### Problem StatementThe 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 ExplanationSince $ BA = AS = AK $, triangle $ BAK $ and triangle $ ASK $ are isosceles. Given $ \angle BSK = 76^\circ $, we need to find $ \angle AKB $.#### Step-by-Step Solution1. 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

Answered: The portable basketball hoop shown is…

The portable basketball hoop shown is made so that BA = AS = AK. The measure of ZBSK is 76°. What is mLAKB? Explain your reasoning. A S

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter6: Ratio, Proportion, And Probability

Section6.5: Similarity And Measurement

Problem 2E

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Question

### 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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