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To solve this problem, let's break down the information provided and use vector addition to find the resultant force, \( F \), and its direction.

### Given Data:
1. **Force 1 (Boat's Propulsion):** 850 lb at an angle of 37°.
2. **Force 2 (Wind Force):** 175 lb at an angle of 321°.

### Steps to Solve:

#### Step 1: Convert Forces into Vector Components
We'll split each force into its horizontal (x-axis) and vertical (y-axis) components using trigonometry.

1. **For the Boat's Force (850 lb at 37°):**
   \[
   F_{1x} = 850 \cos(37^\circ)
   \]
   \[
   F_{1y} = 850 \sin(37^\circ)
   \]

2. **For the Wind Force (175 lb at 321°):**
   The angle 321° is equivalent to \(-39^\circ\) (321° - 360°), making it easier to compute.
   \[
   F_{2x} = 175 \cos(-39^\circ)
   \]
   \[
   F_{2y} = 175 \sin(-39^\circ)
   \]

#### Step 2: Calculate the Components
Using approximate trigonometric values:
1. For the Boat's Force:
   - \( \cos(37^\circ) \approx 0.7986 \)
   - \( \sin(37^\circ) \approx 0.6018 \)
   \[
   F_{1x} = 850 \times 0.7986 \approx 678.81
   \]
   \[
   F_{1y} = 850 \times 0.6018 \approx 511.53
   \]

2. For the Wind Force:
   - \( \cos(-39^\circ) \approx 0.7771 \)
   - \( \sin(-39^\circ) \approx -0.6293 \)
   \[
   F_{2x} = 175 \times 0.7771 \approx 136.00
   \]
   \[
   F_{2y} = 175 \times (-0.6293) \approx -110.13
   \]

#### Step 3: Add the Components
Now, add the corresponding components to find the resultant vector components:

\[
F_x = F_{1x} + F_{2x} = 678.81 + 136.00 \approx 814.81
\]
\[
F_y = F_{1y} + F_{2y} = 511.53 - 110.13 \approx 401.40
\]

#### Step 4: Calculate the Magnitude of the Resultant Force
The magnitude \( F \) of the resultant force is:
\[
F = \sqrt{F_x^2 + F_y^2}
\]
\[
F = \sqrt{(814.81)^2 + (401.40)^2}
\]
\[
F \approx \sqrt{663930.7 + 161122.0} \approx \sqrt{825052.7} \approx 908.3
\]
Rounded to the nearest integer, \( F \approx 908 \) lb.

#### Step 5: Determine the Direction of the Resultant Force
The direction angle \( \theta \) with respect to the positive x-axis is:
\[
\theta = \tan^{-1} \left( \frac{F_y}{F_x} \right)
\]
\[
\theta = \tan^{-1} \left( \frac{401.40}{814.81} \right)
\]
\[
\theta \approx \tan^{-1} (0.4925) \approx 26.2^\circ
\]

### Answer:
- **Magnitude of Resultant Force, \( F \):** Approximately \( 908 \) lb.
- **Direction of Resultant Force:** Approximately \( 26.2^\circ \) from the direction of the boat's propulsion.

Would you like further details on any part of the calculation?

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Here are five related questions for practice:
1. How would the resultant force change if the wind force was increased to 200 lb?
2. What if the boat headed in a different direction, such as 45° instead of 37°?
3. How would we calculate the angle of the resultant force if the wind came from the opposite direction?
4. How can vector addition be applied to solve problems involving three or more forces?
5. How does the angle between two forces impact the magnitude of their resultant force?

**Tip:** When adding vectors, always break them down into their components along each axis; it simplifies both magnitude and direction calculations.

A boat heads 37°, propelled by a force of 850 lb. A wind from 321° exerts a force of 175 lb on the boat. How large is the resultant force F, and in what direction is the boat moving?

Learn how to calculate the resultant force and direction for a boat moving at 37° with an 850 lb propulsion force and a 175 lb wind force from 321°. This trigonometry problem involves vector addition, force components, and Pythagorean theorem. Suitable for students in Grades 10-12.

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