Given AJKL, find mZL. K L. (13x – 9)° (4x + 11)° J (2x + 7)° O 68° O 58° 29° O47

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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**Problem Statement:**

Given \( \triangle JKL \), find \( m\angle L \).

**Diagram Explanation:**

The diagram shows triangle \( JKL \) with:

- \( \angle J \) labeled as \( (2x + 7)^\circ \)
- \( \angle K \) labeled as \( (13x - 9)^\circ \)
- \( \angle L \) labeled as \( (4x + 11)^\circ \)

**Multiple-Choice Answers:**

- 68°
- 58°
- 29°
- 47°

To solve for \( m\angle L \), you need to use the triangle angle sum property, which states that the sum of the angles in a triangle is always 180°:

\[ 
(2x + 7) + (13x - 9) + (4x + 11) = 180 
\]

Simplify and solve for \( x \):

\[ 
2x + 7 + 13x - 9 + 4x + 11 = 180 
\] 
\[ 
19x + 9 = 180 
\] 
\[ 
19x = 171 
\] 
\[ 
x = 9 
\]

Now, substitute \( x = 9 \) back into \( \angle L \):

\[ 
m\angle L = 4x + 11 = 4(9) + 11 = 36 + 11 = 47^\circ 
\]

Therefore, the correct answer is:

- 47°
Transcribed Image Text:**Problem Statement:** Given \( \triangle JKL \), find \( m\angle L \). **Diagram Explanation:** The diagram shows triangle \( JKL \) with: - \( \angle J \) labeled as \( (2x + 7)^\circ \) - \( \angle K \) labeled as \( (13x - 9)^\circ \) - \( \angle L \) labeled as \( (4x + 11)^\circ \) **Multiple-Choice Answers:** - 68° - 58° - 29° - 47° To solve for \( m\angle L \), you need to use the triangle angle sum property, which states that the sum of the angles in a triangle is always 180°: \[ (2x + 7) + (13x - 9) + (4x + 11) = 180 \] Simplify and solve for \( x \): \[ 2x + 7 + 13x - 9 + 4x + 11 = 180 \] \[ 19x + 9 = 180 \] \[ 19x = 171 \] \[ x = 9 \] Now, substitute \( x = 9 \) back into \( \angle L \): \[ m\angle L = 4x + 11 = 4(9) + 11 = 36 + 11 = 47^\circ \] Therefore, the correct answer is: - 47°
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