n the diagram below of AABC, side BC is extended to point D, mLA = 3x + 2, mLB = 4x-1, ACD = x+19. Find mLB. %3D A 3x+2 A+b =ACD x+2 + 2x+15 =X+19 X+19 4x D 5xジ=xt19 -17 5xニメ+2 ーx -× X=2
n the diagram below of AABC, side BC is extended to point D, mLA = 3x + 2, mLB = 4x-1, ACD = x+19. Find mLB. %3D A 3x+2 A+b =ACD x+2 + 2x+15 =X+19 X+19 4x D 5xジ=xt19 -17 5xニメ+2 ーx -× X=2
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
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![## Transcription and Explanation of the Diagram in the Given Geometry Problem
### Problem Statement
In the diagram below of triangle \( \triangle ABC \), side \( BC \) is extended to point \( D \). The measures of the angles are given as follows:
- \( m \angle A = 3x + 2 \)
- \( m \angle LB = 4x - 1 \)
- \( m \angle ACD = x + 19 \)
Find \( m \angle LB \).
### Diagram Description
The diagram shows an extended triangle with the following details:
- Point \( A \), \( B \), \( C \), and \( D \) are labeled with lines connecting them to form the structure of the triangle and the extended line.
- Angle measures are marked on the diagram as follows:
- \( \angle A = 3x + 2 \)
- The external angle \( \angle ACD = x + 19 \)
- The angle \( \angle LB = 4x - 1 \)
### Solution Steps
1. **Equate the Angles**:
The exterior angle \( ACD \) of triangle \( ABC \) is equal to the sum of the opposite interior angles \( A \) and \( B \):
\[
A + B = ACD
\]
\[
3x + 2 + (4x - 1) = x + 19
\]
2. **Simplify the Equation**:
\[
5x + 1 = x + 19
\]
3. **Solve for \( x \)**:
\[
5x - x = 19 - 1
\]
\[
4x = 18
\]
\[
x = \frac{18}{4} = 4.5
\]
4. **Find \( m \angle LB \)**:
Substitute \( x = 4.5 \) into \( 4x - 1 \):
\[
m \angle LB = 4(4.5) - 1 = 18 - 1 = 17
\]
### Conclusion
The measure of \( m \angle LB \) is \( 17 \) degrees.
This problem illustrates the relationship between interior and exterior angles in geometry, highlighting](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff94d9d23-7978-492a-bec9-1b9c11961432%2Fd2457649-9f59-412e-aada-b408121bfa68%2Fjhxbusb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Transcription and Explanation of the Diagram in the Given Geometry Problem
### Problem Statement
In the diagram below of triangle \( \triangle ABC \), side \( BC \) is extended to point \( D \). The measures of the angles are given as follows:
- \( m \angle A = 3x + 2 \)
- \( m \angle LB = 4x - 1 \)
- \( m \angle ACD = x + 19 \)
Find \( m \angle LB \).
### Diagram Description
The diagram shows an extended triangle with the following details:
- Point \( A \), \( B \), \( C \), and \( D \) are labeled with lines connecting them to form the structure of the triangle and the extended line.
- Angle measures are marked on the diagram as follows:
- \( \angle A = 3x + 2 \)
- The external angle \( \angle ACD = x + 19 \)
- The angle \( \angle LB = 4x - 1 \)
### Solution Steps
1. **Equate the Angles**:
The exterior angle \( ACD \) of triangle \( ABC \) is equal to the sum of the opposite interior angles \( A \) and \( B \):
\[
A + B = ACD
\]
\[
3x + 2 + (4x - 1) = x + 19
\]
2. **Simplify the Equation**:
\[
5x + 1 = x + 19
\]
3. **Solve for \( x \)**:
\[
5x - x = 19 - 1
\]
\[
4x = 18
\]
\[
x = \frac{18}{4} = 4.5
\]
4. **Find \( m \angle LB \)**:
Substitute \( x = 4.5 \) into \( 4x - 1 \):
\[
m \angle LB = 4(4.5) - 1 = 18 - 1 = 17
\]
### Conclusion
The measure of \( m \angle LB \) is \( 17 \) degrees.
This problem illustrates the relationship between interior and exterior angles in geometry, highlighting
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