Worksheet for identifying vertex, complementary, and supplementary angles with labeled diagrams.

To solve the vector addition problems presented in the image, we need to break each vector into its x-component and y-component, then add the components separately. Here's a step-by-step explanation for each problem:

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Problem 1:

#### Vectors:
- Vector \( \vec{A} \): Magnitude = 5 units, Angle = 30°
- Vector \( \vec{B} \): Magnitude = 3 units, Angle = 120°

#### Step 1: Resolve each vector into components.
The general formulas for resolving a vector into components are:
\[
A_x = A \cos(\theta), \quad A_y = A \sin(\theta)
\]
\[
B_x = B \cos(\theta), \quad B_y = B \sin(\theta)
\]

For \( \vec{A} \):
\[
A_x = 5 \cos(30^\circ) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \approx 4.33
\]
\[
A_y = 5 \sin(30^\circ) = 5 \cdot \frac{1}{2} = 2.5
\]

For \( \vec{B} \):
\[
B_x = 3 \cos(120^\circ) = 3 \cdot \left(-\frac{1}{2}\right) = -1.5
\]
\[
B_y = 3 \sin(120^\circ) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \approx 2.60
\]

#### Step 2: Add the x-components and y-components.
\[
R_x = A_x + B_x = \frac{5\sqrt{3}}{2} + (-1.5) \approx 4.33 - 1.5 = 2.83
\]
\[
R_y = A_y + B_y = 2.5 + \frac{3\sqrt{3}}{2} \approx 2.5 + 2.60 = 5.10
\]

#### Step 3: Find the magnitude and direction of the resultant vector \( \vec{R} \).
The magnitude \( R \) is given by:
\[
R = \sqrt{R_x^2 + R_y^2}
\]
\[
R = \sqrt{(2.83)^2 + (5.10)^2} = \sqrt{8.01 + 26.01} = \sqrt{34.02} \approx 5.83
\]

The direction \( \theta \) is given by:
\[
\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]
\[
\theta = \tan^{-1}\left(\frac{5.10}{2.83}\right) \approx \tan^{-1}(1.80) \approx 61.0^\circ
\]

#### Final Answer for Problem 1:
\[
\boxed{R \approx 5.83 \text{ units}, \theta \approx 61.0^\circ}
\]

---

Problem 2:

#### Vectors:
- Vector \( \vec{C} \): Magnitude = 4 units, Angle = 210°
- Vector \( \vec{D} \): Magnitude = 6 units, Angle = 300°

#### Step 1: Resolve each vector into components.
For \( \vec{C} \):
\[
C_x = 4 \cos(210^\circ) = 4 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} \approx -3.46
\]
\[
C_y = 4 \sin(210^\circ) = 4 \cdot \left(-\frac{1}{2}\right) = -2
\]

For \( \vec{D} \):
\[
D_x = 6 \cos(300^\circ) = 6 \cdot \frac{1}{2} = 3
\]
\[
D_y = 6 \sin(300^\circ) = 6 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} \approx -5.19
\]

#### Step 2: Add the x-components and y-components.
\[
R_x = C_x + D_x = -2\sqrt{3} + 3 \approx -3.46 + 3 = -0.46
\]
\[
R_y = C_y + D_y = -2 + (-3\sqrt{3}) \approx -2 - 5.19 = -7.19
\]

#### Step 3: Find the magnitude and direction of the resultant vector \( \vec{R} \).
The magnitude \( R \) is:
\[
R = \sqrt{R_x^2 + R_y^2}
\]
\[
R = \sqrt{(-0.46)^2 + (-7.19)^2} = \sqrt{0.21 + 51.69} = \sqrt{51.90} \approx 7.20
\]

The direction \( \theta \) is:
\[
\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]
\[
\theta = \tan^{-1}\left(\frac{-7.19}{-0.46}\right) \approx \tan^{-1}(15.63) \approx 86.5^\circ
\]
Since both \( R_x \) and \( R_y \) are negative, the resultant vector is in the third quadrant. Thus:
\[
\theta = 180^\circ + 86.5^\circ = 266.5^\circ
\]

#### Final Answer for Problem 2:
\[
\boxed{R \approx 7.20 \text{ units}, \theta \approx 266.5^\circ}
\]

---

Problem 3:

#### Vectors:
- Vector \( \vec{E} \): Magnitude = 7 units, Angle = 45°
- Vector \( \vec{F} \): Magnitude = 2 units, Angle = 225°

#### Step 1: Resolve each vector into components.
For \( \vec{E} \):
\[
E_x = 7 \cos(45^\circ) = 7 \cdot \frac{\sqrt{2}}{2} = \frac{7\sqrt{2}}{2} \approx 4.95
\]
\[
E_y = 7 \sin(45^\circ) = 7 \cdot \frac{\sqrt{2}}{2} = \frac{7\sqrt{2}}{2} \approx 4.95
\]

For \( \vec{F} \):
\[
F_x = 2 \cos(225^\circ) = 2 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \approx -1.41
\]
\[
F_y = 2 \sin(225^\circ) = 2 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \approx -1.41
\]

#### Step 2: Add the x-components and y-components.
\[
R_x = E_x + F_x = \frac{7\sqrt{2}}{2} + (-\sqrt{2}) \approx 4.95 - 1.41 = 3.54
\]
\[
R_y = E_y + F_y = \frac{7\sqrt{2}}{2} + (-\sqrt{2}) \approx 4.95 - 1.41 = 3.54
\]

#### Step 3: Find the magnitude and direction of the resultant vector \( \vec{R} \).
The magnitude \( R \) is:
\[
R = \sqrt{R_x^2 + R_y^2}
\]
\[
R = \sqrt{(3.54)^2 + (3.54)^2} = \sqrt{12.53 + 12.53} = \sqrt{25.06} \approx 5.01
\]

The direction \( \theta \) is:
\[
\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
\]
\[
\theta = \tan^{-1}\left(\frac{3.54}{3.54}\right) = \tan^{-1}(1) = 45^\circ
\]

#### Final Answer for Problem 3:
\[
\boxed{R \approx 5.01 \text{ units}, \theta = 45^\circ}
\]

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Final Answers:

1. \( \boxed{R \approx 5.83 \text{ units}, \theta \approx 61.0^\circ} \)
2. \( \boxed{R \approx 7.20 \text{ units}, \theta \approx 266.5^\circ} \)
3. \( \boxed{R \approx 5.01 \text{ units}, \theta = 45^\circ} \)