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It looks like you're working on a trigonometry problem involving solving a triangle. From the first image, the given data is:

- \( a = 63.55 \) mi,
- \( b = 41.46 \) mi,
- \( C = 54.9^\circ \).

This suggests a Law of Cosines or Law of Sines problem. Since you have two sides and one angle (SSA case), there could be one solution, two solutions, or no solution based on the triangle's properties.

### Step 1: Check if the given information forms a valid triangle
We can apply the **Law of Sines** to determine if one or two triangles are possible. The Law of Sines states:

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

### Step 2: Solve for angle \( B \) using the Law of Sines
First, calculate \( B \):

\[
\frac{\sin B}{b} = \frac{\sin C}{c} \Rightarrow \sin B = \frac{b \cdot \sin C}{a}
\]

\[
\sin B = \frac{41.46 \cdot \sin 54.9^\circ}{63.55}
\]

Calculate the value of \( \sin B \) to check if it's within the valid range (i.e., between 0 and 1).

Do you want me to continue solving the values and check the number of solutions? Let me know if you'd like a more detailed breakdown!

Solve the triangle with the given parts: a = 63.55 mi, b = 41.46 mi, C = 54.9°. Determine if there is one solution, two solutions, or no solution.

This problem involves solving a triangle where two sides (a = 63.55 mi, b = 41.46 mi) and one angle (C = 54.9°) are given. The solution uses the Law of Sines and Cosines to determine if one, two, or no solutions exist. Suitable for Grades 10-12 trigonometry students.

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