T Madas Exam Question T Madas 40 14

© T Madas

Exam Question © T Madas

40 14 3 0 1 0 20 1 50 6 10 17 0 0 0 180 50 Breston N Whitepool 0 80 10 110 70 120 0 13 60 075° 90 Falihax 80 100 7 0 14 50 40 1 30 160 20 0 10 17 0 180 0 110 60 50 12 13 0 Nanchester 0 Measure and write down the bearing of Whitepool from Falihax Hester 0 20 kilometres © T Madas

0 80 10 110 70 120 0 13 90 Whitepool 80 1 00 70 110 60 50 12 13 0 0 0 14 50 40 1 30 160 20 0 10 17 0 180 Falihax 136° 60 0 Breston N Nanchester Measure and write down the bearing of Nanchester from Breston Hester 0 20 kilometres © T Madas

N Measure and write down the bearing of Hester from Nanchester Whitepool 0 14 50 0 1 4 0 3 160 20 0 10 17 0 180 Breston 90 80 1 00 70 110 6 5 0 0 12 13 0 0 Nanchester 14 15 0 20 1 0 6 10 17 0 0 0 180 0 80 10 0 11 70 120 0 3 60 1 50 194° 30 40 Falihax Hester 0 20 kilometres © T Madas

N Whitepool 0 14 50 0 1 4 0 3 160 20 0 10 17 0 180 Breston 90 80 1 00 70 110 6 5 0 0 12 13 0 0 Falihax Nanchester 315° 30 40 14 15 0 20 1 0 6 10 17 0 0 0 180 0 80 10 0 11 70 120 0 60 13 50 Hester Measure and write down the bearing of Breston from Nanchester 0 20 kilometres © T Madas

N Whitepool Breston Falihax Nanchester Hester A radio mast is to be built in Nanchecter. The signal can be received within a distance of 30 km. Shade in the diagram the region in which the signal can be received. 0 20 kilometres © T Madas

© T Madas

This is a diagram that uses the scale 1 cm = 10 km. Point C is on bearing of 300° and at a distance of 65 km from point A. 1. Measure and write down the bearing of point A from point B. 2. Mark with a cross the position of point C. N N B 50° A 230° 180° 1 cm = 10 km © T Madas

This is a diagram that uses the scale 1 cm = 10 km. Point C is on bearing of 300° and at a distance of 65 km from point A. 1. Measure and write down the bearing of point A from point B. 2. Mark with a cross the position of point C. N N B 230° 60° 120° A 180° 1 cm = 10 km © T Madas

This is a diagram that uses the scale 1 cm = 10 km. Point C is on bearing of 300° and at a distance of 65 km from point A. 1. Measure and write down the bearing of point A from point B. 2. Mark with a cross the position of point C. N N B C 0 1 2 3 4 5 6 A 7 8 9 230° 1 cm = 10 km 10 © T Madas

This is a diagram that uses the scale 1 cm = 10 km. Point C is on bearing of 300° and at a distance of 65 km from point A. 1. Measure and write down the bearing of point A from point B. 2. Mark with a cross the position of point C. N N B C A 230° 1 cm = 10 km © T Madas

© T Madas

The diagram below shows aeroplane A in a position due west of aeroplane B, with both aeroplanes heading towards airport C. At that instant, the airport C is on a bearing of 045° from A and on a bearing of 330° from B. Mark the position of the airport C. A 80 1 00 70 110 60 50 12 13 0 0 0 80 10 0 N 90 0 14 50 0 1 4 0 3 160 20 0 10 17 0 180 N B © T Madas

2 10 11 12 13 14 15 16 17 18 19 20 The diagram below shows aeroplane A in a position due west of aeroplane B, with both aeroplanes heading towards airport C. At that instant, the airport C is on a bearing of 045° from A and on a bearing of 330° from B. Mark the position of the airport C. N 4 5 6 7 8 9 N 2 3 045° B 0 1 A © T Madas

The diagram below shows aeroplane A in a position due west of aeroplane B, with both aeroplanes heading towards airport C. At that instant, the airport C is on a bearing of 045° from A and on a bearing of 330° from B. Mark the position of the airport C. A 20 10 0 80 1 00 70 110 60 50 12 13 0 0 60 14 15 0 0 160 170 180 30 045° 50 40 N 90 0 80 10 110 70 120 0 13 N B 0 14 50 0 1 4 0 3 160 20 0 10 17 0 180 © T Madas

1 2 The diagram below shows aeroplane A in a position due west of aeroplane B, with both aeroplanes heading towards airport C. At that instant, the airport C is on a bearing of 045° from A and on a bearing of 330° from B. Mark the position of the airport C. 3 4 5 6 7 8 9 10 N 11 N 12 13 14 16 B 17 A 15 045° 18 19 20 21 22 © T Madas 2

The diagram below shows aeroplane A in a position due west of aeroplane B, with both aeroplanes heading towards airport C. At that instant, the airport C is on a bearing of 045° from A and on a bearing of 330° from B. Mark the position of the airport C. C N N 045° A B 330° © T Madas

© 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

© 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

3 DC = 3 x cos 50° 1. 93 Base θ c DC ≈ 1. 93 km N D c cos 50° = c DC cos 50° = AC DC 2. 3 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? © T Madas

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

© T Madas

Boat A is 7. 2 km from a lighthouse H on a bearing of 195°. Boat B is 11. 4 km from H on a bearing of 293°. 1. Calculate the distance between A and B 14. 3 km 2. Calculate the bearing of boat A from boat B N By the cosine rule on AHB d 2 = 7. 22 + 11. 42 – 2 x 7. 2 x 11. 4 x cos 98° d 2 = 51. 84 + 129. 96 – 164. 16 cos 98° B 11. 4 98° 5 . 30 7. 2 14 d d 2 ≈ 204. 65 H c c c N d ≈ 14. 305 km 195° A © T Madas

Boat A is 7. 2 km from a lighthouse H on a bearing of 195°. Boat B is 11. 4 km from H on a bearing of 293°. 1. Calculate the distance between A and B 14. 3 km 2. Calculate the bearing of boat A from boat B 143° 14 5 . 30 d 98° H 195° c 67° sinθ = sin 98° x 7. 2 14. 3 sinθ = 7. 2 sin 98° 14. 3 c 113° 3 θ 0 11. 4 ° By the sine rule on AHB : 7. 2 x 7. 2 B N sinθ ≈ 0. 498 θ ≈ sin-1 (0. 498) θ ≈ c c N 30° A © T Madas

© 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