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 radii OP=OR=OS=OQ to form isosceles triangles, or apply the external angle formula for two secants from T.
Step 1: Mark the equal radii.
Since \(O\) is the center of the circle, \(OP\), \(OQ\), \(OR\), and \(OS\) are all radii of the same circle, so they are equal in length. This means triangles \(OPR\) and \(OQS\) are both isosceles. Step 2: Set up base angles for triangle \(OPR\).
In triangle \(OPR\), since \(OP=OR\), the base angles are equal. Let
\[
\angle OPR=\angle ORP=a
\] Step 3: Set up base angles for triangle \(OQS\).
In triangle \(OQS\), since \(OQ=OS\), the base angles are equal. Let
\[
\angle OQS=\angle OSQ=b
\] Step 4: Use the straight line \(PQ\) through \(O\).
Since \(P\), \(O\), \(Q\) lie on a straight line (as \(PQ\) is a diameter), the three angles \(\angle POR\), \(\angle ROS\), and \(\angle SOQ\) together make a straight angle:
\[
\angle POR+\angle ROS+\angle SOQ=180^\circ
\]
In triangle \(OPR\), \(\angle POR=180^\circ-2a\), and in triangle \(OQS\), \(\angle SOQ=180^\circ-2b\). Step 5: Substitute and simplify.
\[
(180^\circ-2a)+80^\circ+(180^\circ-2b)=180^\circ
\]
\[
440^\circ-2a-2b=180^\circ
\]
\[
a+b=130^\circ
\] Step 6: Use triangle \(PTQ\).
Since \(R\) lies on line \(PT\) and \(S\) lies on line \(QT\), the angle \(\angle RTS\) is the same as \(\angle PTQ\). Also, \(\angle TPQ=\angle OPR=a\) and \(\angle TQP=\angle OQS=b\), since \(O\) lies on segment \(PQ\). In triangle \(PTQ\), the angles sum to \(180^\circ\):
\[
a+b+\angle PTQ=180^\circ
\]
\[
130^\circ+\angle PTQ=180^\circ
\]
\[
\angle PTQ=50^\circ
\] Step 7: Final conclusion.
Since \(\angle RTS=\angle PTQ\),
\[
\boxed{\angle RTS=50^\circ}
\]
Hence, the correct option is (B).
