Problem Analysis

We are tasked with finding the angles $\angle RQS$ and $\angle RTQ$ in the given circle, where:

• Arc $RS$ subtends a central angle of $42^\circ$ at $O$.
• Lines $PR$ and $QS$ are extended to meet at a point $T$.

Step-by-Step Solution

1. Angle at the Circumference

The angle subtended by an arc at the circumference is half of the angle subtended at the center. Thus: $\angle RQS = \frac{\angle ROS}{2} = \frac{42^\circ}{2} = 21^\circ.$

2. Finding $\angle RTQ$

The four points $P, R, Q, T$ form a cyclic quadrilateral since they lie on the circle. For a cyclic quadrilateral, the opposite angles are supplementary: $\angle RQS + \angle RTQ = 180^\circ.$ Substituting $\angle RQS = 21^\circ$, we get: $\angle RTQ = 180^\circ - 21^\circ = 159^\circ.$

Final Answers

1. $\angle RQS = 21^\circ$
2. $\angle RTQ = 159^\circ$

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Related Questions

1. What are the properties of cyclic quadrilaterals and how are they used to solve problems like this?
2. How do we derive the relationship between the central angle and the angle at the circumference of a circle?
3. Can this solution be extended if the angle subtended by the arc changes?
4. What happens if $PR$ and $QS$ intersect outside the circle instead of at $T$?
5. How can this problem be solved using coordinate geometry?

Tip:

When solving problems in circles, always check for relationships involving cyclic quadrilaterals and arcs—they often simplify the calculations!