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SULIT 3472/2

[ Lihat sebelah 3472/2 SULIT

1

1 3472/2 Form Four Additional Mathematics Paper 2 2010

221 hours

PEPERIKSAAN AKHIR TAHUN 2010

TINGKATAN 4

ADDITIONAL MATHEMATICS

Paper 2

221 hours

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. This questions paper consists of three sections : Section A, Section B and Section C. 2. Answer all questions in Section A , four questions from Section B and two questions from Section C. 3. Give only one answer / solution to each question.. 4. Show your working. It may help you to get marks. 5. The diagram in the questions provided are not drawn to scale unless stated. 6. The marks allocated for each question and sub-part of a question are shown in brackets.. 7. A list of formulae is provided on pages 2 to 3. 8. A booklet of four-figure mathematical tables is provided. 9. You may use a non-programmable scientific calculator.

Kertas soalan ini mengandungi 15 halaman bercetak

http://tutormansor.wordpress.com/

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The following formulae may be helpful in answering the questions.The symbols given are the ones commonly used. Rumus-rumus berikut boleh digunakan untuk membantu anda menjawab soalan. . Simbol-simbol yang diberi adalah yang biasa digunakan.

ALGEBRA

1. x =a

acbb2

42

2 am an = a m + n 3 am an = a m - n

4 (am) n = a nm 5 loga mn = log am + loga n

6 loga nm

= log am - loga n

7 log a mn = n log a m

8 logab = ab

c

c

loglog

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan

11 Tn = ar n-1

12 Sn = rra

rra nn

1)1(

1)1(

, (r 1)

13 r

aS

1 , r <1

CALCULUS

1 y = uv , dxduv

dxdvu

dxdy

2 vuy , 2v

dxdvu

dxduv

dxdy

,

3 dxdu

dudy

dxdy

GEOMETRY

1 Distance = 221

221 )()( yyxx

2 Midpoint

(x , y) =

221 xx

,

221 yy

3 22 yxr

4 ř 22 yx

yjxi

4 Area under a curve

= b

a

y dx or

= b

a

x dy

5 Volume generated

= b

a

y 2 dx or b

a

x 2 dy

5 . A point dividing a segment of a line

( x,y) = ,21

nmmxnx

nmmyny 21

6. Area of a triangle =

)()(21

312312133221 yxyxyxyxyxyx


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3

STATISTICS

TRIGONOMETRY

1 x = N

x

2 x =

ffx

3 = N

xx 2)( =

2_2

xN

x

4 =

fxxf 2)(

= 22

xf

fx

5 M = Cf

FNL

m

2

1

6 1000

1 PPI

7 1

11

wIwI

8 )!(

!rn

nPrn

9. !)!(

!rrn

nCrn

10 P(AB)=P(A)+P(B)-P(AB)

11 p (X=r) = rnrr

n qpC , p + q = 1

12 Min(mean) = np 13 npq

14 z = x

1 Arc length , s = r

2 Area of a sector, L = 221 r

3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosek2 A = 1 + cot2 A

6 sin2A = 2 sinAcosA 7 cos 2A = cos2A – sin2 A = 2 cos2A-1 = 1- 2 sin2A

8 tan2A = A

A2tan1

tan2

9 sin (A B) = sinAcosB cosAsinB

10 cos (A B) = cos AcosB sinAsinB

11 tan (A B) = BABA

tantan1tantan

12 C

cB

bA

asinsinsin

13 a2 = b2 +c2 - 2bc cosA

14 Area of triangle = Cabsin21


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Section A

[ 40 Marks ]

Answer all questions from this section.

1. Solve the following simultaneous equations. Give your answers correct to four decimal places.

Selesaikan persamaan serentak berikut. Berikan jawapan anda kepada empat

tempat perpuluhan. x + 2y = 6 2x2 – y2 + xy = 2

[5 marks]

2. Functions g and h are defined by 21)(

xxxg , 2x and

0,3:

xx

axxh where a is a constant.

Fungsi g dan h didefinasikan sebagai 21)(

xxxg , 2x dan

0,3:

xx

axxh dimana a adalah pemalar.

(a) Find 1g Cari 1g (b) Given that 1hg (4) = 6 , find the value of a.

Diberi 1hg (4) = 6 , cari nilai a . [ 6 marks] 3. A set of examination marks x1, x2, x3, x4, x5 has a mean of 4 and standard deviation of 2.3 Suatu set markah peperiksaan x1, x2, x3, x4, x5 mempunyai min 4 dan sisihan piawai 2.3

a) Find / Cari ,

i) the sum of the marks, ∑x

jumlah markah, ∑x ii) the sum of the squares of the marks, ∑ x2. jumlah kuasa dua markah, ∑x 2. [ 2 marks]


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b) Each mark is multiplied by 3 and then 7 is added to it.

Setiap markah itu didarabkan dengan 3 dan ditambah nombor 7 ke dalamnya.

Find / Carikan

i) the mean / min ii) the variance / varian [ 5 marks]

4. The gradient function of a curve is px 2 + hx, where p and h are constants.The curve has turning point at ( 3, - 2 ). The gradient of the tangent to the curve at the point where x = -1 is 12 Fungsi kecerunan satu lengkung ialah px 2 + hx, dengan keadaan p dan h ialah pemalar.Lengkung itu mempunyai titik pusingan pada ( 3, - 2). Kecerunan tangent kepada lengkung itu pada titik x = -1 ialah 12.

a) Find the value of p and h. Carikan nilai p dan h. [ 4 marks ]

b) A curve, y = f(x) with the gradient function, 13

2

xx

dxdy .

Satu lengkung, y = f(x) mempunyai fungsi kecerunan 13

2

xx

dxdy

Find , Cari ,

2

2

dxyd

[2 marks]

5. The straight line y + 2x - 4 = 0 is a tangent to the curve y = x3 + 3x2 – 11x + 9 at a point P. Garis lurus y + 2x - 4 = 0 ialah tangen kepada lengkung y = x3 + 3x2 – 11x + 9 pada titik P. ( a ) Find the gradient of the tangent at point P

Cari kecerunan tangen itu pada titik P. [ 2 marks ]


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( b ) Find the coordinates of point P. Cari koordinit bagi titik P. [ 4 marks ]

( c ) Find the equation of another tangent which is parallel to the tangent at point P.

Cari persamaan bagi tangen yang satu lagi yang selari dengan tangen pada titik P. [ 2 marks ]

6. ( a ) Given log3 m = x and log3 n = y . Express log3 m n in terms of x and y. Diberi log3 m = x dan log3 n = y. Ungkapkan log3 m n dalam sebutan x dan y. [ 2 marks ]

( b ) Solve the equation 2x . 3x = 6 2x – 6 Selesaikan persamaan 2x . 3x = 6 2x – 6 [ 2 marks ] ( c ) Solve the equation 2 log9 x = log3 4

Selesaikan persamaan 2 log9 x = log3 4 [ 4 marks ]

Section B

[ 40 Marks ]

Answer four questions from this Section.

7. A point P moves along the curve of a circle with centre G ( 2,3 ). The arc

passes through A ( -2 , 0 ) and B ( 5 , t ) Titik P bergerak di sepanjang suatu lengkok bulatan yang berpusat G ( 2,3 ).Lengkok

bulatan itu melalui A ( -2,0) dan B ( 5 , t )

(a) Find / Carikan

( i ) the equation of the locus of a point P persamaan locus bagi titik P


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(ii) the values of t

nilai-nilai t [ 6 marks ]

(b) The tangent to the circle at point A intersects the y-axis at point H . Find the area of triangle OAH

Tangen pada bulatan itu di titik A bersilang dengan paksi-y di titik H. Carikan luas bagi segitiga OAH [ 4 marks ]

8. y U ( 6 , 7 ) T ( 0 , 5 ) x V ( p , q ) W The Diagram 8 shows the vertices of a rectangle TUVW on a Cartesian plane Rajah 8 menunjukkan bucu-bucu sebuah segi empat tepat TUVW di atas satah

Cartesian

(a) Find the equation that related p and q by using the gradient of UV Carikan persamaan yang menghubungkan p dan q dengan menggunakan

kecerunan UV [ 3 marks ]

(b) Show that the area of triangle TUV can be expressed as p – 3q + 15 Tunjukkan bahawa luas segitiga TUV boleh diungkapkan sebagai p – 3q + 15

[ 2 marks ]

Diagram 8 / Rajah 8


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(c) Hence, calculate the coordinates of point V,given that the area of rectangle TUVW is 40 unit2

Seterusnya , hitungkan koordinat titik V, diberi luas segi empat tepat TUVW

ialah 40 unit2 [ 3 marks ]

(d) Find the equation of the straight line TW in the intercept form.

Carikan persamaan garis lurus TW dalam bentuk pintasan . [ 2 marks ]

9. The Diagram 9 above shows a sector OPQ of as circle with centre O. Point A lies on OP and point B lies on OQ. Given OB = 2 cm , tan = 1 and

OB : OQ = 1 : 3

Rajah 9 di atas menunjukkan sebuah sektor bulatan OPQ yang berpusat di O. Titik A terletak pada OP dan titik B terletak pada OQ. Diberi OB = 2 cm, tan = 1 dan OB : OQ = 1 : 3. a) Find the value of , in term of . Cari nilai , dalam sebutan .

[ 1 marks ] b) Find the length of OA, in cm. Cari panjang OA, dalam cm.

[ 2 marks ] c) Calculate the perimeter, in cm, of the shaded region Hitung perimeter, dalam cm, kawasan yang berlorek

[ 4 marks ] d) Calculate the area, in cm 2 , of the shaded region. Hitung luas, dalam cm 2 , kawasan yang berlorek.

[ 3 marks ]

Diagram 9 / Rajah 9


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10. A piece of wire of length 120 cm is bent into a shape as shown in the Diagram 10. Segulung wayar sepanjang 120 cm diikat dalam satu bentuk seperti yang ditunjukkan dalam Rajah 10.

a) Express y in term of x. Ungkapkan y dalam sebutan x. [ 2marks ]

b) Hence, shows that the area, A cm 2 , is given by A= 36x(10 - x) Seterusnya tunjukkan luas bagi A ialah A= 36x(10 – x) [ 4marks ]

c) Find the value of x and y for which A is maximum and state its maximum value.

Cari nilai x dan y di mana nilai A ialah maximum dan nyatakan nilai maksimum tersebut. [ 4marks ]

5x cm

5x cm

y cm

y cm

y cm

Diagram 10 / Rajah 10


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11. The marks obtained by a group of students in a test is shown in Table 1. The mean mark of the students is 48.25

Markah yang diperoleh sekumpulan pelajar dalam suatu ujian ditunjukkan dalam

Jadual 1. Min markah pelajar-pelajar ialah 48.25.

Marks Markah

Number of students Bilangan pelajar

20 – 29 8

30 – 39 15

40 – 49 22

50 – 59 p

60 – 69 10

70 – 79 8

Table 11 Jadual 11

(a) Based on the data in Table 11 and without using the graphical method,

Berdasarkan data dalam Jadual 11 dan tanpa menggunakan kaedah graf, Calculate / hitung

(i) the value p, nilai p,

(ii) the median of the distribution.

median untuk taburan itu. [ 6 marks ]

(b) Draw the histogram to represent the data in Table 11 and estimate the mode of mark of the distribution.

Lukis sebuah histogram untuk mewakili data dalam Jadual 11 dan anggarkan markah mod bagi taburan itu.

[ 4 marks ]


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Section C [ 20 Marks ]

Answer two questions from this Section.

12. The Table 12 shows the prices, the price indices and the percentages

of five components, P, Q, R, S and T, used to produce a kind of pliers. Jadual 12 menunjukkan harga, indeks harga dan pemberat untuk lima komponen, P, Q, R, S dan T, yang digunakan untuk membuat sejenis playar.

Price (RM) for the year Harga (RM) pada tahun

Component Komponen

2003

2005

Price index for the year 2005 base on the year

2003 Indeks harga pada tahun 2005 berasaskan tahun

2003

Percentage Pemberat

P

2.50

3.00

120

30

Q

6.00

8.40

x

20

R

9.00

Y

125

10

S

Z

10.40

130

25

T

4.50

6.75

150

15

Table 12 / Jadual 12

a) Find the value of X, Y and Z. Dapatkan nilai X, Y dan Z. [ 3 marks ]

b) Calculate the composite index for the production cost of the pliers in the year 2005 based on the year 2003.

Kirakan indeks komposit untuk kos pembuatan playar pada tahun 2005 berasaskan tahun 2003. [ 3 marks ]

c) The price of each component increased by 20 % from the year 2005 to the year 2007.Given that the production cost of one pliers in the year 2003 was RM 70, calculate the corresponding cost in the year 2007,

Harga bagi setiap komponen meningkat 20 % dari tahun 2005 ke tahun 2007. Diberi bahawa kos pembuatan sebuah playar pada tahun 2003 ialah RM 70, kirakan harga yang sepadan bagi tahun 2007. [ 4 marks ]


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13. Diagram 13 is a bar chart which shows the average weekly number of pairs

of tennis, badminton, soccer and golf sports shoes sold in the year 2005. Rajah 13 menunjukkan carta palang purata minggu bagi sepasang kasut sukan

tenis , badminton , bolasepak dan golf yang dijual pada tahun 2005

Number of pairs / bilangan pasang

100

80

60

40

20

0 Sports / kasut Tennis Badminton Soccer Golf shoes sukan Tenis Badminton Bolasepak Golf Diagram 13 / Rajah 13 Table 13 shows the average price per pair of tennis, badminton, soccer and

golf shoes in the years 2005 and 2007 and the price indices of these sports

shoes in the year 2007 based on the year 2005.

Jadual 13 menunjukkan purata harga bagi sepasang kasut sukan tenis, badminton,

bolasepak dan golf bagi tahun 2005 dan 2007 dan indeks harga bagi kasu-kasut

sukan pada tahun 2007 berdasarkan tahun 2005 sebagai tahun asas.


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Average price per pair ( RM ) Purata harga untuk satu pasang ( RM )

Types of sports shoes

Jenis-jenis kasut sukan

Year 2005 Tahun 2005

Year 2007 Tahun 2007

Price index for the year 2007 based on

the year 2005 Indek Harga 2007

dengan 2005 sebagai tahun asas

Tennis / Tenis x 220.00 110

Badminton /Badminton

150.00 187.50 y

Soccer/ Bolasepak 180.00 189.00 105

Golf /Golf 400.00 z 130

Table 13 Jadual 13

a) Find the value of x, y and z Cari nilai x, y dan z [ 3 marks ]

b) Using the data in Diagram 13 as weightage, calculate the composite index

number of the four types of sports shoes in the year 2007 based on

the year 2005. Dengan menggunakan data dari Rajah 13 sebagai pemberat, kirakan nombor

indeks gubahan bagi keempat-empat jenis kasut sukan pada tahun 2007 dengan

tahun 2005 sebagai tahun asas.

[ 2 marks ]

c) If the monthly sales netted from these four types of sports shoes in January

2005 is RM75 000, calculate the corresponding monthly sales netted in

January 2007 Jika jualan pukal bulanan keempat-empat jenis kasut sukan pada bulan Januari

pada tahun 2005 adalah RM 75 000, hitungkan jualan bulanan bersih yang sepadan

pada tahun 2007.

[ 2 marks ]


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d) The average prices of the sports shoes are expected to rise by 40% from the

year 2005 to the year 2009. Calculate the composite index number for the

year 2009 based on the year 2007. Harga purata kasut-kasut sukan dijangkakan akan naik sebanyak 40% dari tahun

2005 ketahun 2009. Hitungkan nombor indeks gubahan tahun 2009 dengan 2007

sebagai tahun asas.

[ 3 marks]

14.

Diagram 14 above shows a quadrilateral ABCD Rajah 14 di atas menunjukkan sebuah sisi empat ABCD

a) Find the length of AC, in cm Cari panjang AC, dalan cm.

(2 marks ) b) Calculate Hitung i) ADC ii) ACD (4 marks ) c) Calculate the area, in cm 2 , of the quadrilateral ABCD Hitung luas, dalam cm 2 , sisi empat ABCD

(4 marks )

B

A

065

12 cm 12.8 cm

C

D

075


8 cm


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15

15.

In Diagram 15 , sin ADC = 23 where ADC is an obtuse angle.

Di dalam Rajah 15 , sin ADC = 23

dimana ADC ialah sudut cakah.

Find / Hitung ( a ) the length of AC panjang AC [ 4 marks ] ( b ) ABC [ 2 marks ] ( c ) the perpendicular distance from C to the side AB. jarak serenjang dari C ke sisi AB. [ 4 marks ]

END OF QUESTIONS

B A

C D

9.3 cm 5.4 cm

6.6 cm

500



1

3472/2 Form 4 Additional Mathematics Paper 2 2010

PEPERIKSAAN AKHIR TAHUN 2010

TINGKATAN 4

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 11 printed pages.


2

Number

Solution and marking scheme

Submarks

Full marks

1

x = 6 – 2y 2(6 – 2y)2 – y2 + y(6 –2y) = 2

)5(2)70)(5(4)42()42( 2

y

y = 6.108 , y = 2.292 x = -6.216 , x = 1.416 atau setara.

1

1

1

1 1

5

2

( a ) g -1 = g -1(4) = 3 ( b ) = 6

a = 5

2

1 1

2

6

3

a i) 20 ii) 106.45 b i) 19 4(3) + 7 ii) 47.61 2.3 2 (9)

1 1

2 B1

3

B2

7


3

4

a) 9,3 hp

039

0

hpdxdy

@ p – h = 12

b) 2

2

2

2

)13()3)(()2)(13(

xxxx

dxyd

2

2

)13(23

xxx

2, 2

B1

B1

1

1

6

( a ) y + 2x - 4 = 0 y = -2x + 4 Gradient of a tangent = -2

1 1

2

5

( b ) y = x3 + 3x2 -11x + 9

dxdy = 3x2 + 6x - 11

-2 = 3x2 + 6x - 11 3x2 + 6x - 9 = 0 x2 + 2x - 3 = 0 ( x-1)(x+3) = 0 x=1 or x = -3 When x= 1 , y=2 When x=-3 , y=42 Substitut (1,2) into y = -2x + 4 2=2 Substitut (-3,42) into y = -2x + 4 42 ≠10

1

1

1

4


4

Thus the coordinate of point P is ( 1, 2 )

1

( c ) y - 42 = -2 ( x + 3 ) y = -2x + 36

1 1

2

( a ) log3 m√n = log3 m + ½ log3 n = x + ½ y

1 1

2

( b ) 2x 3x = 62x – 6

( 2 X 3 )x = 62x – 6

6x = 62x – 6 x = 6

1

1

2

6

(c ) 2 log9 x = log34 log9 x2 = log34

9log

log

3

23 x = log34

23

23

3loglog x = log34

log3 x2 = log342

x2 = 42 x = 4

1

1

1

1

4

Answer four questions from this section

7

( a ) i. PG = GA 22 )3()2( yx = 22 )03()22( x2+y2 -4x -6x – 12 =0 ii. B (5 , t ) , 52 +t2 -4(5) – 6t -12 = 0 t2 -6t – 7= 0 ( t + 1 )( t – 7 ) = 0 t = - 1 or 7

1 1 1 1 1 , 1


5

( b ) Gradient GA =

2230 =

43

The equation of tangent ; y – 0 = 3

4 ( x + 2 )

y = 3

4 x - 38

x = 0 , y = - 38 T ( 0 , -

38 )

Area of triangle 0AH = ½ x 2 x 38

= 38 unit2

1 1 1 1

10

8

( a ) Gradient TU =

0657

= 31

Gradient UV = - 3 3

67

pq

3p + q -25 = 0………………….(1) ( b ) Area triangle TUV = ½

575060

qp

= ½ )56()730( pqp 15 + p – 3q shown ( c ) Area of TUVW = 40 units2

Area triangle TUV = 20 units2

15 + p – 3q = 20 p = 5 + 3q …………….( 2) sub (2) into ( 1) 3 [ 5 +3q ] + q – 25 = 0 q = 1 , p = 8 Koordinat V =( 8 , 1 )

1 1 1 1 1 1 1 1

10


6

( d ) Gradient TW = UV = -3 Equation TW ; y = -3x + 5 y + 3x = 5

5y 1

53

x

1 1

9

a)

4rad

b) 2 22 2OA = 2.828 cm c) OQ = 3OB = 3(2) = 6 BQ = 6 – 2 = 4 AP = OP – OA = 6 – 2.828 = 3.172 cm

Length of arc PQ = 64

= 4.713 cm

Perimeter of the shaded region = AP + AB + BQ + Arc PQ = 3.172 + 2 + 4 + 4.713 = 13.89 cm d) Area of the shaded region = Area sector OPQ – Area ofOAB

= 21 1(6 ) (2)(2)2 4 2

= 12.14 2cm

1

1 1

1

1

1

1

1,1

1

10

10

a) 5x + 5x + 6x + y + y = 120 y = 60 – 8x ………….. (1 )

b) A = 16 ( ) (6 )(4 )2

x y x x

= 26 12xy x ……………………(2) Substitude (1) into (2)

1 1 1 1

10


7

26 (60 8 ) 12

36 (10 )A x x x

x x

c) 360 72dA xdx

For A is a maximum

0dAdx

360 – 72x = 0 x = 5 when x = 5 then y = 60 – 8(5) = 20 Maximum area = 36(5)(10 – 5) = 900 2cm

1 1 1 1 1 1

11

(a) (i) =48.25

p = 17

(ii) m =

= 47.23

(b) Graph Draw the histogram with the uniform scale – x-axis & y-axis

2 1

2 1

2

10


8

Draw – find mode Mode = 45.5

1 1

Answer two questions from this section

12 a) x = 140, y = RM 11.25 , z = RM8.00 b) 131.50

100

)15(150)25(130)10(125)20(140)30(120

1, 1, 1

3

B2

10


9

a) RM 110.46

80.157

10012050.131

8.15710070

07/05

05

0705

XI

XPPXP

4

B3

B2

13

a) (i) 110100220

xx

x = RM 200 (ii) y = 100

1505.187 x

y = 125 (iii) 130100

400xz

z = RM 520 (b)

WIW

I

280

)40130()60105()80125()100110( xxxx

116.07 c) 07.116100

75000xU

U = RM 87 053.57 (d) 07.116

100140 x

162.5

1 1 1

1 1 1 1 1,1 1

10

14

a)

2 2 2

2 2 0

2( ( )cos12 8 2(12)(8)cos65126.86

11.26

AC AB BC AB BC ABC

AC cm

1 1

10


10

b) i)

0

0 '

sin sin 7511.26 12.8

58 11

ADC

ADC

ii) 0 0 0 '180 75 58 11ACD

= 0 '46 49 c) Area of quadrilateral ABCD = Area of ABC + Area of ACD

0 0 '1 1(12)(8)sin 65 (11.26)(12.8)sin 46 492 2

= 96.05 2

cm

1 1 1 1 1,1,1

1


11

15

Sin ADC = 23

ADC = 60 0 ( a ) AC 2 = (6.6)2 + (5.4)2 -2(6.6)(5.4)cos 60 0 = 43.56 + 29.16 - 71.28(0.5) = 37.08 AC = 6.089 cm

(b) 089.6

sin = 3.950sin 0

= 30.100

(c) 9.99sin

AB = 50sin3.9

AB = 11.96 cm ½ X t X 11.96 = ½ X 6.089 X 9.3 X sin 99.9 t = 4.664 cm OR EC = AC sin 50 = 4.664 cm

1

1

1

1

1

1

1

1

1 1

1 1

10


ujian selaras add.maths t4.melaka kertas … four additional mathematics paper 2 2010 2 2 1 hours...

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