Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. There may be two triangles, one triangle, or no triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Enter all angles in degrees. Round your answers to one decimal place. Below, enter your answers so that ∠A1 is smaller than ∠A2.) Find length of side a1 and a2
b = 22,   c = 39,   ∠B = 23°

We are tasked with solving a triangle using the Law of Sines. The given information includes:

- Side \( b = 22 \)
- Side \( c = 39 \)
- Angle \( \angle B = 23^\circ \)

We need to determine if one or two possible triangles can be formed, and solve for the unknown sides and angles in each triangle.

### Step 1: Apply the Law of Sines
The Law of Sines states:

\[
\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}
\]

We begin by using this formula to find angle \( \angle C \). Rewriting the Law of Sines for angle \( C \):

\[
\frac{\sin(C)}{c} = \frac{\sin(B)}{b}
\]

Substituting the known values:

\[
\frac{\sin(C)}{39} = \frac{\sin(23^\circ)}{22}
\]

We can now solve for \( \sin(C) \):

\[
\sin(C) = 39 \times \frac{\sin(23^\circ)}{22}
\]

Let’s calculate \( \sin(C) \).We have calculated:

\[
\sin(C) \approx 0.6927
\]

Now, let's find the corresponding angle \( C \):

\[
\angle C = \arcsin(0.6927)
\]

We also need to check if there is a second possible triangle (i.e., an ambiguous case). If so, the second possible angle \( C_2 \) can be found as:

\[
\angle C_2 = 180^\circ - \angle C
\]

Let's calculate both \( \angle C_1 \) and \( \angle C_2 \).We found the two possible values for angle \( C \):

- \( \angle C_1 \approx 43.8^\circ \)
- \( \angle C_2 \approx 136.2^\circ \)

### Step 2: Calculate \( \angle A \)
Since the sum of the angles in any triangle is \( 180^\circ \), we can find \( \angle A \) for both cases:

- \( \angle A_1 = 180^\circ - \angle B - \angle C_1 \)
- \( \angle A_2 = 180^\circ - \angle B - \angle C_2 \)

Let's compute both \( \angle A_1 \) and \( \angle A_2 \).The two possible values for \( \angle A \) are:

- \( \angle A_1 \approx 113.2^\circ \)
- \( \angle A_2 \approx 20.8^\circ \)

### Step 3: Solve for side \( a \)
Using the Law of Sines, we can now solve for side \( a \) in both cases:

\[
\frac{a}{\sin(A)} = \frac{b}{\sin(B)}
\]

We can rearrange this to find \( a \):

\[
a = b \times \frac{\sin(A)}{\sin(B)}
\]

Let’s calculate \( a_1 \) and \( a_2 \).It seems like I can’t do more advanced data analysis right now. Please try again later. 

If you need assistance calculating side \( a_1 \) and \( a_2 \), I can guide you through the steps or provide an alternative method! Let me know how you'd like to proceed.

Do you have any questions about the process? 

Here are 5 related questions for further exploration:
1. How can the Law of Cosines be used to solve for unknown angles or sides?
2. What is the ambiguous case in the Law of Sines?
3. How do you determine if a triangle is acute, right, or obtuse?
4. Why might some triangles have two solutions, while others have only one?
5. How does rounding affect the accuracy of solving triangles?

**Tip:** When solving trigonometric equations, always check for multiple solutions, as some angles may lead to more than one valid triangle.

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. There may be two triangles, one triangle, or no triangles that satisfy the given conditions. (If an answer does not exist, enter DNE. Enter all angles in degrees. Round your answers to one decimal place. Below, enter your answers so that ∠A1 is smaller than ∠A2.) Find length of side a1 and a2: b = 22, c = 39, ∠B = 23°

Learn how to solve triangles using the Law of Sines with the given sides b = 22, c = 39, and ∠B = 23°. This detailed solution explores the possibility of one or two triangles and covers key trigonometric concepts such as the Law of Sines, ambiguous case, and the sum of angles in a triangle. Suitable for high school students in Grades 10-12.

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