Vector addition problems using the computational method, requiring calculation of missing sides and angles in right triangles.

Problem: Vector Addition Computational Method

We are tasked with solving for the missing sides and/or angles in each of the given right triangles using vector addition principles. We will use the Pythagorean theorem and trigonometric relationships (sine, cosine, tangent) to solve these problems.

---

Problem 1:

Given:
- \( a = 20 \, \text{m} \)
- \( b = 15 \, \text{m} \)

#### Step 1: Find the hypotenuse \( c \)
Using the Pythagorean theorem:
\[
c = \sqrt{a^2 + b^2}
\]
Substitute the given values:
\[
c = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \, \text{m}
\]

#### Step 2: Find the angle \( \alpha \)
Using the tangent function:
\[
\tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}} = \frac{b}{a} = \frac{15}{20} = 0.75
\]
\[
\alpha = \tan^{-1}(0.75) \approx 36.87^\circ
\]

#### Step 3: Find the angle \( \beta \)
Since \( \alpha + \beta = 90^\circ \):
\[
\beta = 90^\circ - \alpha = 90^\circ - 36.87^\circ \approx 53.13^\circ
\]

Solution for Problem 1:
\[
c = 25 \, \text{m}, \quad \alpha \approx 36.87^\circ, \quad \beta \approx 53.13^\circ
\]

---

Problem 2:

Given:
- \( c = 5.0 \, \text{m/s} \)
- \( \beta = 35^\circ \)

#### Step 1: Find side \( a \)
Using the cosine function:
\[
\cos(\beta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{c}
\]
\[
a = c \cdot \cos(\beta) = 5.0 \cdot \cos(35^\circ)
\]
\[
a \approx 5.0 \cdot 0.8192 \approx 4.096 \, \text{m/s}
\]

#### Step 2: Find side \( b \)
Using the sine function:
\[
\sin(\beta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c}
\]
\[
b = c \cdot \sin(\beta) = 5.0 \cdot \sin(35^\circ)
\]
\[
b \approx 5.0 \cdot 0.5736 \approx 2.868 \, \text{m/s}
\]

#### Step 3: Find angle \( \alpha \)
Since \( \alpha + \beta = 90^\circ \):
\[
\alpha = 90^\circ - \beta = 90^\circ - 35^\circ = 55^\circ
\]

Solution for Problem 2:
\[
a \approx 4.096 \, \text{m/s}, \quad b \approx 2.868 \, \text{m/s}, \quad \alpha = 55^\circ
\]

---

Problem 3:

Given:
- \( a = 10.0 \, \text{m} \)
- \( c = 12.5 \, \text{m} \)

#### Step 1: Find side \( b \)
Using the Pythagorean theorem:
\[
b = \sqrt{c^2 - a^2}
\]
Substitute the given values:
\[
b = \sqrt{12.5^2 - 10.0^2} = \sqrt{156.25 - 100} = \sqrt{56.25} = 7.5 \, \text{m}
\]

#### Step 2: Find angle \( \theta \)
Using the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c} = \frac{7.5}{12.5} = 0.6
\]
\[
\theta = \sin^{-1}(0.6) \approx 36.87^\circ
\]

Solution for Problem 3:
\[
b = 7.5 \, \text{m}, \quad \theta \approx 36.87^\circ
\]

---

Problem 4:

Given:
- \( a = 36 \, \text{m/s} \)
- \( b = 25 \, \text{m/s} \)

#### Step 1: Find the hypotenuse \( c \)
Using the Pythagorean theorem:
\[
c = \sqrt{a^2 + b^2}
\]
Substitute the given values:
\[
c = \sqrt{36^2 + 25^2} = \sqrt{1296 + 625} = \sqrt{1921} \approx 43.83 \, \text{m/s}
\]

#### Step 2: Find angle \( \theta \)
Using the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{b}{a} = \frac{25}{36}
\]
\[
\theta = \tan^{-1}\left(\frac{25}{36}\right) \approx \tan^{-1}(0.6944) \approx 34.75^\circ
\]

Solution for Problem 4:
\[
c \approx 43.83 \, \text{m/s}, \quad \theta \approx 34.75^\circ
\]

---

Problem 5:

Given:
- \( c = 45 \, \text{m} \)
- \( \alpha = 25^\circ \)

#### Step 1: Find side \( a \)
Using the cosine function:
\[
\cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{c}
\]
\[
a = c \cdot \cos(\alpha) = 45 \cdot \cos(25^\circ)
\]
\[
a \approx 45 \cdot 0.9063 \approx 40.78 \, \text{m}
\]

#### Step 2: Find side \( b \)
Using the sine function:
\[
\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c}
\]
\[
b = c \cdot \sin(\alpha) = 45 \cdot \sin(25^\circ)
\]
\[
b \approx 45 \cdot 0.4226 \approx 19.02 \, \text{m}
\]

#### Step 3: Find angle \( \beta \)
Since \( \alpha + \beta = 90^\circ \):
\[
\beta = 90^\circ - \alpha = 90^\circ - 25^\circ = 65^\circ
\]

Solution for Problem 5:
\[
a \approx 40.78 \, \text{m}, \quad b \approx 19.02 \, \text{m}, \quad \beta = 65^\circ
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad c = 25 \, \text{m}, \, \alpha \approx 36.87^\circ, \, \beta \approx 53.13^\circ \\
2. & \quad a \approx 4.096 \, \text{m/s}, \, b \approx 2.868 \, \text{m/s}, \, \alpha = 55^\circ \\
3. & \quad b = 7.5 \, \text{m}, \, \theta \approx 36.87^\circ \\
4. & \quad c \approx 43.83 \, \text{m/s}, \, \theta \approx 34.75^\circ \\
5. & \quad a \approx 40.78 \, \text{m}, \, b \approx 19.02 \, \text{m}, \, \beta = 65^\circ
\end{aligned}
}
\]