1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. The longitude of P is 80°E.

Answer: B

2. Difference in longitude between P and Q= 105° – 10°= 95°

Answer: C

3.

30°150°

P(30°W)O

Q

0°

The longitude of Q is 150°E.

Answer: C

4. The latitude of X is 40°N.

Answer: B

5. Difference in latitude between P and Q = 70° – 25°= 45°

Answer: B

6.

40°40°

0°

K

O

N

S

J(40°N)

The latitude of K is 40°S.

Answer: A

7. The position of M is (40°N, 35°W).

Answer: C

8. A = (50°N, 180°E)B = (50°N, 40°E)C = (0°, 40°E)D = (0°, 0°)

Answer: B

9.

75°

75°0°

N

S

O

Q

P(75°N)

20°

0°

160°O

Q

20°W

The position of Q is (75°S, 160°E).

Answer: B

Paper 2

1. (a), (b), (c)

N

S

Equator

80°

80°E

120°E20°

20°W60°W

40°

GreenwichMeridian

CHAPTER

29 Earth as a SphereCHAPTER

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Mathematics SPM Chapter 29

© Penerbitan Pelangi Sdn. Bhd.

2.

95°60°

85°

10°

M(95°W)O

L(10°W)

K(60°E)

J(85°E)

0°

(a) Difference in longitude between J and K = 85° – 60° = 25°

(b) Difference in longitude between K and L = 60° + 10° = 70°

(c) Difference in longitude between J and M = 85° + 95° = 180°

3. (a) Latitude of point E = 70°S

(b) Latitude of point F = 42°N

(c) Latitude of point G = (90° – 25°)N = 65°N

4.

28°

50°

10° 0°65°

P(65°N)N

S

K(10°N)

M(28°S)

N(50°S)

O

(a) Difference in latitude between P and K = 65° – 10° = 55°

(b) Difference in latitude between K and M = 10° + 28° = 38°

(c) Difference in latitude between P and N = 65° + 50° = 115°

5. N

P

Q

R T

S

40°N

50°E60°W

75°S

0°O

6. Position of H = (56°N, (180° – 135°)E) = (56°N, 45°E)Position of J = (0°, 135°W)Position of K = (30°S, 0°)Position of L = (30°S, 45°E)

7. (a) ∠AOB = 50° AB = 50 × 60 = 3000 nautical miles

(b) ∠BOC = 40° + 40° = 80°

N

S

OA

B C40° 40°

0°50°

BNC = 80 × 60 = 4800 nautical miles

8. (a) Difference in longitude between P and Q = 180° PQ = 180 × 60 = 10 800 nautical miles

(b) TR = PQ × cos 60° = 10 800 × cos 60° = 5400 nautical miles

9. (a) PQ = x × 60

x = 3600–––––60

= 60

(b) RQ = y × 60

y = 1500–––––60

= 25

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Mathematics SPM Chapter 29

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Paper 1

1. Longitude of K= (180° – 65°)W= 115°W

Answer: D

2. N

S

O

FG

H(20°S, 15°E)

20°

40° 0°

Latitude of G = (40° – 20°)N = 20°NLongitude of G = (20° – 15°)W = 5°WHence, the position of G is (20°N, 5°W).

Answer: D

3. Longitude of G = (180° – 70°)E = 110°E

Answer: C

4. Longitude of L = (180° – 65°)W = 115°W

Answer: C

5. B is due east of P. The difference in longitude between P and B= 10° + 20°= 30°

Answer: B

6. Latitude of X = (90° – 70°)N = 20°NLongitude of X = 105°W

Hence, the position of X is (20°N, 105°W).

Answer: D

7. Latitude of Q = (70° – 60°)N = 10°N

Longitude of Q = (20° + 100°)E = 120°EHence, the position of Q is (10°N, 120°E).

Answer: A

8.

0°

P

Q

O

N

S

30°

The latitude of Q = (10° + 30°)N= 40°N

Answer: B

Paper 2

1. N

S

EquatorO

B

C

A(58°N, 20°E)

58°32°

(a) Longitude of B = (180° – 20°)W = 160°W

(b) Position of C = (58°S, 160°W)

(c) Distance from A to B via N = (32 × 2) × 60 = 3840 nautical miles

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Mathematics SPM Chapter 29

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(d) (i) AD = 600 × 5.5 = 3300 nautical miles

D

A(20°E)

3300 nm

58°N

y °

O

(ii) y × 60 × cos 58° = 3300 y = 3300–––––––––––

60 × cos 58° = 103.79

Longitude of D = (103.79° + 20°)E = (123.79°)E = 123°47′E

Hence, the position of D is (58°N, 123°47′E).

2. (a)

0°

N

S

O

F(55°N, 108°W)

62°W

G

H

J

Longitude of H = (180° – 108°)E = 72°E

(b) N

S

O

F H

0°55°

35°

55°

35°

Shortest distance from F to H = FNH = (35 + 35) × 60 = 4200 nautical miles

(c) Distance from F to G = (108° – 62°) × 60 × cos 55° = 1583.07 nautical miles

(d) (i) Distance = Speed × Time

= 640 × 8 12

= 5440 nautical miles

(ii) N

S

O

G

J

0°55°

90°40ʹ

∠GOJ = 5440–––––60

= 90°409

Latitude of J = (90°409 – 55°)S = 35°409S

3. (a)

32°O

N

S

F(32°S, 12°E)

G(p°N, 12°E)

D (32°S, 65°W)

H

0°

Latitude of H = 32°N Longitude of H = (180° – 12°)W = 168°W Hence, the location of H is (32°N, 168°W)

(b)

p°32°

N

G

O0°

F

S

3480 nm

∠GOF = 348060

p° + 32° = 58° p° = 58° – 32° p = 26

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Mathematics SPM Chapter 29

© Penerbitan Pelangi Sdn. Bhd.

(c) Distance from D to F = (65 + 12) × 60 × cos 32° = 3917.98 nautical miles

(d) Total distance travelled from D to F and to G = 3917.98 + 3480 = 7397.98 nautical miles

Average speed = 7397.9812

= 616.5 knots

4.

40°

100°E

40°E

40°N

O

N

S

F

H

D

G

0°

(a) Latitude of G = 40°S Longitude of G = (180° – 100°)W = 80°W Position of G = (40°S, 80°W)

(b) Distance from D to N = (90 – 40) × 60 = 3000 nautical miles

(c) ∠DOH = 432060

= 72°

0°

D(40°N)

H

O 72°

Latitude of H = (72° – 40°)S = 32°S

(d) Distance from D to F = (100 – 40) × 60 × cos 40° = 2757.76 nautical miles Timeofflight

= 2757.76500

= 5.52 hours

5.

x°

y°E

N

S

X

C

D

B

40° + y °

30°

4800 nm

A(30°S, 40°W)

30° 0° O

(a) Longitude of X = (180° – 40°)E = 140°E Position of X = (30°N, 140°E)

(b) (i) AB = 4800 (30 + x) × 60 = 4800

30 + x = 4800–––––60

30 + x = 80 x = 80 – 30 = 50

(ii) BD = 3857 (40 + y) × 60 × cos 50° = 3857

(40 + y) = 3857–––––––––60 cos 50°

y = 100 – 40 = 60

(c) Total distance travelled from A to D = AB + BD = 4800 + 3857 = 8657 nautical miles

Average speed = 8657–––––10.5

= 824 knots

6. (a) Position of H = (58°N, (95 – 90)°W) = (58°N, 5°W)

(b)

x°

O

N

S

K

H1800 nm

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Mathematics SPM Chapter 29

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x = 1800–––––60

= 30

Latitude of K = (58° – 30°)N = 28°N

(c) JH = 95 × 60 × cos 58° = 3021 nautical miles

(d) Average speed

= 18004

= 450 knots

7. (a) Laitude of Q = 65°N Longitude of Q = (180° – 100°)W = 80°W

The location of Q = (65°N, 80°W)

(b) ∠UOT = 372060

= 62°

The longitude of T = (80° – 62°)W = 18°W

(c) Distance QR = 3720 × cos 65° = 1572.14 nautical miles

(d) Distance TR = 65 × 60 = 3900 nautical miles

Totaltimeofflight

= 3900 + 1572.14530

= 10.32 hours

Paper 1

1. Difference in longitude between M and N= 20° + 60°= 80°

Answer: C

2.

O35° 40° 0°

L(35°N)

N

S

M(40°S)

Difference in latitude between L and M= 35° + 40°= 75°

Answer: D

3. Position of D = (25°N, (180° – 40°)E) = (25°N, 140°E)

Answer: D

4.

O50°

50°

N

S

0°

N(50°S, 70°W)

M(50°N, 110°E)

Answer: C

5.

O

80°

N

S

0°

F(80°S, 42°W)

G(80°N, 138°E)

Answer: A

6. Latitude of H=(90°−20°)N = 70°NPosition of H = (70°N, 100°W)

Answer: D

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Mathematics SPM Chapter 29

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7.

40°

S

N

0°U(0°, 0°E)

V

O

Position of V = (0°, 40°E)

Answer: B

8. Latitude of R = (90° – 40°)S = 50°S

Longitude of R=(95°−80°)E = 15°EHence, position of R is (50°S, 15°E).

Answer: C

9.

30° 20°

S

N

0°

J(50°N, 80°E)

K

O

Position of K = (20°N, 80°E)

Answer: A

10.

50° 30°

S

N

0°

P(50°N, 15°W)

Q

O

Position of Q = (30°S, 15°W)

Answer: C

11. Position of Q = (65°N, (180° – 110°)W) = (65°N, 70°W)

Answer: B

12. Difference in longitude between P and Q= 20° + 80°= 100°

Answer: D

13. Latitude of P=(80°−45°)N = 35°NLongitude of R=(180°−60°)W = 120°WHence, position of R = (35°N, 120°W)

Answer: D

Paper 2

1. (a) N

10°ES

P(55°N, 80°W)

0°

R

Q

O

Longitude of R=(180°−80°)E = 100°E

(b)

0°

N

S

55°55°

35° 35°

O

P R

PNR–––––PSR = 35 × 2 × 60–––––––––––––––––––

[360−(35×2)]×60

= 70––––290

= 7–––29

(c) QP = (80 + 10) × 60 × cos 55° = 3097 nautical miles

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Mathematics SPM Chapter 29

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(d)

0°55°

N

S

x°O

Q

T

Letthelatitudeofthefinallocation,T = x°S QT = 6480 (55 + x) × 60 = 6480

55 + x = 6480–––––60

x =108−55 = 53 Hence,thelatitudeofthefinallocationis53°S.

2. (a)

O

Q(56°S, 150°E)

56°0°

N

S

P(56°N, 30°W)

Longitude of Q=(180°−30°)E = 150°E

Hence, position of Q = (56°S, 150°E)

(b)

O

P

56° 0°

S

N

Shortest distance from P to S = (56 + 90) × 60 = 8760 nautical miles

(c)

O

P

R

56°

1980 nm

0°

N

S

x°

PR = 1980 x × 60 = 1980

x = 1980–––––60

= 33

Hence, latitude of R=(56°−33°)N = 23°N

(d)

O

V(26°E)

P(30°W)

56°N

0°

30°26°

Difference in longitude between P and V = 30° + 26° = 56°

Timeofflight

= PV–––––speed

= 56 × 60 × cos 56°––––––––––––––––600

= 3.13 hours

3.

O34°

B(34°N, 152°E)

N

S

A(34°N, 28°W)

0°

(a) (i) Distance from A to B along the common parallel of latitude

= 180 × 60 × cos 34° = 8954 nautical miles

(ii) ANB=2(90−34)×60 = 2 × 56 × 60 = 6720 nautical miles

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Mathematics SPM Chapter 29

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(b) (i) TimeofflightofaeroplaneP

= 6720–––––600

= 11.2 hours

(ii) 34°N

BO

C

Ax °

Let aeroplane Q reached point C when aeroplane P reached point B.

AC = 6720 x × 60 × cos 34° = 6720

x = 6720–––––––––––60 × cos 34°

= 135.1

Longitude of point C =(135.1°−28°)E = 107.1°E

4.

S

5220 nm

55°W 10°W

36°N

N

0°O

Q

R

T

P

x °

(a) Longitude of R=(180°−55°)E = 125°E

(b) QP=(55−10)×60×cos36° = 2184 nautical miles

(c) Let ∠POT = x° PT = 5220 60 × x = 5220

x = 5220–––––60

= 87

Hence, latitude of T =(87°−36°)S = 51°S

(d) TimeofflightfromP to R

= PSR––––––speed

= 180 × 60––––––––750

= 14.4 hours

5.

y°0°

V

N

S

20°E

O

R

Q

P

60°E

35°S

4740 nm

(a) Longitude of R =(180°−60°)W = 120°W

(b) QP=(60−20)×60×cos35° = 1966 nautical miles

(c) Let ∠POV = y° PV = 4740 60 × y = 4740

y = 4740–––––60

= 79

Hence, latitude of V =(79°−35°)N = 44°N

(d) TimeofflightfromP to R

= PNR––––––speed

= 180 × 60––––––––864

= 12.5 hours

6.

40°0° 30°

D(40°N, 120°E)

F

N

S

O

G

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Mathematics SPM Chapter 29

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(a) (i) Latitude of F =(70°−40°)S = 30°S

(ii) Longitude of G =(180°−120°)W = 60°W

(b) DF = 70 × 60 = 4200 nautical miles FG = 180 × 60 × cos 30° = 9353 nautical miles Hence, total distance travelled = 4200 + 9353 = 13 553 nautical miles

7.

35°E105°E

0°

F

P

N

S

O

QR

T

42°N

79°N

(a)

70°35°

0°

35°

42°N

R(105°E)

Q(35°E)P

O

Given PQ = QR, hence, longitude of P = 35°W

(b) (i) RQ = 70 × 60 × cos 42° = 3121 nautical miles

(ii) Shortest distance from R to T =(79−42)×60 = 2220 nautical miles

(c) Timeofflight

= total distance––––––––––––speed

= 3121 + 2220–––––––––––500 = 10.68 hours

8.

0°

N

S

x°O

P

Q

TR

37°W

67°N

4800 nm

(a) PR = 67 × 60 = 4020 nautical miles

(b) Let ∠ROT = x° RT = 4800 x × 60 = 4800

x = 4800–––––60

= 80

Hence, the longitude of T =(80°−37°)E = 43°E

(c) (i) PQ = 4800 × cos 67° = 1876 nautical miles

(ii) Timeofflight= 1876–––––600

= 3.13 hours

9.

0°

S

N

40°O

AB

C

50°E30°W

40°S

(a) Longitude of C =(180°−30°)E = 150°E

(b) (i) AB = (30 + 50) × 60 × cos 40° = 3677 nautical miles

(ii) ∠AOC=180°−(2×40°) = 100° ASC = 100 × 60 = 6000 nautical miles

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Mathematics SPM Chapter 29

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(c) Distance travelled = 10 × 570 = 5700 nautical miles

D

A

O40°

0°

N

S

5700 nm θ

Let the difference in latitude between A and the new point, D, be θ.

AD = 5700 θ × 60 = 5700

θ = 5700–––––60

= 95

Hence, latitude of the new point =(95°−40°)N = 55°N

10. G

O

F

P

x°

65°N

20°E

N

S

65°

0°

4800 nm

(a) Longitude of G =(180°−20°)W = 160°W

(b) Let ∠FOP = x° FP = 4800 x × 60 = 4800

x = 4800–––––60

= 80

Hence, latitude of P =(80°−65°)S = 15°S

(c) (i) FG = 180 × 60 × cos 65° = 4564 nautical miles

(ii) ∠GOF=180°−(2×65°) = 50°

Distance GNF = 50 × 60 = 3000 nautical miles

11.

x°S

5°E

0°

PQ R

O

10°W

N

S

(a) Difference in longitude between P and R = 10° + 5° = 15°

(b) x°S

7.5°P(10°W)

Q

O

R(5°E)

Longitude of Q =(10°−7.5°)W = 2.5°W = 2°30′W

(c) PQ = 7.5 × 60 × cos 40° = 344.7 nautical miles

(d) Timeofflight

= 344.7–––––400

= 52 minutes The time when the aeroplane landed at Q is 1252.

12. (a)

S

N

60°

P(60°N, 40°W)

O

Q(60°N, 140°E)

0°

(b) Shortest distance from P to Q = PNQ =(180−60×2)×60 = 3600 nautical miles

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(c) Let the two aeroplanes meet at X after t hours.

180° – x°x°

X

OP(40°W) Q(140°E)

60°N

PX–––XQ

= x–––––––180−x

600t–––––400t

= x–––––––180−x

3—2 = x–––––––180−x

3(180−x) = 2x 540−3x = 2x 540 = 5x x = 108

PX = 108 × 60 × cos 60° = 3240 nautical miles 600t = 3240

t = 3240–––––600

= 5.4

Hence, longitude of X =(108°−40°)E = 68°E

13. 140°W 40°E

Z(60°S, 15°E)X

CY

Q

O

q°

P

y°

0°

N

S

x°N

(a) Let ∠POQ = y° PNQ = 3000 y × 60 = 3000

y = 3000–––––60

= 50

Common latitude of P and Q

= 180°−50°–––––––––2

N

= 65°N Hence, the value of x is 65.

(b) (i) Difference in latitude between X and Y

= 2400–––––60

= 40°

(ii) Let ∠XCZ = q° XZ = 2100 q × 60 × cos 60° = 2100

q = 2100–––––––––––60 × cos 60°

= 70 Longitude of X =(70°−15°)W = 55°W

Hence, the position of X is (60°S, 55°W).

(iii) Total distance from Z to X and to Y = 2100 + 2400 = 4500 nautical miles

Averagespeedoftheflight

= 4500–––––10

= 450 knots

14.

20°W 40°E

30°S

60°N

A B

O

CD

0°60°30°

N

S

(a) (i) Distance from B to C = (60 + 30) × 60 = 5400 nautical miles

(ii) Distance from A to B = (20 + 40) × 60 × cos 30° = 3118 nautical miles

(b) TimeofflightofaeroplaneX from B to A

= 3118–––––400

= 7.795 hours

Distance travelled by aeroplane Y from D when aeroplane X reached A

= 600 × 7.795 = 4677 nautical miles

Hence, distance of aeroplane Y from A when aeroplane X reached A

=5400−4677 = 723 nautical miles

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Mathematics SPM Chapter 29

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15.

50°E

56°N

40°W

CD

B

A

O0°

N

S

(a) (i) Distance from A to B = (40 + 50) × 60 = 5400 nautical miles

(ii) Distance from C to D = 5400 × cos 56° = 3020 nautical miles

(b) (i) Total distance from A to C and to D = (56 × 60) + 3020 = 6380 nautical miles

(ii) Averagespeedoftheflight

= 6380–––––12

= 531.7 knots