Vector Components online exercise for - Free Printable

Let's solve each of the three vector component problems step by step. Each problem involves finding the horizontal and vertical components of a vector given its magnitude and direction angle (measured from the positive x-axis).

We use the following formulas:

- Horizontal component (x):
$$
x = |v| \cdot \cos(\theta)
$$

- Vertical component (y):
$$
y = |v| \cdot \sin(\theta)
$$

Where:
- $ |v| $ is the magnitude of the vector,
- $ \theta $ is the direction angle in degrees.

---

Problem 1:

- Magnitude = 15 m
- Direction angle = 25°

#### Horizontal component:
$$
x = 15 \cdot \cos(25^\circ)
$$
Using a calculator:
$$
\cos(25^\circ) \approx 0.9063
$$
$$
x = 15 \cdot 0.9063 = 13.5945 \, \text{m} \approx 13.6 \, \text{m}
$$

#### Vertical component:
$$
y = 15 \cdot \sin(25^\circ)
$$
$$
\sin(25^\circ) \approx 0.4226
$$
$$
y = 15 \cdot 0.4226 = 6.339 \, \text{m} \approx 6.34 \, \text{m}
$$

✔ Answer for Problem 1:
- Magnitude = 15 m
- Direction angle = 25°
- $ x = 15 \cdot \cos(25^\circ) \approx 13.6 \, \text{m} $
- $ y = 15 \cdot \sin(25^\circ) \approx 6.34 \, \text{m} $

---

Problem 2:

- Magnitude = 40 m
- Direction angle = 10°

#### Horizontal component:
$$
x = 40 \cdot \cos(10^\circ)
$$
$$
\cos(10^\circ) \approx 0.9848
$$
$$
x = 40 \cdot 0.9848 = 39.392 \, \text{m} \approx 39.4 \, \text{m}
$$

#### Vertical component:
$$
y = 40 \cdot \sin(10^\circ)
$$
$$
\sin(10^\circ) \approx 0.1736
$$
$$
y = 40 \cdot 0.1736 = 6.944 \, \text{m} \approx 6.94 \, \text{m}
$$

✔ Answer for Problem 2:
- Magnitude = 40 m
- Direction angle = 10°
- $ x = 40 \cdot \cos(10^\circ) \approx 39.4 \, \text{m} $
- $ y = 40 \cdot \sin(10^\circ) \approx 6.94 \, \text{m} $

---

Problem 3:

- Magnitude = 32 m
- Direction angle = 45°

#### Horizontal component:
$$
x = 32 \cdot \cos(45^\circ)
$$
$$
\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071
$$
$$
x = 32 \cdot 0.7071 = 22.6272 \, \text{m} \approx 22.6 \, \text{m}
$$

#### Vertical component:
$$
y = 32 \cdot \sin(45^\circ)
$$
$$
\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071
$$
$$
y = 32 \cdot 0.7071 = 22.6272 \, \text{m} \approx 22.6 \, \text{m}
$$

✔ Answer for Problem 3:
- Magnitude = 32 m
- Direction angle = 45°
- $ x = 32 \cdot \cos(45^\circ) \approx 22.6 \, \text{m} $
- $ y = 32 \cdot \sin(45^\circ) \approx 22.6 \, \text{m} $

---

✔ Final Answers Summary:

| Problem | Magnitude | Angle | x-component | y-component |
|--------|-----------|-------|-------------|-------------|
| 1 | 15 m | 25° | 13.6 m | 6.34 m |
| 2 | 40 m | 10° | 39.4 m | 6.94 m |
| 3 | 32 m | 45° | 22.6 m | 22.6 m |

> Note: All angles are measured from the positive x-axis (standard position), so these components are in the first quadrant — both x and y are positive.

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