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Let's go through the questions step by step to solve the right triangles using trigonometric functions based on the information provided.

### 1. Solve for the missing sides in right triangles:
We will use basic trigonometric ratios (sine, cosine, and tangent).

For a right triangle, the following trigonometric identities apply:

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

Let’s solve each part:

#### a. \( \beta = 25^\circ \), \( c = 15 \, \text{in} \), find \( a \)
We know:
- \( \beta \) is the angle opposite to side \( b \),
- \( a \) is the adjacent side to \( \beta \),
- \( c \) is the hypotenuse.

We use the cosine function to solve for \( a \):

\[
\cos(25^\circ) = \frac{a}{15} \implies a = 15 \cdot \cos(25^\circ)
\]

#### b. \( \alpha = 33.5^\circ \), \( b = 21.9 \, \text{mi} \), find \( a \)
In this case, we use the tangent function because we know \( \alpha \) and \( b \):

\[
\tan(33.5^\circ) = \frac{b}{a} \implies a = \frac{b}{\tan(33.5^\circ)} = \frac{21.9}{\tan(33.5^\circ)}
\]

#### c. \( \beta = 75^\circ 10' \), \( b = 96.5 \, \text{km} \), find \( c \)
First, convert \( 75^\circ 10' \) to decimal degrees:
\[
75^\circ 10' = 75 + \frac{10}{60} = 75.1667^\circ
\]

We can use the sine function to solve for \( c \):
\[
\sin(75.1667^\circ) = \frac{b}{c} \implies c = \frac{b}{\sin(75.1667^\circ)} = \frac{96.5}{\sin(75.1667^\circ)}
\]

### 2. Solve for the missing angles and sides:
Use the same trigonometric methods for these.

#### a. \( \alpha = 30^\circ \), \( c = 21 \, \text{ft} \)
Using the sine function to find \( b \):
\[
\sin(30^\circ) = \frac{b}{c} \implies b = 21 \cdot \sin(30^\circ)
\]

#### b. \( \beta = 65^\circ \), \( c = 8.55 \, \text{mm} \)
Using the cosine function to find \( a \):
\[
\cos(65^\circ) = \frac{a}{c} \implies a = 8.55 \cdot \cos(65^\circ)
\]

I'll compute these answers now.Here are the computed solutions for the given problems:

### 1. Solving for the missing sides in the triangles:
- **1a.** \( \beta = 25^\circ \), \( c = 15 \, \text{in} \), \( a \approx 13.59 \, \text{in} \)
- **1b.** \( \alpha = 33.5^\circ \), \( b = 21.9 \, \text{mi} \), \( a \approx 33.09 \, \text{mi} \)
- **1c.** \( \beta = 75^\circ 10' \), \( b = 96.5 \, \text{km} \), \( c \approx 99.83 \, \text{km} \)

### 2. Solving for the remaining sides in the triangles:
- **2a.** \( \alpha = 30^\circ \), \( c = 21 \, \text{ft} \), \( b \approx 10.50 \, \text{ft} \)
- **2b.** \( \beta = 65^\circ \), \( c = 8.55 \, \text{mm} \), \( a \approx 3.61 \, \text{mm} \)

Would you like further clarification or details on any step?

### Related Questions:
1. How do you use the sine, cosine, and tangent functions in non-right triangles?
2. Can you explain how to convert angles from degrees, minutes, and seconds to decimal form?
3. What is the importance of rounding in trigonometric calculations?
4. How can you verify the accuracy of calculated sides in a triangle?
5. How do you determine the remaining angle in a right triangle if two angles are known?

**Tip:** Always keep your calculator in "degree mode" when working with angles in degrees.

Solve the right triangles using the given angles and side lengths based on Figure 5 in the image.

Learn how to solve right triangles using sine, cosine, and tangent functions. This problem covers angle measures in decimal degrees and includes examples like β = 25°, c = 15 in, α = 33.5°, b = 21.9 mi, and more. Detailed step-by-step solutions provided.

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