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: Use the angle in a semicircle property.
Since \(\overline{PQ}\) is a diameter, any point on the circle, such as \(R\) or \(S\), forms a right angle with the two ends of the diameter. So \(\angle PRQ = 90^{\circ}\) and \(\angle PSQ = 90^{\circ}\).
Step 2: Set up variable angular positions for R and S.
Let \(O\) be the centre with \(P\) and \(Q\) at the two ends of the diameter. Let \(\angle QOS = \varphi\), the angle radius \(OS\) makes with \(OQ\). Since \(\angle ROS = 80^{\circ}\) and R lies further from Q than S, as seen in the figure, \(\angle QOR = \varphi + 80^{\circ}\).
Step 3: Convert these central angles to inscribed angles at P and Q.
The inscribed angle theorem says an inscribed angle is half of the central angle standing on the same arc.
At P, the angle \(\angle RPQ\) stands on arc RQ, whose central angle is \(\angle ROQ = \varphi + 80^{\circ}\), so
\[ \angle RPQ = \frac{\varphi + 80^{\circ}}{2} \]
At Q, the angle \(\angle SQP\) stands on arc SP, whose central angle is \(\angle SOP = 180^{\circ} - \varphi\) (because \(\angle POQ = 180^{\circ}\)), so
\[ \angle SQP = \frac{180^{\circ} - \varphi}{2} = 90^{\circ} - \frac{\varphi}{2} \]
Step 4: Apply the angle sum property in triangle PTQ.
Since \(R\) lies on segment \(PT\) and \(S\) lies on segment \(QT\), the angle \(\angle RTS\) is the same as the angle \(\angle PTQ\) of triangle \(PTQ\). The three angles of this triangle, \(\angle TPQ = \angle RPQ\), \(\angle TQP = \angle SQP\), and \(\angle PTQ\), must sum to \(180^{\circ}\).
\[ \angle PTQ = 180^{\circ} - \left(\frac{\varphi}{2} + 40^{\circ}\right) - \left(90^{\circ} - \frac{\varphi}{2}\right) \]
\[ \angle PTQ = 180^{\circ} - 40^{\circ} - 90^{\circ} - \frac{\varphi}{2} + \frac{\varphi}{2} \]
\[ \angle PTQ = 50^{\circ} \]
The \(\varphi\) terms cancel out completely, so the answer does not depend on exactly where R and S sit on the circle, only on the fact that \(\angle ROS = 80^{\circ}\).
Final Answer:
\(\angle RTS = 50^{\circ}\).
\[ \boxed{50^{\circ}} \]
