To solve this problem, we need to use the Law of Sines, which states:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Given:

• $a = 38$
• $c = 42$
• $\angle A = 31^\circ$

We’ll follow these steps:

Step 1: Calculate $\angle C$ and $\angle B$ for each possible triangle

Since we’re given two sides and an angle opposite one of them ($a$ and $\angle A$), there may be two solutions (an ambiguous case) or just one, depending on whether we can form two triangles with the given values.

Calculating $\angle C$

Using the Law of Sines: $\frac{\sin C}{c} = \frac{\sin A}{a}$ Rearranging to solve for $\sin C$: $\sin C = \frac{c \cdot \sin A}{a} = \frac{42 \cdot \sin 31^\circ}{38}$ Calculate $\sin C$ to determine if it yields one or two possible angles.

Step 2: Find $\angle B$ and Side $b$ for Each Triangle

Once we have $\angle C$ (or both possible values if there’s an ambiguous case), we can find:

1. $\angle B = 180^\circ - \angle A - \angle C$
2. Side $b$ using: $b = \frac{a \cdot \sin B}{\sin A}$

I'll compute these values step-by-step.The solutions for the possible triangles are as follows:

1. First Triangle (larger $B_1$):

   • $\angle B_1 = 114.3^\circ$
   • $\angle C_1 = 34.7^\circ$
   • $b_1 = 67.2$

2. Second Triangle (smaller $B_2$):

   • $\angle B_2 = 3.7^\circ$
   • $\angle C_2 = 145.3^\circ$
   • $b_2 = 4.8$

If you have any questions or need more details, feel free to ask!

Here are some related questions you may find helpful:

1. How does the Law of Sines work in non-right triangles?
2. When does the ambiguous case for the Law of Sines apply?
3. What other methods can solve oblique triangles?
4. How do you verify if the angles and sides form a valid triangle?
5. How can we use the Law of Cosines in similar problems?

Tip: Always check if the computed sine values lie between -1 and 1, as values outside this range indicate that no triangle exists.