Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
![Title: Solving Circle Geometry Problems
---
**Problem 13:**
Given:
- On circle \( R \), \( m\overset{\frown}{QT} = m\overset{\frown}{TS} = 120^\circ \).
- Find \( m\angle QRS \) and \( m\angle QTS \).
**Solution Explanation:**
1. **Understanding the Given Information:**
- Two arcs, \( \overset{\frown}{QT} \) and \( \overset{\frown}{TS} \), both measure \( 120^\circ \).
- The circle is centered at point \( R \).
2. **Identifying Key Angles and Arcs:**
- Since the sum of the angles around the center is \( 360^\circ \), this means that \( \overset{\frown}{QS} \), the remaining arc, will be:
\[
360^\circ - (120^\circ + 120^\circ) = 120^\circ
\]
- Thus, we have three arcs: \( QT = 120^\circ \), \( TS = 120^\circ \), and \( QS = 120^\circ \).
3. **Finding \( m\angle QRS \):**
- \( \angle QRS \) is an inscribed angle that intercepts \( \overset{\frown}{QS} \).
- Inscribed angles are half the measure of the arc they intercept.
- So,
\[
m\angle QRS = \frac{1}{2} \times 120^\circ = 60^\circ
\]
4. **Finding \( m\angle QTS \):**
- Similarly, \( \angle QTS \) intercepts the same \( \overset{\frown}{QS} \).
- Thus,
\[
m\angle QTS = \frac{1}{2} \times 120^\circ = 60^\circ
\]
**Conclusion:**
- \( m\angle QRS = 60^\circ \)
- \( m\angle QTS = 60^\circ \)
**Visual Explanation:**
In the provided diagram (Figure not drawn to scale), points \( Q \), \( T \), and \( S \) lie on the circumference of the circle with \( R \) as](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c8a549a-ce13-4c73-8e46-ff62e8d935c2%2F93968df0-eab1-41dc-9766-ad08f0048b58%2Fvxxnpw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Title: Solving Circle Geometry Problems
---
**Problem 13:**
Given:
- On circle \( R \), \( m\overset{\frown}{QT} = m\overset{\frown}{TS} = 120^\circ \).
- Find \( m\angle QRS \) and \( m\angle QTS \).
**Solution Explanation:**
1. **Understanding the Given Information:**
- Two arcs, \( \overset{\frown}{QT} \) and \( \overset{\frown}{TS} \), both measure \( 120^\circ \).
- The circle is centered at point \( R \).
2. **Identifying Key Angles and Arcs:**
- Since the sum of the angles around the center is \( 360^\circ \), this means that \( \overset{\frown}{QS} \), the remaining arc, will be:
\[
360^\circ - (120^\circ + 120^\circ) = 120^\circ
\]
- Thus, we have three arcs: \( QT = 120^\circ \), \( TS = 120^\circ \), and \( QS = 120^\circ \).
3. **Finding \( m\angle QRS \):**
- \( \angle QRS \) is an inscribed angle that intercepts \( \overset{\frown}{QS} \).
- Inscribed angles are half the measure of the arc they intercept.
- So,
\[
m\angle QRS = \frac{1}{2} \times 120^\circ = 60^\circ
\]
4. **Finding \( m\angle QTS \):**
- Similarly, \( \angle QTS \) intercepts the same \( \overset{\frown}{QS} \).
- Thus,
\[
m\angle QTS = \frac{1}{2} \times 120^\circ = 60^\circ
\]
**Conclusion:**
- \( m\angle QRS = 60^\circ \)
- \( m\angle QTS = 60^\circ \)
**Visual Explanation:**
In the provided diagram (Figure not drawn to scale), points \( Q \), \( T \), and \( S \) lie on the circumference of the circle with \( R \) as
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