Raylene who was involved in an
orienteering competition. The first two
legs of her triangular course are:
• 1.2 km on a true bearing of 200 °
• 2.3 km on a true bearing of 320 ° .
To find the shortest way back to the
start/finish line, Raylene must
calculate the distance and bearing of
the final leg.
1 Draw a diagram showing the first
two legs of the course.
2 Consider the first leg of the course.
Calculate the distance south and the
distance west of the starting point
travelled in the first leg.
3 Now, considering only the second
leg of the course, calculate the
distance travelled north and west
during this leg.
4 Calculate how far west of the start
point Raylene is at the end of the second leg.
5 Calculate how far north or south of the start point Raylene is at the end of the
second leg.
6 Use Pythagoras’ theorem to find the distance back to the start/finish line.
7 Find the bearing on which Raylene must run to get back to the start/finish line.
8 The organisers made an error in calculating the bearing on which to run the
second leg. It was the intention of the organisers that the final leg of the course
be due east. Calculate the bearing on which the second leg should have been run
so that the final leg is due east.
what is your question about this assignment?
1 Draw a diagram showing the first
two legs of the course.
2 Consider the first leg of the course.
Calculate the distance south and the
distance west of the starting point
travelled in the first leg.
3 Now, considering only the second
leg of the course, calculate the
distance travelled north and west
during this leg.
4 Calculate how far west of the start
point Raylene is at the end of the second leg.
5 Calculate how far north or south of the start point Raylene is at the end of the
second leg.
6 Use Pythagoras’ theorem to find the distance back to the start/finish line.
7 Find the bearing on which Raylene must run to get back to the start/finish line.
8 The organisers made an error in calculating the bearing on which to run the
second leg. It was the intention of the organisers that the final leg of the course
be due east. Calculate the bearing on which the second leg should have been run
so that the final leg is due east.
--- These are the questions. I've done question 1 but it confuses me in question 2. Please help.
question 2. If you did this on graph paper you have the coordinates of the two points, use the distance formula.
what are the answers to these questions?
1. To draw a diagram showing the first two legs of the course, you can use a ruler and protractor. Start by drawing a line segment to represent the first leg of 1.2 km on a true bearing of 200°. Then, from the endpoint of the first leg, draw another line segment to represent the second leg of 2.3 km on a true bearing of 320°. The two line segments should form a triangle.
2. To calculate the distance south and the distance west of the starting point traveled in the first leg, use trigonometry. Since the true bearing is given, you can break it down into its components: the angle relative to the north direction (bearing) and the angle relative to the east direction. In this case, the bearing is 200°, which means the angle relative to the north direction is 20° (180° + 20°) and the angle relative to the east direction is 70° (90° - 20°).
To find the distance south, use sine: distance south = 1.2 km * sin(20°) ≈ 0.41 km.
To find the distance west, use cosine: distance west = 1.2 km * cos(20°) ≈ 1.13 km.
3. Now, considering only the second leg of the course, calculate the distance traveled north and west during this leg. Again, break down the true bearing of 320° into its components. The angle relative to the north direction is 40° (360° - 320°) and the angle relative to the east direction is 70° (90° - 40°).
To find the distance north, use sine: distance north = 2.3 km * sin(40°) ≈ 1.48 km.
To find the distance west, use cosine: distance west = 2.3 km * cos(40°) ≈ 1.76 km.
4. To calculate how far west of the start point Raylene is at the end of the second leg, add the distance west traveled in the first leg (1.13 km) to the distance west traveled in the second leg (1.76 km). The total distance west from the start point is approximately 2.89 km.
5. To calculate how far north or south of the start point Raylene is at the end of the second leg, subtract the distance south traveled in the first leg (0.41 km) from the distance north traveled in the second leg (1.48 km). The difference is approximately 1.07 km north of the start point.
6. To find the distance back to the start/finish line, we can use Pythagoras' theorem. The distance back to the start/finish line is the hypotenuse of the right triangle formed by the total distance west (2.89 km) and the total distance north (1.07 km) calculated in the previous steps.
Using the Pythagorean theorem, the distance back to the start/finish line is:
distance = √(2.89 km^2 + 1.07 km^2) ≈ 3.11 km.
7. To find the bearing on which Raylene must run to get back to the start/finish line, use trigonometry. The bearing is the angle relative to the north direction.
The bearing = arctan(distance west / distance north)
= arctan(2.89 km / 1.07 km) ≈ 70.2°.
So, Raylene must run on a bearing of approximately 70.2° to get back to the start/finish line.
8. To calculate the bearing on which the second leg should have been run so that the final leg is due east, we need to find the difference between the desired bearing (90°) and the current bearing (320°) of the second leg.
The difference in bearing = desired bearing - current bearing
= 90° - 320°
= -230°.
Since the difference is negative (-230°), Raylene needs to run 230° to the right (clockwise) from the current bearing of the second leg.
The bearing on which the second leg should have been run is:
corrected bearing = current bearing + difference in bearing
= 320° + (-230°)
= 90°.
So, the second leg should have been run on a bearing of 90° to make the final leg due east.