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We are given a circle with the following conditions:

1. \( QS \) is a diameter.
2. \( m \overarc{QR} = 46^\circ \)
3. \( m \angle QRT = 44^\circ \)

We need to find the following angles:

- (a) \( m \angle RQS \)
- (b) \( m \angle SRT \)

Let's break down the problem step by step.

### 1. \( m \overarc{QR} = 46^\circ \)
The arc \( \overarc{QR} \) subtends the central angle at \( O \). Since \( QS \) is the diameter, the entire circle subtends a \( 180^\circ \) arc on one side.

### 2. \( m \angle RQS \) (Part (a))
- Since \( QS \) is the diameter, \( \angle RQS \) is an inscribed angle subtended by the arc \( \overarc{QR} \).
- An inscribed angle is half the measure of the arc it subtends.
  
Therefore,
\[
m \angle RQS = \frac{m \overarc{QR}}{2} = \frac{46^\circ}{2} = 23^\circ
\]

### 3. \( m \angle SRT \) (Part (b))
- \( \angle SRT \) is an exterior angle formed outside the triangle \( QRT \).
- The exterior angle is equal to the sum of the opposite interior angles. Therefore,
\[
m \angle SRT = m \angle RQS + m \angle QRT = 23^\circ + 44^\circ = 67^\circ
\]

### Final Answers:
- (a) \( m \angle RQS = 23^\circ \)
- (b) \( m \angle SRT = 67^\circ \)

Would you like further explanation on any part? 

Here are some related questions:
1. What is the relationship between inscribed and central angles in a circle?
2. How do you calculate the measure of an arc in a circle?
3. What is the property of an angle subtended by a diameter in a circle?
4. How do you calculate exterior angles in polygons?
5. How would the problem change if \( QS \) were not a diameter?

**Tip:** Remember, inscribed angles in a circle are always half the measure of the intercepted arc.

In the circle below, QS is a diameter. Suppose mQR = 46° and m∠QRT = 44°. Find the following:
(a) m∠RQS = ?
(b) m∠SRT = ?

Learn how to find the inscribed and exterior angles in a circle where QS is the diameter, mQR = 46°, and m∠QRT = 44°. This problem covers key concepts such as the inscribed angle theorem and the exterior angle theorem, suitable for students in Grades 9-10.

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