Solve all triangles where A = 44°, a = 11in, b = 9in (round all values to 1 decimal place) B= C = C O in
Solve all triangles where A = 44°, a = 11in, b = 9in (round all values to 1 decimal place) B= C = C O in
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
![### Solving Triangles with Given Conditions
**Problem Statement:**
Solve all triangles where the following conditions are given:
- Angle \( A = 44^\circ \)
- Side \( a = 11 \text{ in} \)
- Side \( b = 9 \text{ in} \)
*Note: Round all values to one decimal place.*
**Solution:**
To solve for the missing angles and side, use the Law of Sines.
#### 1. Using the Law of Sines
The Law of Sines states:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
Given:
- \( a = 11 \text{ in} \)
- \( b = 9 \text{ in} \)
- \( A = 44^\circ \)
#### 2. Solving for \( B \)
\[
\frac{11}{\sin 44^\circ} = \frac{9}{\sin B}
\]
Calculate \( \sin 44^\circ \):
\[
\sin 44^\circ \approx 0.694
\]
Substitute and solve for \( \sin B \):
\[
\frac{11}{0.694} = \frac{9}{\sin B}
\]
\[
15.85 = \frac{9}{\sin B}
\]
\[
\sin B = \frac{9}{15.85}
\]
\[
\sin B \approx 0.568
\]
Find \( B \):
\[
B \approx \sin^{-1}(0.568)
\]
\[
B \approx 34.7^\circ
\]
#### 3. Solving for \( C \)
Since the sum of angles in a triangle is \( 180^\circ \):
\[
C = 180^\circ - A - B
\]
\[
C = 180^\circ - 44^\circ - 34.7^\circ
\]
\[
C \approx 101.3^\circ
\]
#### 4. Solving for \( c \)
Using the Law of Sines again:
\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\]
Substitute known values:
\[
\frac{11}{0.694](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb77dd63d-90e0-4a84-b222-b885a237b41a%2F70c8e6e4-2a6d-4335-9b8c-0d458d164ca6%2Fornh1bn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Triangles with Given Conditions
**Problem Statement:**
Solve all triangles where the following conditions are given:
- Angle \( A = 44^\circ \)
- Side \( a = 11 \text{ in} \)
- Side \( b = 9 \text{ in} \)
*Note: Round all values to one decimal place.*
**Solution:**
To solve for the missing angles and side, use the Law of Sines.
#### 1. Using the Law of Sines
The Law of Sines states:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
Given:
- \( a = 11 \text{ in} \)
- \( b = 9 \text{ in} \)
- \( A = 44^\circ \)
#### 2. Solving for \( B \)
\[
\frac{11}{\sin 44^\circ} = \frac{9}{\sin B}
\]
Calculate \( \sin 44^\circ \):
\[
\sin 44^\circ \approx 0.694
\]
Substitute and solve for \( \sin B \):
\[
\frac{11}{0.694} = \frac{9}{\sin B}
\]
\[
15.85 = \frac{9}{\sin B}
\]
\[
\sin B = \frac{9}{15.85}
\]
\[
\sin B \approx 0.568
\]
Find \( B \):
\[
B \approx \sin^{-1}(0.568)
\]
\[
B \approx 34.7^\circ
\]
#### 3. Solving for \( C \)
Since the sum of angles in a triangle is \( 180^\circ \):
\[
C = 180^\circ - A - B
\]
\[
C = 180^\circ - 44^\circ - 34.7^\circ
\]
\[
C \approx 101.3^\circ
\]
#### 4. Solving for \( c \)
Using the Law of Sines again:
\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\]
Substitute known values:
\[
\frac{11}{0.694
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