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
The portable basketball hoop shown in the diagram is constructed such that \( BA = AS = AK \). The measure of \( \angle BSK \) is \( 76^\circ \). Determine the measure of \( \angle AKB \). Provide your reasoning.
### Diagram Explanation
- The diagram features a basketball hoop and net setup.
- Point \( B \) is at the top end of the pole where the backboard is attached.
- Point \( A \) is on the hoop's post.
- Point \( S \) is on the base of the hoop vertically below point \( A \).
- Point \( K \) is on the base, horizontally offset from point \( S \).
- Lines \( BA \) and \( AK \) are drawn such that they form a triangle \( BAK \) with point \( A \) as the vertex.
- Line \( AS \) is a vertical line, and \( K \) lies on the base of the basketball hoop.
### Mathematical Explanation
Since \( BA = AS = AK \), triangle \( BAK \) and triangle \( ASK \) are isosceles. Given \( \angle BSK = 76^\circ \), we need to find \( \angle AKB \).
#### Step-by-Step Solution
1. First, recognize that point \( A \) is vertically aligned with point \( S \).
2. Because \( BA = AK \), triangle \( BAK \) is isosceles with \( AB = AK \). The analysis must focus on triangle properties.
3. The angles \( \angle BAK \) and \( \angle AKB \) within triangle \( BAK \) are identical because opposite sides ( \( BA \) and \( AK \) ) are equal.
4. This is also reflected in angle relations with \( \angle BSA \) and \( \angle ASK \). Given \( BA = AS = AK \), both angles \( \angle BAS \) and \( \angle ASK \) are equal.
5. Given \( \angle BSK = 76^\circ \):
\[
\angle BSA + \angle ASK = 180^\circ - 76^\circ = 104^\circ
\]
6. Hence, \( \angle BSA = \angle ASK = 52^\circ \).
Now, angle \( \angle AKB \) will](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbde1279-10a5-4128-9c22-ecc62dd16890%2F5a4cd9cf-fe54-4058-a95a-d996adc21927%2F8bllzrc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
The portable basketball hoop shown in the diagram is constructed such that \( BA = AS = AK \). The measure of \( \angle BSK \) is \( 76^\circ \). Determine the measure of \( \angle AKB \). Provide your reasoning.
### Diagram Explanation
- The diagram features a basketball hoop and net setup.
- Point \( B \) is at the top end of the pole where the backboard is attached.
- Point \( A \) is on the hoop's post.
- Point \( S \) is on the base of the hoop vertically below point \( A \).
- Point \( K \) is on the base, horizontally offset from point \( S \).
- Lines \( BA \) and \( AK \) are drawn such that they form a triangle \( BAK \) with point \( A \) as the vertex.
- Line \( AS \) is a vertical line, and \( K \) lies on the base of the basketball hoop.
### Mathematical Explanation
Since \( BA = AS = AK \), triangle \( BAK \) and triangle \( ASK \) are isosceles. Given \( \angle BSK = 76^\circ \), we need to find \( \angle AKB \).
#### Step-by-Step Solution
1. First, recognize that point \( A \) is vertically aligned with point \( S \).
2. Because \( BA = AK \), triangle \( BAK \) is isosceles with \( AB = AK \). The analysis must focus on triangle properties.
3. The angles \( \angle BAK \) and \( \angle AKB \) within triangle \( BAK \) are identical because opposite sides ( \( BA \) and \( AK \) ) are equal.
4. This is also reflected in angle relations with \( \angle BSA \) and \( \angle ASK \). Given \( BA = AS = AK \), both angles \( \angle BAS \) and \( \angle ASK \) are equal.
5. Given \( \angle BSK = 76^\circ \):
\[
\angle BSA + \angle ASK = 180^\circ - 76^\circ = 104^\circ
\]
6. Hence, \( \angle BSA = \angle ASK = 52^\circ \).
Now, angle \( \angle AKB \) will
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