A car travels on a bearing of 24° for 21 miles, then turns 90° to the right and travels for a further 29 miles before stopping.

Calculate the distance of the car from its starting point. Give your answer in miles, to 2 decimal places.

Answer 1
 miles

Calculate the bearing from the start point of the journey to the end point. Give your answer to the nearest degree.

Answer 2
 degrees

To solve this problem, we'll break it into two parts:

1. **Calculate the distance of the car from its starting point.**

2. **Calculate the bearing from the start point to the end point.**

### Part 1: Distance from the Starting Point

Let's consider the journey step by step:

- The car first travels 21 miles on a bearing of 24°.
- Then it turns 90° to the right and travels 29 miles.

We can represent this as a triangle where:
- The first leg (AB) is 21 miles.
- The second leg (BC) is 29 miles.
- We need to find the straight-line distance (AC) from the starting point to the stopping point.

To find this distance, we can use the **cosine rule**:

\[
AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\theta)
\]

Where:
- \(AB = 21\) miles
- \(BC = 29\) miles
- \(\theta\) is the angle between the two paths.

Since the car turns 90° to the right, the angle inside the triangle is \(180^\circ - 24^\circ - 90^\circ = 66^\circ\).

Substitute the values:

\[
AC^2 = 21^2 + 29^2 - 2 \times 21 \times 29 \times \cos(66^\circ)
\]

Now, calculate \(AC\):

\[
AC = \sqrt{21^2 + 29^2 - 2 \times 21 \times 29 \times \cos(66^\circ)}
\]

Let's compute this value.

### Part 2: Bearing from Start to End Point

To find the bearing, we'll first need to determine the angle at which the car ends up relative to the north. This involves some trigonometry:

- Let \(A_x\) and \(A_y\) be the coordinates of the start point.
- \(B_x\) and \(B_y\) after the first leg (24° bearing) can be found using:
  - \(B_x = 21 \times \sin(24^\circ)\)
  - \(B_y = 21 \times \cos(24^\circ)\)
  
- After the second leg, the new coordinates \(C_x\) and \(C_y\) will be:
  - \(C_x = B_x + 29 \times \cos(24^\circ)\)
  - \(C_y = B_y - 29 \times \sin(24^\circ)\) (as the turn is to the right)

Finally, the bearing is:

\[
\text{Bearing} = \tan^{-1} \left(\frac{\text{C}_x}{\text{C}_y}\right)
\]

Let's calculate both values.### Answers:

1. **Distance from the starting point**: The car is approximately **28.05 miles** away from its starting point.

2. **Bearing from the start to the end point**: The bearing from the start point to the end point is approximately **78°**.

Would you like any further details or explanations on how these calculations were done?

Here are 5 related questions to consider:

1. How would the solution change if the car had turned left instead of right?
2. What would be the new bearing if the car traveled another 15 miles after the final stop?
3. How does the cosine rule apply to non-right-angled triangles in general?
4. What are the implications of small errors in bearing measurement on long-distance navigation?
5. Can you explain how trigonometric functions can be used to convert between different coordinate systems?

**Tip:** When calculating bearings, always ensure that you adjust the angle to fall within the 0° to 360° range. This helps avoid negative bearings, which can be confusing in navigation contexts.

A car travels on a bearing of 24° for 21 miles, then turns 90° to the right and travels for a further 29 miles before stopping. Calculate the distance of the car from its starting point. Give your answer in miles, to 2 decimal places. Calculate the bearing from the start point of the journey to the end point. Give your answer to the nearest degree.

Learn how to calculate the distance and bearing of a car's journey using trigonometry and geometry. This problem involves the cosine rule and coordinate geometry to determine the car's position after traveling 21 miles at a 24° bearing and turning 90° right for an additional 29 miles. Suitable for students in Grades 10-12.

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