Vector addition diagrams for Practice A and Practice B showing magnitude and direction of resultant vectors.

Let's solve both Practice A and Practice B step by step. These are vector addition problems involving displacement vectors in different directions.

We'll use the Pythagorean theorem to find the magnitude of the resultant vector and trigonometry (specifically tangent) to find the direction.

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🔹 Practice A

Vectors:
- 10 km, North
- 5 km, West

This means:
- North is the positive y-direction.
- West is the negative x-direction.

So we can represent these as components:
- North: +10 km in the y-direction → $ y = 10 $
- West: -5 km in the x-direction → $ x = -5 $

Now, we find the resultant vector R:

#### ✔ Magnitude of R:
$$
R = \sqrt{x^2 + y^2} = \sqrt{(-5)^2 + (10)^2} = \sqrt{25 + 100} = \sqrt{125} \approx 11.18 \text{ km}
$$

#### ✔ Direction (angle θ from the positive x-axis or reference):
But note: since the vector is in the second quadrant (negative x, positive y), we need to be careful with the angle.

Let’s find the angle $ \theta $ that the resultant makes with the positive x-axis, but we’ll usually report it relative to a cardinal direction.

Let’s compute the angle between the resultant and the north direction (since north is dominant).

Alternatively, let’s calculate the angle $ \theta $ between the resultant and the west or north.

From the diagram:
- The angle $ \theta $ is between the resultant and the vertical (north) direction.
- Opposite side = 5 km (west)
- Adjacent side = 10 km (north)

So:
$$
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{10} = 0.5
\Rightarrow \theta = \tan^{-1}(0.5) \approx 26.6^\circ
$$

So the direction is 26.6° west of north.

✔ Answer for Practice A:
- Magnitude: ≈ 11.18 km
- Direction: 26.6° west of north

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🔹 Practice B

Vectors:
- 30 km, West
- 40 km, South

Again:
- West → negative x-direction → $ x = -30 $
- South → negative y-direction → $ y = -40 $

So both components are negative — this is in the third quadrant.

#### ✔ Magnitude of R:
$$
R = \sqrt{(-30)^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ km}
$$

#### ✔ Direction:
We want the angle $ \theta $ between the resultant and the south direction (as shown in the diagram), or relative to the southwest.

From the diagram:
- The angle $ \theta $ is between the resultant and the south direction.
- Opposite side = 30 km (west)
- Adjacent side = 40 km (south)

So:
$$
\tan(\theta) = \frac{30}{40} = 0.75 \Rightarrow \theta = \tan^{-1}(0.75) \approx 36.9^\circ
$$

So the direction is 36.9° west of south.

✔ Answer for Practice B:
- Magnitude: 50 km
- Direction: 36.9° west of south

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📌 Summary:

| Practice | Resultant Magnitude | Direction |
|--------|----------------------|----------|
| A | ~11.18 km | 26.6° west of north |
| B | 50 km | 36.9° west of south |

These match the diagrams provided, where:
- In A, the red arrow (R) is diagonal pointing northwest.
- In B, the red arrow points southwest.

Let me know if you'd like the answers in vector notation or compass bearings!