Dave walked from point A to point B due North East for a distance of 20m. He immediately changed direction and walked 50m from B to C on a bearing of 110 degrees.
i. Calculate the distance AC
ii. Find the bearing of C from A
See previous post.
To calculate the distance AC, we can use the concept of vector addition. We can break down the different parts of Dave's journey.
1. Dave walks from point A to point B due North East for a distance of 20m.
- This means Dave walks 20m in a direction that is a combination of North and East, but we need to find the components of this movement. Since it's due North East, we can split it into equal North and East components. This gives us two right-angled triangles, with each side equal to 20m.
- Using the Pythagorean theorem, we can calculate the length of each component.
- The North component is 20 cos 45° (since cos 45° = adjacent/hypotenuse = North/20m).
- The East component is 20 sin 45° (since sin 45° = opposite/hypotenuse = East/20m).
- Simplifying these equations gives us:
- North = 20 cos 45° = 20 * √2/2 = 10√2 m
- East = 20 sin 45° = 20 * √2/2 = 10√2 m
2. Dave changes direction and walks 50m from B to C on a bearing of 110 degrees.
- This means Dave walks 50m in a specific direction measured clockwise from North. To find the components of this movement, we can use trigonometry.
- The North component is 50 sin 110° (since sin 110° = opposite/hypotenuse = North/50m).
- The East component is 50 cos 110° (since cos 110° = adjacent/hypotenuse = East/50m).
- Simplifying these equations gives us:
- North = 50 sin 110° ≈ -37.46 m (Note the negative sign as the component is in the opposite direction to the North)
- East = 50 cos 110° ≈ -23.12 m (Note the negative sign as the component is in the opposite direction to the East)
3. To find the distance AC, we need to add the North and East components together and find the magnitude of the resulting vector.
- The North component is 10√2 m.
- The East component is -23.12 m.
- So, the resultant vector can be calculated using the Pythagorean theorem:
- AC = √(North^2 + East^2)
= √((10√2)^2 + (-23.12)^2)
= √(200 + 535.6544)
= √735.6544
≈ 27.1 m
Therefore, the distance AC is approximately 27.1 meters.
To find the bearing of C from A, we can use trigonometry and bearing conventions.
4. The bearing of C from A is the angle between the North direction and the line connecting A to C, measured clockwise.
- To find this bearing, we can use the tangent function.
- The tangent of the bearing angle is given by the ratio of the East component to the North component. Therefore:
- Tan bearing angle = East/North
- Tan bearing angle = -23.12 / 10√2
- Simplifying the equation gives us:
- Tan bearing angle = -1.63
- Bearing angle ≈ tan^(-1)(-1.63)
- Using a calculator, we find that the bearing angle is approximately -58.68°.
5. However, bearings are usually given as angles between 0° and 360°, with 0° representing North. Since we obtained a negative result, we need to convert it to a positive angle.
- Adding 180° to the negative angle will give us the positive equivalent of the bearing angle.
- So, the bearing of C from A is approximately 180° - 58.68° = 121.32°.
Therefore, the bearing of C from A is approximately 121.32°.