Bearing (SPM)

28 Bearing

Paper 1

Paper 1 1.

1.

Bearing of P from Q = 040°

North

Answer: B

P 70°

2.

Bearing of E from from F = = 120° Answer: C

75°

35°

R

3.

Q

75°

North

∠PRQ =

180° − 70 ° − 35 ° = 75 °

Bearing of R from Q = 180° + 75° = 255°

N

75°

M

Answer: D

Bearing of M from from N = 180° + 75° = 255°

2.

North

E

Answer: C

80°

4. North

50°

P

50°

G

F 30°

∠ EFG

180° − 80 ° = –––––––––– 2 = 50°

Bearing of G from F = = 050°

Q

Answer: A

Bearing of P from Q = 360° − 30 ° = 330°

3. North

Answer: D

120° X

North

120°

Y

Answer: D

1

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Penerbitan Pelangi Sdn. Bhd.

Mathematics

4.

SPM

Chapter 28

H 25° North

Paper 1

North 35°

70° 50° 50°

G 300°

1. North

F

P

∠ HFG

= 180° – 70° – 25° = 85° Bearing of H from from F = = 360° − 35 ° = 325°

70°

Q

Answer: D

Bearing of P from Q = 360° − 70 ° = 290° 5.

North

Answer: D

North

105° E F

2.

North

105° M

Bearing of E from from F = 180° − 105 ° = 075°

130°

N

Answer: B Answer: B 6.

North

3.

242°

North

Z

62° Y

70°

North

F

62° 34°

E

70°

X

Bearing of E from from F = = 070°

∠ XZY = =

242° – 180° = 62°

Answer: B

Bearing of Y from from X = 62° + 34° = 096°

4.

North E

Answer: A

45° West

7.

F

North

P

Bearing of E from from F = = 360° − 45 ° = 315°

30°

Answer: D

Q

Answer: C

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2

Mathematics

5.

9.

North

E

Chapter 28

North

F (East)

28°

SPM

080°

Q 50°

50°

P 150°

G

R

Bearing of G from E = 90 ° + 28° = 118° Answer: B

Bearing of Q from R = 050° Answer: D

6.

North

10. North 75°

Q 75°

P

125° M

Bearing of P from Q = 180° + 75° = 255°

55°

55°

N

Answer: C

7.

Bearing of M from from N = = 360° − 55 ° = 305°

North

E

Answer: C

160°

40°

20°

11.

40°

North

60°

F K

G

30°

Bearing of G from F = 40 ° + 60° = 100°

° 0 3 1

30°

J

Answer: B H

8.

∠ JHK =

180° − 130 ° − 30 ° = 20°

North 130°

L

Answer: A

50° M

50°

50°

N

Bearing of N from from M = = 180° + 50° = 230° Answer: C

3

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Mathematics

SPM

12.

Chapter 28

15.

North

R

North

P

North

30° 30°

P

60° 30°

60°

Q

20 km

∠PRQ =

180° − 30 ° − 60 ° = 90 °

30° 80°

Answer: D

Q

70°

16.

R

North

G

∠PQR =

180° − 30 ° − 70 ° = 80°

40°

Bearing of R from Q = 360° − (80 ° + 30°) = 250°

40°

F 50°

Answer: D

E

13.

Bearing of G from F = 360° − 40 ° = 320°

North

180° + 40°

E

Answer: D

= 220° 40°

17.

North

North

F

Q

Answer: D

x

20° 40°

14. 40° North

Q

30°

P

x = 180° − 20°

100°

= 160° Answer: C

40°

R

∠PRQ

180° − 100 ° = ––––––––––– 2 = 40°

Bearing of R from Q = 180° − 10 ° = 170° Answer: B

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60°

R

30° 10°

P

60°

4

Bearing (SPM)

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