4) A ship sails 100km on a bearing 170 degrees and then a further 120km on bearing 265 degrees.
a) How far is the ship from its starting point?
b) On what bearing must the ship sail so that it returns to its starting point

To solve the problem, we can treat the ship's movement as a vector problem in the Cartesian plane.

### Step 1: Resolve the vectors into their components
The ship's journey can be divided into two legs:

1. **First leg:** 100 km on a bearing of 170 degrees.
   - \( \text{Eastward component} = 100 \times \cos(170^\circ) \)
   - \( \text{Northward component} = 100 \times \sin(170^\circ) \)

2. **Second leg:** 120 km on a bearing of 265 degrees.
   - \( \text{Eastward component} = 120 \times \cos(265^\circ) \)
   - \( \text{Northward component} = 120 \times \sin(265^\circ) \)

### Step 2: Calculate the total displacement
The total displacement is the sum of the eastward and northward components of both legs:

- \( \text{Total eastward component} = 100 \times \cos(170^\circ) + 120 \times \cos(265^\circ) \)
- \( \text{Total northward component} = 100 \times \sin(170^\circ) + 120 \times \sin(265^\circ) \)

### Step 3: Find the distance from the starting point
The distance \( D \) from the starting point can be found using the Pythagorean theorem:

\[
D = \sqrt{(\text{Total eastward component})^2 + (\text{Total northward component})^2}
\]

### Step 4: Determine the return bearing
The bearing \( \theta \) that the ship must sail to return to its starting point is found using the inverse tangent:

\[
\theta = \tan^{-1}\left(\frac{\text{Total eastward component}}{\text{Total northward component}}\right)
\]

### Let's compute these values:

#### First leg (100 km at 170 degrees):
- \( \cos(170^\circ) = -\cos(10^\circ) \) (since 170° is in the second quadrant)
- \( \sin(170^\circ) = \sin(10^\circ) \)

\[
\text{Eastward component} = 100 \times \cos(170^\circ) \approx 100 \times -0.9848 \approx -98.48 \text{ km}
\]
\[
\text{Northward component} = 100 \times \sin(170^\circ) \approx 100 \times 0.1736 \approx 17.36 \text{ km}
\]

#### Second leg (120 km at 265 degrees):
- \( \cos(265^\circ) = -\sin(5^\circ) \) (since 265° is in the third quadrant)
- \( \sin(265^\circ) = -\cos(5^\circ) \)

\[
\text{Eastward component} = 120 \times \cos(265^\circ) \approx 120 \times -0.0872 \approx -10.464 \text{ km}
\]
\[
\text{Northward component} = 120 \times \sin(265^\circ) \approx 120 \times -0.9962 \approx -119.544 \text{ km}
\]

#### Total displacement:
- **Eastward:** \( -98.48 + (-10.464) = -108.944 \text{ km} \)
- **Northward:** \( 17.36 + (-119.544) = -102.184 \text{ km} \)

#### Distance from the starting point:
\[
D = \sqrt{(-108.944)^2 + (-102.184)^2} \approx \sqrt{11866.68 + 10441.54} \approx \sqrt{22308.22} \approx 149.36 \text{ km}
\]

#### Bearing to return to starting point:
\[
\theta = \tan^{-1}\left(\frac{-108.944}{-102.184}\right) \approx \tan^{-1}(1.067) \approx 47.15^\circ
\]
The bearing is measured clockwise from the north, so to return to the starting point:
\[
\text{Return bearing} = 180^\circ + 47.15^\circ \approx 227.15^\circ
\]

### Final Answers:
a) The ship is approximately **149.36 km** from its starting point.

b) The ship must sail on a bearing of approximately **227.15 degrees** to return to its starting point.

Would you like any details, or do you have any questions?

Here are 5 related questions:
1. How do different bearings affect the eastward and northward components of a journey?
2. How can vector addition be used to determine the final position of an object?
3. What is the significance of the Pythagorean theorem in navigation problems?
4. How do you calculate the bearing between two points given their coordinates?
5. What are the applications of trigonometry in navigation?

**Tip:** Always break down vector problems into their components for easier calculation and visualization.

A ship sails 100km on a bearing 170 degrees and then a further 120km on bearing 265 degrees. a) How far is the ship from its starting point? b) On what bearing must the ship sail so that it returns to its starting point?

Learn how to calculate the distance of a ship from its starting point and determine the return bearing using vector addition and trigonometry. This detailed step-by-step solution applies the Pythagorean Theorem and trigonometric functions, making it ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation