Question

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 ________.

• A. \(40^\circ\)
• B. \(50^\circ\)
• C. \(60^\circ\)
• D. \(80^\circ\)

Hint:

Use radii OP=OR=OS=OQ to form isosceles triangles, or apply the external angle formula for two secants from T.

Answer 1

The Correct Option is B

Step 1: Identify the two secants from external point $T$.
$T$ lies outside the circle, and two secants are drawn from it: one through $R$ and $P$ (with $R$ nearer to $T$), and one through $S$ and $Q$ (with $S$ nearer to $T$).
Step 2: Recall the external angle formula.
When two secants are drawn from a point outside a circle, the angle between them equals half the difference of the two arcs they cut off:
\[ \angle T=\frac{1}{2}\left(\text{arc }PQ-\text{arc }RS\right) \]
where arc $PQ$ is the far arc between the far points $P$ and $Q$, and arc $RS$ is the near arc between the near points $R$ and $S$.
Step 3: Find arc $RS$.
The central angle $\angle ROS=80^\circ$ directly gives the near arc $RS$:
\[ \text{arc }RS=80^\circ \]
Step 4: Find arc $PQ$.
Since $PQ$ is a diameter, it splits the circle into two equal arcs of $180^\circ$ each. Points $R$ and $S$ both lie on the same semicircle, the one nearer to $T$, so the far arc $PQ$ (the one not containing $R$ and $S$) is the other semicircle:
\[ \text{arc }PQ=180^\circ \]
Step 5: Substitute into the formula.
\[ \angle RTS=\frac{1}{2}(180^\circ-80^\circ)=\frac{1}{2}(100^\circ)=50^\circ \]
Step 6: Conclude.
\[ \boxed{\angle RTS=50^\circ} \]