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2. 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°

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11. 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°

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14. 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°

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17. 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 40 2 d 2 =+ 30 2 c 1600+ 900d 2 = c d 2d 2 = 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 = θ ≈ 37° 0.75 c c c 37° 53°

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° 53° 50 miles 053° 53° 233°

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23. 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 3 050° A4B C D

24. © T Madas Base N 3 050° A4B C D sinθ = Opp Hyp c sin50° = DA 3 c = 3 c ≈ 2.30 km x sin50° 2.3 Point B is 6.3 km east of the base

25. © T Madas Base N 3 050° A4B 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 A4B C D 2.3 θ cos50° = DC AC c cos50° = DC 3 c = 3 c ≈ 1.93 km x cos50° 1.93 3

27. © T Madas Base N A4B C D 2.3 θ 1.93 tan -1 tanθ = Opp adj tanθ = 6.3 1.93 θ ≈θ ≈ θ ≈ 73° 3.264 c c c B is at a bearing of 073° from the base 73°

28. © T Madas Base N A4B 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 A4B C D 2.3 1.93 73° N 6.3 2 d 2 =+ 1.93 2 c 39.69+ 3.72d 2 = c d 2d 2 = 43.41 c d =43.41 By Pythagoras Theorem: c d ≈ 6.6 km d B is 6.6 km away from the base

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32. 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 4 060° A5B C 30° d 150°

33. © T Madas Base N 060° A5B C 30° d 150° – 40 x 5 d 2d 2 = 5252 + 4242 – 2x 4x cos150° By the cosine rule on ABC c d 2d 2 = 25 + 16 cos150° c d 2d 2 ≈ 75.64 c d ≈ 8.7 km 4 8.7

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 4 060° A5B C 30° 150° 8.7 8.7 km θ

35. © T Madas Base N 4 060° A5B C 30° 150° 8.7 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 4 060° A5 C 30° 150° 8.7 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° 17° B is at a bearing of 077° from the base 257° B 77°

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