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To solve this problem, we need to use vector addition and trigonometry. The ship makes two legs of a journey, each in a specific direction. We’ll break down each leg into its east and north components and then use the Pythagorean theorem to determine how far it is from its original position.

### Step 1: First leg of the journey
The ship sails 105 km in the direction **E32°S**, which means 32 degrees south of east. We need to break this down into:
- East component = \(105 \times \cos(32^\circ)\)
- South component = \(105 \times \sin(32^\circ)\)

### Step 2: Second leg of the journey
Next, the ship sails 200 km in the direction **N10°E**, which means 10 degrees east of north. We break this down into:
- North component = \(200 \times \cos(10^\circ)\)
- East component = \(200 \times \sin(10^\circ)\)

### Step 3: Calculate net displacement
Now, we sum the components:
- Total east component = (East from first leg) + (East from second leg)
- Total north component = (North from second leg) - (South from first leg)

### Step 4: Use Pythagoras’ theorem
The distance from the original position is the magnitude of the resultant displacement vector, calculated as:
\[
\text{Distance} = \sqrt{(\text{Total east component})^2 + (\text{Total north component})^2}
\]

I will now compute these values.The ship is approximately **187.86 km** away from its original position.

Would you like further details on the calculations, or do you have any additional questions?

Here are some related questions for deeper understanding:

1. How are vector components calculated in trigonometry?
2. What is the significance of the angles in navigation problems?
3. Can you explain the Pythagorean theorem and its use in determining distance?
4. What would the direction of the resultant displacement vector be?
5. How would the result change if the angles were different?

**Tip**: In navigation problems, always break the journey into vector components to simplify calculations involving distance and direction.

A ship sails for 105 km in a direction E32°S. It then sails for another 200 km in the direction N10°E. How far is it from its original position?

Solve a ship navigation problem using vector addition and trigonometry. The ship sails 105 km in the direction E32°S and 200 km in the direction N10°E. Learn to break down distances into components and calculate the resultant displacement using the Pythagorean theorem. Suitable for Grades 9-12.

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