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?

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.