https://file.aiquickdraw.com/mathbot/file/1728100797013wcsd49nq.jpg

We are given the following information:

- \(\overline{QS}\) is the diameter of the circle.
- \(\overline{QT}\) is tangent to the circle at point \(Q\).
- The angle \(\angle QRS = 68^\circ\).
- We need to find the angles \(\angle RQT\) and \(\angle RQS\).

### Step-by-Step Solution:

#### (a) Finding \(\angle RQT\):
- Since \(\overline{QT}\) is tangent to the circle at \(Q\), by the tangent-secant theorem, the angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the opposite segment. Hence, \(\angle RQT = \angle QRS\).
- From the given, \(\angle QRS = 68^\circ\).
- Thus, \(\angle RQT = 68^\circ\).

#### (b) Finding \(\angle RQS\):
- Since \(\overline{QS}\) is the diameter of the circle, \(\angle RQS\) is an inscribed angle subtended by the diameter, meaning it must be a right angle.
- Therefore, \(\angle RQS = 90^\circ\).

### Final Answer:
- (a) \(\angle RQT = 68^\circ\)
- (b) \(\angle RQS = 90^\circ\)

Would you like more details or clarification on any step?

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Here are 5 related questions to expand on the topic:

1. What is the relationship between an inscribed angle and the arc it subtends?
2. How does the tangent-secant theorem apply to other types of angles in a circle?
3. What are the properties of angles in a semicircle?
4. How can you use the tangent-radius property in other problems?
5. What are the different theorems involving tangents and circles?

**Tip:** Always remember that the angle subtended by a diameter in a circle is a right angle (90°).

En el siguiente círculo, QS es un diámetro y QT es tangente en Q. Suponiendo que m∠QR = 68°. Hallar lo siguiente:
(a) ∠RQT
(b) ∠RQS

Solve this geometry problem involving a circle with a diameter (QS) and a tangent (QT) at point Q. Using the tangent-secant theorem and the inscribed angle theorem, learn how to calculate ∠RQT and ∠RQS when ∠QR = 68°.

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