Solving problems with bearings
Apply trigonometry and Pythagoras' theorem to find distances and bearings in single- and two-leg journey problems.
Worked examples
Finding a north component from a single bearing
Straightforward
Problem
A ship sails on a bearing of for 100 km. How far north of its starting point is the ship? Give your answer correct to 1 decimal place.
1
Draw a diagram and identify the right-angled triangle.
Draw a north-south reference line at the starting point. The ship's path makes a angle with north (measured clockwise). Drop a perpendicular from the end of the path to the north-south line. This forms a right-angled triangle with the angle at the starting point, the 100 km path as the hypotenuse, and the north component as the adjacent side.
2
Write the trigonometric ratio for the north component.
The north component is adjacent to the angle and the hypotenuse is 100 km, so:
3
Calculate the north component.
Answer
The ship is km north of its starting point.
Finding straight-line distance after a two-leg journey
Moderate
Problem
A ship sails 80 km on a bearing of and then 60 km on a bearing of . Find the straight-line distance from the starting point.
1
Find the north and east components of Leg 1.
Leg 1: km, bearing .
2
Find the north and east components of Leg 2.
Leg 2: km, bearing .
3
Find the total north and east displacements.
4
Expanding:
Apply Pythagoras' theorem to find the straight-line distance.
Expanding:
Answer
The ship is km from its starting point.
Finding distance and back-bearing for a two-leg journey
Challenging
Problem
A boat sails 30 km due west and then 40 km due south. Find the straight-line distance from the starting point and the true bearing of the starting point from the boat's final position.
1
State the total displacements.
The boat travels 30 km west (east component km) and 40 km south (north component km). From the starting point, the boat's final position is 30 km west and 40 km south.
2
Find the straight-line distance using Pythagoras' theorem.
3
Find the direction from the boat back to the start.
From the boat, the start is 30 km east and 40 km north. This direction is north-east, so the bearing is between and .
4
The bearing of the start from the boat is approximately .
Calculate the angle from north towards east.
The bearing of the start from the boat is approximately .
Answer
The straight-line distance is km; the starting point is on a bearing of from the boat.
Practise
Q1·Straightforward
A ship sails on a bearing of for 60 km. How far north of its starting point has the ship sailed? Give your answer correct to 1 decimal place.
Explanation
The bearing is , so the direction is east of north. The north component is adjacent to the angle in the right-angled triangle:
Q2·Straightforward
A plane flies on a bearing of for 300 km. How far east of its starting point is the plane? Give your answer to the nearest kilometre.
Explanation
The east component is :
Note: because sine is positive in the second quadrant.
Note: because sine is positive in the second quadrant.
Q3·Straightforward
A hiker walks 9 km due north and then 12 km due east. How far is the hiker from the starting point?
Explanation
The north leg (9 km) and east leg (12 km) form a right angle. By Pythagoras' theorem:
Q4·Straightforward
A ship sails on a bearing of until it is 42 km north of its starting point. How far has the ship sailed? Give your answer correct to 1 decimal place.
Explanation
Bearing means the direction is west of north (since ). In the right-angled triangle, the north component (42 km) is adjacent to the angle and the distance sailed is the hypotenuse:
Q5·Moderate
A ship sails 80 km on a bearing of , then 60 km on a bearing of . How far north of the start is the ship? Give your answer correct to 1 decimal place.
Explanation
North component of each leg:
Total north km. The ship is km north of the start.
Total north km. The ship is km north of the start.
Q6·Moderate
A bushwalker hikes 50 km on a bearing of and then 40 km on a bearing of . How far west of the starting point is the bushwalker? Give your answer correct to 1 decimal place.
Explanation
East component of each leg:
Total east km. The negative sign means the bushwalker is km west of the start.
Total east km. The negative sign means the bushwalker is km west of the start.
Q7·Moderate
A ship sails 60 km due north and then 80 km due east. How far is the ship from its starting point?
Explanation
Total north displacement: 60 km. Total east displacement: 80 km. These form the two shorter sides of a right-angled triangle:
Q8·Moderate
A plane flies 150 km on a bearing of and then 200 km on a bearing of . How far is the plane from its starting point?
Explanation
The plane travels 150 km due west and 200 km due south. These legs are perpendicular:
Q9·Challenging
A ship sails 90 km due north and then 120 km due east. What is the straight-line distance from its starting point?
Explanation
Total north: 90 km. Total east: 120 km.
Note: is a multiple of the Pythagorean triple.
Note: is a multiple of the Pythagorean triple.
Q10·Challenging
A ship sails 90 km due north and then 120 km due east. What is the true bearing of the starting point from the ship's final position? Give your answer to the nearest degree.
Explanation
The ship is 90 km north and 120 km east of the start, so from the ship the start is 90 km south and 120 km west. The angle from south towards west:
The bearing from the ship to the start is in the SW sector:
The bearing from the ship to the start is in the SW sector:
Q11·Challenging
A ship sails 40 km on a bearing of and then 30 km on a bearing of . Find the straight-line distance from the starting point.
Explanation
Leg 1 (, 40 km): km, km.
Leg 2 (, 30 km): km, km.
Total: , .
Leg 2 (, 30 km): km, km.
Total: , .
Q12·Challenging
A boat sails 30 km due west and then 40 km due south. Find the true bearing of the starting point from the boat's final position. Give your answer to the nearest degree.
Explanation
The boat is 30 km west and 40 km south of the start, so from the boat the start is 30 km east and 40 km north.
Angle from north towards east:
Since the direction is north-east, the bearing is in the first quadrant:
Angle from north towards east:
Since the direction is north-east, the bearing is in the first quadrant: