Memo 2

NATIONAL SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P2 FEBRUARY/MARCH 2013 MEMORANDUM

MARKS: 150

This memorandum consists of 21 pages.

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2 NSC – Memorandum

DBE/Feb.–Mar. 2013

NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent Accuracy applies in ALL aspects of the marking memorandum. QUESTION 1 1.1

Scatter plot of exchange rate versus oil price 82 81 80

 any 4 points correctly plotted  any 9 points correctly plotted  all points correctly plotted

79 78 77

Oil price (in $)

76 75 74 73 72 71 70 69 68 67 66 65 6.7

6.8

6.9

7

7.1

7.2

7.3

7.4

7.5

7.6

7.7

Exchange rate (in R/$) 1.2

1.3

1.4 1.5

As the exchange rate (R/$) increases the oil price ($) decreases. OR There is a negative correlation between the exchange rate and oil price. 852,6 Mean = 12 = 71,05 Standard deviation is: σ = 4,09 2 standard deviations from the mean = 71,05+ 2(4,09) = 79,23 The public will be concerned in December 2010

7.8

(3)  reason (2)  852,6  71,05 (2)  4,09 (2)  79,23  Dec 2010 (2) [11]

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QUESTION 2 2.1

 94 – 68  26

Range of Peter’s scores is 94 – 68 = 26

(2) 2.2

 76

Vuyani’s minimum score is 76

(1) 2.3

Vuyani was more consistent during the year because the range of his scores is more clustered about the median value OR the range and inter-quartile range are smaller than Peters.

 Vuyani  reason (2) [5]

QUESTION 3 3.1

Cumulative Frequency Graph 160 150

 plotting points at cumulative frequencies

140 130

Cumulative frequency

120 110

 plot against upper limits

100 90 80

 grounded at (0 ; 0)

70 60 50 40

 smooth curve

30 20 10 0 0

10

20

30

40

50

60

Percentage interval

3.2.1

3.2.2

(85 ; ± 144 ) ± 144 learners that scored below 85% (Accept: 144 to 146) Q 1 = 25 or 27 or 26 Q 3 = 61 or 62 or 64 Interquartile range = 36 or 35 or 38

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70

80

90

100

(4)

 (85 ; ± 144)  ± 144 learners (2)  lower quartile  upper quartile  IQR (3) [9]

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QUESTION 4 4.1

m AD =

y 2 − y1 x 2 − x1

 substitution

7 − (−3) = 1 − (−4) =2

2 (2)

4.2

 m AD = 2

AD//BC m AD = m BC = 2

 substitute into formula

y − y1 = m( x − x1 ) y − (−8) = 2( x − (−2) ∴ y = 2x − 4

 y = 2x − 4 (3)

4.3

4.4

At F: y = 0 0 = 2x – 4 x=2 F(2 ; 0) D is translated C according to the rule: D( x ; y ) → C ( x + 2 ; y − 5) A must also be translated according to this rule to B/. ∴A(1 ; 7) → B/ (3 ; 2)

y=0 x=2 (2) x=3 y=2 (2)

OR x=3 y=2

x B / = −2 + (1 + 4) = 3 y B / = −8 + (7 + 3) = 5 4.5

(2)

m BC = 2 tan θ = 2 β θ = 63,43° D(–4 ; –3) − 8 − (−3) 5 =− m DC = − 2 − (−4) 2 C(–2 ; –8) 5 tan β = − 2 β = 180° − 68,20° = 111,80° α = 111,80° − 63.43° = 48,37°

θ F(2 ; 0)

 63,43°  tan β = −

5 2

 111,8°  48,37° (4)

OR

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DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 CF = (−2 − 2) 2 + (−8 − 0) 2 = 80 Subst in cosformula cos α subject 0,6643... 48,37° (4)

DF = (2 + 4) 2 + (0 + 3) 2 = 45 29 + 80 − 45 cos α = 2( 29 )( 80 ) = 0,6643...

α = 48,37° OR DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 DB = (3 + 4) 2 + (2 + 3) 2 = 74 BC = (3 + 2) 2 + (2 + 8) 2 = 125 29 + 125 − 74 cos α = 2( 29 )( 125 ) = 0,6643...

Subst in cosformula cos α subject 0,6643... 48,37° (4)

α = 48,37° 4.6

DC = (−4 + 2) 2 + (−3 + 8) 2 = 29 CF = (−2 − 2) 2 + (−8 − 0) 2 = 80 1 .DC.CF. sin α 2 1 = ( 29 )( 80 ) sin 48,37° 2 = 18 units 2

 substitution into formula  29  substitution into formula  80

Area ∆DCF =

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 substitution into the area rule  18 (6)

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OR 6

F

(1) Area=9

3

 establishing rectangle and area

D 8 5

(3) Area =5

(2) Area = 16  relationship of areas  (1) = 9 (2) = 16 (3) = 5 18 units2 (6)

2 C 4 Area ∆DCF = Area of rectangle – (1) – (2) – (3) = 48 – 9 – 5 – 16 = 18 sq units

OR

G(–4 ; 0)

H(–2 ; 0)

F (2 ; 0)

 drawing perpendiculars

D (–4 ; –3)

C (–2 ; –8) Area CDF = Area CHF + Area CDGH - Area DGF 1 1 1 × 4 × 8 + 2 × (3 × 8) − × 6 × 3 2 2 2 = 16 + 11 − 9 = 18 =

 relationship of areas  16  11  9 18 units2 (6) [19]

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QUESTION 5 5.1.1

x 2 + y 2 + 2x + 6 y + 2 = 0 x 2 + 2 x + 1 + y 2 + 6 y + 9 = −2 + 10 ( x + 1) 2 + ( y + 3) 2 = 8 M (−1 ; − 3)

5.1.2 5.2

radius of circle C1 = 8

x 2 + ( x − 2) 2 + 2 x + 6( x − 2) + 2 = 0

 ( x + 1) 2 + ( y + 3) 2 = 8  –1  –3 (3)  8 (1)  substitution

x 2 + x 2 − 4 x + 4 + 2 x + 6 x − 12 + 2 = 0 2x 2 + 4x − 6 = 0 x 2 + 2x − 3 = 0 (x + 3)( x − 1) = 0 x = −3 or x ≠ 1 y = −3 − 2 = −5

 standard form  factors  value of x  value of y (5)

∴ D(−3 ; − 5) OR

( x + 1) 2 + ( y + 3) 2 = 8 subst. y = x − 2

 substitution

( x + 1) 2 + ( x − 2 + 3) 2 = 8 ( x + 1) 2 + ( x + 1) 2 = 8 x 2 + 2x − 3 = 0 (x + 3)( x − 1) = 0 x = −3 or x ≠ 1 y = −3 − 2 = −5 OR ( x + 1) 2 + ( y + 3) 2 = 8 subst. y = x − 2

 standard form  factors  value of x  value of y (5)

 substitution

( x + 1) 2 + ( x − 2 + 3) 2 = 8 ( x + 1) 2 + ( x + 1) 2 = 8

(x + 1)2

= x +1 = x= y=

4 ±2 −3 or x ≠ 1 −3 − 2 = −5

 simplification  square root of both sides  value of x  value of y

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PM makes 45° with the x-axis.

8 = 22 + 22 Therefore: x D = x M − 2 = −1 − 2 = −3


 8 = 2 2 + 2 2  value of x  value of y (5)

y D = −3 − 2 = −5 5.3

MD ⊥ DB

(tangent ⊥ radius)

MB 2 = MD 2 + DB 2

(Pythagoras)

= ( 8 ) 2 + (4 2 ) 2 = 40 MB is the radius of C 2

MB = 40

 tangent ⊥ radius  substitution into Pythagoras



40

(3) 5.4

5.5

( x + 1) + ( y + 3) = 40 2

2

(

)

Distance from 2 5 ; 0 to centre =

(2

)

5 + 1 + (0 + 3) 2

2

= 6,24

(

)

6,24 < 6,32 40 Distance from 2 5 ; 0 to centre < radius of circle. 2 5 ; 0 lies inside the circle.

(

)

(

)

 LHS  RHS (2) substitution into distance formula  6,24 6,24 < 6,32  conclusion (4) [18]

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QUESTION 6 6.1.1

A( – 5 ; 3) A / (−5 + 4 ; 3 − 3) = (−1 ; 0)

6.1.2

A (−5 ; − 3)

–1 0 (2)  –5  –3

/

(2) 6.2.1

/

Scale factor of enlargement is

/

KM 15 3 = = KM 10 2

/

KM KM 3  2

/



OR 3  3 K (−4 ; 2) → K / (−6 ; 3) = K /  × −4 ; × 2  2  2 3 Scale factor is 2

 y 

 3  3  × −4 ; × 2  2  2 3  2 (2) 3  x 2 3  y 2 (2) 9  2  3 (2)

6.2.2

3 3 ( x ; y) →  x ; 2 2

6.2.3

3 3 P/  ×3 ; 2×  2 2 9  = P /  ; 3 2 

6.2.4

a=1

a=1

6.2.5

K (4 ; − 2)

 4  –2

6.2.6

K K =5

K K = 5  K / M /// = 15

(2)

//

(2) ///

/

K / M /// = 15 K / K /// 5 1 = = / /// 15 3 K M

///



/

1 3 (3) [17]

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QUESTION 7 b –a

7.1

K / (b ; − a )

7.2

K (b cos θ − a sin θ ; − a cos θ − b sin θ ) //

OR K (a cos(90 + θ ) + b sin(90° + θ ) ; b cos(90° + θ ) − a sin(90° + θ ) //

= K // (−a sin θ + b cos θ ; − b sin θ − a cos θ )

7.3

T // (−(−4) sin θ + (−2) cos θ ; − (−2) sin θ − (−4) cos θ ) = T // (4 sin θ − 2 cos θ ; 2 sin θ + 4 cos θ )

OR

T // (−2 cos θ − (−4) sin θ ; − (−4) cos θ − (−2) sin θ ) = T // (−2 cos θ + 4 sin θ ; 4 cos θ + 2 sin θ )

7.4

2 3 + 1 = 4 sin θ − 2 cos θ ........(1) 3 − 2 = 2 sin θ + 4 cos θ ........(2) (2) × 2 : 2 3 − 4 = 4 sin θ + 8 cos θ ....(3) (1) − (3) : 5 = −10 cos θ 1 − = cos θ 2 ∴ θ = 180° − 60° = 120° OR

2 3 + 1 = 4 sin θ − 2 cos θ ........(1) 3 − 2 = 2 sin θ + 4 cos θ ........(2) (1) × 2 : 4 3 + 2 = 8 sin θ − 4 cos θ ....(3) (2) + (3) : 5 3 = 10 sin θ

(2)  b cos θ − a sin θ  − a cos θ − b sin θ (2)  4 sin θ − 2 cos θ  2 sin θ + 4 cos θ (2)  4 sin θ − 2 cos θ  2 sin θ + 4 cos θ (2)  substitution to form equation  substitution to form equation  5 = −10 cos θ 1  − = cos θ 2  120° (5)  substitution to form equation  substitution to form equation  5 3 = 10 sin θ

3 = sin θ 2 ∴ θ = 180° − 60° = 120°

3 = sin θ 2  120°



(5) OR

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mOT =

mOT / =


1 ˆT = 1 ⇒ tan XO 2 2 ˆ XOT = 206,565...° 3−2

⇒ tan XOˆ T // =

2 3 +1 ˆ T = −3,434° XO

3−2 2 3 +1

= −0,06....


ˆT =  tan XO

1 2

 206.565...°  – 0,06...

90° + θ = 209,99...° ≈ 210° θ = 120°

 – 3.434°  120° (5) OR

(TT/)2 = OT2 + (OT/)2 – 2(OT)(OT/).cos (90° + θ) 40 + 20 3 = 40 − 40. cos(90° + θ ) 3 cos(90° + θ ) = − 2 90° + θ = 150° θ = 60°

(TT/)2 = 40 + 20 3 substitution in cos-rule simplification 150° 60° (5) [11]

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QUESTION 8 8.1

1 − sin 2 θ + 3 − cos 2 θ

 simplification

= 4 − (sin 2 θ + cos 2 θ ) =3

3 (2) OR

cos θ + 3 − cos θ =3 2

 substitution with identity 3

2

(2) 8.2

4

sin 150°

.2

 rewrite using reduction formula  substituting special angles  simplification

3 tan 225°

= 4 sin 30°.2 3 tan 45° =

(2 ) .2 1 2 2

3

= 16 =4

4 (4) OR

1 sin 150° = 2 tan 225° = 1

1 2  tan 225° = 1

 sin 150° =

4 sin 150° 2 3 tan 225° 1

= 4 2 23 = 2.2

3

= 16 =4

8.3

cos 2 x(sin 2 x + cos 2 x) 1 − sin x 2 cos x . (1) = 1 − sin x (1 − sin 2 x) = 1 − sin x (1 + sin x)(1 − sin x) = 1 − sin x =1 + sin x

LHS =

= RHS

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1

 42 = 2 4 (4)  factorisation 1  1 − sin 2 x  factors

(4)

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8.4


cos 3θ = cos(2θ + θ ) = cos 2θ . cos θ − sin 2θ . sin θ = (2 cos θ − 1). cos θ − 2 sin θ . cos θ . sin θ 2

= 2 cos 3 θ − cos θ − 2 sin 2 θ . cos θ = 2 cos 3 θ − cos θ − 2(1 − cos 2 θ ). cos θ = 2 cos θ − cos θ − 2 cos θ + 2 cos θ 3

3


expansion  2 cos 2 θ − 1  2 sin θ . cos θ  1− cos 2 θ

= 4 cos 3 θ − 3 cos θ

8.5

cos 3θ = 4 cos θ − 3 cos θ

(4)

3

cos 3(20°) = 4 cos 3 (20°) − 3 cos(20°) 1 = 4 x 3 − 3x 2 8x 3 − 6 x − 1 = 0

θ = 20° cos 60° =

1 2 (2) [16]

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QUESTION 9 9.1

cos160°. tan 200° 2 sin( −10°) (− cos 20°)(tan 20°) = 2(− sin 10°)

 − cos 20°  tan 20°  − sin 10°

 sin 20°  (− cos 20°)  cos 20°   = − 2 sin 10° 2 sin 10° cos10° = 2 sin 10° = cos10° 9.2.1



 2 sin 10° cos10°  cos10°

LHS = cos( x + 45°). cos( x − 45°) = (cos x. cos 45° − sin x sin 45°)(cos x cos 45° + sin x sin 45°) = cos 2 x. cos 2 45° − sin 2 x. sin 2 45°   2  2  − sin 2   or cos 2 = cos x   2    2    1 1 = cos 2 x − sin 2 x 2 2 1 = (cos 2 x − sin 2 x) 2 1 = cos 2 x 2 2

sin 20° cos 20°

2

2

2

 1  x  − sin 2  2

2  1   x    2  

(6)  expand cos( x + 45°)  expand cos( x − 45°)  substitute special angles  simplification

(4) OR 2 cos α cos β = cos(α + β ) + cos(α − β ) 1 cos α cos β = (cos(α + β ) + cos(α − β ) ) 2 Let α = x + 45° and β = x − 45° ∴ cos( x + 45°) cos( x − 45°)

 deriving identity

1 (cos(( x + 45° + x − 45°) + cos( x + 45° − x + 45°) ) 2 1 = (cos 2 x + cos 90°) 2 1 = cos 2 x 2

 substitution

=

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 simplification

(4)

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9.2.2


cos( x + 45°) cos( x − 45°) has a minimum when


1 cos 2 x has a 2

minimum. The minimum value of cos 2 x is –1

cos 2 x = −1 2 x = 180°

 minimum value of –1 2x = 180°  x = 90°

x = 90°

(3) [13]

QUESTION 10 10.1

Range = [−1 ; 1]

 [−1 ; 1] (2)

10.2

3  3  f  x  = sin 2 x  2  2  = sin 3 x 360° ∴ Period = = 120° 3

 sin 3x 120° (2)

OR 3  3  f  x  = sin 2 x  2  2  = sin 3 x = sin(3 x + 360°) = sin 3( x + 120°) ∴ Period = 120°

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 sin 3x

120° (2)

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10.3 1.5

f

1

(30° ; 1) 0,87

0.5

(90° ; 0,5) -180

-120

-60

60

 x intercepts  turning points  shape

-0.5

(-180° ; -0,87)

g

-1

(4)

(-150° ; -1) -1.5

10.4

(−180° ; − 90°) or (−60° ; 0°)

 > – 180°  < –90°  > –60°  < 0°

OR − 180° < x < −90° or − 60° < x < 0°

(4)  translation 30°  to the left

10.5

y = sin 2(x + 30°) ∴ translation of 30° to the left

10.6

sin 2 x = cos( x − 30°) sin 2 x = sin[90° − ( x − 30°)] = sin(120° − x) 2 x = 180° − (120° − x) + 360°k 2 x = 120° − x + 360°k ; k ∈ Z or 2 x − x = 60° + 360°k 3 x = 120° + 360°k x = 60° + 360°k ; k ∈ Z x = 40° + 120°k ; k ∈ Z

(2)  using co-function  2 x = 120° − x + 360°k  x = 40° + 120°k  2 x = 180° − (120° − x) + 360°k  x = 60° + 360°k k ∈Z (6)

OR sin 2 x = cos( x − 30°) cos(90° − 2 x) = cos( x − 30°) 90° − 2 x = x − 30° + 360°k or 90° − 2 x = 360° − ( x − 30°) + 360°k

− 3 x = −120° + 360°k x = 40° − 120°k ; k ∈ Z

− x = 300° + 360°k x = −300° − 360°k ; k ∈ Z

∴ x = 40° + 120°k or x = 60° + 360°k ; k ∈ Z

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 

cos(90° − x) = cos( x − 30°) 90° − 2 x = x − 30°

+ 360°k  x = 40° − 120°k  90° − 2 x = 360° − ( x − 30°) + 360°k  x = −300° − 360°k k ∈Z (6) [20] Please turn over

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QUESTION 11 11.1

AB = sin 2 x r AB = r sin 2 x

AB = sin 2 x r  AB = r sin 2 x



(2) 11.2

ˆ C = 90° + x AK

 AKˆ C = 90° + x (1)

11.3 C

 sine rule

x x

 substitution

r

B

90°- x 90°+ x K

A

 making AK subject of the formula  cos x

In ΔAKC : ˆ C sin ACˆK sin AK = AC AK sin(90° + x) sin x = r AK r sin x r sin x AK = = sin(90° + x) cos x AK 2 = AB 3  r sin x     cos x  = 2 r sin 2 x 3 sin x 2 cos x = 2 sin x cos x 3 sin x 1 2 × = cos x 2 sin x cos x 3 1 2 = 2 2 cos x 3 4 cos 2 x = 3

 2sinx.cosx



 cos x =

3 cos x = 2 x = 30°

3 2

 x = 30° OR

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1 2 cos 2 x

(8)

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C x x r

90°- x 90°+ x K Using the sine-formula in ∆CBK and ∆CKA: sin x sin(90° − x) sin x sin(90° + x) = = and KA AC BK BC BK KA ∴ = BC AC 1 2 ∴ = BC r 1 ∴ BC = r 2 BC 12 r 1 ∴ cos 2 x = = = 2 AC r ∴ 2 x = 60° ∴ x = 30° B

A

sin x sin(90° − x) = BK BC sin x sin(90° + x) =  KA AC BK KA =  BC AC 1 2 =  BC r 1  BC = r 2 1  cos 2 x = 2  2x = 60° 

 x = 30°

OR

(8)

C x x r

90°- x 90°+ x K c

B ∆CBK: KC =

90°- 2 x 2c

A

c sin x

sin x sin(90° − 2 x) sin(90° − 2 x). sin x = = 2c KC c

∆CKA:

1 = sin 30° 2 90° - 2x = 30° x = 30°

∴ sin (90° - 2x) = ∴

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 KC =

c sin x

sin x sin(90° − 2 x) = 2c KC substitution 1  sin (90° - 2x) = 2 90° - 2x = 30°  x = 30° 

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(8)

OR

C x x

r

B

90°- x 90°+ x K c

90°- 2 x 2c

A

∆CBK:

3c = 2 sin x. cos x r 3c .......................(1) ∴ r sin x = 2 cos x sin 2 x =

∆CKA: 2c r = sin x cos x ∴ r sin x = 2c cos x ..........................(2) Equate (1) and (2): 3c 2c. cos x = 2 cos x 3 ∴ cos 2 x = 4 3 ∴ cos x = 2 x = 30° ∴ OR

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3c r  2 sin x. cos x 3c  r sin x = 2 cos x  sin 2 x =

2c r = sin x cos x  r sin x = 2c cos x 

equating

 cos x = 30°

3 2

(8)

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AK 2 = =2 KB 1 2=

1 2 1 2

AK .BC BK .BC

multiplying by 12 BC  area of triangles

area AKC area ABC 1 rCK sin x = 12 2 BC .CK sin x =

 area formula in triangles r =2  BC BC 1 =  r 2 1  cos 2 x = 2  2x = 60°

r BC BC 1 ∴ = r 2 =

∴ cos 2 x =

1 2

∴ 2 x = 60° ∴ x = 30°

 x = 30° (8) OR C x x r

B

c

K

By the Internal Bisector Theorem: CB BK 1 = = CA KA 2 1 cos 2 x = 2 2 x = 60° x = 30°

2c

A  For stating Internal Bisector Theorem CB BK 1 = =  CA KA 2

 cos 2 x =

1 2

 2x = 60°  x = 30° (8)

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OR D x

r

C x x r x B

0°- x 90°+ x K

Produce BC to D and draw CK parallel to DA. CAˆ D = KCˆ A and BCˆ K = Dˆ ∴ DC = CA = r ∴ ∆BKC ||| ∆BAD BK BC = =3 ∴ BA BD ∴ BD = 3BC = BC + r

∴ BC =

1 r 2

∴ cos 2 x = ∴ 2 x = 60° ∴ x = 30°

1 2

r 1 = r 2

A

 DC = CA = r  ∆BKC ||| ∆BAD BK BC = =3  BA BD BD = BC + r 1  BC = r 2 1  cos 2 x = 2 2x = 60° 30° (8) [11] TOTAL: 150

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