canoeist paddles 3.8 km in the north 63° east direction and then 4.2 km in the north 35 west direction determine the length of a direct path to the canoe final position and the nearest 10th of colter?

To determine the length of the direct path from the starting position to the final position of the canoeist, we can break down the given problem using vector addition and trigonometry. Let's label the vectors as follows:

- The first leg of the journey is 3.8 km in the direction North 63° East.
- The second leg is 4.2 km in the direction North 35° West.

We'll now resolve these vectors into their north and east components.

### Step 1: Resolve vectors into components

#### 1st vector: 3.8 km at North 63° East
- North component: \( 3.8 \cos(63^\circ) \)
- East component: \( 3.8 \sin(63^\circ) \)

#### 2nd vector: 4.2 km at North 35° West
- North component: \( 4.2 \cos(35^\circ) \)
- West component: \( 4.2 \sin(35^\circ) \) (since it's to the west, this will be a negative value for the east-west axis)

### Step 2: Add up the north and east components
Now, let's sum the north components and east components of both vectors.

### Step 3: Calculate the magnitude of the resultant vector
The direct path (resultant vector) will be the hypotenuse of a right triangle formed by these summed components. Using the Pythagorean theorem:

\[
\text{Direct path} = \sqrt{(\text{Total north component})^2 + (\text{Total east component})^2}
\]

I will calculate this now.Let me retry the calculation manually to find the direct path of the canoeist.

### Step 1: Components of the first leg (North 63° East)
- North component: \( 3.8 \cos(63^\circ) \approx 3.8 \times 0.454 \approx 1.7252 \, \text{km} \)
- East component: \( 3.8 \sin(63^\circ) \approx 3.8 \times 0.891 \approx 3.3858 \, \text{km} \)

### Step 2: Components of the second leg (North 35° West)
- North component: \( 4.2 \cos(35^\circ) \approx 4.2 \times 0.819 \approx 3.4398 \, \text{km} \)
- West component: \( 4.2 \sin(35^\circ) \approx 4.2 \times 0.574 \approx 2.4108 \, \text{km} \)

Since this is in the west direction, the east component will be negative.

### Step 3: Sum of components
- Total north component: \( 1.7252 + 3.4398 = 5.165 \, \text{km} \)
- Total east component: \( 3.3858 - 2.4108 = 0.975 \, \text{km} \)

### Step 4: Magnitude of the resultant vector
Now, applying the Pythagorean theorem to these total components:

\[
\text{Direct path} = \sqrt{(5.165)^2 + (0.975)^2} \approx \sqrt{26.672 + 0.9506} \approx \sqrt{27.6226} \approx 5.26 \, \text{km}
\]

### Final Answer:
The length of the direct path to the canoe's final position is approximately **5.3 km** to the nearest tenth of a kilometer.

Let me know if you'd like further details or have any questions!

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Here are 5 related questions:
1. How would the calculation change if the second leg were longer?
2. What if the direction of travel were South 35° West instead?
3. How can we find the angle of the direct path relative to North?
4. How do we account for wind or current in vector calculations?
5. Can the same method be used for 3D vectors, such as in air navigation?

**Tip:** When resolving vectors, always be mindful of the direction to avoid sign errors in components (e.g., East is positive, West is negative).

Canoeist paddles 3.8 km in the north 63° east direction and then 4.2 km in the north 35° west direction. Determine the length of a direct path to the canoe's final position to the nearest tenth of a kilometer.

Learn how to determine the length of a canoeist's direct path after paddling in two different directions using vector addition and the Pythagorean theorem. This problem explores trigonometry and vector components, making it suitable for Grades 10-12.

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