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In the diagram below, AB BC, and mZBCD = 115°. Find mZB. D. 115°

In the diagram below, AB BC, and mZBCD = 115°. Find mZB. D. 115°

Elementary Geometry For College Students, 7e
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:** - \( AB \cong BC \) - \( m \angle BCD = 115^\circ \) **Find:** - \( m \angle B \) ### Diagram Explanation: According to the given problem, we have a geometric figure, specifically a triangle \( ABC \) with an additional line segment \( CD \) extended from point \( C \). Here is a detailed description of the diagram: - Triangle \( ABC \) has points \( A \), \( B \), and \( C \). - Segment \( AB \) is congruent to segment \( BC \). - Point \( D \) lies on a straight line extending from \( C \), creating an external angle \( \angle BCD = 115^\circ \). ### Approach: To find \( m \angle B \) in the triangle \( ABC \): 1. **Identifying the given information:** - Since \( AB \cong BC \), triangle \( ABC \) is an isosceles triangle with \( AB = BC \). - The external angle \( \angle BCD \) can be used to find the internal angle at vertex \( C \) of the triangle \( \angle ACB \). 2. **Using the properties of external angles:** - The external angle of a triangle (\( \angle BCD \)) is equal to the sum of the two non-adjacent internal angles of the triangle. - Therefore, \( \angle BCD = \angle CAB + \angle ACB \). 3. **Calculating the internal angles:** - Let \( \angle ACB = x \). Since \( AB \cong BC \), \( \angle CAB \) also equals \( x \). - Hence, \( \angle BCD = x + x \). - So, \( 115^\circ = 2x \). - Solving for \( x \), \( x = 57.5^\circ \). 4. **Finding \( m \angle B \):** - The internal angles of a triangle sum to \( 180^\circ \). - Therefore, \( m \angle B = 180^\circ - ( \angle ACB + \angle CAB ) \). - \( m \angle B = 180^\circ - (57.5^\circ + 57.5^\circ) \).
Transcribed Image Text:### Problem Statement: **Given:** - \( AB \cong BC \) - \( m \angle BCD = 115^\circ \) **Find:** - \( m \angle B \) ### Diagram Explanation: According to the given problem, we have a geometric figure, specifically a triangle \( ABC \) with an additional line segment \( CD \) extended from point \( C \). Here is a detailed description of the diagram: - Triangle \( ABC \) has points \( A \), \( B \), and \( C \). - Segment \( AB \) is congruent to segment \( BC \). - Point \( D \) lies on a straight line extending from \( C \), creating an external angle \( \angle BCD = 115^\circ \). ### Approach: To find \( m \angle B \) in the triangle \( ABC \): 1. **Identifying the given information:** - Since \( AB \cong BC \), triangle \( ABC \) is an isosceles triangle with \( AB = BC \). - The external angle \( \angle BCD \) can be used to find the internal angle at vertex \( C \) of the triangle \( \angle ACB \). 2. **Using the properties of external angles:** - The external angle of a triangle (\( \angle BCD \)) is equal to the sum of the two non-adjacent internal angles of the triangle. - Therefore, \( \angle BCD = \angle CAB + \angle ACB \). 3. **Calculating the internal angles:** - Let \( \angle ACB = x \). Since \( AB \cong BC \), \( \angle CAB \) also equals \( x \). - Hence, \( \angle BCD = x + x \). - So, \( 115^\circ = 2x \). - Solving for \( x \), \( x = 57.5^\circ \). 4. **Finding \( m \angle B \):** - The internal angles of a triangle sum to \( 180^\circ \). - Therefore, \( m \angle B = 180^\circ - ( \angle ACB + \angle CAB ) \). - \( m \angle B = 180^\circ - (57.5^\circ + 57.5^\circ) \).
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