Question
A boat sails for for 30km on a bearing of 145degree then 10km on a bearing of 250degree what is the distance covered from the starting point and the bearing of the harbour from the ship and with the aid of diagram
Answers
GPT-4o mini
To solve this problem, we can break it down into vector components and use a diagram to illustrate the positions of the boat after each leg of its journey.
### Step 1: Understand the Bearings
- **Bearing:** The angle measured clockwise from the north direction.
### Step 2: Draw the Diagram
1. Start from the origin point (A).
2. Draw the first leg of the journey (30 km on a bearing of 145 degrees):
- To get the vector components, we can use trigonometry.
- The bearing of 145 degrees is measured clockwise from north (0 degrees). The angle from the positive x-axis (east) is \( 145 - 90 = 55 \) degrees counterclockwise, so:
- Vector decomposition:
- \( x_1 = 30 \cos(145^\circ) \)
- \( y_1 = 30 \sin(145^\circ) \)
3. Draw the second leg of the journey (10 km on a bearing of 250 degrees):
- The angle from the positive x-axis (east) is \( 250 - 90 = 160 \) degrees counterclockwise:
- Vector decomposition:
- \( x_2 = 10 \cos(250^\circ) \)
- \( y_2 = 10 \sin(250^\circ) \)
### Step 3: Calculate the Components
Using trigonometric functions:
1. For the first leg:
- \( x_1 = 30 \cos(145^\circ) \approx 30 \times -0.8192 \approx -24.576 \text{ km} \)
- \( y_1 = 30 \sin(145^\circ) \approx 30 \times 0.5736 \approx 17.208 \text{ km} \)
2. For the second leg:
- \( x_2 = 10 \cos(250^\circ) \approx 10 \times -0.1736 \approx -1.736 \text{ km} \)
- \( y_2 = 10 \sin(250^\circ) \approx 10 \times -0.9848 \approx -9.848 \text{ km} \)
### Step 4: Total Components
Now, combine the x and y components:
- Total x-component:
\[
x_{total} = x_1 + x_2 \approx -24.576 - 1.736 \approx -26.312 \text{ km}
\]
- Total y-component:
\[
y_{total} = y_1 + y_2 \approx 17.208 - 9.848 \approx 7.36 \text{ km}
\]
### Step 5: Calculate the Distance from the Starting Point
Using the Pythagorean theorem:
\[
d = \sqrt{(x_{total})^2 + (y_{total})^2}
\]
\[
d = \sqrt{(-26.312)^2 + (7.36)^2} \approx \sqrt{692.547 + 54.049} \approx \sqrt{746.596} \approx 27.36 \text{ km}
\]
### Step 6: Calculate the Bearing to the Harbour
To find the bearing from the ship to the harbour, we can use the arctangent function to find the angle below the negative x-axis:
- Angle from the positive x-axis is given by:
\[
\theta = \tan^{-1} \left( \frac{y_{total}}{x_{total}} \right) = \tan^{-1} \left( \frac{7.36}{-26.312} \right)
\]
Calculating this gives:
\[
\theta \approx \tan^{-1} \left( -0.28 \right) \approx -15.62 \text{ degrees}
\]
Since the x-component is negative and y-component is positive, we are in the second quadrant. The bearing can be calculated as:
\[
\text{Bearing} = 180^\circ + |\theta| \approx 180^\circ + 15.62^\circ \approx 195.62^\circ
\]
### Summary of Results
- The distance from the starting point to the final point is approximately **27.36 km**.
- The bearing of the harbour from the ship is approximately **195.62 degrees**.
### Diagram
To create a diagram, plot:
1. Point A (starting point).
2. From A, draw a line towards the direction of the bearing 145° for 30 km.
3. From the end of this line, draw another line towards 250° for 10 km.
4. Mark the final position as point B and draw a line back to point A and label distances and bearings accordingly.
This can be done using a compass for angles or a graphical tool.
### Step 1: Understand the Bearings
- **Bearing:** The angle measured clockwise from the north direction.
### Step 2: Draw the Diagram
1. Start from the origin point (A).
2. Draw the first leg of the journey (30 km on a bearing of 145 degrees):
- To get the vector components, we can use trigonometry.
- The bearing of 145 degrees is measured clockwise from north (0 degrees). The angle from the positive x-axis (east) is \( 145 - 90 = 55 \) degrees counterclockwise, so:
- Vector decomposition:
- \( x_1 = 30 \cos(145^\circ) \)
- \( y_1 = 30 \sin(145^\circ) \)
3. Draw the second leg of the journey (10 km on a bearing of 250 degrees):
- The angle from the positive x-axis (east) is \( 250 - 90 = 160 \) degrees counterclockwise:
- Vector decomposition:
- \( x_2 = 10 \cos(250^\circ) \)
- \( y_2 = 10 \sin(250^\circ) \)
### Step 3: Calculate the Components
Using trigonometric functions:
1. For the first leg:
- \( x_1 = 30 \cos(145^\circ) \approx 30 \times -0.8192 \approx -24.576 \text{ km} \)
- \( y_1 = 30 \sin(145^\circ) \approx 30 \times 0.5736 \approx 17.208 \text{ km} \)
2. For the second leg:
- \( x_2 = 10 \cos(250^\circ) \approx 10 \times -0.1736 \approx -1.736 \text{ km} \)
- \( y_2 = 10 \sin(250^\circ) \approx 10 \times -0.9848 \approx -9.848 \text{ km} \)
### Step 4: Total Components
Now, combine the x and y components:
- Total x-component:
\[
x_{total} = x_1 + x_2 \approx -24.576 - 1.736 \approx -26.312 \text{ km}
\]
- Total y-component:
\[
y_{total} = y_1 + y_2 \approx 17.208 - 9.848 \approx 7.36 \text{ km}
\]
### Step 5: Calculate the Distance from the Starting Point
Using the Pythagorean theorem:
\[
d = \sqrt{(x_{total})^2 + (y_{total})^2}
\]
\[
d = \sqrt{(-26.312)^2 + (7.36)^2} \approx \sqrt{692.547 + 54.049} \approx \sqrt{746.596} \approx 27.36 \text{ km}
\]
### Step 6: Calculate the Bearing to the Harbour
To find the bearing from the ship to the harbour, we can use the arctangent function to find the angle below the negative x-axis:
- Angle from the positive x-axis is given by:
\[
\theta = \tan^{-1} \left( \frac{y_{total}}{x_{total}} \right) = \tan^{-1} \left( \frac{7.36}{-26.312} \right)
\]
Calculating this gives:
\[
\theta \approx \tan^{-1} \left( -0.28 \right) \approx -15.62 \text{ degrees}
\]
Since the x-component is negative and y-component is positive, we are in the second quadrant. The bearing can be calculated as:
\[
\text{Bearing} = 180^\circ + |\theta| \approx 180^\circ + 15.62^\circ \approx 195.62^\circ
\]
### Summary of Results
- The distance from the starting point to the final point is approximately **27.36 km**.
- The bearing of the harbour from the ship is approximately **195.62 degrees**.
### Diagram
To create a diagram, plot:
1. Point A (starting point).
2. From A, draw a line towards the direction of the bearing 145° for 30 km.
3. From the end of this line, draw another line towards 250° for 10 km.
4. Mark the final position as point B and draw a line back to point A and label distances and bearings accordingly.
This can be done using a compass for angles or a graphical tool.