Solve each triangle ABC that exists.
Aequals44.5degrees     aequals8.2 m      bequals10.7 m
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer boxes within the choice.
A.
There is only one possible solution for the triangle.
The measurements for the remaining angles B and C and side c are as follows.
Bequals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
Cequals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
cequals
  
enter your response here
​(Round to the nearest tenth as​ needed.)
B.
There are two possible solutions for the triangle.
The measurements for the solution with the longer side c are as follows.
Upper B 1equals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
Upper C 1equals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
c 1equals
  
enter your response here
​(Round to the nearest tenth as​ needed.)
The measurements for the solution with the shorter side c are as follows.
Upper B 2equals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
Upper C 2equals
  
enter your response heredegrees
​(Round to the nearest tenth as​ needed.)
c 2equals
  
enter your response here
​(Round to the nearest tenth as​ needed.)
C.
There are no possible solutions for this triangle.

This problem involves solving a triangle given two sides (\( a = 8.2 \, \text{m} \), \( b = 10.7 \, \text{m} \)) and one angle (\( A = 44.5^\circ \)). We can solve this triangle using the Law of Sines, as it relates angles to their opposite sides. However, we must first check whether there is one solution, two solutions, or no solution. This is known as the ambiguous case (SSA case) in trigonometry.

### Step 1: Use the Law of Sines
The Law of Sines is expressed as:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
We can solve for angle \( B \) first. Rewriting the Law of Sines for \( B \):
\[
\sin B = \frac{b \sin A}{a}
\]
Substitute the known values:
\[
\sin B = \frac{10.7 \cdot \sin 44.5^\circ}{8.2}
\]

### Step 2: Calculate \( \sin B \)
First, calculate \( \sin 44.5^\circ \), and then solve for \( \sin B \):
\[
\sin 44.5^\circ \approx 0.6986
\]
\[
\sin B = \frac{10.7 \cdot 0.6986}{8.2} \approx 0.9114
\]

### Step 3: Check for possible solutions
Since \( \sin B \approx 0.9114 \), and the sine of an angle can only be between -1 and 1, this indicates that angle \( B \) exists.

Now, we find \( B \) using the inverse sine function:
\[
B = \sin^{-1}(0.9114) \approx 65.6^\circ
\]

Since \( \sin B \) is positive, there are two possible solutions for \( B \) because \( \sin(\theta) = \sin(180^\circ - \theta) \). Thus:
- The first possible value of \( B \) is \( 65.6^\circ \).
- The second possible value of \( B \) is \( 180^\circ - 65.6^\circ = 114.4^\circ \).

### Step 4: Check the sum of angles
If \( B = 65.6^\circ \), then angle \( C \) can be found by using the fact that the sum of angles in a triangle is \( 180^\circ \):
\[
C = 180^\circ - A - B = 180^\circ - 44.5^\circ - 65.6^\circ = 69.9^\circ
\]

If \( B = 114.4^\circ \), then:
\[
C = 180^\circ - A - B = 180^\circ - 44.5^\circ - 114.4^\circ = 21.1^\circ
\]

### Step 5: Solve for side \( c \)
Using the Law of Sines again:
\[
\frac{c}{\sin C} = \frac{a}{\sin A}
\]
Thus:
\[
c = \frac{a \cdot \sin C}{\sin A}
\]

For the first case, where \( C = 69.9^\circ \):
\[
c = \frac{8.2 \cdot \sin 69.9^\circ}{\sin 44.5^\circ} \approx \frac{8.2 \cdot 0.9362}{0.6986} \approx 10.98 \, \text{m}
\]

For the second case, where \( C = 21.1^\circ \):
\[
c = \frac{8.2 \cdot \sin 21.1^\circ}{\sin 44.5^\circ} \approx \frac{8.2 \cdot 0.3595}{0.6986} \approx 4.22 \, \text{m}
\]

### Final Answer:
There are **two possible solutions** for the triangle:

1. **First solution**:
   - \( B = 65.6^\circ \)
   - \( C = 69.9^\circ \)
   - \( c = 11.0 \, \text{m} \)

2. **Second solution**:
   - \( B = 114.4^\circ \)
   - \( C = 21.1^\circ \)
   - \( c = 4.2 \, \text{m} \)

Would you like further details or clarifications?

### Related Questions:
1. What are the conditions for having two solutions in a triangle with SSA?
2. How does the Law of Sines apply to ambiguous cases in trigonometry?
3. Can there ever be no solution for a triangle with SSA? Under what conditions?
4. How do you use the Law of Cosines to solve triangles?
5. What are the possible ranges for the sine of an angle in a triangle?

### Tip:
When solving SSA triangles, always check the possibility of two solutions by considering both \( \theta \) and \( 180^\circ - \theta \).

Solve each triangle ABC that exists. A = 44.5 degrees, a = 8.2 m, b = 10.7 m. Select the correct choice below and, if necessary, fill in the answer boxes within the choice.

Solve triangle ABC with A = 44.5 degrees, a = 8.2 m, and b = 10.7 m. Learn how to apply the Law of Sines to find two possible solutions for angles B, C, and side c. Suitable for Grades 10-12, covering trigonometry concepts and the ambiguous case (SSA).

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