T Madas A bearing is a way of

A bearing is a way of defining direction as an angle measured from due North in a clockwise direction N Bearings are always written in 3 digits 063° B A What is the bearing of B from A ? 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 ? © T Madas

A bearing is a way of defining direction as an angle measured from due North in a clockwise direction N B A 243° What is the bearing of A from B ? 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 ? © T Madas

A bearing is a way of defining direction as an angle measured from due North in a clockwise direction N N 063° A B 243° What is the bearing of A from B ? 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 ? © T Madas

A bearing is a way of defining direction as an angle measured from due North in a clockwise direction N N 063° A B 243° 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° © T Madas

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

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

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° to get the bearing “backwards”: we try to get an angle between 0° and 360° by adding or subtracting 180°. © T Madas

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

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

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

What is the bearing of K from H ? What is the bearing of P from Q ? 047° 292° P K H 68° Q 133° What is the bearing of Z from O ? What is the bearing of A from B ? 223° 132° B 137° O 48° Z A © 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 B d A 30 40 © T Madas

d 2 = 402 + 302 d 2 = 1600 + 900 d 2 = 2500 d = 50 miles c c By Pythagoras Theorem: B d A 30 40 © 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 50 miles 2. The bearing of port B as measured from port A 3. The bearing of port A as measured from port B B d ? A 30 θ 40 © T Madas

c 30 40 θ = tan-1 0. 75 c tanθ = c Opp tanθ = adj θ ≈ 37° B 3° d ? 5 A 30 θ 37° 40 © 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 50 miles 2. The bearing of port B as measured from port A 053° 3. The bearing of port A as measured from port B 233° B 3° d 5 A 53 ° 30 37° 40 © 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. N A D 050° Base 4 B 3 C © T Madas

DA 3 DA = 3 x sin 50° 2. 3 050° Base c DA ≈ 2. 30 km N D c sin 50° = c Opp sinθ = Hyp 3 A 4 B Point B is 6. 3 km east of the base C © 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 ? 6. 3 km 2. The bearing of B as measured from the base. 3. The bearing of the base as measured from B. N 2. 3 D 050° Base θ A 4 B 3 C © T Madas

c 6. 3 1. 93 θ ≈ tan-1 3. 264 θ ≈ 73° N 2. 3 A 4 B 1. 93 D c tanθ = c Opp tanθ = adj θ Base 73° C B is at a bearing of 073° from the base © 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 ? 6. 3 km 2. The bearing of B as measured from the base. 073° 3. The bearing of the base as measured from B. 253° N N 1. 93 D 2. 3 A 4 B 73° Base 73° C B is at a bearing of 073° from the base © T Madas

Final question for all you wimps! How far is point B from the base?

d 2 = 6. 32 + 1. 932 d 2 = 39. 69 + 3. 72 d 2 = 43. 41 d ≈ 6. 6 km N 1. 93 D Base 73° C 2. 3 A 4 c c By Pythagoras Theorem: N B d B is 6. 6 km away from the base © 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. N A 5 B 30° 150° 060° Base 4 d C © T Madas

d 2 = 52 + 42 – 2 x 5 x 4 x cos 150° d 2 = 25 + 16 – 40 cos 150° d 2 ≈ 75. 64 c c c By the cosine rule on ABC d ≈ 8. 7 km N A 5 B 30° 150° 060° Base 4 8 d. 7 C © 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? 8. 7 km 2. The bearing of B as measured from the base. 3. The bearing of the base as measured from B. N A 5 B 30° 150° 060° Base θ 4 8. 7 C © T Madas

8. 7 sinθ = 5 sin 150° 8. 7 sinθ ≈ 0. 287 θ ≈ sin-1 (0. 287) N θ ≈ A 5 c 5 = sin 150°x 5 c c 5 x sinθ c By the sine rule on ABC : 17° B 30° 150° 060° Base 4 θ 17° 8. 7 C © 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? 8. 7 km 2. The bearing of B as measured from the base. 077° 3. The bearing of the base as measured from B. 257° N A 5 30° 150° 060° Base C B 77° 4 17 ° 8. 7 B is at a bearing of 077° from the base © T Madas