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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Help plz I need it and thankyou
![### Problem Statement
In triangle \(\Delta GHI\), the side lengths are given as follows: \( g = 9.8 \) cm, \( h = 1.6 \) cm, and \( i = 9.2 \) cm. Find the measure of \(\angle I\) to the nearest degree.
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### Explanation:
To solve for the measure of \(\angle I\) in \(\Delta GHI\), you can use the Law of Cosines, which states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
In this case, let \( g \), \( h \), and \( i \) correspond to sides \( a \), \( b \), and \( c \) respectively. Then:
\[ i^2 = g^2 + h^2 - 2gh \cos(\angle I) \]
Solving this will give you the measure of \(\angle I\).
Here's the step-by-step process to find \(\angle I\):
1. Substitute the given values into the equation:
\[ 9.2^2 = 9.8^2 + 1.6^2 - 2 \cdot 9.8 \cdot 1.6 \cdot \cos(\angle I) \]
2. Simplify and solve for \(\cos(\angle I)\):
\[ 84.64 = 96.04 + 2.56 - 31.36 \cdot \cos(\angle I) \]
\[ 84.64 = 98.6 - 31.36 \cdot \cos(\angle I) \]
\[ 31.36 \cdot \cos(\angle I) = 98.6 - 84.64 \]
\[ 31.36 \cdot \cos(\angle I) = 13.96 \]
\[ \cos(\angle I) = \frac{13.96}{31.36} \]
\[ \cos(\angle I) \approx 0.445 \]
3. Finally, use the inverse cosine function to find \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb41ca57-8d09-44f6-8eaf-3cf66a6bbae8%2F414cb08c-093b-4fd1-963f-2adfb777dae8%2Ffyne73p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
In triangle \(\Delta GHI\), the side lengths are given as follows: \( g = 9.8 \) cm, \( h = 1.6 \) cm, and \( i = 9.2 \) cm. Find the measure of \(\angle I\) to the nearest degree.
### Answer Submission
**Answer: [ ]**
**Submit Answer**
---
_Privacy Policy_ | _Terms of Service_
Copyright © 2021 DeltaMath.com. All Rights Reserved.
---
### Explanation:
To solve for the measure of \(\angle I\) in \(\Delta GHI\), you can use the Law of Cosines, which states:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
In this case, let \( g \), \( h \), and \( i \) correspond to sides \( a \), \( b \), and \( c \) respectively. Then:
\[ i^2 = g^2 + h^2 - 2gh \cos(\angle I) \]
Solving this will give you the measure of \(\angle I\).
Here's the step-by-step process to find \(\angle I\):
1. Substitute the given values into the equation:
\[ 9.2^2 = 9.8^2 + 1.6^2 - 2 \cdot 9.8 \cdot 1.6 \cdot \cos(\angle I) \]
2. Simplify and solve for \(\cos(\angle I)\):
\[ 84.64 = 96.04 + 2.56 - 31.36 \cdot \cos(\angle I) \]
\[ 84.64 = 98.6 - 31.36 \cdot \cos(\angle I) \]
\[ 31.36 \cdot \cos(\angle I) = 98.6 - 84.64 \]
\[ 31.36 \cdot \cos(\angle I) = 13.96 \]
\[ \cos(\angle I) = \frac{13.96}{31.36} \]
\[ \cos(\angle I) \approx 0.445 \]
3. Finally, use the inverse cosine function to find \
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