Www.ketam.pja.My -Nota 2 Matematik Tingkatan 4 Dan 5 SPM

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Form 4 – Chapter 1 Standard FormPassport To Success

(Fully–worked Solutions)

Paper 1 1 27.035 = 27.0 (3 sig. fig.)

Answer: B

2 4.23 × 10–4 = 0.0004.23 = 0.000423

Answer: B

3 429 000————–1.5 × 10–2

= 4.29 × 105—————1.5 × 10–2

= 4.29——1.5

× 105——10–2

= 2.86 × 105 – (–2)

= 2.86 × 107

Answer: B

4 2.35 × 108 – 2.48 × 107

= 2.35 × 108 – 0.248 × 101 × 107

= 2.35 × 108 – 0.248 × 108

= (2.35 – 0.248) × 108

= 2.102 × 108

Answer: D

3 � 5

Suc Math SPM (Passport).indd 4Suc Math SPM (Passport).indd 4 10/7/2008 3:17:36 PM10/7/2008 3:17:36 PM

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Form 4 – Chapter 2 Quadratic Expressions and EquationsPassport To Success


Paper 1 1 3h(1 – h) + (h – 1)2

= 3h – 3h2 + h2 – 2h + 1= –2h2 + h + 1

Answer: D

Paper 2

2 3m = 8—m

– 10

3m = 8 – 10m————–m

3m2 = 8 – 10m 3m2 + 10m – 8 = 0(3m – 2)(m + 4) = 0

m = 2—

3 or –4

3 (9p – 1)2 = 9p2

81p2 – 18p + 1 = 9p2

72p2 – 18p + 1 = 0(12p – 1)(6p – 1) = 0

p = 1—–12

or 1—6


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P

Q

R

Set R

Set (P � Q)’

Union with

PQ

R

PQ

R

PQ

R

Form 4 – Chapter 3 SetsPassport To Success


Paper 1 1 A = {4, 9}

Set A has 2n = 22 = 4 subsets.The subsets are {4}, {9}, {4, 9}, { }.

Answer: D

2

4 ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

P = {6, 11, 16} Q = {1, 2, 3, 4, 6, 8, 12}

P ’ � Q = {1, 2, 3, 4, 8, 12}n(P ’ � Q) = 6

Answer: B

Paper 2 5 (a)

(P � Q) � R

(b)

Uniting (P � Q)’ and R, we have (P � Q)’ � R, as shown in the following Venn diagram.

Answer: A

3

In the above diagram,(a) the shaded region represents the set

(Q � R)’, and(b) the shaded region represents the set

P. The intersection of (a) and (b) is the set

that is required by the shaded region of the question i.e. (Q � R)’ � P.

Answer: C

ξP

Q

R

ξK T

15 5 30

20

n(K ) – n(K � T )= 20 – 5= 15

It is given that n(K � T ) = 5.

n(T ) – n(K � T )= 35 – 5= 30

n(ξ) – n(K � T )= 70 – (15 + 5 + 30)= 20


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Form 4 – Chapter 4 Mathematical ReasoningPassport To Success


Paper 2 1 (a) 3m2 + 5m – 2 = 0 is not a statement.

(b) Premise 1: All sets which contain n elements have 2n subsets.

Premise 2: Set A contains 3 elements. Conclusion: Set A has 23 subsets.

(c) 1 = 2(1)3 – 1 15 = 2(2)3 – 1 53 = 2(3)3 – 1 127 = 2(4)3 – 1 The nth term is 2n3 – 1, n = 1, 2, 3, 4, …

2 (a) ‘Some quadratic equations have two distinct roots.’

(b) If x � 3, then x < 8.

The converse of the above statement is ‘If x � 8, then x � 3.’

The converse is false.

(c) Premise 1: If set M is a subset of set N, then M � N = M.

Premise 2: M � N ≠ M Conclusion: Set M is not a subset

of set N.

3 (a) (i) 15 ÷ 3 = 5 and 72 = 14 is false.

(ii) 24 is a multiple of 6 or 1—7

> 1—5

is true.

(b) Premise 1: If the side a rhombus is 5 cm, then its perimeter is 20 cm.

Premise 2: The side of rhombus P is 5 cm.

Conclusion: The perimeter of rhombus P is 20 cm.

(c) 5x � 10 if and only if x � 2. Implication 1: If 5x � 10, then x � 2. Implication 2: If x � 2, then 5x � 10.

This is because we cannot determine its truth value.

The given argument is a type 1 argument.

Premise 1: All P is Q.Premise 2: R is P.Conclusion: R is Q.

whereP : ‘3 elements’Q : ‘have 23 subsets’R : ‘Set A’

A quadratic equation may have ‘two distinct roots’, ‘two equal roots’ or ‘no roots’.

When x � 8, x = 7, 6, 5, 4,… but x = 7, 6, 5 and 4 is not less than 3.


Premise 1: If p, then q.Premise 2: Not q.Conclusion: Not p.

wherep : ‘set M is a subset of set N ’q : ‘M � N = M ’

‘15 ÷ 3 = 5’ is true.‘72 = 14’ is false.‘true’ and ‘false’ is ‘false’.

‘24 is a multiple of 6’ is true.

‘ 1—7

� 1—5

’ is false.

‘true’ or ‘false’ is ‘true’.


Premise 1: If p, then q.Premise 2: p is true.Conclusion: q is true.

where

p : ‘The side of rhombus P is 5 cm.’q : ‘The perimeter of rhombus P is 20 cm.’


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Form 4 – Chapter 5 The Straight LinePassport To Success


Paper 1 1 2x + 5y = 7 5y = –2x + 7

y = – 2—5

x + 7—5

∴ m = – 2—5

Answer: A

2 3x + 6y + 5 = 0 6y = –3x – 5

y = – 1—2

x – 5—6

∴ c = – 5—6

Answer: B

Paper 2 3

(a) mDE

= mGF

= – 1—3

The equation of DE is y = mx + c, i.e.

y = – 1—3

x + c

Since DE passes through point D(–2, 3), x = –2 and y = 3.

3 = – 1—3

(–2) + c

c = 3 – 2—3

= 7—3

Hence, the equation of DE is

y = – 1—3

x + 7—3

.

At the x-axis, y = 0.

0 = – 1—3

x + 7—3

0 = –x + 7 x = 7 ∴ x-intercept = 7

(b) G is point (–2, 0). The equation of GF is y = mx + c, i.e.

y = – 1—3

x + c

Since GF passes through point G(–2, 0), x = –2 and y = 3.

0 = – 1—3

(–2) + c

c = – 2—3

Hence, the equation of GF is

y = – 1—3

x – 2—3

3y = –x – 2

y

O x

D(–2, 3)

G(–2, 0)E

F


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Form 4 – Chapter 6 Statistics IIIPassport To Success


Paper 2

1 (a) Distance (km) Midpoint (x) Tally f fx Class boundaries

21 – 30 25.5 2 51.0 20.5 – 30.5

31 – 40 35.5 4 142.0 30.5 – 40.5

41 – 50 45.5 11 500.5 40.5 – 50.5

51 – 60 55.5 10 555.0 50.5 – 60.5

61 – 70 65.5 8 524.0 60.5 – 70.5

71 – 80 75.5 4 302.0 70.5 – 80.5

81 – 90 85.5 1 85.5 80.5 – 90.5

Σf = 40 Σfx = 2160

(b) x = Σfx

—––Σf

= 2160—–––40

= 54 km

(c) (i), (ii)

(c) (i) Q3 = 75

(ii) The third quartile means 3—4

of the

students (i.e. 30 students) have marks of 75 and below.

2 MarksUpper

boundaryTally f

Cumulative frequency

20 – 29 29.5 0 0

30 – 39 39.5 4 4

40 – 49 49.5 5 9

50 – 59 59.5 7 16

60 – 69 69.5 10 26

70 – 79 79.5 7 33

80 – 89 89.5 5 38

90 – 99 99.5 2 40

20.5

12

10

8

6

4

2

0

Fre

quen

cy

30.5 40.5 50.5 60.5 70.5 80.5 90.5

Distance (km)

40

35

30

25

20

15

10

5

029.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5

Cum

ulat

ive

freq

uenc

y

Marks75


Bab 7 tidak ada

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Form 4 – Chapter 8 Circles IIIPassport To Success


Paper 1 1

∠ABD = ∠ADE = 54°

∠CBD = 180° – ∠ABD = 180° – 54° = 126°

∠BDC = 180° – ∠DBC——————–2

= 180° – 126°—————–2

= 27°

t° = ∠BDC = 27°

Answer: A

2

∠PQB = ∠PBA = 50°

∠PQO + ∠OQB = 50° 20° + ∠OQB = 50° ∠OQB = 30°

∠OBQ = ∠OQB = 30°

∠BOQ = 180° – ∠OBQ – ∠OQB = 180° – 30° – 30° = 120°

y° = ∠BOQ———–

2 = 120°

——2

= 60°

x° = ∠QPB = 60°

∴ x° + y° = 60° + 60° = 120°

Answer: D

A

Angle in the alternate segment

Angles on a straight line

BD = BC


B

C

D

E

126°27°

54°

27°

t° 54°Angle in the alternate segment

OB = OQ

Angles in a triangle

A B C

P

Q

y°20°

30°120°

30°50° x°

O

The angle subtended by an arc at the centre of a circle is twice the angle at the circumference.



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Form 4 – Chapter 9 Trigonometry IIPassport To Success


Paper 1 1

sin x° = 5—9

sin ∠RSQ = 5—9

RQ——SQ

= 5—9

10——SQ

= 5—9

5SQ = 90 SQ = 18

∴ TQ = SQ – ST = 18 – 13 = 5 cm

In �UTQ, based on the Pythagorean triples, TU = 12 cm.

tan y° = –tan ∠TQU = – 12—–5

Answer: D

2

∠RSQ is the basic angle which corresponds to the obtuse ∠VSQ (x°). sin x° is positive because x° is an angle in the second quadrant.

P

QR

S

T

U

x°

13 cm

12 cm

13 cm5 cm

10 cm

V

y°

cos θ = –cos ∠PQR = – 8—–10

= – 4—5

Answer: A

3 The graph of y = cos x for 0° � x � 180° is as shown below.

Answer: B

4 cos θ = –0.4226 Basic ∠ = 65°

∴ θ1 = 180° – 65° = 115° ∴ θ2 = 180° + 65° = 245°

Answer: A

∠POR is the basic angle which corresponds to θ. cos θ is negative because θ is an angle in the third quadrant.

y

O R x

θ

10 6

8

T

P(8, 6)

∠TQU is the basic angle which corresponds to the obtuse ∠PQU (y°). tan y° is negative because y° is an angle in the second quadrant.

y

1

090°

180°x

–1

cos θ is negative in the second and third quadrants.

S

T

A

C

65°

65°

θ2

θ1


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Form 4 – Chapter 10 Angles of Elevation and DepressionPassport To Success


Paper 1 1

In �RTP,

tan 40° = x—–

15

x = 15 × tan 40° = 12.586 m ∴ RS = 2x = 2 × 12.586 = 25.17 m

Answer: C

2

In �PQR,

tan 40° = x—–

18

x = 18 × tan 40° = 15.10 m

∴ Height of tree = 15.10 + (3 – 1) = 17.10 m

Answer: B

P

Q

R

S

T40°

40°

x m

15 m

15 m

Angle of elevation

Angle of depression

P

3 m

Q

R40°

x m

1 m

18 m


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Form 4 – Chapter 11 Lines and Planes in 3–DimensionsPassport To Success


Paper 1 1

The line is KN.The normal is KJ.The orthogonal projection is JN.

The angle between the line KN and the plane NMJ is the angle between the line KN and its orthogonal projection (JN), i.e. ∠KNJ.

Answer: A

2

The line of intersection of the planes NCM and QBC is MC.

∠NMC is a right angle on the plane NCM. ∠QMC is a right angle on the plane QBC.

Hence, the angle between the planes NCM and QBC is ∠NMQ.

Answer: A

Paper 2 3

The line of intersection of the planes ABM and ABCD is AB.

∠BAM is a right angle on the plane ABM. ∠BAD is a right angle on the plane ABCD.

Hence, the angle between the planes ABM and ABCD is ∠MAD.

Let N be the midpoint of AD. In �ANM,

tan ∠MAN = 8—5

∠MAN = 57.99° (or ∠MAD = 57.99°)

J

K L

M

N

Orthogonal projection

Normal

A B

CD

P Q

M

N

A

B C

D

EF

G H

M

N

10 cm

8 cm

12 cm

5 cm 5 cm

A

M

N5 cm

8 cm


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53 52 51 50

2 3 4 05

∴ 23405 = 2 × 53 + 3 × 52 + 4 × 51 + 0 × 50

But it is given that: 23405 = 2 × 53 + 3 × 52 + y × 51 + 0 × 50

Hence, by comparison, y = 4.

Answer: D

4

∴ 15710 = 2358

But it is given that 15710 = 2k58. Hence, by comparison, k = 3.

Answer: C

8 157

8 19 –5

8 2 –3

0 –2

Form 5 – Chapter 1 Number BasesPassport To Success


Paper 1 1

Answer: A

2

Answer: D

3

1 1

1 1 0 0 12

+ 1 1 1 0 12

1 1 0 1 1 02

12 + 12 = 10212 + 12 + 12 = 112

100 000 1112

421 421 421

4 0 78


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Form 5 – Chapter 2 Graphs of Functions IIPassport To Success


Paper 1 1 The general shape of the cubic graph

y = –2x3 – 9 is

The y–intercept of y = –2x3 – 9 is –9. Hence, the graph of y = –2x3 – 9 is as

shown below.

Answer: D

Paper 2 2 For y � –2x + 4, shade above the straight

line y = –2x + 4 and it should be a solid line.

For y � x + 1, shade below the straight line

y = x + 1 and it should be a solid line. For x � 4, shade to the left of the straight

line x = 4 and it should be a dashed line.

The region which satisfies all the given inequalities is as follows.

3 (a) When x = –1, y = 8 – (–1)3 = 9 When x = 1.5, y = 8 – (1.5)3 = 4.625(b)

(c) From the graph, (i) when x = 1.25, y = 6 (ii) when y = 8.4, x = –0.75

(d) y = 8 – x3

+ 0 = x3 – x – 8 y = –x

From the graph, the value of x which satisfies the equation x3 – x – 8 = 0 is the x-coordinate of the point of intersection of the curve y = 8 – x3 and the straight line y = –x, i.e. x = 2.15.

y

O x

–9

6

4

2

O

–2

–4

y

x–2 2 4

y = x

+ 1

y = –2x + 4

10

8

6

4

2

8.4

–1.0–0.75

–0.5 0.5 1.0

1.25

1.5 2.0 2.5

2.15x

y

y = 8 – x 3

y = –x–2

–4

–6

–8

0

Graph drawn

Given equation

This is the equation of the straight line which has to be drawn.


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Form 5 – Chapter 3 Transformations IIIPassport To Success


Paper 2 1

(a) Draw a line segment to join the points A and P. Construct a perpendicular bisector of the line segment AP. …➀

Draw a line segment to join the points BQ. Construct a perpendicular bisector of the line segment BQ. …➁

(i) The centre of rotation is the point of intersection of the perpendicular bisectors of ➀ and ➁, i.e. (3, 5).

(ii) The angle of rotation is 90° (anticlockwise).

(b) (i) A(1, 3) ⎯→ A’(5, 0) ⎯→ A’’(0, 5)

(ii) A(1, 3) ⎯→ A’(3, 1) ⎯→ A’’(7, –2)

(c)

(i) �PQR is transformed to �KQ’R under transformation V, i.e. reflection in the straight line y = 6.

(ii) �KQ’R is transformed to �KLM under transformation W, i.e. enlargement with centre (5, 9) and a scale factor of 2.

Centre of rotation

G H

H Gy

6

4

2

O 2x

A

B C

P Q

R

4 6

8

6

4

2

O 2 4 6 8x

K L

P Q

R

M

y

y = 6

Q ′


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Form 5 – Chapter 4 MatricesPassport To Success


Paper 1

1 –2 h� � 3 5 – 2 0 2� � –3 k

= –2 0� � 9 7

–2 h� � 3 5 – 0 4� � –6 2k

= –2 0� � 9 7

–2 h – 4� � 9 5 – 2k = –2 0� � 9 7

h – 4 = 0 h = 4

∴ h + k = 4 – 1 = 3

Answer: C

5 – 2k = 7 5 – 7 = 2k –2 = 2k k = –1

2 2A – 1 0� � –1 4 = 5 2� � 9 8

2A = 5 2� � 9 8 + 1 0� � –1 4

2A = 6 2� � 8 12

A = 1—2

6 2� � 8 12

A = 3 1� � 4 6

Answer: D

3 (k 4) 2 0� � –k 7 = (14 28)

(k(2) + 4(–k) k(0) + 4(7)) = (14 28) (–2k 28) = (14 28) –2k = 14 k = –7

Answer: D

Paper 2

4 (a) PQ = 1 0� � 0 1

PP –1 = 1 0� � 0 1

∴ Q = P –1

= 1—————–—3(3) – (–4)(2)

3 4� � –2 3

= 1—–

17 3 4� � –2 3

But it is given that Q = 1—k

3 h� � –2 3.

Hence, by comparison, k = 17 and h = 4.

(b) 3x – 4y = –5 2x + 3y = 8

The matrix equation is

3 –4� � 2 3 x� � y

= –5� � 8

x� � y =

1—–17

3 4� � –2 3 –5� � 8

x � � y =

1—–17

17� � 34

x� � y = 1� � 2

∴ x = 1, y = 2

5 (a) Let A = 2 –1� � –6 4

A–1 = 1—————–—–2(4) – (–1)(–6)

4 1� � 6 2

= 1—2

4 1� � 6 2

= 2 1—

2� � 3 1

But it is given that A–1 = 2 h� � 3 1.

Hence, by comparison, h = 1—2

.

(b) 2m – n = 6 –6m + 4n = –20

The matrix equation is

2 –1� � –6 4 m� � n

= 6� � –20

m� � n =

2 1—2� � 3 1

6� � –20

m� � n =

2 × 6 – 1—2 (20)

3 × 6 – 1(20)� � m� � n

= 2� � –2

∴ m = 2, n = –2

Px� �y =

–5� �8

P –1Px� �y = P –1 –5� �8

Ix� �y = P –1 –5� �8

x� �y = P –1 –5� �8

Am� �n =

6� �–20

A–1Am� �n = A–1 6� �–20

Im� �n = A–1 6� �–20

m� �n = A–1 6� �–20


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Form 5 – Chapter 5 VariationsPassport To Success


Paper 1 1 y ∝ x3

y = kx3, where k is a constant When x = 3, y = 9, 9 = k(3)3

k = 9—–

27

k = 1—3

∴ y = 1—3

x3

When x = k and y = 8—3

,

8—3

= 1—3

k3

k3 = 8 k = 2

Answer: A 2 Q ∝

1——3 R

Q = k——R

1—3

, where k is a constant

Q = kR 1–— 3

Answer: B

3 s ∝ m—n

s = km—–n

, where k is a constant

When m = 2 and n = 8, s = 1—2

1—2

= k(2)—––

8 4k = 8 k = 2

∴ s = 2m—–n

When s = 25 and m = 50,

25 = 2(50)—–––

n

n = 100—––25

n = 4

Answer: C


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Chapter 6 Gradient and Area Under a GraphPassport To Success


Paper 2 1

(a) Rate of change of speed from nth s to

18th s = – 5—2

m s–2

–10 – 0�———–�18 – n

= – 5—2

5(18 – n) = 20 90 – 5n = 20 5n = 90 – 20 5n = 70

n = 70—–5

n = 14

(b) (i) Length of time the particle travels at a uniform speed

= n – 7 = 14 – 7 = 7 s

(ii) Average speed in the first 7 s

= Total distance——————–

Total time =

Area P + Area Q——————–—

Total time

1—2

(8 + 16)(4) + 1—2

(16 + 10)(3) = —————————————— 7 =

87—–7

= 12 3—7

m s–1

Negative gradient

Speed (m s–1)

16

108

Time (s)O4 7 n 18

P Q

Horizontal part of the graph

Distance (m)

d

400

O5 11 22

Time (min)

2

(a) The length of time Normala stops for a rest

= 11 – 5 = 6 minutes

(b) Speed in the first 5 minutes

= 400—––

5 = 80 m min–1

(c) Average speed = 30 m min–1

Total distance——————–

Total time = 30

d—–22

= 30

d = 660

Horizontal part of the graph

Gradient


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Form 5 – Chapter 7 Probability I & IIPassport To Success


Paper 1 1 Let S – Sample space A – Event that the card drawn is a factor

of 48

A = {6, 12, 24, 16, 3, 4, 8} = {3, 4, 6, 8, 12, 16, 24} n(A) = 7

∴ P(A) = n(A)

—–––n(S)

= 7—–12

Answer: D

2 Let M – Event that a male fish is chosen F – Event that a female fish is chosen S – Sample space

P(F) = 1 – 5—–

12 =

7—–12

n(F)

—–––n(S)

= 7—–

12

35

—–––n(S)

= 7—–

12

n(S) = 12—–7

× 35

= 60

Answer: A

3 Let R – Event of drawing a red pen B – Event of drawing a blue pen H – Event of drawing a black pen S – Sample space

P(B) = 2—5

× 50 = 20

n(H) = n(S) – n(R) – n(B) = 50 – 18 – 20 = 12

Answer: A

4 Let M – Event that a male student is chosen F – Event that a female student is chosen H – Event that a student carrying a

handphone is chosen S – Sample space

n(S) = n(M) + n(F) = 24 + 16 = 40

n(H) = P(H) × n(S) = 5—8

× 40 = 25

Hence, the number of male students who carry handphones

= Total number of students who carry handphones – Number of female students who carry handphones

= 25 – 7 = 18

Answer: D

Paper 2 5 (a) Let R – Event that a red cube is drawn Y – Event that a yellow cube is drawn P(RR or YY)

= � 2—5

× 4—–10 � + � 3—

5 ×

6—–10 �

= 13—–25

(b)

P(RR or YY) = P(RR) + P(YY)

= � 2—5

× 5—–11 � + � 3—

5 ×

7—–11 �

= 31—–55

2R3Y

Jar

5R6Y

Bowl

4R7Y

Bowl

R

Y

2—5

3—5

R

Y

R

Y

RR

RY

YR

YY

5—11

6—114—11

7—11

Outcomes


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Form 5 – Chapter 8 BearingPassport To Success


Paper 1 1

∠TRS = ∠TSR = 42°

∠VTR = ∠TRS = 42°

∴ Bearing of R from T = 180° + 42° = 222°

Answer: C

2

∠N1QR = 50°

∠N2RQ = 180° – 50° = 130°

∴ Bearing of P from R = 360° – (130° + 20°) = 210°

Answer: C

TS = TR

RS //VT and alternate angles are equal.

It is given that the bearing of R from Q is 050°.

QN1//RN2 and the sum of interior angles is 180°.

P

QR

N1

N2

50°20°

130°

Bearing of P from R

N2

T

V

N1

SR

42°

42°42°

Bearing of R from T


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0.36 hours= 0.36 × 60= 22 minutes

∠MOP = 180° because MP is the diameter of the earth.

N

O

S

PQ

M

42°

55°W10°E

42°ST

Form 5 – Chapter 9 Earth as a SpherePassport To Success


Paper 1 1

Since the longitude of point H is 35°E, ∠GOH = 35°.

Since the difference in longitude between point F and point H is 100°, ∠FOH = 100°.

∴ ∠GOF = 100° – 35° = 65° Therefore, the longitude of point F is 65°W.

Hence, the longitude of point J is(180 – 65)°E = 115°E

Answer: B

2

Since the difference in latitude between point D and point F is 90°, then

∠BOF = 90° – 50° = 40°

Therefore, ∠AOH = 40° because FOH is the diameter of the earth.

Hence, the latitude of point H is 40°N.

Answer: A

Paper 2 3

(a) Longitude of point P = (180 – 55)°E = 125°E

(b) Distance of MT = (55 + 10) × 60 × cos 42° = 2898.3 n.m.

(c) Distance of MQ = 4740 n.m. ∠MOQ × 60 = 4740

∠MOQ =

4740———

60 = 79°

Hence, the latitude of point Q = (79 – 42)°N = 37°N

(d) Time = Distance of MNP——————–—–

Speed

=

180 × 60————–

660

= 16.36 hours = 16 hours 22 minutes

N

S

F

G

O

H

J65° 35°

0° 35°E

N

S

A B

H

F

D

O

50°N

40°40°

40°S


Weblink 23

Paper 2 1 (a)

(b) (i)

Form 5 – Chapter 10 Plans and ElevationsPassport To Success


U/T

(ii)

T/SM/N J/R

2 cm3 cm

7 cm

U/P L/V K/QPlan

L/A/M

V/N

B/J

2 cm

3 cm

5 cm6 cm

P/S Q/C/R

Elevation as viewed from X

L B/A J/M3 cm 4 cm

V/U N/T1 cm

5 cm

Q/P C R/S

Elevation as viewed from Y

1 cm


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