NORTH (N)

SOUTH (S)

WEST (W)

EAST (E)

North-east

North-west

South-east

South-west

45 ̊

North-west

South-east

South-west

North-east

NORTHWEST

EASTSOUTH

North does not always

have to point upright.

o Written in three-digits form (000� to 360� )o For angle less than 100� , add a zero in front

to write the bearing in three-digits.o Measure in angle in degrees clockwise from

the line pointing north.

55 ̊

A

N

B

Bearing of A from B is 055 ̊.

48 ̊ 32’

P

North

Q

48 ̊ 32’ = 48.5 ̊ Bearing of P from

Q is 048.5 ̊

a)b) Nort

h

Q42 ̊

P

South

180 ̊

180 ̊ + 42 ̊ =222 ̊Bearing of P from Q

is 222 ̊

Angles in degrees and minutes have to be converted

to degrees, correct to one decimal place.

Angles in degrees and minutes have to be converted

to degrees, correct to one decimal place.

North

Q43 ̊ 12’

P

South

43 ̊ 12’=43.2 ̊360 ̊ - 43.2 ̊ =316.8 ̊

Bearing of P from Q is 316.8 ̊

1̊ = 60’

2.060

12'12

An archaeologist knows that a treasure is buried at

a bearing of 150 ̊ from an old oak tree. He wants to draw on a map to help him remember its precise location.

150 � 150 �

Oak tree

North

Treasure

In each of the following, find the bearing of point P from point Q given the bearing of point Q from point P is (a) 067 ̊ (b) 307 ̊

P

N

67 ̊ Q

N

67 ̊

Bearing of point P from point Q

180 ̊+ 67̊̊ = 247̊̊

P

N

307 ̊Q

N

Ө

Bearing of point P from point Q

Ө = 307̊ ̊ - 180 ̊= 127̊ ̊

In the diagram, point P is due east of point Q. Find the bearing of (a) R from Q (b) R from P

Bearing of R from Q= 90 ̊ +67 ̊=157 ̊

N

N

N

N

Bearing of R from P= 360 ̊ - 90 ̊ - 78 ̊ = 192 ̊

783567180QPR

N

50 ̊

(a) Bearing of P from Q 180 ̊ + 50 ̊ =230̊

N

44 ̊

44 ̊

(b) Bearing of P from Q 224 ̊ - 180 ̊ =44̊

N

134 ̊

(c) Bearing of P from Q 180 ̊ + 134 ̊ = 314̊

A man in a boat saw a lighthouse at a bearing 056 . ̊What was the bearing of the boat as seen by a guard in the lighthouse? If a camper on the beach saw the boat at a bearing of 290 ,what would the bearing of the camper be ̊from the boat?

56 �

Lighthouse

Boat56 �

Lighthouse

Boat

56 �

180 �

N

N

N

Bearing of the lighthouseFrom the boat =056 ̊

Bearing of the boat from The lighthouse=180 ̊+ 56 ̊

=236 ̊

Boat

Camper

N

290 �

N

110 �

70 �

Bearing of the camper from the boat =180 ̊-7̊0 ̊=110 ̊

34 ̊

B

A

NN

124 ̊

C

N34 ̊ 56 ̊

56 ̊

AB=20kmBC=50km

2.6820

50tan

20

50tan)(

1

BAC

BAC

BACa

68.2 ̊

Bearing of C from A 34 ̊ + 68.2 ̊ =102.2 ̊

kmACAC

ACb

85.532900

5020)( 222

A fisherman set out from a fishing village A to fishing ground B, 20 km away at a bearing of 034̊. After the catch, the fishermanwent to sell his fish in town C, 50 km from B at a bearing of 124̊. Find(a) the bearing of C from A (b) Distance AC

1. The bearing of Ben’s house from Ahmad’s house is 075 � whereas the bearing of Chin’s house from Ahmad’s house is 105 ̊ . Given that Ben’s house and Chin’s house are both 150 m from Ahmad’s house, find (a) the bearing of Ben’s house from Chin’s house. (b) the distance between Ben’s and Chin’s house.

2. In the treasure hunt, a scout has to move 400 m at a bearing of 146 ̊ from the base camp to find a key. After getting the key, he has to move 300 m at a bearing of 056 ̊ to a treasure chest. Find(a)the bearing of the treasure chest from the base camp.(b)the distance of the treasure chest from the base camp.

Ahmad

N

Ben

Chin

75 ̊

150 m

105 ̊ 30 ̊

150 m

(a)Bearing of Ben’s house from Chin’s house.

000�

(b) Distance between Ben’s and Chin’s house.

x

x

mxxxxx

x

65.7726457.772

)8229.38(228229.38

15sin150150

15sin

Distance between Ben’s and Chin’s house =77.65 m

Basecamp

N

Key

400 m

146 ̊

N

Treasurechest

300 m56 ̊

146 ̊

34 ̊

(a)Bearing of the treasure chest from the base camp.

9.364

3tan

400

300tan

1

x

x

x

Bearing of the treasure chestfrom the base camp = 146̊ - 36.9̊ = 109.1 ̊

Basecamp

N

Key

400 m

146 ̊

N

Treasurechest

300 m56 ̊

146 ̊

34 ̊

(b) Distance of the treasure chest from the base camp.

Use Pythagoras theorem

Distance of the treasure chest from the base camp = 500 m

3.