The document discusses bearings and how to calculate them. A bearing is defined as an angle measured clockwise from true north. Examples are given of calculating the bearing between two points using trigonometry. Related concepts like calculating the reverse bearing by adding or subtracting 180 degrees are also covered. Several multi-step word problems are worked through demonstrating how to find distances and bearings between points when given travel directions or legs.

1. © T Madas

2. © T Madas
A bearing is a way of defining direction as an angle
measured from due North in a clockwise direction
N
A
B
063°
In other words if you are at A and facing North, by
how many degrees must you turn in a clockwise
direction until you are facing B ?
What is the bearing of B from A ?
Bearings are always
written in 3 digits

3. © T Madas
A bearing is a way of defining direction as an angle
measured from due North in a clockwise direction
In other words if you are at B and facing North, by
how many degrees must you turn in a clockwise
direction until you are facing A ?
What is the bearing of A from B ?
N
A
B
243°

4. © T Madas
A bearing is a way of defining direction as an angle
measured from due North in a clockwise direction
In other words if you are at B and facing North, by
how many degrees must you turn in a clockwise
direction until you are facing A ?
What is the bearing of A from B ?
N
B
243°
N
063°
A

5. © T Madas
A bearing is a way of defining direction as an angle
measured from due North in a clockwise direction
N
B
243°
N
063°
A
The bearing of B from A is 063°
The bearing of A from B is 243°
A bearing is an angle between 0° and 360°

6. © T Madas
How do these bearings relate to each other?
N
B
243°
N
063°
A

7. © T Madas
243°
180°
How do these bearings relate to each other?
N
B
N
063°
A
Alternate Angles
63°
to get the bearing “backwards”:
we try to get an angle between 0° and 360° by adding
or subtracting 180°.

8. © T Madas
to get the bearing “backwards”:
we try to get an angle between 0° and 360° by adding
or subtracting 180°.
to get the bearing “backwards”:
we try to get an angle between 0° and 360° by adding
or subtracting 180°.
The bearing of Norwich from London is 042°
∴ The bearing of London from Norwich is 222°
The bearing of Birmingham from Dover is 300°
∴ The bearing of Dover from Birmingham is 120°

9. © T Madas

10. © T Madas

11. © T Madas
What is the bearing of X from Y ?
X
120°
060°
What is the bearing of P from Q ?
Q
P
110°
290°
110°
What is the bearing of W from O ?
O
W
130°
230°
What is the bearing of A from B ?
A
B
50°
130°
Y

12. © T Madas
What is the bearing of K from H ?
K
H
133°
047°
What is the bearing of P from Q ?
Q
P
68°
292°
68°
What is the bearing of Z from O ?
O
Z
137°
223°
What is the bearing of A from B ?
A
B
48°
132°
48°

13. © T Madas

14. © T Madas
What is the bearing of X from Y ?
X
120°
060°
What is the bearing of P from Q ?
Q
P
110°
290°
110°
What is the bearing of W from O ?
O
W
130°
230°
What is the bearing of A from B ?
A
B
50°
130°
Y

15. © T Madas
What is the bearing of K from H ?
K
H
133°
047°
What is the bearing of P from Q ?
Q
P
68°
292°
68°
What is the bearing of Z from O ?
O
Z
137°
223°
What is the bearing of A from B ?
A
B
48°
132°
48°

16. © T Madas

17. © T Madas
A ship left port A heading for port B.
It sailed due East for 40 miles.
It then sailed due North for 30 miles.
Find:
1. The distance between the two ports
2. The bearing of port B as measured from port A
3. The bearing of port A as measured from port B
A
B
40
30
d

18. © T Madas
A
B
40
30
d
402
d2
= + 302
c
1600 + 900
d2 =
c
d2
= 2500
c
d = 2500
By Pythagoras Theorem:
c
d = 50 miles

19. © T Madas
A
B
40
30
d
A ship left port A heading for port B.
It sailed due East for 40 miles.
It then sailed due North for 30 miles.
Find:
1. The distance between the two ports
2. The bearing of port B as measured from port A
3. The bearing of port A as measured from port B
?
θ
50 miles

20. © T Madas
A
B
40
30
d
?
θ
tan-1
tanθ =
Opp
adj
tanθ =
30
40
θ =
θ ≈ 37°
0.75
c
c
c
37°

21. © T Madas
A
B
40
30
d
A ship left port A heading for port B.
It sailed due East for 40 miles.
It then sailed due North for 30 miles.
Find:
1. The distance between the two ports
2. The bearing of port B as measured from port A
3. The bearing of port A as measured from port B
37°
50 miles
053°
233°

22. © T Madas

23. © T Madas
A soldier walked from his base for 3 km on a bearing of
050° to a point A.
He then walked a further 4 km due east to a point B.
Find:
1. How far east of the base is point B ?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
Base
N
050°
A 4 B
C
D

24. © T Madas
Base
N
050°
A 4 B
C
D
sinθ =
Opp
Hyp
c
sin50° =
DA
3
c
DA = 3
c
DA ≈ 2.30 km
x sin50°
2.3
Point B is 6.3 km east of the base

25. © T Madas
Base
N
050°
A 4 B
C
D
2.3
A soldier walked from his base for 3 km on a bearing of
050° to a point A.
He then walked a further 4 km due east to a point B.
Find:
1. How far east of the base is point B ?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
θ
6.3 km

26. © T Madas
Base
N
A 4 B
C
D
2.3
θ
cos50° =
DC
AC
c
cos50° =
DC
3
c
DC = 3
c
DC ≈ 1.93 km
x cos50°
1.93

27. © T Madas
Base
N
A 4 B
C
D
2.3
θ
1.93
tan-1
tanθ =
Opp
adj
tanθ =
6.3
1.93
θ ≈
θ ≈ 73°
3.264
c
c
c
B is at a bearing of 073° from the base
73°

28. © T Madas
Base
N
A 4 B
C
D
2.3
1.93
B is at a bearing of 073° from the base
A soldier walked from his base for 3 km on a bearing of
050° to a point A.
He then walked a further 4 km due east to a point B.
Find:
1. How far east of the base is point B ?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
73°
6.3 km
073°
N
73°
253°

29. © T Madas
Final question for all
you wimps!
How far is point B from
the base?

30. © T Madas
Base
N
A 4 B
C
D
2.3
1.93
73°
N
6.32
d2
= + 1.932
c
39.69 + 3.72
d2 =
c
d2
= 43.41
c
d = 43.41
By Pythagoras Theorem:
c
d ≈ 6.6 km
d
B is 6.6 km away from the base

31. © T Madas

32. © T Madas
A soldier walked from his base for 4 km on a bearing of
060° to a point A.
He then walked a further 5 km due east to a point B.
Find:
1. The distance of point B from the base?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
Base
N
060°
A 5 B
C
30°
d
150°

33. © T Madas
Base
N
060°
A 5 B
C
30°
d
150°
– 40
x 5
d 2
= 52
+42 – 2 x 4 x cos150°
By the cosine rule on ABC
c
d 2
= 25 + 16 cos150°
c
d 2
≈ 75.64
c
d ≈ 8.7 km

34. © T Madas
A soldier walked from his base for 4 km on a bearing of
060° to a point A.
He then walked a further 5 km due east to a point B.
Find:
1. The distance of point B from the base?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
Base
N
060°
A 5 B
C
30° 150°
8.7 km
θ

35. © T Madas
Base
N
060°
A 5 B
C
30° 150°
θ
By the sine rule on ABC :
sinθ
5
sin150°
8.7
=
c
5sin150°
8.7
sinθ =
c
x 5
5 x
0.287
sinθ ≈
c
sin-1 (0.287)
θ ≈
c
17°
θ ≈

36. © T Madas
Base
N
060°
A 5
C
30° 150°
A soldier walked from his base for 4 km on a bearing of
060° to a point A.
He then walked a further 5 km due east to a point B.
Find:
1. The distance of point B from the base?
2. The bearing of B as measured from the base.
3. The bearing of the base as measured from B.
8.7 km
077°
B is at a bearing of 077° from the base
257°
B
77°

37. © T Madas