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To solve this triangle, we’ll use the **Law of Sines** because we have two sides and a non-included angle (SSA case). Here's the solution process:

### Given:
- \( a = 31 \)
- \( b = 25 \)
- \( B = 34^\circ \)

### Step 1: Find Angle \( A \) Using the Law of Sines
The Law of Sines states:
\[
\frac{\sin A}{a} = \frac{\sin B}{b}
\]
Substitute the values:
\[
\frac{\sin A}{31} = \frac{\sin 34^\circ}{25}
\]

Calculate \( \sin 34^\circ \) (rounded to four decimal places):
\[
\sin 34^\circ \approx 0.5592
\]
So:
\[
\frac{\sin A}{31} = \frac{0.5592}{25}
\]

Solving for \( \sin A \):
\[
\sin A = \frac{31 \times 0.5592}{25} \approx 0.6932
\]

Now, we calculate \( A \):
\[
A = \arcsin(0.6932) \approx 43.9^\circ
\]

### Step 2: Determine if There's a Second Possible Solution for \( A \)
In an SSA case, there could be two possible triangles (the ambiguous case). We also check for:
\[
A' = 180^\circ - 43.9^\circ \approx 136.1^\circ
\]

If \( A' \) were true, then \( A' + B = 136.1^\circ + 34^\circ = 170.1^\circ \), which is possible in a triangle (sum < 180°). Therefore, we have **two possible solutions** for \( A \).

### Step 3: Calculate Angle \( C \) for Each Case
1. **First Solution**: \( A = 43.9^\circ \)
   \[
   C = 180^\circ - A - B = 180^\circ - 43.9^\circ - 34^\circ = 102.1^\circ
   \]

2. **Second Solution**: \( A' = 136.1^\circ \)
   \[
   C' = 180^\circ - A' - B = 180^\circ - 136.1^\circ - 34^\circ = 9.9^\circ
   \]

### Step 4: Find Side \( c \) Using the Law of Sines
For each case:

1. **First Solution**:
   \[
   \frac{c}{\sin C} = \frac{a}{\sin A} \Rightarrow c = \frac{31 \cdot \sin 102.1^\circ}{\sin 43.9^\circ}
   \]
   Calculate \( \sin 102.1^\circ \approx 0.974 \) and \( \sin 43.9^\circ \approx 0.6941 \):
   \[
   c \approx \frac{31 \cdot 0.974}{0.6941} \approx 43.5
   \]

2. **Second Solution**:
   \[
   c' = \frac{31 \cdot \sin 9.9^\circ}{\sin 136.1^\circ}
   \]
   Calculate \( \sin 9.9^\circ \approx 0.172 \) and \( \sin 136.1^\circ \approx 0.6941 \):
   \[
   c' \approx \frac{31 \cdot 0.172}{0.6941} \approx 7.7
   \]

### Summary of Solutions
1. **First Solution**:
   - \( A \approx 43.9^\circ \)
   - \( C \approx 102.1^\circ \)
   - \( c \approx 43.5 \)

2. **Second Solution**:
   - \( A' \approx 136.1^\circ \)
   - \( C' \approx 9.9^\circ \)
   - \( c' \approx 7.7 \)

Would you like any further details on this solution, or have any questions?

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### Related Questions:
1. How can we apply the Law of Cosines in a similar SSA scenario?
2. What conditions determine whether a triangle has no solution in SSA cases?
3. How would the solution change if angle \( B \) were larger than 90 degrees?
4. Can you have multiple solutions for a triangle in other configurations besides SSA?
5. What other methods exist for solving non-right triangles?

### Tip:
In SSA cases, always check for the ambiguous case to see if two solutions are possible.

Consider a triangle ABC like the one below. Suppose that b = 25, a = 31, and B = 34°. Solve the triangle.

Learn how to solve a triangle given sides a = 31, b = 25, and angle B = 34°. This problem demonstrates the use of the Law of Sines in an SSA configuration, covering possible ambiguous cases and showing step-by-step calculations for angles and side lengths. Ideal for Grades 10-12.

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