) is the diameter of a circle with center \(O\). Two points \(R\) and \(S\) are chosen on the circle such that \(\angle ROS = 80^{\circ}\). When \(\overline{PR}\) and \(\overline{QS}\) are extended, they meet at \(T\). The value of \(\angle RTS\) is .
Show Hint
Use the right angle from the diameter at R, find angle RQS as half of angle ROS, then apply the triangle angle sum in triangle QRT.
Step 1: Mark the right angle that the diameter creates.
Since $PQ$ is a diameter, any point on the circle sees it at a right angle, so $\angle PRQ = 90^\circ$. Point $T$ lies on ray $PR$ extended beyond $R$, so $R$ sits between $P$ and $T$ on a straight line. That makes $\angle QRT$ the supplement of $\angle QRP$ along that straight line:
\[ \angle QRT = 180^\circ - \angle QRP = 180^\circ - 90^\circ = 90^\circ \]
Step 2: Find angle RQT using the inscribed angle theorem.
Point $S$ lies on ray $QS$ extended to $T$, so $Q$, $S$, $T$ sit on one straight line, which means $\angle RQT$ is the same angle as $\angle RQS$.
$\angle RQS$ is an inscribed angle standing on chord $RS$, viewed from point $Q$ on the circle. By the inscribed angle theorem, an inscribed angle is half of the central angle standing on the same chord:
\[ \angle RQS = \frac{1}{2} \angle ROS = \frac{1}{2}(80^\circ) = 40^\circ \]
So $\angle RQT = 40^\circ$.
Step 3: Apply the angle sum property in triangle QRT.
Triangle $QRT$ has vertices $Q$, $R$, $T$, with interior angles $\angle RQT = 40^\circ$ (Step 2), $\angle QRT = 90^\circ$ (Step 1), and $\angle RTQ$, the angle we want. The three angles of any triangle add up to $180^\circ$:
\[ \angle RTQ = 180^\circ - 40^\circ - 90^\circ = 50^\circ \]
Step 4: Identify angle RTQ with angle RTS.
Since $S$ lies on segment $QT$, the ray $TS$ is the same ray as $TQ$, so $\angle RTS$ and $\angle RTQ$ are the exact same angle.
