Vector A has a magnitude of 8.0 at an angle of 60º from the x-axis. Vector B has a magnitude of 6.0 at an angle of -30º from the x-axis. Determine the resultant vector.

To find the resultant vector \( \vec{R} = \vec{A} + \vec{B} \), we'll break each vector into its x and y components and then add these components.

### Step 1: Find the components of vector \( \vec{A} \).
- Magnitude \( |\vec{A}| = 8.0 \)
- Angle \( \theta_A = 60^\circ \)

The components of \( \vec{A} \) are:
\[
A_x = |\vec{A}| \cos(\theta_A) = 8.0 \cos(60^\circ)
\]
\[
A_y = |\vec{A}| \sin(\theta_A) = 8.0 \sin(60^\circ)
\]

Substituting the values:
\[
A_x = 8.0 \times \frac{1}{2} = 4.0
\]
\[
A_y = 8.0 \times \frac{\sqrt{3}}{2} \approx 8.0 \times 0.866 = 6.928
\]

So, the components of \( \vec{A} \) are:
\[
\vec{A} = (4.0, 6.928)
\]

### Step 2: Find the components of vector \( \vec{B} \).
- Magnitude \( |\vec{B}| = 6.0 \)
- Angle \( \theta_B = -30^\circ \)

The components of \( \vec{B} \) are:
\[
B_x = |\vec{B}| \cos(\theta_B) = 6.0 \cos(-30^\circ)
\]
\[
B_y = |\vec{B}| \sin(\theta_B) = 6.0 \sin(-30^\circ)
\]

Substituting the values:
\[
B_x = 6.0 \times \frac{\sqrt{3}}{2} \approx 6.0 \times 0.866 = 5.196
\]
\[
B_y = 6.0 \times \left(-\frac{1}{2}\right) = -3.0
\]

So, the components of \( \vec{B} \) are:
\[
\vec{B} = (5.196, -3.0)
\]

### Step 3: Add the components to find the resultant vector \( \vec{R} \).
\[
R_x = A_x + B_x = 4.0 + 5.196 = 9.196
\]
\[
R

Learn how to calculate the resultant vector of vectors A and B with magnitudes 8.0 and 6.0 at angles 60º and -30º from the x-axis respectively. This problem involves vector addition using components and trigonometric functions. Suitable for high school students.

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