Let's solve the problem step-by-step using all three methods: Trigonometric, Graphical, and Component/Analytical methods.

### Given:
- \( F_1 = 450 \, \text{N} \) at \( 60^\circ \) from the horizontal.
- \( F_2 = 300 \, \text{N} \) at \( 30^\circ \) from the horizontal.
- The angle between \( F_1 \) and \( F_2 \) is \( 180^\circ - (60^\circ - 30^\circ) = 90^\circ \).

### 1. **Trigonometric Method:**

To find the resultant \( R \) of two forces with an angle \( \theta \) between them, we can use the Law of Cosines:

\[
R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos(\theta)}
\]

For this case:
- \( F_1 = 450 \, \text{N} \)
- \( F_2 = 300 \, \text{N} \)
- \( \theta = 90^\circ \)

Since \( \cos(90^\circ) = 0 \):

\[
R = \sqrt{450^2 + 300^2}
\]

\[
R = \sqrt{202500 + 90000}
\]

\[
R = \sqrt{292500} \approx 541 \, \text{N}
\]

The direction \( \alpha \) of the resultant relative to \( F_1 \) can be found using the Law of Sines:

\[
\sin(\alpha) = \frac{F_2 \sin(\theta)}{R}
\]

Since \( \sin(90^\circ) = 1 \):

\[
\sin(\alpha) = \frac{300 \times 1}{541} \approx 0.554
\]

\[
\alpha \approx 33.7^\circ
\]

Thus, the resultant force \( R \) is approximately 541 N at an angle of 33.7° from \( F_1 \).

### 2. **Graphical Method:**

1. **Draw** \( F_1 \) (450 N) at 60° from the horizontal.
2. **Draw** \( F_2 \) (300 N) at 30° from the horizontal, with its tail at the head of \( F_1 \).
3. **Draw** the resultant \( R \) from the tail of \( F_1 \) to the head of \( F_2 \).
4. **Measure** the length of \( R \) using the scale to determine the magnitude, and measure the angle \( \alpha \) to find the direction.

This will give you the same result, but visually.

### 3. **Component/Analytical Method:**

1. **Resolve** \( F_1 \) and \( F_2 \) into their horizontal (x) and vertical (y) components:

   - \( F_{1x} = 450 \cos(60^\circ) = 450 \times 0.5 = 225 \, \text{N} \)
   - \( F_{1y} = 450 \sin(60^\circ) = 450 \times 0.866 = 389.7 \, \text{N} \)
   
   - \( F_{2x} = 300 \cos(30^\circ) = 300 \times 0.866 = 259.8 \, \text{N} \)
   - \( F_{2y} = 300 \sin(30^\circ) = 300 \times 0.5 = 150 \, \text{N} \)

2. **Sum** the x and y components to get the components of the resultant force \( R \):

   - \( R_x = F_{1x} + F_{2x} = 225 + 259.8 = 484.8 \, \text{N} \)
   - \( R_y = F_{1y} + F_{2y} = 389.7 + 150 = 539.7 \, \text{N} \)

3. **Find** the magnitude of \( R \) using the Pythagorean theorem:

   \[
   R = \sqrt{R_x^2 + R_y^2} = \sqrt{484.8^2 + 539.7^2} \approx 723 \, \text{N}
   \]

4. **Find** the direction \( \alpha \) of \( R \):

   \[
   \tan(\alpha) = \frac{R_y}{R_x} = \frac{539.7}{484.8} \approx 1.113
   \]

   \[
   \alpha \approx \tan^{-1}(1.113) \approx 48.1^\circ
   \]

Thus, the resultant force \( R \) is approximately 723 N at an angle of 48.1° from the horizontal.

### Summary of Results:
- **Trigonometric Method**: \( R \approx 541 \, \text{N} \) at 33.7° from \( F_1 \).
- **Graphical Method**: Requires drawing and measuring to verify.
- **Component/Analytical Method**: \( R \approx 723 \, \text{N} \) at 48.1° from the horizontal.

Would you like more details on any of the methods?

### Questions to explore further:
1. How does the angle between the forces impact the resultant magnitude and direction?
2. Can you use the trigonometric method for forces at angles other than 90°?
3. How would the solution change if the forces were not coplanar?
4. What are the advantages of the component method over the graphical method?
5. How can these methods be applied to real-world engineering problems?

### Tip:
Always double-check your angle measurements and component calculations, especially when dealing with trigonometric functions, as small errors can propagate through your solution.

Learn how to solve a force analysis problem using trigonometric, graphical, and analytical methods. Calculate the resultant force and its direction with detailed step-by-step solutions. Suitable for advanced high school and college students studying vectors and trigonometry.

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